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Supporting Information

This file contains all supporting information including answers on the questions of all 20 sections.

Chapter 1. Alkaloids 3

1.1 Pseudopelletierine from the root-bark of the pomegranate tree 3 1.2 Colchicine from the seeds of the autumn crocus 17 1.3 Capsaicin from Kenyan "African Bird’s Eye Chilies" 21

Chapter 2. Coloured Compounds 33

2.1 Thymoquinone from the oil of the seeds of black caraway 33 2.2 Berberine chloride from the bark of the common barberry 39 2.3 Carminic acid from dried cochineal 45 2.4 Safflomin A from flowers of the safflower 47 2.5 Chlorophyll a from deep frozen spinach leaves 57

Chapter 3. Carbohydrates and glycosides 74

3.1 Raffinose from the seeds of blue lupins 74 3.2 Fraxin from the shredded bark of the ash tree 83 3.3 Stevioside from the dried leaves of Stevia rebaudiana 93

Chapter 4. Terpenoids 109

4.1 Linalool from Brazilian rosewood oil 109 4.2 Camphor from camphor tree oil 120 4.3 Cantharidin from Spanish fly Lytta vesicatoria 124 4.4 Artemisinin from the dried leaves of the annual mugwort 126 4.5 Diosgenin from an extract of the roots of Mexican yams 136 4.6 Friedelin from cork from the bark of the cork-oak 143 4.7 Boswellic acid from frankincense, the resin of the Arabian olibanum tree 152

Chapter 5. Aromatic Compounds 158

5.1 Sinensetin from cold-pressed orange oil 158 5.2 Rosmarinic acid from the dried leaves of lemon balm 160

Supporting information Pseudopelletierine

Chapter 1 1.1 Pseudopelletierine

Fig. S1.1-1 Structure of pseudopelletierine

Fig. 1.1-2 Excerpt from the title page of the "Universal-Lexicon der practischen Medizin und Chirurgie" (Universal Lexicon of Medicine and Surgery), 5th Volume, H. Frank'sche Verlagsexpedi- tion, Leipzig 1838; for the quotation in text see page 13 of the main book

3 3 Supporting information Pseudopelletierine

Piccinini's Degradation of Pseudopelletierine to Suberic Acid

Fig. S1.1-3 Piccinini's degradation of pseudopelletierine to suberic acid [10] Synthesis of Pseudopelletierine

Fig. S1.1-4 Robinson-Schöpf synthesis of pseudopelletierine by a double Man- nich reaction

4 5 Supporting information Pseudopelletierine

Biosynthesis of Pelletierine, N-Methylpelletierine and Pseudopelletierine

Fig. S1.1-5 Biosynthesis of pelletierine, N-methylpelletierine and pseudopelletierine. The biosynthesis of the Punica granatum alkaloids [21] shows several similarities to the laboratory synthesis. It begins with the amino acid lysine, which is initially cyclized to Δ1-piperideine. The iminium ion derived from this adds subsequently to the enolate of the activated acetonedicarboxylic acid (Fig. S1.1-5). Parallel to the decarboxylation, the precursor of the N-methylpelletierine can be oxidized to a further iminium ion, which via intramolecular enolate addition and decarboxylation forms pseudopelletierine. Willstätter's Synthesis of Cyclooctatetraene Willstätter's degradation of pseudopelletierine to COT begins with the reduction of the carbonyl group by the Bouveault-Blanc method and subsequent acid catalysed dehydration. The second double bond is introduced by Hofmann elimination with opening of the N-bridge. With this step the first difficulty occurs. On heating the quaternary ammonium hydroxide under normal pressure, instead of the desired 5-dimethylamino-1,3-cyclooctadiene the isomeric 1-dimethylamino compound is formed. The reaction only proceeds as desired, when the reaction is carried out under vacuum. A further Hofmann elimination leads to 1,3,5-cyclooctatriene. Its reaction with

Br2 gives the dibromo-adduct. Without having a proof, Willstätter assumed, that a 1,6-addition occurs. The debromination of the dibromo-compound

gives a product C8H8 that, however, on catalytic hydrogenation takes up less than 8 H-atoms. The detour over an exchange of both bromine atoms by dimethylamino substituents and further Hofmann eliminations proved to be successful (Fig. 1.1-16).

4 5 Supporting information Pseudopelletierine

1. How is 1-dimethylamino-1,3-cyclooctadiene formed under "normal" conditions by the first Hofmann exhaustive methylation? The formation of this product during the distillation under atmospheric pressure can be considered to be a further reaction, i.e. a thermally allowed superfacial 1,5-H-shift, of the formed primary product. Under mild conditions the desired primary product can be obtained.

2. Willstätter's assumption, the 1,6-addition of Br2, was later confirmed by Cope [1]. The adduct is identical to an independently synthesised 1,6-dibromo-2,4-cyclooctadiene. After the elucidation of the structure, the question of the stereochemistry remains. We can only speculate. If the addition of bromine takes place via a bromonium intermediate by transfer of Br+ to the ends of the conjugated triene, then the two bromine atoms must be in a trans-arrangement.

3. Why does the double dehydrobromination of 1,6-dibromo-2,4-cy-

clooctadiene with quinolone as a base give a C8H8-product that has less than four easily hydrogenated double bonds instead of COT? Will- stätter assumed that the high temperature required for the dehydrobromination produced a mixture of bicyclo[4.2.0]octa-2,4,7-triene and tricyclo[4.2.0.02.5]octa-3,7-diene.

Today we know, that the bicyclic compound is contained to 0.01% in COT in a valence tautomeric equilibrium. Independently synthesised bicyclo[4.2.0]octa-2,4,7-triene isomerizes at 0°C with a half-life of 14 minutes to COT [S2]. A further ring closure to the tricyclic diene is improbable,

because of ring strain. Therefore, both C8H8-isomers can be eliminated as products of the base catalysed dehydrobromination. Willstätter mentioned, that the smell of the hydrocarbon was reminiscent of benzene and reported the boiling point at 737 torr to be 142.8 – 143.8°C and the specific gravity to be 0.912 g/mL. These properties agree with those of styrene (bp = 146°C, d = 0.909 g/mL), its smell is strongly reminiscent of benzene. The formation of styrene from 1,6-dibromocycloocta-2,4-diene and COT from 1,6-diaminocycloocta-2,4-diene must depend upon different mecha- nisms of the β-elimination.

6 7 Supporting information Pseudopelletierine

The properties of the Br-substituents in the allylic positions as good leaving groups point to an E1-type mechanism for the two dehydrobromination steps. In the second elimination step, a cation is formed that corresponds to the protonated COT and therefore exists in the form of the homotropylium ion. To explain the formation of styrene, we suggest, that the homotropylium ion is in equilibrium with a low concentration of the bicyclo[4.2.0]-octa-2,4-di- en-7-yl cation, which can be deprotonated to styrene, as shown. Hofmann eliminations are typical E2-eliminations with a certain E1cB-character, i.e. carbenium ions are avoided. Therefore, the stepwise cleavage of two trimethylamine molecules occurs as follows: For the sake of simplification we have up to now usually shown cyclooctatetraene as a planar eight-membered ring. In reality, the spacial arrangement of COT is tub-shaped [S3], with the conse- quence, that neighbouring double bonds build a dihedral angle of 56°. Because of the poor overlapping of neighbouring p-orbitals, we are dealing with fixed alternating double and single bonds.

If cyclooctatetraene were planar and the π-system could therefore be delocalized, then according to the Hückel Rule COT would be antiaromatic and not aromatic (the number of electrons is 4n and not 4n+2). However, it is neither, because the antiaromaticity is avoided by the non-planar structure and it exists as a polyolefine (annulene).

6 7 Supporting information Pseudopelletierine

Apart from the already mentioned valence isomerism to bicyclo[4.2.0]-octa-2,4,7-triene COT demonstrates two further dynamic processes that can be most easily explained by planar D8h or respectively D4h structures as transition states. This refers to the shift of the double bond and the ring inversion [S4]. Because of the only slight overlapping of for example the p-orbitals of C-1 and C-8, the shifts of the double bonds cannot be obtained by a simple relocation of the π-bonds. However, it could succeed via a flattening to a delocalized π-system (TS(bs)). With this approach, the ring inversion requires the formation of a transition state TS(ri) with localized π-bonds. Both processes can be observed independently of each other, e.g. with dynamic NMR spectroscopy of cyclooctatetraene with a suitable substitution. Be- cause of the antiaromatic transition state, the activation energy of the shift of the double bond is higher. Pelletierine, "Isopelletierine", N-Methylpelletierine Our article focuses on pseudopelletierine, the main alkaloid of the pomegranate tree. However, pelletierine and N-methylpelletierine were responsible for the pharmaceutical and medical interest. The history of these two alkaloids with their apparently simple structures is a textbook example of how difficult and time-consuming structure elucidation could be a century ago. Therefore, we want now to briefly peruse the situation and so complete this overview of the chemistry of the pomegranate tree. In 1917 the research group of Hess began a series of reports on the structure of pelletierine, isopelletierine and N-methyl(iso)pelletierine. The subject was approached with great synthetic effort and with the deployment of the methods for structural elucidation, which were available at that time. Tragically, Hess reached erroneous conclusions. The reason for this is not understandable, even after a thorough study of the many publications. In the following scheme, the structures suggested by Hess are compared with those that were later proved to be correct.

Pelletierine Isopelletierine Methyl(iso)pelletierine Hess was convinced, that the pelletierines in Tanret (–) C H NO rac. C H NO (–) C H NO 8 15 8 15 9 17 the pomegranate tree were racemates. Conse- O O quently, he had to con- correct (±)-Pelletierine N clude, that pelletierine structure N H and isopelletierine were CH3 structural isomers. Pel- letierine was assumed to be 3-(2-piperidyl)pro- Hess O panal and isopelletierine N N N H H to be 2-propionylpiperi- H O CH3 O dine.

According to this, isopelletierine and N-methylisopelletierine should be structurally identical to the products of oxidation of conhydrine (a hemlock alkaloid) or respectively N-methylconhydrine.

8 9 Supporting information Pseudopelletierine

As this was not confirmed, Hess tried a more daring interpretation. He assumed, that the triply coordinated N-atom of the piperidone ring was a chiral centre [S6]. In modern terms, he postulated a stability against inversion at the N-atom. Together with the chiral centre at C-2 one arrives at two di- astereomeric pairs of enantiomers. One of these should be the racemic conhydrinone or N-methylconhydrinone, the other the racemic isopelletierine or respectively N-methylisopelletierine. This explanation fitted at the time, in which unsolved cases of isomerism were not seldom explained by a so- called asymmetric N-atom. Meisenheimer was sceptical about this interpretation, since he had dis- proved other cases of supposed stereoisomerism at a trivalent N-atom, and showed that conhydrinone and isopelletierine are structural isomers [10]. Concurrently he was the first to elucidate the true structure of isopelletierine and since this, as we shall soon see, is the racemate of (-)-pelletierine, also its structure. The reaction scheme used by Meisenheimer is shown in Fig. S1.1-6.

Fig. S1.1-6 Proof of structure for isopelletierine and N-methylpelletierine by Meisenheimer

In a detailed experimental investigation [S7] Hess later corrected his suggested structure, following Meisenheimer's proof. The previous papers of Hess's research group can be found in [S6]. Later, syntheses of isopelletierine followed that made use of the newly developed Li-organic chemistry [S8]. A totally different concept that does not start from a pyridine derivative and in principle corresponds to the biosynthesis is due to Schöpf [S9, S10] (Fig. S1.1-7). The conditions used resem- ble as closely as possible those in the cell.

Fig. S1.1-7 Synthesis of N-methylisopelletierine from 4-methylaminopentanal and acetonedicarboxylic acid

8 9 Supporting information Pseudopelletierine

An important step forward was made by a research group in Vienna by the separation of synthetic isopelletierine and N-methylisopelletierine in their enantiomers with (+)-6,6'-dinitro-2,2'-diphenic acid or (+)-tartaric acid [S11]. The racemization of the pure enantiomer in weakly alkaline solution could be followed from the change in optical rotation. In acidic medium, no racemization was observed. The catalysis by bases is illustrated in the reaction scheme shown in Fig. S1.1-8. In the first step, an enolate is formed by deprotonation that delivers an achiral intermediate by retro-Michael addition. On reversal both enantiomers are formed in equal amounts. This delivered an explanation, why in the older literature pelletierine was sometimes described as optically active and sometimes as inactive. Apparently, the biosynthesis delivers an optically active product that through the extraction and work-up partially or totally racemizes under the conditions normally used for alkaloids. Tanret, who discovered the alkaloids, used mild methods for the isolation and therefore retained in part the optical activity.

Fig. S1.1-8 Base catalysed racemization It is now clear, that pelletierine and isopelletierine have the same constitu- of optically active pelletierine (R=H, tional structure. Isopelletierine is the racemate of pelletierine. pelletierine; R=CH3, N-methylisopelletierine) The subsequently drawn conclusion, "Pelletierine in the literature is isopelletierine", is not exactly incorrect, but unfortunate. It is more logical to relinquish the term "isopelletierine", which has in the meantime been accepted. However, the name isopelletierine can also be found in current monographs and textbooks, when really pelletierine is meant. In the section on the GC-MS analysis of the alkaloid extract from the pomegranate tree, an experiment with chiral GC-phases is described, which demonstrated the dependency of the racemization of the alkaloid fraction on the conditions used for preparation. The investigation showed that pelletierine has a high susceptibility for racemization, whereas N-methylpelletierine is more stable. This unequivocal finding is contrary to the report of Galinowsky et al. [S11], that N-methylpelletierine is more easily racemized. The importance of pelletierine as an intermediate in the biosynthesis of more complex alkaloids leads today to a focussing in synthesis work on the development of enantioselective methods [S13-S15].

GC-MS Analysis For an analytical overview, we stirred the coarsely hacked bark for 24 hours in dichloromethane/conc. ammonia, to set the bases free. The dichloromethane phase (Extract 1) was extracted with dilute sulphuric acid, the acidic extract made alkaline and re-extracted with dichloromethane (Extract 2). Fig. S1.1-11 in the article shows the gas chromatogram of the alkaloid mixture, obtained in this way. According to this, there are essentially three alkaloids present, which according to the mass spectra have M+•-ions with m/z 141, 155 and 153 in the ratio 25:12:63. The main alkaloid is pseudopelletierine 10 11 Supporting information Pseudopelletierine

+• (C9H15NO, M 153), the next most abundant substance is either pelletierine +• or isopelletierine (C8H15NO, M 141) and the third alkaloid must be meth- +• ylpelletierine (C9H17NO, M 155). In the base line of the chromatogram several other alkaloids are hidden that only occur in trace amounts and represent <1% of the total amount of alkaloids. In Extract 1 we found an arte- fact (M+• 152), which on further preparation was converted back to pseudopelletierine, and large amounts of friedelin [S16]. The EI spectra of pelletierine and N-methylpelletierine, which were obtained from the GC-MS analysis described, are simple and unequivocally interpreted.

Fig. S1.1-9 EI fragmentation of pelletierine and N-methylpelletierine

Apart from pseudopelletierine, pelletierine and N-methylpelletierine one finds in the chromatogram further alkaloids, which occur in trace amounts (<1%), including norpseudopelletierine (M+• 139), and sedridine. In a detailed investigation, Neuhöfer et al. [S17] also identified the pyrrolidine alkaloids norhygrine and hygrine.

The GC-MS investigation of Extract 1 shows, that apart from pelletierine, N-methylpelletierine and pseudopelletierine a further alkaloid with a molecular ion at m/z 152 (relative intensity 59%) and a [M-H]+-peak at m/z 151 (14%) is present. This compound disappears again on further preparation of the extract. According to the mass spectrum, it is iminopseudopelletierine, which is formed by the reaction of the carbonyl group with ammonia. The mass spectrum shows in analogy to pseudopelletierine the fragments m/z 82 (23%), 94 (5%), 95 (6%), 96 (60%), 110 (23%) and in addition m/z 109 (5%). In the case of pseudopelletierine this last peak falls together with the m/z 110 peak.

10 11 Supporting information Pseudopelletierine

At m/z 111, 84, and 83 ions occur that have no corresponding peak in the mass spectrum of pseudopelletierine. The reason for this could be, that apart from the ionisation of the amino-nitrogen of the iminocompound, which leads to the ions shown above, the imino group is ionized, which opens a further fragmentation path (Fig. S1.1-10).

Fig. S1.1-10 Fragment formation after ionisation of the imino group of iminopseudopelletierine In the root-bark of the pomegranate tree, the alkaloids occur as salts of tannic acids and other organic acids. Using a standard procedure the alkaloids are released from the acids by stirring the hacked bark in a mixture of dichloromethane and conc. ammonia. The fact that the optically active pelletierine from the biosynthesis is prone to base catalysed racemization has caused much confusion in the structure elucidation, as reported. In the following, experiments are described that demonstrate the degree of racemization as a function of the conditions used for the preparation. If the alkaloid extract after stirring for 24 hours with conc. ammonia (see gas chromatogram in Fig. S1.1-11) is analysed on a chiral stationary phase, it can be seen, that the racemization of pelletierine has already progressed far (ratio of enantiomers 57:43, 14% ee). For N-methylpelletierine the amount of the not naturally occurring enantiomer has only reached 8% (84% ee) (Fig. S1.1-11a). After stirring for a further 24 hours with conc. ammonia the pelletierine is fully racemized. For N-methylpelletierine the enantiomeric excess is still 48% (Fig. S1.1-11b).

Fig. S1.1-11a /11b Gas chromatographic separation of the alkaloid extract from cortex punica granatum on a chiral phase (1 = pelletierine, 2 = N-methylpelletierine, 3 = pseudopelletierine). (a) after 24 h and (b) after 48 h contact with ammonia (for details

see text). GC conditions: Instrument: HP5890 chromatograph; injector: 240 °C; detector: FID, 250°C; carrier gas: H2; column: 13 m×0.25 mm; film thickness 0.13 μm (30 % 2,3-diacetyl-6-TBDMS-β-cyclodextrin in PS086); temperature: 80 °C.

12 13 Supporting information Pseudopelletierine

The extraction of the root-bark with saturated NaHCO3 solution instead of conc. ammonia led to an almost complete racemization of pelletierine (enantiomeric ratio 53:47, 6% ee). We obtained the greatest excess of the biosynthetic enantiomers of pelletierine (74% ee) under the following con-

ditions: the root-bark was stirred with 5% H2SO4, to release the alkaloid as its soluble sulphate from the tannic acids. The yellow-orange filtrate was made alkaline with conc. ammonia, whereby the colour changed to brown. The subsequent extraction with dichloromethane was complicated by the formation of an emulsion, so that the contact with ammonia was longer than intended (about 1 h). This was sufficient to cause a partial racemization (enantiomeric ration 87:13)

Literature [S1] A. C. Cope, C. L. Stevens, F. A. Hochstein "Cyclic Polyolefins V. Preparation of Bromocyclooctadiene and 1,3,5-Cyclooctatriene from 1,5-Cyclooctadiene" J. Am. Chem. Soc. 1949, 72, 2510–2514. [S2] E. Vogel, H. Kiefer, R. Roth "Bicyclo[4.2.0]octa-2,4,7-trien" Angew. Chem. 1964, 76, 432–433. [S3] J. Bordner, R. G. Parker, R. H. Stanford "The Crystal Structure of Octamethylcyclooctatetraene" Acta Cryst. Sect. B 1972, 28, 1069– 1075. [S4] L. A. Paquette "The Current View of Dynamic Change within Cy- clooctatetraene" Acc. Chem. Res. 1993, 26, 57–62. [S5] E. Vaupel, "Krieg der Chemiker" Chem. Unserer Zeit 2014, 48, 460– 475. [S6] K. Hess, R. Grau, "Neue Umwandlungen von Conhydrin und Methylisopelletierin, V. Mitteilung zur Frage des asymmetrischen dreiwertigen Stickstoffatoms" Liebigs. Ann. Chem. 1925, 441, 101– 137. [S7] K. Hess, O. Littmann "Methylisopelletierin und 1-[α-N-Methyl-piperidyl]-propan-2-on" Liebigs Ann. Chem. 1932, 494, 7–17. [S8] J. P. Wibaut, J. I. de Jong "Pyridine and Quinoline Derivatives. LXXIX. Synthesis of Ketones by the Action of 2-Picolyllithium on Nitriles" Rec. Trav. Chim. Pay-Pas 1949, 68, 485–490. [S9] C. Schöpf, G. Lehmann "Die Synthese des Tropinons, Pseudopelle- tierins, Lobelanins, und verwandter Alkaloide unter physiologischen Bedingungen" Liebigs Ann. Chem. 1935, 518, 1−37. [S10] F. Galinowsky, A. Wagner, R. Weiser "Die Umesterung von N-Me-

thyl-α-pyrrolidon und N-Methyl-α-piperidon mit LiAlH4 zu den ω-Methylaminaldehyden" Monatsh. Chem. 1951, 82, 551−559. [S11] F. Galinowsky, G. Bianchetti, O. Vogl "Über die Racemisierung des Isopelletierins und Methylisopelletierins" Monatsh. Chem. 1953, 84, 1221−1227. [S12] G. Habermehl, P. E. Hamman, H. C. Krebs "Naturstoffchemie" 2. Auflage, Springer Verlag Berlin2002 , S. 158.

12 13 Supporting information Pseudopelletierine

[S13] H. Takahata, M. Kubota, S. Takahashi, T. Momose "A New Asymetric Entry to 2-Substituted Piperidines. A Concise Synthesis of (+)-Coni- ine, (−)-Pelletierine, (+)-δ-Coniceine, and (+)-Epidihydropinidine" Tetrahedron Asymmetry, 1996, 7, 3047−3054. [S14] E. C. Carlson, L. K. Rathbone, H. Yang, N. D. Collett, R. G. Carter "Improved Protocol for Asymmetric Intramolecular Heteroatom Mi- chael Addition Using Organocatalysis: Enantioselective Synthesis of Homoproline, Pelletierine, and Homopipeolic Acid" J. Org. Chem. 2008, 73, 5155−5158. [S15] W.-H. Chion, G.-T. Chen, C.-L. Kao, Y.-K. Gao "Synthesis of (−)-Pelletierine and (−)-Homopipeolic Acid" Org: Biomol. Chem. 2012, 10, 2510−2520. [S16] R. Seupel, A Roth, K. Steinke, D. Sicker, H.-U.-Siehl, K.-P. Zeller, S. Berger "Nicht verkorkst, Friedelin aus Kork" Chem. Unserer Zeit 2014, 49, 60–72. [S17] H. Neuhöfer, L. Witte, M. Gorunovic, F.-C. Czygan "Alkaloids in the bark of punica granatum L. (pomegranate) from Jugoslavia" Phar- mazie 1993, 48, 389 – 391.

Questions and Answers A. Eating a pomegranate is not so easy and often ends with coloured stains on the table cloth, shirt or blouse. Suggest how these can be removed. Our suggestion is, to dab the coloured stain with a swab that has been soaked with an aqueous solution of sodium dithionite. The stain disappears immediately. Subsequently rinse with clear water. The strong reducing

agent Na2S2O4 reduces the delphinidin to a colourless leuko compound, which can be washed out with water.

In this way, also other reducible vegetable dyes can be removed. The reduction interrupts the extended, conjugated chromophore system. The leuko compound does nor absorb in the visible range and is therefore colourless.

B. What is to be understood by the term "alkaloid"? Do all alkaloids have a common structural element? Nitrogen containing organic compounds from the secondary metabolism are termed alkaloids, which are produced mainly but not exclusively by plants. They often have a strong physiological effect on humans and animals, which can be used pharmaceutically. A high dose of many alkaloids is toxic or even deadly. The structure

14 15 Supporting information Pseudopelletierine

element common to all alkaloids is at least one N-atom in the structure. It is often part of a heterocycle but this is not compulsory. The apothecary C. F. W. Meißner from Halle coined the term alkaloid in 1819, because the basicity of the first "alkali-similar" phytochemicals (e.g. morphine and strychnine) was a conspicuous property. However, it is not necessary for an alkaloid to be basic (see Question C). To date over 10,000 alkaloids are known.

C. Which alkaloid was the first to be isolated and from what? Give examples for basic and non-basic alkaloids. The first alkaloid was morphine (earlier also called morphium) isolated from the drug opium of the opium poppy. This was first achieved in 1804 by the German pharmacist F. W. Sertürner in Paderborn. For example, nicotine, atropine, coniine, scopolamine, cocaine, quinine, papaverine, codeine, strychnine and the particularly poisonous aconitine are basic. Non-basic alkaloids are e.g. caffeine, colchicine and capsaicin. Advisable to know: basic amino acids and peptides, such as the hormone insulin, are not classified as alkaloids, as they are products of the primary metabolism.

D. Which alkaloid was first isolated on an industrial scale? From what was it isolated and for what was it used? It was quinine from the bark of the cinchona. Several chemists have thereby earned their merits. The first isolation in pure form was performed by F. F. Runge, who was also the first to isolate caffeine. The first isolation on an industrial scale was done by the pharmacist F. Koch, who from 1823 extracted quinine from cinchona bark. From 1824, the famous French chemists Pelletier and Caventou delivered quinine on a large scale for the French colonial troops. The main application for quinine is as an antimalarial drug but it is also used against muscle cramps. Annually about 500 tonnes are obtained for pharmaceutical purposes by extraction. F. Koch himself suffered from malaria but was infected not in the tropics but in Germany in the lower Rhine region. During the Biedermeier time malaria was endemic in that region as well as in Cologne. Koch had an enormous motivation to find an antidote. Recommended reading: S. Streller, K. Roth "Eine Rinde erobert die Welt – Von der Apotheke an die Bar" Chem. Unserer Zeit 2012, 46, 228–247.

E. Which plants gained strategic importance in World War 2, because of the alkaloids it contains? They were plants of the genus cinchona. For the allies obtaining sufficient material from these trees that grow in Middle and South America was of strategic importance, since the war against the Japanese troops in the Pacific was fought in tropical and sub-tropical areas and hundreds of thousands of soldiers needed a stable supply of quinine to combat malaria. This aspect of the use of quinine is now almost entirely forgotten.

F. Explain the yellow colour of cyclooctatetraene. The yellow colour of cyclooctatetraene cannot be explained by the presence of four conjugated double bonds, since their π-orbitals hardly overlap. However, the face to face lying π-orbitals can interact. This leads to a splitting of the π energy levels and a bathochromic shift of the long wave-

length ππ*-transition. ππ* (ethene), λmax = 165

nm, log ε = 4.2; π2π3* (COT), ), λmax = 285 nm, log ε = 2.3. The yellow colour is a result of the

extension of the end absorption of the π2π3*-transition into the visible region. The yellow colour of COT pales on cooling to low temperatures and returns, when it is warmed again. As an explanation for this phenomenon, it is assumed, that the long wavelength end of the absorption spectrum is

14 15 Supporting information Pseudopelletierine

due to transitions from thermally excited vibrational states of COT in the electronic ground state. At higher temperatures, these levels are more densely populated. A. Langseth, S. Brodersen "Temperature Effect on the Absorption Spectrum of Cyclooctatetraene" Acta Chim. Scand. 1949, 3, 778–782.

G. The sharp doublet from H-2/4α is strongly reminiscent of a similarly situated proton of a bicyclic compound discussed in this book. Which one? H-3a of camphor, see Chapter 4.2.

16 16 Colchicine

1.2 Colchicine

Fig. S1.2-1 Structure of colchicine

Biosynthesis of colchicine according to scheme 1 in ref. [1]

Fig. S1.2-2 Biosynthesis of colchicine

17 Supporting information

IR Spectrum in KBr

100 95 90 85 80 75 70 65 60 55

%T 50 45 T% 40 35 30 25 20 15 10 5

4000 3500 3000 2500 2000 1500 1000 500 WavenumbersWavenumbers (cm-1)(cm–1) Fig. S1.2-3 IR spectrum of colchicine The relatively sharp NH-band at 3210 cm–1 is obvious. The CH-stretching region around 3000 cm–1 shows sp2 and sp3 groups. The structure of the bands in the double bond region between 1700 and 1500 cm–1 appears too complex for an exact assignment, however, Siddiqui et al. [14] accom- plished this in a computationally aided analysis of the IR spectrum.

Additional NMR Spectra at 600 MHz in CDCl3 COSY Spectrum 14 13 15 17 18 NH 8 12 11 4 7 5 5 6 6 δH/ppm 6 17 6 55

13 15 18 14 7

4 11 8 12 NH

δH/ppm

Fig. S1.2-4 COSY spectrum of colchicine

The COSY spectrum is a textbook example (Fig. S1.2-4). Beginning with the NH signal at δH = 8.12, we ﬁ nd via

the COSY cross signal the proton H-7 at δH = 4.66. From here the cross peaks lead to one of the protons H-6 at δH= 1.95 in the base of the methyl group signal. The vicinal partners of H-6 and H-6', can be determined from the COSY

spectrum at δH = 2.35. The signal is linked to H-5 and H-5' at δH = 2.39 and δH = 2.56.

18 Colchicine

13C-APT NMR Spectrum 17 15 15 H3C 14 16 H3CO 13 5 6 O 3 4 4a 11 4 7 17 7 N 14 2 1 12b 7a 8 H H3CO 12 13 12a 12 8 H3CO 9 11 10 O 18 OCH3

7a 2 12a 12b 9 16 10 3 1 4a 6 5

δC/ppm

Fig. S1.2-5 APT 13C spectrum of colchicine The APT 13C NMR spectrum is a typical 13C spectrum of an alkaloid, in which all 22 signals are distributed over the entire range of chemical shifts. A direct assignment is only possible for a few resonances. The most deshielded signal at δC = 179.5 can easily be attributed to the α/β-unsaturated C=O carbonyl group of the tropolone ring, followed by the signal of the C=O group of the acetamido group C-16 at δC = 170.1. The individual assignment of almost all other signals depends on correlations from the HSQC and HMBC spectra. Only the resonance of the methyl group

C-17 at δC = 22.8 is obvious.

Questions and Answers A. Why is colchicine not basic? The N-atom is bound in the form of an acetylamino group, i.e. in the form of the amide of a carboxylic acid, which suppresses the basicity of the N-atom, as can be explained by a mesomeric formula. The effect is analogous to the properties of a peptide bond.

B. What is the quintessence of the isolation procedure of basic alkaloids such as nicotine, cytisine or galanthamine? Basic alkaloids such as nicotine, cytisine or galanthamine are isolated by the usual acid extraction process, by which the alkaloid is protonated with a mineral acid and selectively extracted into the aqueous phase from which it can then be released by the addition of a base and extracted into an organic phase.

C. How can the relatively high solubility of colchicine in water be explained, when it contains no functional groups such as OH or COOH? The structural reason is the same as that, which leads to the aqueous solubility of the pentamethoxyﬂ avone sinensetin (cf. chapter 5.1). The six O-atoms in the four methoxy groups of colchicine and the two carbonyl groups are acceptors for hydrogen bonds to the solvent water and the NH-function is a donor for such a hydrogen bond. These multiple hydrogen bonds lead to the solubility in water.

D. What do such structures as benzene, the cyclopentadienyl anion and the tropylium cation have in common? All three satisfy with their cyclic conjugated π-electrons the Hückel rule (number of π-electrons = 2n+1) and are therefore aromatic. 19 Supporting information E. What effect causes the increase in the vicinal coupling constant of unsaturated seven-membered rings compared with aromatic six-membered rings? Compared to a six-membered ring the C-C-H bond angle is larger.

F. Apart from the cleavage of a methyl radical, explained in Fig. 1.2-23, the radical cation with m/z 312 makes • • use of the elimination of H (m/z 311) and CH3O (m/z 281) to lose its radical character. Suggest how this fragmentation can occur. Both processes can be explained by the formation of a methoxy substituted dibenzotropyli- um ion. The formation of m/z 311 can easily be understood. It is more difﬁ cult to ﬁ nd a plausible explanation for the methoxy cleavage to give m/z 281. Since no labelling experiments have been made, it is not known, which methoxy group is involved. Apart from this, the degradation cannot be a simple cleavage of an sp2-C-O bond, because a σ-radical centre would be formed that cannot combine with the already present π-radical. However, a diradical cation would be unfavourable, to be compete with other fragmentation paths of the m/z 312 ion. A solution to this is offered by a hydrogen shift from C-5 to one of the methoxy substituted carbon atoms prior to the fragmentation. A C-O-homolysis can then occur from the then sp3-hybridised C-atom without a problem. If the methoxy group on C-2 is eliminated (which has not yet been proved) the following scheme illustrates the process:

G. Discuss the formation of the ions with m/z 356, m/z 328 and m/z 43 (second most intensive fragment). The acetyl cation (m/z 43) is usually formed by α-cleavage. The formation of m/z 356 could be understood as an analogous process with the retention of charge on the N-atom. However, there are justiﬁ able reasons to doubt this, because a divalent N-cation would be formed. To avoid such a species, a hydrogen transfer from a C-atom to the N-atom might be assumed. An α-cleavage of the rearranged radical cation would then produce a stable iminium ion. The ion with m/z 328 is consequently formed by CO-elimination.

20 Supporting information Capsaicin

1.3 Capsaicin

Fig. S1.3-1 Structure of capsaicin

NMR spectra of dihydrocapsaicin at 400 MHz in CDCl3

Fig. S1.3-2 1H NMR spectrum of dihydrocapsaicin (2)

Fig. S1.3-3 APT 13C spectrum of dihydrocapsaicin (2) 21 21 Supporting information Capsaicin

Assignment table for the 1H NMR δ [ppm] Multiplicity Integral Correlation COSY Assignment signals of dihydrocapsaicin (2) 6,85 d 1 H H-6' (3J = 8,0 Hz) H-2' H-5'

6,80 d 1 H H-5' H-6' (4J = 1,9 Hz) H-2'

6,75 dd 1 H H-2' (4J = 1,9 Hz) H-5' H-6' (3J = 8,0 Hz)

5,76 bs 1 H H-7' H-9'

4,35 d 2 H H-9' H-7'

3,87 s 3 H H-8'

2,20 t 2 H H-3 (3J = 7,6 Hz) H-2

1,64 p 2 H H-2 (3J = 7,2 Hz) H-4 H-3

1,50 sept. 1 H H-7 H-9 (3J = 6,7 Hz) H-8

1,30 m 2 H H-3 H-5 H-4

1,27 m 2 H H-4 H-6 H-5

1,26 m 2 H H-5 H-7 H-6

1,13 m 2 H H-8 H-7

0,85 d 6 H H-8 (3J = 6,7 Hz) H-9

Assignment table for the 13C NMR δ [ppm] Type of Correlation Correlation HMBC Assignment signals of dihydrocapsaicin (2) carbon HSQC

173,2 Cq H-2; H-3; H-7' 1

146,9 Cq H-5'; H-8' 3'

145,3 Cq H-2' ;H-6' 4'

130,5 Cq H-2', H-6'; H-7' 1' 121,0 CH H-6' H-7' 6'

114,5 CH H-5' H-7' 5'

110,9 CH H-2' H-6'; H-7' 2'

56,1 CH3 H-8' H-7' 8'

43,7 CH2 H-7' H-2'; H-6' 7'

39,1 CH2 H-7 H-6; H-9 7

37,0 CH2 H-2 H-3 2

29,8 CH2 H-5 H-6; H-7 5

29,5 CH2 H-4 H-3 4 28,1 CH H-8 H-7; H-9 8

27,4 CH2 H-6 H-5; H-7; H-8 6

26,0 CH2 H-3 H-2; H-4 3

22,8 CH3 H-9 H-7; H-8 9

22 23 Supporting information Capsaicin

Details of the X-Ray Crystallographic Analysis and Crystallographic Data of Capsaicin: CCDC 1033836 contains the supplementary crystallographic data to the X-ray structure analysis. The data can be obtained, free of charge, from the Cambridge Crystallographic Data Centre at www.ccdc.cam.uk/data. Single crystals suitable for the X-ray structure analysis were obtained from the capsaicin (1) recrystallized from MTBE/n-pentane. A crystal was fixed to a glass needle with perfluoroether oil and measured on an Xcaliburdiffrac- tometer (Oxford Instruments) in a stream of cooled nitrogen at 130 K. The structure was solved by direct methods and refined by the F2 full matrix least squares procedure (SHELX 97) (see G. M. Sheldrick,Acta Cryst. 2008, A64, 112–122). All non-hydrogen atoms were anisotropically refined. The calculated positions of hydrogen were refined using the "Riding modus". Exceptions were the H-atoms of the atoms O2 and N1. These positions, which are important for hydrogen bonding, were taken from a difference Fourier function and isotropically refined.

Crystal data and structure refinement for x1909

Identification code x1909

Empirical formula C18H27NO3 Formula weight 305.41 Temperature 130(2) K Wavelength 0.71073 Å Crystal system Monoclinic Space group P2(1)/c Unit cell dimensions a = 12.1733(3) Å a= 90° b = 14.7993(3) Å b= 94.142(2)° c = 9.4782(2) Å g = 90° Volume 1703.10(7) Å3 Z 4 Density (calculated) 1.191 Mg/m3 Absorption coefficient 0.080 mm–1 F(000) 664 Crystal size 0.35 × 0.10 × 0.05 mm3 Theta range for data collection 2.97 to 26.00°. Index ranges –15<=h<=14, –18<=k<=18, –11<=l<=11 Reflections collected 12925 Independent reflections 3345 [R(int) = 0.0348] Completeness to theta = 26.00° 99.9% Absorption correction None Max. and min. transmission 0.9960 and 0.9725 Refinement method Full-matrix least-squares on F2 Data / restraints / parameters 3345 / 0 / 207 Goodness-of-fit on F2 1.043 Final R indices [I>2sigma(I)] R1 = 0.0400, wR2 = 0.0859 R indices (all data) R1 = 0.0595, wR2 = 0.0928 Largest diff. peak and hole 0.223 and –0.197 e.Å-3

22 23 Supporting information Capsaicin

Atomic coordinates ( x 104) and equivalent isotropic displacement parameters (Å2x 103) for x1909 U(eq) is defined as one third of the trace of the orthogonalized Uij tensor.

x y z U(eq) C(1') –1107(1) 3334(1) 9484(1) 20(1) C(1) 1228(1) 2202(1) 10068(1) 18(1) C(2) 2356(1) 2130(1) 10821(2) 22(1) C(2') –371(1) 3928(1) 8924(1) 21(1) C(3') –751(1) 4684(1) 8180(2) 21(1) C(3) 3275(1) 2434(1) 9921(2) 21(1) C(4) 3232(1) 3435(1) 9564(2) 22(1) C(4') –1880(1) 4859(1) 7987(1) 21(1) C(5) 4185(1) 3727(1) 8713(2) 22(1) C(5') –2607(1) 4267(1) 8538(2) 23(1) C(6') –2225(1) 3510(1) 9283(2) 22(1) C(6) 4147(1) 4698(1) 261(2) 23(1) C(7) 4253(1) 4973(1) 6955(2) 24(1) C(7') –724(1) 2487(1) 10283(2) 24(1) C(8) 4284(1) 5923(1) 6401(2) 26(1) C(9) 4054(2) 6634(1) 7495(2) 36(1) C(10) 3503(1) 6024(1) 5079(2) 32(1) C(8') 1048(1) 5148(1) 7652(2) 29(1) N(1) 423(1) 2494(1) 10829(1) 21(1) O(1) 1054(1) 1977(1) 8812(1) 23(1) O(3) –2278(1) 5603(1) 7265(1) 26(1) O(2) –103(1) 5317(1) 7594(1) 27(1)

24 25 Supporting information Capsaicin

Bond lengths [Å] and angles [°] for x1909 C(1')-C(6') 1.385(2) C(9)-H(9C) 0.9800 C(1')-C(2') 1.3883(19) C(10)-H(10A) 0.9800 C(1')-C(7') 1.5198(19) C(10)-H(10B) 0.9800 C(1)-O(1) 1.2403(16) C(10)-H(10C) 0.9800 C(1)-N(1) 1.3302(18) C(8')-O(2) 1.4203(17) C(1)-C(2) 1.5049(19) C(8')-H(8'1) 0.9800 C(2)-C(3) 1.5221(19) C(8')-H(8'3) 0.9800 C(2)-H(2A) 0.9900 C(8')-H(8'2) 0.9800 C(2)-H(2B) 0.9900 N(1)-H(1N) 0.854(16) C(2')-C(3') 1.385(2) O(3)-H(3O) 0.89(2) C(2')-H(2') 0.9500 C(6')-C(1')-C(2') 118.90(13) C(3')-O(2) 1.3675(17) C(6')-C(1')-C(7') 119.11(12) C(3')-C(4') 1.398(2) C(2')-C(1')-C(7') 121.97(13) C(3)-C(4) .5192(19) O(1)-C(1)-N(1) 121.73(13) C(3)-H(3A) .9900 O(1)-C(1)-C(2) 121.55(13) C(3)-H(3B) 0.9900 N(1)-C(1)-C(2) 116.70(12) C(4)-C(5) 1.5231(19) C(1)-C(2)-C(3) 113.29(11) C(4)-H(4A) 0.9900 C(1)-C(2)-H(2A) 108.9 C(4)-H(4B) 0.9900 C(3)-C(2)-H(2A) 108.9 C(4')-O(3) 1.3660(17) C(1)-C(2)-H(2B) 108.9 C(4')-C(5') 1.375(2) C(3)-C(2)-H(2B) 108.9 C(5)-C(6) 1.498(2) H(2A)-C(2)-H(2B) 107.7 C(5)-H(5A) 0.9900 C(3')-C(2')-C(1') 120.38(13) C(5)-H(5B) 0.9900 C(3')-C(2')-H(2') 119.8 C(5')-C(6') 1.388(2) C(1')-C(2')-H(2') 119.8 C(5')-H(5') 0.9500 O(2)-C(3')-C(2') 125.39(13) C(6')-H(6') 0.9500 O(2)-C(3')-C(4') 114.23(12) C(6)-C(7) 1.319(2) C(2')-C(3')-C(4') 120.37(13) C(6)-H(6) 0.9500 C(4)-C(3)-C(2) 113.48(12) C(7)-C(8) 1.503(2) C(4)-C(3)-H(3A) 108.9 C(7)-H(7) 0.9500 C(2)-C(3)-H(3A) 108.9 C(7')-N(1) 1.4536(18) C(4)-C(3)-H(3B) 108.9 C(7')-H(7'1) 0.9900 C(2)-C(3)-H(3B) 108.9 C(7')-H(7'2) 0.9900 H(3A)-C(3)-H(3B) 107.7 C(8)-C(9) 1.517(2) C(3)-C(4)-C(5) 112.32(12) C(8)-C(10) 1.525(2) C(3)-C(4)-H(4A) 109.1 C(8)-H(8) 1.0000 C(5)-C(4)-H(4A) 109.1 C(9)-H(9A) 0.9800 C(3)-C(4)-H(4B) 109.1 C(9)-H(9B) 0.9800 C(5)-C(4)-H(4B) 109.1

24 25 Supporting information Capsaicin

C(1')-C(6') 1.385(2) C(9)-H(9C) 0.9800 C(1')-C(2') 1.3883(19) C(10)-H(10A) 0.9800 C(1')-C(7') 1.5198(19) C(10)-H(10B) 0.9800 C(1)-O(1) 1.2403(16) C(10)-H(10C) 0.9800 C(1)-N(1) 1.3302(18) C(8')-O(2) 1.4203(17) C(1)-C(2) 1.5049(19) C(8')-H(8'1) 0.9800 C(2)-C(3) 1.5221(19) C(8')-H(8'3) 0.9800 C(2)-H(2A) 0.9900 C(8')-H(8'2) 0.9800 C(2)-H(2B) 0.9900 N(1)-H(1N) 0.854(16) C(2')-C(3') 1.385(2) O(3)-H(3O) 0.89(2) C(2')-H(2') 0.9500 C(6')-C(1')-C(2') 118.90(13) C(3')-O(2) 1.3675(17) C(6')-C(1')-C(7') 119.11(12) C(3')-C(4') 1.398(2) C(2')-C(1')-C(7') 121.97(13) C(3)-C(4) .5192(19) O(1)-C(1)-N(1) 121.73(13) C(3)-H(3A) .9900 O(1)-C(1)-C(2) 121.55(13) C(3)-H(3B) 0.9900 N(1)-C(1)-C(2) 116.70(12) C(4)-C(5) 1.5231(19) C(1)-C(2)-C(3) 113.29(11) C(4)-H(4A) 0.9900 C(1)-C(2)-H(2A) 108.9 C(4)-H(4B) 0.9900 C(3)-C(2)-H(2A) 108.9 C(4')-O(3) 1.3660(17) C(1)-C(2)-H(2B) 108.9 C(4')-C(5') 1.375(2) C(3)-C(2)-H(2B) 108.9 C(5)-C(6) 1.498(2) H(2A)-C(2)-H(2B) 107.7 C(5)-H(5A) 0.9900 C(3')-C(2')-C(1') 120.38(13) C(5)-H(5B) 0.9900 C(3')-C(2')-H(2') 119.8 C(5')-C(6') 1.388(2) C(1')-C(2')-H(2') 119.8 C(5')-H(5') 0.9500 O(2)-C(3')-C(2') 125.39(13) C(6')-H(6') 0.9500 O(2)-C(3')-C(4') 114.23(12) C(6)-C(7) 1.319(2) C(2')-C(3')-C(4') 120.37(13) C(6)-H(6) 0.9500 C(4)-C(3)-C(2) 113.48(12) C(7)-C(8) 1.503(2) C(4)-C(3)-H(3A) 108.9 C(7)-H(7) 0.9500 C(2)-C(3)-H(3A) 108.9 C(7')-N(1) 1.4536(18) C(4)-C(3)-H(3B) 108.9 C(7')-H(7'1) 0.9900 C(2)-C(3)-H(3B) 108.9 C(7')-H(7'2) 0.9900 H(3A)-C(3)-H(3B) 107.7 C(8)-C(9) 1.517(2) C(3)-C(4)-C(5) 112.32(12) C(8)-C(10) 1.525(2) C(3)-C(4)-H(4A) 109.1 C(8)-H(8) 1.0000 C(5)-C(4)-H(4A) 109.1 C(9)-H(9A) 0.9800 C(3)-C(4)-H(4B) 109.1 C(9)-H(9B) 0.9800 C(5)-C(4)-H(4B) 109.1

26 27 Supporting information Capsaicin

H(4A)-C(4)-H(4B) 107.9 C(8)-C(10)-H(10B) 109.5 O(3)-C(4')-C(5') 119.25(13) H(10A)-C(10)-H(10B) 109.5 O(3)-C(4')-C(3') 121.65(13) C(8)-C(10)-H(10C) 109.5 C(5')-C(4')-C(3') 119.10(13) H(10A)-C(10)-H(10C) 109.5 C(6)-C(5)-C(4) 114.47(12) H(10B)-C(10)-H(10C) 109.5 C(6)-C(5)-H(5A) 108.6 O(2)-C(8')-H(8'1) 109.5 C(4)-C(5)-H(5A) 108.6 O(2)-C(8')-H(8'3) 109.5 C(6)-C(5)-H(5B) 108.6 H(8'1)-C(8')-H(8'3) 109.5 C(4)-C(5)-H(5B) 108.6 O(2)-C(8')-H(8'2) 109.5 H(5A)-C(5)-H(5B) 107.6 H(8'1)-C(8')-H(8'2) 109.5 C(4')-C(5')-C(6') 120.44(13) H(8'3)-C(8')-H(8'2) 109.5 C(4')-C(5')-H(5') 119.8 C(1)-N(1)-C(7') 122.13(13) C(6')-C(5')-H(5') 119.8 C(1)-N(1)-H(1N) 119.4(10) C(1')-C(6')-C(5') 120.80(13) C(7')-N(1)-H(1N) 118.4(10) C(1')-C(6')-H(6') 119.6 C(4')-O(3)-H(3O) 110.2(14) C(5')-C(6')-H(6') 119.6 C(3')-O(2)-C(8') 117.36(11) C(7)-C(6)-C(5) 124.13(14) C(7)-C(6)-H(6) 117.9 C(5)-C(6)-H(6) 117.9 C(6)-C(7)-C(8) 128.60(14) C(6)-C(7)-H(7) 115.7 C(8)-C(7)-H(7) 115.7 N(1)-C(7')-C(1') 115.10(12) N(1)-C(7')-H(7'1) 108.5 C(1')-C(7')-H(7'1) 108.5 N(1)-C(7')-H(7'2) 108.5 C(1')-C(7')-H(7'2) 108.5 H(7'1)-C(7')-H(7'2) 107.5 C(7)-C(8)-C(9) 113.59(12) C(7)-C(8)-C(10) 110.37(12) C(9)-C(8)-C(10) 111.02(13) C(7)-C(8)-H(8) 107.2 C(9)-C(8)-H(8) 107.2 C(10)-C(8)-H(8) 107.2 C(8)-C(9)-H(9A) 109.5 C(8)-C(9)-H(9B) 109.5 H(9A)-C(9)-H(9B) 109.5 C(8)-C(9)-H(9C) 109.5 H(9A)-C(9)-H(9C) 109.5 H(9B)-C(9)-H(9C) 109.5 C(8)-C(10)-H(10A) 109.5

26 27 Supporting information Capsaicin

Anisotropic displacement parameters (Å2x 103) for x1909 The anisotropic displacement factor exponent takes the form: –2p2[ h2a*2U11 + ... + 2 h k a* b* U12 ]

U11 U22 U33 U23 U13 U12 C(1') 23(1) 20(1) 18(1) –2(1) 2(1) –1(1) C(1) 23(1) 13(1) 19(1) 2(1) 2(1) –1(1) C(2) 23(1) 22(1) 20(1) 3(1) –1(1) 0(1) C(2') 18(1) 24(1) 22(1) 0(1) 2(1) 2(1) C(3') 21(1) 20(1) 22(1) –2(1) 4(1) –3(1) C(3) 21(1) 21(1) 20(1) 1(1) 0(1) 1(1) C(4) 24(1) 21(1) 21(1) 0(1) 2(1) 1(1) C(4') 22(1) 21(1) 19(1) –1(1) 0(1) 1(1) C(5) 22(1) 24(1) 20(1) 0(1) 0(1) –1(1) C(5') 17(1) 29(1) 23(1) 0(1) 0(1) –1(1) C(6') 21(1) 24(1) 21(1) 0(1) 2(1) –5(1) C(6) 23(1) 24(1) 22(1) –2(1) 0(1) –4(1) C(7) 21(1) 27(1) 23(1) –1(1) 2(1) –2(1) C(7') 20(1) 24(1) 27(1) 3(1) 4(1) –1(1) C(8) 26(1) 29(1) 24(1) 5(1) 3(1) –4(1) C(9) 55(1) 25(1) 27(1) 1(1) –4(1) –9(1) C(10) 39(1) 33(1) 24(1) 1(1) 1(1) 3(1) C(8') 21(1) 28(1) 40(1) 6(1) 10(1) –1(1) N(1) 23(1) 24(1) 17(1) 0(1) 0(1) 0(1) O(1) 26(1) 24(1) 18(1) –2(1) 0(1) 1(1) O(3) 22(1) 25(1) 31(1) 7(1) 0(1) 3(1) O(2) 20(1) 24(1) 38(1) 10(1) 6(1) 0(1)

28 29 Supporting information Capsaicin

Hydrogen coordinates (x104) and isotropic displacement pa- Torsion angles [°] for x1909 rameters (Å2x103) for x1909

x y z U(eq) O(1)-C(1)-C(2)-C(3) 41.78(17) H(2A) 2489 1494 11110 26 N(1)-C(1)-C(2)-C(3) –140.16(13) H(2B) 2379 2503 11690 26 C(6')-C(1')-C(2')-C(3') 0.2(2) H(2') 398 3815 9051 25 C(7')-C(1')-C(2')-C(3') 178.85(13) H(3A) 3993 2297 10433 25 C(1')-C(2')-C(3')-O(2) 179.40(13) H(3B) 3228 2083 9030 25 C(1')-C(2')-C(3')-C(4') 0.1(2) H(4A) 3252 3789 10452 27 C(1)-C(2)-C(3)-C(4) 65.63(16) H(4B) 2528 3570 9016 27 C(2)-C(3)-C(4)-C(5) 177.79(12) H(5A) 4884 3619 9288 27 O(2)-C(3')-C(4')-O(3) 0.25(19) H(5B) 4190 3342 7858 27 C(2')-C(3')-C(4')-O(3) 179.61(13) H(5') –3377 4378 8407 28 O(2)-C(3')-C(4')-C(5') –179.79(12) H(6') –2736 3107 9661 27 C(2')-C(3')-C(4')-C(5') –0.4(2) H(6) 4040 5143 8961 27 C(3)-C(4)-C(5)-C(6) 176.69(12) H(7) 4318 4509 6274 28 O(3)-C(4')-C(5')-C(6') –179.59(13) H(7'1) –1191 2401 11086 29 C(3')-C(4')-C(5')-C(6') 0.4(2) H(7'2) –844 1961 9645 29 C(2')-C(1')-C(6')-C(5') –0.2(2) H(8) 5047 6033 6112 32 C(7')-C(1')-C(6')-C(5') –178.87(13) H(9A) 4089 7236 7069 54 C(4')-C(5')-C(6')-C(1') –0.2(2) H(9B) 4605 6589 8297 54 C(4)-C(5)-C(6)-C(7) –129.86(15) H(9C) 3318 6537 7823 54 C(5)-C(6)-C(7)-C(8) –176.67(14) H(10A) 3531 6646 4731 48 C(6')-C(1')-C(7')-N(1) –161.50(13) H(10B) 2750 5881 5310 48 C(2')-C(1')-C(7')-N(1) 19.8(2) H(10C) 3725 5608 4347 48 C(6)-C(7)-C(8)-C(9) –6.2(2) H(8'1) 1417 5646 7197 44 C(6)-C(7)-C(8)-C(10) –131.66(17) H(8'3) 1184 4582 7156 44 O(1)-C(1)-N(1)-C(7') 5.9(2) H(8'2) 1335 5098 8641 44 C(2)-C(1)-N(1)-C(7') H(1N) 574(12) 2661(10) 11683(18) 23(4) –172.15(12) H(3O) –1727(18) 5935(15) 6990(20) 65(7) C(1')-C(7')-N(1)-C(1) –88.74(16) C(2')-C(3')-O(2)-C(8') 5.5(2) C(4')-C(3')-O(2)-C(8') –175.22(12)

Hydrogen bonds for x1909 [Å and °]

N(1)-H(1N)...O(1)#1 0.854(16) 2.128(17) .9803(16) 176.0(15) O(3)-H(3O)...O(1)#2 0.89(2) 1.93(2) 2.7603(14) 156(2) O(3)-H(3O)...O(2) 0.89(2) 2.22(2) 2.6772(15) 112.1(17)

Symmetry transformations used to generate equivalent atoms: #1 x,–y+1/2,z+1/2 #2 –x,y+1/2,–z+3/2

28 29 Supporting information Capsaicin

Further comments on the EI-mass spectrum

Fig. S1.3-5 Formation of the ions with m/z 268, 248, 208 and 206 by degradation of the carbon chain of capsaicin

• • • The cleavage of C3H7 , C4H9 and C7H13 radicals cannot be understood, if the molecular ion does not rearrange. The following suggestion is based on the radical cation "A" that is formed in the first step of the McLafferty rearrangement (Fig. 1.3-30). As a homoallyl radical, "A" can equilibrate with the cyclopropylmethyl radical "B" and the isomeric homoallyl radical "C". Over all, it is a radical induced 1.2-vinyl-shift, which in contrast to the analogous 1.2-H-shift and 1.2-alkyl-shift is well known. From the equilibrium by way of a radical H-shift the radical cations "D" and "E" are formed. As shown in Fig. S1.3-5, the fragments with m/z 262 and 248 can be derived from "D" and m/z 208 from "E".

It is not easy to find a plausible explanation for the appearance of m/z 206. An elimination of 2H from m/z 208 comes into question, however, there is no reason for it in the ion that is directly formed from "E". This ion unites in itself a nucleophilic aromatic part with an electrophilic Michael acceptor. Our attempt to explain the formation of m/z 206 starts with an intramolecular Michael addition. The adduct that is formed has a good reason to eliminate

H2, because by rearomatisation the benzyl cation with m/z 206 can be formed.

Questions and Answers A. Why are the capsaicinoids not basic alkaloids? The N-atom of the amide substructure in capsaicin is not basic, because the n-electron pair on the nitrogen atom is involved in the delocalisation of the n- and π-electrons in a common molecular orbital of the three atoms of the amide function (N-C=O) and is therefore not available for acid-base reactions. The delocalisation of electrons over more than 2 atoms is referred to as mesomerism or resonance. In the MO theory this is represented by molecular orbitals. In valence bond (VB) theory mesomerism is illustrated by VB resonance structures drawn as localized Lewis structures separated by the resonance arrow.

30 31 Supporting information Capsaicin

The delocalisation of the π- and n-electrons leads to π-electron density in the C-N-bond and therefore to an increased barrier of rotation around this bond and, corresponding to their different electronegativities, to a partial negative charge on the O-atom and a partial positive charge on the N-atom.

B. Apart from capsicum species, such as chilli, pepper and ginger are regarded as further pungent spices. The question arises if there is possibly a common structural element in those components that are found to be pungent. This can actually be recognized, if the structures of capsaicin, piperine (from pepper), gingerol (from ginger) and shogaol (a product of dehydration of gingerol) are compared. Find the structures of the first three named substances in the literature and name the common feature.

The common structural element is the phenyl ring substituted in the 3 and 4 position with O-substituents, whereby the O-atoms can belong to a 1,3-dioxolane ring, which is just as typical for natural products as OH or

OCH3 groups. All molecules possess a hydrophobic hydrocarbon chain. It can be saturated or unsaturated, the amide function is not necessary and it can be differently positioned, as a comparison of piperine and capsaicin shows. Further functions are possible as demonstrated by the components of ginger. Only the substituted arene and the hydrophobic hydrocarbon chain seem to be essential. This apparently allows all these substances to attach to the same receptor.

C. What is to be understood by the term oleoresin? From what and how are oleoresins obtained and for what are they used? An oleoresin is an extract that is obtained from vegetable materials. Oleoresins can be used to colour or flavour foods or as an additive in other products such as ointments. Organic solvents are used for the extraction. These can be non-polar, such as n-hexane or polar, such as methanol or ethanol. Overcritical carbon dioxide can be used in some cases. Oleo refers to the essential oils, which are often present, resin to the smeary, pasty part, which can consist of waxes, fats, non-polar dyes and similar things. Typical raw materials for the production of oleoresins are for example diverse species of capsicum (for colour and pungency), pepper, nutmeg and its flower (macis), ginger, coriander and cardamom.

D. The NOESY spectrum shows the diagonal signals in black and the NOE cross signals in red, to demonstrate the different phases. The exchange signals between the NH-group and the OH-group have the same phase as the diagonal signals. Why?

30 31 Supporting information Capsaicin

NOESY cross signals arise from dipolar relaxation of different spins through space. Their Lamor frequency remains unchanged, only the intensity is altered. During an EXSY experiment, which is formally identical to the NOESY experiment, the spins of dynamically exchanging atoms lose their identity as a result of the exchange. In a system of two spins, spin 1 is averaged with spin 2 and vice versa. The cross signals therefore have the same phase as the diagonal signals. The rate of exchange and many other things make the conditions more complicated (see A. D. Bain "Chemical Exchange in NMR" Progr. NMR Spectrosc. 2003, 43, 63–103).

E. Of which named reaction in organic chemistry does the suggested explanation for the formation of the ion with m/z 137 in Fig. 1.3-25 remind you? In the hydroxamic acid-isocyanate degradation, the so-called Lossen rearrangement, the formation of an acyl- nitrenium ion is avoided by the migration of the residue R. An isocyanate is formed:

+• In the case of the M -ion of capsaicin the role of the leaving group H2O is taken over by the 4-hydroxy-3-methoxybenzyl radical.

• F. An ion with m/z 122 that can only be explained by the elimination of CH3 from m/z 137 can be found 15 a.m.u. below the signal with m/z 137. What is the problem? The 4-hydroxy-3-methoxybenzyl radical (m/z 137) is an ion with an even number of electrons that should be maintained during fragmentation reactions, so that no electron pair needs to be separated (Even Electron Rule). Exceptions to this rule are possible, e.g. when by the elimination of a radical a particularly stable ionic species is formed. This could be the case here.

32 32 Thymoquinone

Chapter 2 2.1 Thymoquinone

Fig. S2.1-1 Structure of thymoquinone

Comment: An alternative process of purification is also conceivable, by which the material after the 2nd chromatographic step is dried for a short time under oil pump vacuum. This is followed by vacuum sublimation under membrane pump vacuum in a water-cooled vacuum sublimator (15 mbar pressure, gentle external heating). Preparation of the Semiquinone Anion Radical

1. DMF (50 mL) dried over molecular sieve is filled into a N2-flask fitted

with a rubber septum and saturated with N2 for 30 minutes

2. Thymoquinone (20 mg) or KO2 (20 mg) is each given into one of two

N2-flasks (25 mL) fitted with rubber septa. Using a syringe2 N -saturat-

ed DMF (5 mL) is filled into each of the flasks, whereby the KO2 does not quite dissolve. 3. An aliquot (0.5 mL) of each of the solutions prepared under 2. are

injected together into a N2-flask (10 mL) with rubber septum. After 5 minutes a green solution forms.

4. The reaction solution (0.1 mL) is injected under N2 into an ESR tube and the spectrum immediately recorded.

Fast Atom Bombardment Mass Spectra The importance of redox processes for thymochinone is apparent in the negative and positive FAB mass spectra. In the FAB ion source negative ions are formed usually by deprotonation and the positive ions by the addition of a proton. Unintentional traces of metal cations can also lead to the appearance of [M+Na]+ and similar quasi-molecular ions. A deprotonation of thymoquinone can be discounted. Instead, the bombardment with fast xenon atoms using either a 3-nitrobenzyl alcohol or a glycerine matrix leads to electron capture resulting in the formation of the anion radical M−• (m/z 164, see Fig. S2.1-1). Interestingly, using the same matrices and the mass spectrometer in the positive mode the corresponding cation radical M+• (m/z 164) appears with a strong intensity. Also to be found are all the fragments that are known from the EI mass spectrum that can be MF 33 Supporting information

derived from the M+• of thymoquinone (see Fig. S2.1-2). Not surprisingly the proton addition ion [M+H]+ (m/z 165) can be found as well.

Fig. S2.1-2 FAB mass spectrum of thymoquinone in negative mode

Fig. S2.1-3 FAB mass spectrum of thymoquinone in positive mode

34 Thymoquinone

On bombardment with Xe-atoms thymoquinone forms M−• ions (electron capture, reduction) as well as M+• ions (electron donation, oxidation). The first thought might be, that these are processes involving electron exchange between thymoquinone and the molecules of the matrix. In principle this could be true when 3-nitro-benzyl alcohol is the matrix, since this molecule is able to donate electrons (formation of M−• from thymoquinone) as well as accept electrons (formation of M+• from thymoquinone) and thereby to be converted to the radical ion A or B respectively. The formation of a radical ion from glycerine can be discounted, so that contrary to the experimental finding with this matrix no M+• ion should be formed by electron transfer from thymoquinone to the matrix. From the fact, that thymoquinone shows exactly the same behaviour in both matrices, it must be concluded, that the electron exchange occurs between two mole- Fig. S2.1-4 Matrix radicals from 3-nitro- cules of thymoquinone (Fig. 2.1-5) benzyl alcohol

Fig. S2.1-5 Ion formation from thymoquinone in the FAB ion source

Questions and Answers

A. Why is thymoquinone yellow? Thymoquinone contains two C,C and two C,O double bonds that are conjugated with each other, so that a new chromophore is formed. Neither an isolated C,C double bond nor a carbonyl group alone absorb radiation in the visible range. The π and π* orbitals combine by conjugation to give a new set of π/π* orbitals, in which the highest π orbital is higher and the lowest π* orbital is lower in energy. This causes a bathochromic shift of the π→π* transition. The n(p) orbitals of the O-atoms are unaffected by the conjugation. However, because of the lowering of the π* orbital the "forbidden" n→π* transition also experiences a bathochromic shift and occurs in the visible region. Despite its small extinction coefficient this transition is responsible for the yellow colour of the quinone. To see this transition in the UV-Vis spectrum, ε must be plotted on a logarithmic scale. The combination of the π/π* orbitals of two neighbouring carbonyl groups is sufficient, to lead to a yellow colour for the n→π* transition, as can be seen for diacetyl (butan-2,3-dione), the yellow liquid flavouring of butter.

34 35 Supporting information

B. What is the origin of the name of the class of substances known as quinones? The name quinone is derived from quinic acid, which is an important intermediate in the biosynthesis of aromatics via the shikimate pathway. It is responsible for example for the aromatic amino acids phenylalanine, tryptophan and tyrosine. Quinones are the oxidation products of aromatics. Quinic acid itself can be oxidized to 1,4-benzoquinone, which is not aromatic but fully conjugated. As in benzene all 6 C-atoms are sp2 hybridized. Quinoidal systems can exist in a para or ortho form, i.e. analogue to either 1,4- or 1,2-benzoquinone.

The isolation and spectroscopy of the quinone lawsone from the leaves of henna (chap. 3.1) and of shikimic acid from star anise (chap. 6.1) have been described in "Classics in Spectroscopy – Isolation and Structure Elucidation of Natural Products" (S. Berger, D. Sicker, Wiley-VCH 2009).

C. Both substances of the redox system p-benzoquinone – hydroquinone have an electronic stabilisation. How does this occur? In this conjugated redox system p-benzoquinone is the oxidant and hydroquinone the reductant. Hydroquinone as an aromatic substance possesses a considerable energetic stabilization that is expressed by the term "resonance energy". The structural reason is the cyclic conjugated aromatic system, which contains six π-electrons and obeys the Hückel Rule. For p-benzoquinone this is not the case. However, the conjugation of both C=C bonds with both carbonyl groups has a stabilizing effect (cf. Question A)

D. Usually in an azeotropic distillation in the laboratory an entrainer is used to remove water and not as in the case described here, that water itself is used as the entrainer. What is the name of the apparatus used? Give an example of an entrainer and examples of its use. The laboratory apparatus is called a water separator. There are two types depending upon if the entrainer has a higher or lower density than water and thus if in the condensed azeotrope water forms the upper or lower layer. Toluene as a typical entrainer requires a different water separator than chloroform. Water separators are for example suitable to dry solvents such as toluene or chloroform to a certain residual water content. The solvent is distilled until no more water separates in the water separator. This fraction is discarded. The solvent remaining after the distillation is not totally free of water but has a low water content. However, a typical use of a water separator in the laboratory is in equilibrium reactions such as esterifications and condensations, in which the equilibrium can be shifted entirely to the side of the products, if one of the products, namely water, is continuously removed.

E. Which of the following substances are steam volatile and which are not? Amygdalin, anethole, camphor, carvone, glucose, hesperidin, lactose, limonene, menthol, 2-nitrophenol, 4-nitrophenol, patchouli alcohol, raffinose, saccharose, cinnamaldehyde. Give reasons for your answer. Steam volatile are: anethole, camphor, carvone, limonene, menthol, 2-nitrophenol, patchouli alcohol, and cinnamaldehyde. All of these compounds are hydrophobic and have a sufficiently high vapour pressure, i.e. a sufficiently low molecular mass. They therefore fulfil the requirements explained in the text. Not steam volatile are: amygdalin, glucose, hesperidin, lactose, raffinose, saccharose and 4-nitrophenol. All these substances are so hydrophilic that the water as the solvent forms hydrogen bonds to the OH-groups of the carbohydrates and retains them in the aqueous phase. This applies to the pure carbohydrates such as the monosaccharide glucose, the disaccharides lactose and saccharose and the trisaccharide raffinose as well as to the glycosides amygdalin

36 Thymoquinone

and hesperidin (cf. "Classics in Spectroscopy – Isolation and Structure Elucidation of Natural Products" (S. Berger, D. Sicker, Wiley-VCH 2009) chap. 4.3 and 4.4), in which each aglycon is bound by an O-glycosidic bond to a disaccharide. An interesting case is the comparison of the two isomers of nitrophenol. In 2-nitrophenol the O-atom of the

OH-group and one O-atom of the nitro group (NO2) are linked by an intramolecular hydrogen bond. This intramolecular hydrogen bond prevents the OH-group from hydrogen bonding to water, as is possible for 4-nitrophenol. In comparison to 4-nitrophenol 2-nitrophenol is non-polar and hydrophobic. 4-Nitrophenol is therefore hydrophilic and not steam volatile. In contrast 2-nitrophenol is sufficiently hydrophobic to be steam volatile. The effect of the different polarity of both molecules can be seen clearly by TLC on silica gel 60.

2-Nitrophenol is the less polar molecule and has a higher Rf-value than 4-nitrophenol. The structural difference can be used for a selective separation of 2-nitrophenol and substituted analogues that share this structure from mixtures with other nitrophenols. This applies to 5-methoxy-2-nitrophenol, 2-nitroresorcinol or 3-nitro- catechol.

F. Why are ortho-benzoquinones always darker coloured than para-benzoquinones? Ortho-benzoquinones in contrast to para-benzoquinones have a longer π-conjugation system, whereas the para-benzoquinones are cross-conjugated. Therefore, the ortho-benzoquinones absorb at a longer wavelength.

G. In a high resolution scan the following exact masses were recorded: m/z 164.085, 136.082 and 108.092. With-

in experimental error these agree with the molecular formulae C10H12O2 (164.084), C9H12O (136.088) and

C8H12 (108.094). What structural information can you derive from this? The number of double bond equivalents can be calculated from the molecular formula of the molecular ion. This means the number of double bonds (degree of unsaturation) and rings that are present in the molecular structure. In a saturated, open chain hydrocarbon with n C-atoms the number of H-atoms is 2n+2. This does

not change, if heteroatoms with a valency of 2 are present (e.g. ethane C2H6, ethanol C2H6O). Each ring and

each double bond reduces the number of H-atoms by 2. The double bond equivalents of a compound CnHmOx is therefore given by: r + d = (2n + 2 – m)/2 r = number of rings, d = number of double bonds, n = number of C-atoms, m = number of H-atoms

For C10H12O2 five double bond equivalents can be calculated (to calculate r + d when other elements with other valencies are present see Literature [18] p. 305). From the loss in mass it follows that twice CO (Δm = 27.995)

and not perhaps C2H4 (Δm = 28.038) has been eliminated. Under consideration of the colour of the analyte this is already a good indication, that it could be a benzoquinone (one ring, 4 double bonds, 2 carbonyl groups, yellow).

H. Is it absolutely necessary, that in each of the two decarbonylation steps induced by electron impact (Fig. 2.1- 27) a new σ-bond is formed?

+ + The formation of [M-2CO] and [M-2CO-2CH3] ions merely proves, that a new σ-bond has been formed. If the formation of a new C-C-bond did not occur in one of the two steps, then [M-2CO]+ ions with the structure A or B would be formed, depending on if the missing bond formation occurred during the elimination of C1 or C-2 as CO. However, in these ions the charge and radical would be localised in orbitals, whose direction is

36 37 Supporting information

orthogonal to the π-system. In the formulation shown in Fig. 2.1-27 the charge and radical belong to a delocalized π-system. Therefore in the literature about the mass spectrometry of quinones the [M-2CO]+ ion is described by the formula C or D, which should both express the same thing.

I. Make a mechanistic proposal for the formation of m/z 68 and 96 from m/z 136 (Fig. 2.1-28).

38 Supporting information Berberine chloride

2.2 Berberine chloride

Fig. S2.2-1 Structure of berberine chloride

COSY Spectrum at 400 MHz in DMSO-d6 2 15 16 14 4 8 13 11 12

δH / ppm 6 5 5

16, 15

14 12, 11

δH / ppm

Fig. S2.2-2 COSY spectrum of berberine chloride

ESI(+) Mass Spectrum

Fig. S2.2-3 ESI-MS/MS spectrum of berberine chloride at 46,8 eV

39 39 Supporting information Berberine chloride

The ESI(+) mass spectrum of berberine chloride shows the berberine cation at m/z 336 with a high intensity. No further ions occur. In the MS/MS spectrum (Fig. S2.2-3) recorded with a collision energy of 46.8 eV, one observes the fragmentation ions summarized in Tab. S1, which are found to a large extent in the EI mass spectrum. The determination of high resolution masses leaves no doubt, that the ions with m/z 320 and 306 are formed from m/z 336 by the elimination of methane or ethane respectively. Both ions eliminate CO to give m/z 292 and 278. The fragment at m/z 304 is the result of

the cleavage of CH3OH from m/z 336 and m/z 318 is obviously a product of dehydrogenation from m/z 320. A mechanistic rationalisation of the process described is shown in Fig. S2.2-4 and S2.2-5. If the collision energy is increased to 66.2 eV, the ions with m/z 320, 306 and 292 clearly decrease in intensity, whereas the ions with m/z 318 and 278 become more intense. In addition, fragments with low values of m/z are formed (Fig. S2.2-6).

m/z Molecular Formula Experimental Calculated

336 C20H18NO4 336.1231 336.1230

320 C19H14NO4 320.0921 320.0923

318 C19H12NO4 318.0765 318.0766

306 C18H12NO4 306.0761 306.0765

Tab. S1 Fragment ions from the collision 304 C19H14NO3 304.0972 304.0968 a activation of m/z 336 in the ESI-MS/MS 292 C H NO 292.0972 292.0974 experiment of berberine chloride 18 14 3

278 C17H12NO3 278.0815 278.0817 a Collision energy 46.2 eV

Fig. S2.2-4 Collision induced cleavage of methane from the berberine cation und subsequent processes

40 Supporting information Berberine chloride

Fig. S2.2-5 Collision induced cleavage of methanol and ethane from the berberine cation

Fig. S2.2-6 ESI-MS/MS spectrum of berberine chloride at 66,2 eV

Fig. S2.2-7 Formation of the radical cation m/z 263 by ring closure and elimina- • tion of CH3 from m/z 278 40 41 Supporting information Berberine chloride

+• The ion C16H9NO3 at m/z 263.0580 (calc. 263.0577) deserves particular attention, because of its uneven mass number and the nitrogen rule it must be a radical cation. It can only occur by the cleavage of a radical from an ion of higher mass with an even number of electrons. Its formation is therefore in conflict with the even electron rule, from which exceptions are possible in well-grounded cases. An investigation of the differences in mass make +• the C16H9NO3 -ion with m/z 263.0580 (calc. 263.0577) the probable pre-

cursor, which by elimination of CH3 becomes m/z 263. Fig. S2.2-7 shows a mechanistic proposal that includes as intermediates a phenonium ion and a benzyl cation that after ring closure forms a five membered ring with an

attached CH3 group.

Fig. S2.2-8 ESI-MS/MS spectrum of berberine chloride at 86,2 eV A further increase of the collision energy to 86.2 eV initiates further degra- +• dation steps that cannot all be dealt with here (Fig. S2.2-8). The C14H9N -ion with m/z 191.0731 (calc. 191.0730) is conspicuous, because of its relatively high intensity. Again, it is a radical cation. We assume, that it is formed via +• the C15H9NO2 -ion with m/z 235.0628 (calc. 235.0606) as an intermediate +• from C16H9NO3 by successive elimination of CO and CO2 (Fig. S2.2-9).

Fig. S2.2-9 Proposal for the formation of m/z 191 at higher collision energy

42 Supporting information Berberine chloride

Questions and Answers

A. What is to be understood by the term heterocycles? What in organic chemistry is a heteroatom? In organic chemistry all atoms except C and H are regarded to be heteroatoms. Heterocycles are cyclic chemical compounds with at least two different chemical elements as ring atoms. Simple examples are pyridine, thiophene and furan. In organic chemistry heterocycles contain apart from C at least one further element in the ring.

B. Which structural relationship do the two nudicaulines have to each other, and what is that of their aglycones? The two aglycones are enantiomers, since the mirror images are not superposable. In contrast, the two nudicalines are diastereomers, because the sugar residues are not mirror images, but are the same in both compounds.

C. Which structural element is responsible for the colour of the nudicalines? How many π-electrons are conjugated? What is the strongest chromophore? Are there any auxochromes? The system marked in blue causes the colour. It is conjugated and contains 16 π-electrons. The strongest chromophore is the C=N double bond, the C=C double bonds are weaker chromophores. There is only one auxochrome, which leads to a bathochromic shift (i.e. to longer wavelength) of the absorption maximum. That is the OH-group that is conjugated to the chromophore and increases by the +M-effect the release of electrons into the chromophore system.

D. Consider the systematic name for berberine chloride: 5,6-dihydro-9,10-dimethoxy-benzo[g]-1,3-benzodioxol- o[5,6-a]quinolizinium chloride (1:1). What is the meaning of the expressions in square brackets? How do they come about? This name for berberine chloride is formed according to the so-called Hantzsch-Widman nomenclature for heterocycles . Firstly, the basic component is determined. That is the quinolizinium cation, since it contains an N-atom, which according to these rules takes priority. This substructure is marked red in the figure below. For the basic component there is a mandatory numbering system that is shown here in red, however, it is not related with the final numbering of the whole molecule. The fused heterocyclic component is also numbered according to the convention (shown in black). The atoms common to both rings are defined by numbers (here in black) and letters (here in blue), whereby the sequence of the numbers must correspond in direction to the lettering in the basic component. Consequently, the benzene ring substituted with the two methoxy groups is annulated on the g-side of the quinolizinium ion and therefore becomes benzo[g], while the 1,3-benzodioxol- ane is annulated on the a-side of the quinolizinium ion via its own as 5,6-bond defined edge and is therefore [5,6-a]. Thereby, the direction of viewing should be the same, which is shown in the figure by violet or green pairs of arrows. Finally, the whole molecule is numbered, without regard to the components. The numbering starts at the O-atom with the highest priority and progresses so that the N-atom receives the lowest possible number. At this point, for the first time, it is taken into account, that it is a dihydroquinolizinium ion. As can be seen the 5,6-dihydro prefix has absolutely nothing to do with the 5,6-description in the square brackets! The latter refer only to the annulated side. Furthermore, it is mandatory, to write the letters in the brackets in italic.

42 43 Supporting information Berberine chloride

This seems complicated to you? How do we deal with it in our profession? No university teaches nomenclature in detail. However, there are good books on nomenclature. A well tried method is, to find a concrete problem and ask the question: how does it get that name? That is what I want to know! That is what the author did in this case.

E. Are the two protons on C-2 not diastereotopic and consequently should they not represent an AB-spin system?

The molecule has no C2-axis, therefore the two protons on C-2 in principle have different chemical shifts. Since viewed from C-2 the molecule is symmetric until the C-atoms 5 and 13a, the difference in chemical shift should be very small. In addition, the geminal spin coupling between the protons on C-2 in a compound with a five-membered ring with two oxygen atoms at the 1,3-position is very small. Consequently, in general, one observes singlets in such compounds.

F. Is there an alternative possibility to the path proposed in the main text, for the formation of the most intensive ion (m/z 321) in the EI spectrum of berberine chloride?

In the inlet system a thermally induced SN2 reaction could occur, in which the neutral molecule D would be the leaving group.

44 Supporting information Carminic acid

2.3 Carminic acid

Fig. S2.3-1 Structure of carminic acid

Questions and Answers A. Why is carminic acid chemically so stable, although it is a glycoside? Carminic acid is a C-glucoside and not like most glycosides an O-glycoside. The bond between the aglycone and the saccharide in a C-glycoside is a C-C bond. This bond is comparatively stable and is not sensitive to hydrolysis, as is the C-O bond of an O-glycoside e.g. ruberythric acid. It is possible to boil a solution of carminic acid without decomposition.

B. Why is carminic acid a mordant dye? What is a mordant substance? What is the colouring principle? A solution of a multivalent metal cation (Al3+, Fe3+, Cr3+, Cu2+ inter alia) that can form a stable complex between this central ion, the functional groups of the dye and the fibre is referred to as a mordant. Fibres that are well suited to mordant dyeing are wool and silk, which have strongly nucleophilic groups such as –SH and

–NH2 that can bind to the metal atom. A typical structural element of such dyes is catechol (1,2-dihydroxyben- zene), which is found as a substructure in alizarin and also in carminic acid and many other naturally occurring dyes (e.g. in quercetin found in the skin of onions). The typical procedure for dyeing consists of bating the fibre (e.g. with a solution of alum), steaming and colouring with a solution of the dye. An example of this is the colouring of wool for carpets with red or yellow dyes.

C. Explain why carminic acid is an indicator, whose colour changes between orange (acid) and violet (alkaline). An indicator is a weak organic acid or base, the neutral form of which has a different colour than the ionic form. For carminic acid this means, that the neutral and the protonated form, which exists in acidic media, has an orange-red appearance, whereas the deprotonated anionic form, which exists in alkaline media (as carboxylate or phenolate ion), undergoes a shift to a longer wavelength (red shift). The anions have a stronger electron donor effect on the chromophore system causing a considerable bathochromic shift. Visually the colour is experienced as scarlet to violet. If the system is made acidic the colour change is reversible, as expected for an indicator.

D. Why does the HSQC spectrum make it clear, that carminic acid is not an O-glycoside? In the HSQC spectrum the signal from H-1' at 4.71 ppm correlates with the signal from a carbon atom at 77.2 ppm. The latter cannot therefore be the anomeric signal of an O-glycoside, the characteristic shift for C-1' would be 100 ppm.

E. Which OH-group of the molecule (excluding the carboxylate group) is the most acidic and how can this be determined? The OH-groups of C-3, C-5 and C-8 are possibly bound by intramolecular hydrogen bonds. The OH-group of C-6 remains as a free phenolic function. Such determinations are attainable by the titration of the 13C-spectra.

45 45 Supporting information Carminic acid

F. Why is there no description of an EI mass spectrum given in the foregoing article? A condition for the electron impact ionization, as for all types of ionization that occur in the gas phase, is that a sufficiently high vapour pressure of the intact substance is reached. With very polar molecules or macromolecules this is not possible. In these cases techniques are used, in which ions are transferred directly from the condensed phase (e.g. field desorption, fast atom beam bombardment) or from sprayed solutions (e.g. electrospray ionization) into the gas phase.

G. Why do ESI mass spectra contain so few or as in the case of carminic acid no fragments? This is partly due to the fact that the ions that reach the gas phase have very little excess energy. In the case of molecules in the gas phase that are bombarded with 70 eV this is quite different. More important is the fact that in contrast to electron impact ionization no radical cations M+• are formed. This is because the unpaired electrons are reactive, but "quasi molecular ions", which being ions with a closed shell are intrinsically stable. In the negative mode, in which [M–H]–-ions are formed, we are dealing with for example mesomerism stabilized carboxylate, phenolate and phosphonate ions etc. In the positive mode relatively stable onium ions [M+H]+ are formed.

46 Supporting information Safflomin A

2.4 Safflomin A

Fig. S2.4-1 Structure of safflomin A

Characteristics of Carthamin

IUPAC Name: (2Z,6S)-6-β-D-Glu- copyranosyl-2-[[(3S)-3-β-D-glucopyranosyl-2,3,4-trihydroxy-5-[(2E)-3-(4- hydroxyphenyl)-1-oxo-2-propen-1- yl]-6-oxo-1,4-cyclohexadien-1-yl] methylen]-5,6-dihydroxy-4-[(2E)-3-(4-hydroxyphenyl)-1-oxo-2-propen-1-yl]-4-cy- clohexen-1,3-dion Synonyms: C.I. Natural Red 26, carthamine, safflower red, –1 C43 H42O22 MW 910.78 g×mol Fig. S2.4-2 The since 1979 valid structu- CAS Registry Number: 36338-96-2 ral formula of carthamin as C-glucoside, CAS RN 36338-96-2 About the Attempts to Isolate Pure Carthamin In October 2016, there were no physico-chemical data for the substance cited in Scifinder®. Preliminary Remark to Carthamin – What is the Ma- jor Difference to Safflomin A? Our experiments were based on the already mentioned work of Schlieper and Kuroda. Until 1979 the structural formula of Fig. S2.4-3 was considered to be correct for carthamin. That changed with the work of Obara and Onodera from 1979. From the NMR spectrum of the Ca-salt of carthamin they determined, that there was one missing and one additional signal, which led to the present structural formula (see Fig. S2.4-2) that shows carthamin not as an O-glucoside but as Fig. S2.4-3 Outdated, until 1979 accep- a double C-glucoside. ted formula for carthamin as O-glucoside H. Obara, J. Onodera "Structure of Carthamin" Chem. Lett. 1979, 201–204. What is the driving force behind the quest after pure carthamin? For a long time, there has been a lack of NMR data for carthamin. There have been no 2D NMR measurements. Only a few papers from Obara and Onodera (vide infra) and a paper from Takahashi from 1982 exist, in which NMR spectroscopic investigations of carthamin are reported. The papers show, that the major problem is the poor solubility of carthamin. Obara and

Onodera made measurements on the calcium salt of carthamin in DMSO-d6 and because they assumed, that the false structure in Fig. S2.4-3 was correct, they compared the spectra with the reference compound of Fig. S2.4-4. Fig. S2.4-4: Reference compound of Obara et al. Takahashi (vide infra) used a mixture of pyridine-d5 and ethanol-d6 (95/5) as a solvent and measured at 400 MHz. On the basis of the spectra that they 47 47 Supporting information Safflomin A

obtained, both groups defined the structure in Fig. S2.4-2 as the correct structure of carthamin. However, it must be noted, that the quality of the spectra measured more than 30 years ago do not satisfy today’s standards and that today a complete structural elucidation includes 2D NMR spectra. However, after the problem of the solubility is solved, will it be possible to close this gap, provided pure carthamin is available. There are therefore two problems to solve, purity and measurement. H. Obara, J. Onodera "Carthamin and Isocarthamin" Chem. Lett. 1978, 643– 644. and Y. Takahashi "Constitution of two coloring matters in the flowers Fig. S2.4-5: Commercially available petals of carthamus tinctorius" Tetrahedron Lett. 1982, 23, 5163–5166. safflower Finally, 1996 the following paper appeared in Korean language from authors ofSouth Korea: J. B. Kim, M. H. Cho, T.-R. Hahn, Y.-S. Paik, Young-Sook "Purification and structural identification of carthamin from Carthamus tinctorius" Han'guk Nonghwa Hakhoechi 1996, 39, 501–505. The translated name of the journal is Agricultural Chemistry and Biochem- istry. In this work, there are two lists of peaks for 1H and 13C NMR of carth-

amin measured in DMSO-d6, which confirm the structure shown in Fig. S2.4-2. Unfortunately the spectra are not shown directly, so that an impres- Fig. S2.4-6: Safflower washed with sion of the purity cannot be obtained. The abstract in Scifinder® contains warm water that contains almost no more the description: "After removing the yellow pigments by washing Cartha- safflomin A. The flowers appear to be mus flowers with water and methanol, the red pigment was extd. with 0.5 much redder. M Na2CO3 and pptd. with aq. citrate. The pigment was further purified by cellulose adsorption and Sephadex LH-20 column chromatog. The purified

red pigment was decompd. at about 300°C, and its Rf value on silica gel in

BAW (n-BuOH:HOAc:H2O; 4:1:5) was 0.56." Unfortunately, we could not reproduce the underlined, very general instructions. However, it would surely be worthwhile, to read this publication in English and then to reproduce it. The greatest challenge in the isolation of carthamin is its generally poor solubility. This holds for water and for organic, with water immiscible solvents. In the literature, a poor solubility in water and ethanol is repeatably reported, cf. J. Falbe, U. Zorll, B. Fugmann, Römpp Lexikon, Thieme, Stutt- gart, 1992-1999. Fig. S2.4-7: Cotton wool coloured with carthamin On the other hand, after the deprotonation of carthamin, it shows a very good solubility in aqueous, alkaline media, which was the basis for the initially mentioned isolation by Schlieper, even if it does not yield pure carthamin. The good solubility of the red dye as an anion is confirmed by the intensive coloration of the aqueous, alkaline medium, however, this method has a grave disadvantage. The formation of inorganic salts by dissolution in

bases, especially solutions of Na2CO3, and their subsequent neutralization with acids represents a major obstacle for further processing. The transfer of carthamin to an organic solvent that is immiscible with water, is unfortunately not possible. For that, it is too polar. If processing is continued in aqueous medium, the salts are dragged through every step and finally crys- tallize out with carthamin, without the carthamin being separated. Faced Fig. S2.4-8: after decolouration with soda with this, two fundamental solutions (literally) are apparent. Either the in- troduction of these ions during the extraction must be completely avoided,

48 49 Supporting information Safflomin A

by improving the solubility of carthamin in the pure solvent, or the ions must be removed from the aqueous medium. We recommend avoiding every use of cellulose, e.g. in the form of cotton wool or filter paper, as long as it is not intentionally used as a carrier. These materials are intensively coloured by carthamin, which is not easily released. Of course, this proves that carthamin is known to be a good dye for cotton, but for the isolation, it is a disaster. For the chromatography this means, that only columns with a glass frit and without cotton wool (since it is cellulose) and for filtration, where possible, a fine metal sieve, since filter paper is also cellulose and would bind carthamin, should be used. Cf. Fig S2.4-7 and Fig. S2.4-8. About some experiments to isolate carthamin The material basis was the initially mentioned "fat-free, washed safflower" obtained from the flowers of the safflower after treatment with diethyl ether and much warm water. It is shown in Fig. S2.4-6 and appears to be much redder than the commercially available flowers of safflower. Attempt to isolate carthamin by lyophilisation of a neutralized soda extract In this experiment the procedure of Schlieper (vide supra) was modified. The washed safflower was placed in a glass beaker and so much water add-

ed, that the mixture could be well stirred. Solid Na2CO3 was now added until the pH was about 11, whereby the vegetable material lost its colour and simultaneously the solution became red. Only a small amount of soda was required for this, in contrast to the amount given in the literature. Subsequent- ly the discoloured flowers were pressed in a metal kitchen sieve and the filtrate collected. The orange-red solution obtained was neutralized with very little concentrated hydrochloric acid, which caused a change of the colour to pink. Instead of the usual rotary evaporator, the lyophilisation equipment (CHRIST® Alpha 1-2) was used. A brown powder-like residue remained in the flask that in contact with water immediately became red again (see Fig. S2.4-9). An NMR spectrum and an ESI mass spectrum were recorded. Whereas, in the NMR spectrum none of the Fig. S 2.4-9: Left: Raw carthamin residue, wet with water, rechts: dry. signals present could be assigned to carthamin, in the mass spectrum, apart from numerous impurities, a peak with the exact mass of carthamin appeared. Because of this information, that carthamin was present in the residue, work was continued. A classical chromatographic separation on silica gel followed. The choice of the eluent was a challenge. Whereas for TLC the mixture n-butanol/acetic acid/water in the ratio 4:1:2 proved to be successful, it could not be used for column chromatography, because during the chromatography, the eluent esterified. The n-butyl acetate formed, because of its high boiling point, could not be removed completely from the fractions collected. Since a sep- 48 49 Supporting information Safflomin A

aration of the raw extract with usual eluents, such as n-hexane/ethyl acetate or petroleum ether/chloroform, was not successful, the n-butanol was replaced by n-propanol. The boiling point of n-propyl acetate of 102°C is low enough, so that it could be completely removal under oil-pump vacuum making the separation possible. From the beginning, a red front, which was visible to the naked eye ran quickly behind the solvent front. These fractions were collected, united and the solvent removed on the rotary evaporator and finally dried under fine vacuum. The weakly red residue was investigated by NMR spectroscopy, whereby no signals from carthamin were detected.

Extraction of the Flowers of the Safflower to Obtain Carthamin by the Method of Schlieper 35 g fat-free, washed safflower were stirred with 450 mL 15% soda solution in a 2 L glass beaker for 30 minutes. The dark brown flowers were removed and the dark red filtrate neutralized by the stepwise addition of about 70 mL glacial acetic acid with stirring. Because of the strong foaming, the reaction was carried out in a tall 2 L glass beaker. Subsequently three cosmetic cotton wool balls, each 2 g, were added to the solution for 2 h. The dye was absorbed onto the cellulose, the balls were coloured deep red (see Fig. S2.4-7) as described by Schlieper for cotton material. The cotton wool

was copiously washed with water and placed in 500 mL 5% Na2CO3 solution. After 2 h the effectively discoloured cotton wool was removed (Fig. S2.4-8) and pressed out. The dark red solution (pH = 10-11) was gradually neutralized (pH = 6) by the addition of 30 g citric acid monohydrate with stirring. The liquid was centrifuged, the precipitates obtained united and the process repeated. It was assumed, that through the absorption on cellulose and release a selective purification occurs. The residues were washed with water and the residual solvent removed under reduced pressure. 325.0 mg of a magenta coloured precipitate were obtained. A 1H NMR spectrum failed to show any definite signals for carthamin. Despite this, a purification by column chromatography was attempted. Conditions: Dimensions : d = 45 mm, l = 230 mm Stationary Phase : Silica gel 60, 0,035 – 0,070 mm, Merck Mobile Phase : n-butanol/acetic acid/water (4:1:2,v/v/v) Volume of fractions: ca. 10 mL Number of fractions: 42

Fig. S2.4-10 Attempt at separation by co- After investigation of the fractions by TLC, the fractions 1-28 and 29-37 lumn chromatography of raw carthamin were respectively united and the solvent removed under reduced pressure. on silica gel 60 (Eluent n-butanol/acetic After drying in vacuum: acid/water 4:1:2 v/v) Sample 1 (fractions 1-28) 320.0 mg Sample 2 (fraction 29-37) 181.1 mg were obtained as colourless (!), sweet smelling residues. Both samples were investigated by NMR spectroscopy (see Fig. S2.4-10 and S2.4-11). Carth- amin is not detectable, at about 8.5 ppm an impurity appears.

50 51 Supporting information Safflomin A

Fig. S2.4-11 and S2.4-12 1H-NMR spectra of both samples after column chromatography at 400 MHz

Extraction of the Flowers of the Safflower to Obtain Carthamin According to Kuroda About 20 g fat-free, washed safflower were shaken with 200 mL pyridine in an Erlenmeyer flask at 60°C on a water bath, the plant material was removed and the solvent evaporated under reduced pressure. The process was repeated with 6 portions of distilled pyridine. The extracts were united (Fig. S2.4-12). On the final extraction the solvent was not entirely removed, about 100 mL of the red liquid were left. The solution was filtered and 150 mL distilled water were added and left cool overnight. On filtration 60.7 mg of a brown waxy precipitate were obtained, it was discarded. The solvent was removed under reduced pressure from the remaining solution to yield 3.1 g of a red residue. 50 mL chloroform were added to the residue and the solution filtered. Sub- sequently the solution was centrifuged to obtain 375.1 mg of a brown-red precipitate that was investigated by NMR spectroscopy (Fig. S13).

Fig. S2.4-13: Pyridine extract of carthamin

Fig. S2.4-14 Pyridine raw extract of

safflower measured in DMSO-d6

The spectrum is not indicative of a marked content of carthamin. Attempt to Isolate Carthamin with Ion Exchangers As already mentioned, the anorganic salts that are formed during the neutralization represent a problem. Whereas in the preceding experiments these reactions could not be avoided, the alkaline solution should now be neu-

50 51 Supporting information Safflomin A

tralized in another way, namely by the use of ion exchangers, with which we have had good experience in the isolation of other natural products (e.g. shikimic acid, carminic acid). Firstly, it was attempted to replace the acid that was used to neutralize the alkaline solution by Amberlite® IR 120. Fat-free, washed safflower was placed in a glass beaker with just so much distilled water, that continuous stirring was possible and solid sodium car- bonate was added, to bring the pH to about 11. As described for the previous extractions the plant material discoloured and simultaneously a red coloration of the solution was observed. The flowers were removed after some time over a fine metal sieve and the filtrate collected in a glass beaker. The strong acid cation exchanger Amberlite® IR120 (CAS RN 68441-33-8, Sigma Aldrich) was now used to neutralize the solution. The resin bound cations are protons, which should be exchanged against the sodium ions that come from the soda. The idea was as follows: the carthamin is deprotonated and is present in the solution as the counter ion to Na+. Through the cation exchanger, the sodium ion should be exchanged against a proton, the Fig. S2.4-15: Residue (containing carthamin) after neutralization with amberlite® alkali ions are therefore removed from the solution and carthamin returns to IR 120 its neutral form. Before use, the ion exchange resin was thoroughly washed with several litres of distilled water, until the wastewater showed no colour. Because the amount of ion exchanger needed for neutralization is difficult to determine in advance, for safety's sake a volume of 250 mL was prepared in this way. The washed cation exchanger was added in small amounts with control of the pH-value to the deep red extract. As with the neutralization with acid, on reaching a neutral pH-value a change in colour in the direction pink was recognized. To avoid the use of filter paper the ion exchanger beads had to be removed with the metal sieve. The water was removed from the now pink solution on the rotary evaporator. Because of the carbonic acid contained in solution,

which decomposes to CO2 and H2O, a strong foaming develops, however, with skill and patience it is relatively easy to command control. A promis- ing, pink residue remained on the wall of the flask (Fig. S2.4-15). In its 1H NMR spectrum (Fig. S2.4-16), for the first time, signals could be detected in the aromatic region that agree with the data available for carthamin in the literature. S. Yue, Y. Tang, S. Li, J.-A. Duan "Chemical and biological properties of quinochalcone C-glycosides from the florets of Carthamus tinctorius"Mol - ecules (Basel, Switzerland) 2013, 18, 15220–15254.

Fig. S2.4-16: Excerpt from the 1H NMR spectrum of sample AS 35 from the isolation experiment for carthamin with the ion exchanger Amberlite® IR 120. The large peaks at 8.72, 7.58 und 7.21 ppm

come from the solvent pyridine-d5.

52 53 Supporting information Safflomin A

At the same time, in the mass spectrum a peak for the molecular ion was detected with an intensity that up to now had not been reached (Fig. S2.4- 17). The peaks with m/z 806.0, 1034.0 and 1633.0 belong to the calibration mixture. Furthermore, in this spectrum apart from numerous impurities that cannot be determined a peak with m/z 909.2 appears. Calculation of the

molecular formula for this peak gives C43H41O22 as the most probable. This corresponds exactly to the molecular formula for carthamin in the negative mode ([M-H+]).

Fig. S2.4-17 Excerpt of the ESI (-) mass spectrum of the sample AS 35 from the isolation experiment for carthamin with the ion exchanger Amberlite® IR 120.

The result of this analysis allows the conclusion, that for the first time in this work, carthamin was definitely detected in an isolated residue, even if not in a pure form. This indicated, that we were on the right path with this method. What was needed was a method to purify the precipitate obtained. Since the carthamin spot in the TLC could be seen with the naked eye to run as a sharp red band, it seemed reasonable, instead of the usual column chromatography to try the purification by preparative TLC (TLC plate SIL G-25, 20×20cm, 0.25 mm layer silica gel, Macherey-Nagel). Because the n-butanol or n-propanol containing eluents, which were used until now, esterify, a better eluent mixture was sought. Finally, the acetic acid was

successfully replaced by ethanol. This led to a clearly reduced Rf-value for the red band, however, for the preparative TLC this was not important. Part of the raw extract was loaded onto a plate and developed with n-butanol/ ethanol/water (4:1:2). Because of the two different types of gel on the plate, the chromatograph- ically inactive "collecting gel" below the start zone and the separating gel above it, a focussed thin red starting zone and coloured band were achieved. However, after the development, the plate was dried in air and the red colour disappeared entirely. How could that happen? The zone with the supposedly carthamin band was scratched from the plate and transferred to a flask, in which it was attempted, to dissolve the carthamin with pyridine. However, no red colouring of the pyridine was observed and the NMR spectrum showed none of the peaks, which in the raw extract had been allocated to carthamin. Therefore, this "purification" proved to be unsuccessful.

52 53 Supporting information Safflomin A

Based on the findings of all the experiments that had been carried out, a final isolation experiment was undertaken. Apart from the cation exchanger Am- berlite® IR 120 the anion exchanger Amberlite® IR 400 would come into use. After the neutralization of the soda solution with an acidic ion exchanger, instead of removing the solvent, as had been done up to now, it should be attempted to bind the dye out of the solution as an anion on the basic anion exchanger. An advantage seemed to be, that this process should be so specific, that it would only happen with carthamin and the impurities would remain in solution. Subsequently the dye could be released (competitively displaced) and hopefully pure carthamin would be obtained. To begin, the fat-free, washed safflower was placed in a glass beaker with sufficient water and stirred. Then enough solid soda was added to bring the pH-value to 11. As expected the alkaline solution became coloured and simultaneously the flowers became pale. Now the solution was neutralized with the cationic exchanger Amberlite® IR 120 and the ion exchanger removed over a sieve. A chromatography column with a glass frit was now filled to a height of 20 cm with the ion exchanger Amberlite® IR 400 in the chloride form (CAS Nummer 60177-39-1, Sigma Aldrich) and washed in the column with several litres of distilled water. Subsequently, in the same way, the coloured soda solution was loaded onto the column and the output from the column collected. Already with the first run it could be recognized, that the beads of the ion exchanger took up the colour. The intensity of the colour became less in going from top to bottom of the column. Simultaneously, the solution leaving the column lost its pink colour and was now a clear yellow. This process was repeated until no further visible change of the solution or the column was discernible. Finally, the Amberlite® IR 400 in the column was washed with distilled water, whereby the splendid colouring could be recognized even better (Fig. S2.4-18) A further step in the experimental plan was to recover the red dye from the column. Here, unexpectedly large problems were encountered. First the attempt was made with acid, in the form of 25% acetic acid, to dissolve the red dye by percolation. Neither in the solution nor in the beads was there a change in colour. A higher concentration and a change to hydrochloric acid brought the same result. To avoid finding a great number of different Fig. S2.4-18 Amberlite® IR 400 with solvents in the column, tests were conducted with small amounts of the bound carthamin coloured beads contained in glass beakers. Thereby, neither 50% formic acid nor pyridine nor trimethylamine brought about a colour change. Soda effected a discoloration, whereby the solution itself remained colourless. After the beads were separated from the soda solution and the solution was neutralized with a few drops of hydrochloric acid the expected red colour was not obtained. However, if a drop of acid was given to the beads that had been separated and washed, these immediately became scarlet. The dye was still bound to the ion exchanger and the attempt to release it from the beads was unsuccessful.

54 55 Supporting information Safflomin A

Fig. S2.4-19 and S2.4-20 Amberlite® IR 400-beads, left: unloaded, right: charged with carthamin.

That is the end of the story of our attempts to obtain pure carthamin. Maybe, by taking heed of the literature and our information, you will be able to do better.

Questions and Answers A. Which parts of the structure of safflomin A contribute to the colour and which do not? What are the chromophores in safflomin A and carthamin and what are the auxochromes? The glucose residues do not contribute to the colour, because they do not contain chromophores. Chromo- phores are functional groups with a double bond. C=C bonds are weak and C=O strong chromophores. One chromophore alone cannot cause colour, but a conjugated π-system is required. In safflomin A 8 double bonds are conjugated but in carthamin 17. Therefore, even less energy rich light can be absorbed and carthamin has a darker colour: pink-red. Auxochrome groups, as +M substituents that release electrons into the conjugated system, have a colour darkening effect. In both cases, these are the OH-groups on the quinol or benzene rings, but not on the glucose rings.

B. Which C-substructure would you use, to come to a dimer that is bridged over a C-atom, as shown in the scheme "synthesis of dimeric non-glycosidic carthamin analogues"? S. Sato et al. solved this problem in 2005. The quinochalcone substructure shown in the middle of the scheme is allowed to react with:

- 2,3,4,5,6-penta-O-acetyl-aldehydo-D-glucose/NaOMe, which leads to the compound top left in Fig. 4.2-9 and after removal of the protective groups with NaOMe to the compound below - glyoxylic acid /NaOMe, which leads to the carboxylic acid top right in Fig. 4.2-9, from which by oxidative

decarboxylation with very dilute KMnO4 solution the carthamin analogue bottom right becomes available, as it also would be with - triethyl orthoformate/NaH.

C. What is a "drying oil"? To what use can the properties of a drying oil be put? Fatty oils containing multiply unsaturated fatty acids are referred to as "drying oils". Chemically these are triglycerides, i.e. threefold esters of glycerine with fatty acids such as linoleic or linolenic acids. To these belong: linseed oil, walnut oil, poppy-seed oil, safflower oil, fish oil, amongst others. By "drying" is meant not the evaporation of a possibly contained solvent but the formation of a quite hard but elastic layer, after the application of an oil paint, by reaction with oxygen in the air. This causes the formation of hydroperoxides and a radically induced cross-linking, which converts the monomer triglyceride to a polymeric, hydrophobic network. Drying oils, particularly linseed oil, were widely used in the manufacture of the slow drying oil paints, which have been used in Europe for decorative purposes since the 12th century. Such paints are a suspension of a 54 55 Supporting information Safflomin A

coloured pigment in a drying oil. These dry or harden slowly to give a glossy, hardwearing surface that is water-repellent and washable. Today oil paints are used in finishing and protecting wood in buildings or exposed metal surfaces such as ships and bridges. Such paints are best used in summer, since they "dry" i.e. harden faster. In cold weather, it can take weeks, until the surface is no longer sticky. By the use of siccatives the oxidative drying process can be accelerated. For this, various metal salts can be used. Artists still use oil paints. Linseed oil was also used, because of its drying properties for the manufacture of the decorative materials Linoleum and Lincrusta, that were invented in 1860 by the British chemist Frederick Walton.

D. What is to be understood by an "essential fatty acid"? Is linoleic acid a ω-3 fatty acid? What is meant by ω-3? Essential fatty acids are required for the metabolism in the body, but cannot be synthesized by the body itself. Therefore, they must be taken in with nutrition. The most well-known essential fatty acids are linoleic acid ((9Z,12Z)-octadeca-9,12-dienoic acid) and linolenic acid ((9Z,12Z,15Z)-octadeca-9,12,15-trienoic acid), which received their names from their presence in linseed oil (genus Linum L.). Linolenic acid is a ω-3 fatty acid, but linoleic acid is not. ω-3 means, that counting from the end of the C-chain of the carboxylic acid, the third bond must be a C=C double bond. This coding seems to be rather esoteric and proved to have an excep- tional marketing effect on laypersons – as intended.

E. The signals of the C-atoms 2' and 3' show the typical sequence of an α/β-unsaturated carbonyl system. The signals of the corresponding protons, H-2' and H-3' behave conversely. Discuss this! The resonance structure demonstrates, that the partial positive charge can be delocalized as far as the OH- group on C-4". Thereby, a greater electron density is created at the position 3', to which the protons react more sensitively.

F. The strong deshielding of the proton signals from H-2''' in comparison to H-2'''' is very noticeable. Is there an explanation for this? The glucose residue''' is bound to C-6, which is part of the dieneone system, whereas the glucose residue'''' is bonded to the sp3 hybridized C-4. Probably, because of the anisotropy of the π-system both H-1''' and H-2''' show a clear deshielding.

G. Very often the intensity of [M-H]−-ions can be increased by the addition of formic acid to the dissolved analyte. This applies to the measurement of the ESI(-) mass spectrum of safflomin A. Give an explanation for the paradox, that the yield of deprotonated product is increased by the addition of formic acid. Water undergoes cathodic reduction, to form hydrogen and hydroxide ions, on the spray capillary, which has a high negative potential. Thereby, OH-ions enter the escaping drops and are responsible for the deprotonation. The formic acid added supports the electrochemical process by increasing the electrical conductivity.

56 56 Supporting information Chlorophyll a

2.5 Chlorophyll a

Fig. S2.5-1 Structure of chlorophyll a

Numbering of Chlorophyll a Three different numbering systems exist, according to Fischer [S1], IUPAC [S2] and Scifinder®, which is used in this article.

Fischer 1 2 3 4 5 6 7 8 9 IUPAC 2 3 7 8 12 13 17 18 131 Scifinder 8 9 13 14 18 19 3 4 20 Fischer 10 11 12 13 14 15 16 17 18 IUPAC 132 1 4 6 9 11 14 16 19 Scifinder 21 7 10 12 15 17 22 2 5 Fischer 1a 2a 2b α 3a 4a 4b β 5a IUPAC 21 31 32 5 71 81 82 10 121 Scifinder 81 91 92 11 131 141 142 16 181 Fischer 10a 10b 7a 7b 7c 8a δ P1 P2 IUPAC 133 134 171 172 173 181 20 Scifinder 211 212 31 32 33 41 6 Fischer P3 P4 P5 P6 P7 P8 P9 P10 P11 IUPAC Scifinder Fischer P12 P13 P14 P15 P16 P3a P7a P11a P15a IUPAC Scifinder Fischer γ IUPAC 15 Scifinder 1

57 57 Supporting information Chlorophyll a

A Look into the Leaf: Chlorophyll and Photosynthesis Let us take a look at the role of chlorophyll in photosynthesis, without going as deeply as in [1]. Through photosynthesis energy rich substances are produced from substances of lower energy by the agency of light energy. This can occur in plants, algae and some bacteria. By the assimilation of carbon dioxide carbohydrates, our source of energy, are produced in the chloroplasts. The overall equation of the reaction is:

6 H2O + 6 CO2 + 2872 kJ light energy → C6H12O6 + 6 O2 It is a redox reaction, in which carbon dioxide is reduced to carbohydrates

and the oxygen in water is oxidized to elemental oxygen. O in H2O delivers the necessary electrons for the reduction and water is the source of the oxygen in the air. The photosynthesis, as the basis of life, takes place on an enormous scale. About 150 × 109 tonnes of vegetable biomass are produced each year (calculated as dry mass). About the same amount of oxygen is formed! All the oxygen that we breathe is of biological origin. Before this reaction developed, there was no oxygen in the atmosphere. Putting it into an imaginable form: a large deciduous tree on a summer day assimilates about 10 m3 carbon dioxide and sets 10 m3 oxygen free. That is the daily oxygen requirement for three humans. In addition, 12 kg of biomass are produced. Photosynthesis offers a biochemical synthesis that performs by the energy being supplied by light. The agent for this wonder is chlorophyll. The distribution of molecules of chlorophyll in a leaf is not random, such as the distribution of raisins in a cake, but perfectly organized as part of a cellular machinery that collects energy, transfers it and transforms it into chemical energy. How does this machinery work, roughly described from an external view? There are two types of photosynthesis, the anoxygenic (without the produc-

tion of O2) and the oxygenic (with the formation of O2). In an evolutionary context, the anoxygenic variant came first, 3.5 × 109 years ago; the oxygenic form came later, about 3 × 109 years ago. Chlorophylls exist that long! Of interest is, that only the existence of oxygen made the formation of an ozone layer possible. In principle three steps occur: - The absorption of electromagnetic radiation (light) by a pigment such as chlorophyll - The transformation of radiation energy into chemical energy. Here- by, energy rich adenosine triphosphate (ATP), a carrier and short-term store of biochemically useable energy, is synthesized. - The biosynthesis of organic substances that serve as raw materials and a source of energy for all organisms in the food chain, including those without a photosynthesis of their own. The first two steps are called light reactions. For plants these occur in the photosynthesis system I (PS I) and photosynthesis system II (PS II). The third step is independent of light. The chlorophylls of plants, which absorb light, are referred to in this context as "chromophores". If they are bound in chromoproteins, they are called pigments (both terms are not unambiguous). Light excites the conjugated

58 Supporting information Chlorophyll a

π-system. However, not all regions of the visible spectrum are absorbed. Blue and red light is preferably absorbed by Chl a and Chl b, leading to a "green gap" that we can see as the green of leaves. Excited chlorophyll can donate an electron to an electron acceptor. A positively charged chlorophyll radical (Chl•+) remains. Finally, the donated electron returns over an electron transport chain back to the chlorophyll. Thereby, the transport of protons through the membrane (proton pump) is effected. Light energy is therefore converted by chemiosmosis (movement of ions across semipermeable membranes) to an electrical and osmotic potential. The PS II offers a strong oxidizing agent for the oxidation of water. The PS I delivers a strong reducing agent for the reduction of NADP+. Thus, light cleaves water, producing

O2 and the strong reducing agent NADPH:

+ + 2 H2O + 2 NADP + light energy → O2 + 2 NADPH/H Energy rich ATP is synthesised from ADP and phosphate:

3 ADP + 3 Pi (+ADP synthase) → 3 ATP + 3 H2O This gives overall a chlorophyll mediated light reaction:

+ + 2 H2O + 2 NADP + 3 ADP + 3 Pi → O2 + 2 NADPH/H + 3 H2O The structural arrangement of chlorophyll in light collecting complexes, with which we want to conclude this "View into the Leaf", is fascinating. Light collecting complexes are to be found in the inner thylakoid membrane of the chloroplasts, in which photosynthesis occurs. They absorb light and direct this energy to the reaction centre of the light reaction shown above. Apart from chlorophylls, also carotinoids and xanthophylls are present as chromophores that absorb but can also emit light. In the course of evolution, light collecting complexes have formed as antennas, that makes them much more effective than single pigments. Because the cross section of antennas that are arranged around a reaction centre is larger and the absorption area is broader.

Liquid-Liquid-Chromatography (LLC) The two phases of a two-phased solvent system that is prepared by mixing at least two solvents are used as the stationary and mobile phases. One of the two phases, either the upper or lower phase, is held stationary in a spe- cially constructed, rotating column by a centrifugal field (stationary phase), while the other, the mobile phase, is pumped through. The actual separation is based on the partition of the targeted component between the two liquid phases. LLC is at present the only chromatographic technique, for which the user must prepare both the mobile and stationary phases in one step, since they cannot be chosen independently of each other. The choices for the composition of the solvent mixture are almost innumerable, making this separation technique very versatile and allowing custom-made stationary and mobile phases for each separation problem. The fact, that the stationary phase is a liquid without a solid carrier, brings considerable advantages. The total volume of the stationary solid phase, which normally takes up 60-80% of the total volume of the column, is available for the separation. This makes a very high loading of the column and a very high separating capacity possible. Additionally, the absence of irreversible absorption means, that the

58 59 Supporting information Chlorophyll a

sample can be completely regained and that the separation is highly repro- ducible. The size of the apparatus varies between 5 mL and 18 L and there are two basic designs, hydrodynamic and hydrostatic. Over the years, in addition to the original, hydrodynamic apparatus with two axes a hydrostatic construction with a single axis has been established; we used such an apparatus. Apart from the principle of construction, the modern apparatuses differ only very slightly from the first prototypes. The main characteristics of the hydrostatic equipment are the single axis drive of the rotor with the resulting constant centrifugal field and the typical construction of the chambers and channels (Fig. S2.5-2) [14]. In the chambers, which are joined to each other by narrow channels, the mobile phase flows through the retained stationary phase and the desired exchange of substances takes place. Depending on the manufacturer, the chambers have different geometries and can occur either single or paired in the form of "twin-cells" (Fig. S2.5-2a). The chambers are milled into a disc of PTFE or stainless steel, which enables a modular construction. Depending on the capacity needed, a number of similar discs can be placed over each other and the chambers of two adjacent discs joined. Thereby, the last chamber of each disc is joined to the first chamber of the following disc. To facilitate this linkage, two holes are bored in the ring plate that are used to physically separate neighbouring discs (Fig. S2.5-2 b). A rotor can be put together from a number of discs and their neighbouring ring plates in modular fashion and one or more of these can be contained in the apparatus (Fig. S2.5-2c).

b) c)

Fig. S2.5-2 Basic construction of a disc with chambers and channels b) ring plate of Teflon® c) modular construction of a CCC column from discs and ring plates

60 Supporting information Chlorophyll a

The basic construction and the mode of operation of the hydrodynamic column is displayed in detail in Fig. S2.5-3

Fig. S2.5-3 Hydrodynamic Apparatus a) construction and planetary gears and b) schematic representation of the mixing and settling-out zones in position I, II, III and IV during a complete rotation of the coil around the central axis of rotation as a coiled column (top) and in its uncoiled state (bottom) [S5-S9].

A tube, made of an inert substance, e.g. Teflon® or stainless steel and open at both ends, is wrapped spirally around a coil carrier. In contrast to the hydrostatic construction, the two open ends can be connected to the periphery of the column (pump and detector) without a rotating seal [S7, S8]. In the hydrodynamic construction the column therefore exists in the form of a single tube wrapped around a coil carrier that accelerates not only around its own axis but also around a central rotary axis. Since the angular speed and direction of rotation is the same around both axes of rotation, a centrifugal field results that changes periodically in its intensity and direction. This varying force field leads to alternating mixing and settling-out zones along the column (Fig. S2.5-2b). While in the settling-out zones a phase separation occurs, in the mixing zones the desired exchange of substances takes place. The mixing zones take up about a quarter of the total volume of the column and are found near the centre of rotation. They make one complete turn per revolution of the column. Similar to an Archimedean screw, which transports water against gravity, the mobile phase is pumped through the column, while the stationary phase is retained. For the counter current chromatography (CCC) experiments the following solvents were used: n-heptane (EMSURE, purity 99%), ethanol (LiChro- solv®, purity 99.9% and acetonitrile (LiChrosolv®, purity 99.9%) from Merck KGaA (Darmstadt, Germany). Millipore 18 MOhm water was used. Each column consists of ten stainless steel plates in series, each containing 90 twin cells (Fig. S2.5-3a). To carry out the separation in this project, two of the four HPLC pumps (maximum flow rate 50 mL×min–1) shown in Fig. S2.5-3 were used, one to fill the column with the stationary phase and the other to pump the mobile phase during the separation. After the second 60 61 Supporting information Chlorophyll a

column a UV detector (ECOM DAD600 2WL 200-600 nm, Prague, Czech Republic) was installed. Fractions were collected with a fraction collector (LS 5600, Armen Instrument, France).

Additional NMR Spectra at 700 MHz in THF-d8

Equilibrium of Epimeres 16a 11a 6a

16a' 11a' 6a'

δH / ppm

Fig. S2.5-4 Epimeric equilibrium of chlorophyll a and a'. Trace a: initial spectrum of isolated chlorophyll a' in THF-d8 15 min after dissolution; trace b: spectrum after 30 minutes at 295 K; trace c: spectrum after 2 h; trace d: spectrum after 18 h, the same result as in Fig. 2.5-8 of the main text was obtained.

THF 212 181 131 81

1 P1 14

P2 4 3

δH / ppm

Fig. S2.5-5 Expansion of the 1H NMR spectrum of chlorophyll a in olefinic/aliphatic region

62 Supporting information Chlorophyll a

In Fig. S2.5-5 at δH = 5.19 we can see a triplet of an olefinic proton δ( C =

119.7) that is therefore assigned to P2. Two multiplets follow at δH = 4.50

(1H, δC = 50.2) and 4.47 (2H, δC = 61.5). In the COSY spectrum the former

couples with the doublet of the methyl group at δH = 1.74 and is therefore assigned to H-4. In the COSY spectrum, the latter shows a link to the signal of the proton P2 and therefore belongs to the proton P1. The next signal at

δH = 4.14 (1H, δC = 51.8) couples with a multiplet at δH = 2.39 and is therefore attributed to H-3.

2 H2O 14 P16

P3a P5------P14 P7a

P11a

31,2

P4 P15 31 32

δH / ppm Fig. S2.5-6 Expansion of the 1H NMR spectrum of chlorophyll a in aliphatic region

A multiplet from two protons at δH = 3.79 (δC = 20.3) couples with the signal

of a methyl group at δH = 1.73 (δC = 18.1) that is extensively covered by the

signal from the solvent THF-d8. These resonances are assigned to the ethyl

group on C-14. Four signals from methyl groups follow as singlets at δH =

3.72/ δC = 52.2; δH = 3.60/ δC = 12.6; δH = 3.34/ δC =12.6 and δH = 3.29/ δC = 11.2. From its 13C chemical shift the first can be assigned to the methoxy group H-212. According to the chemical shifts of the other three, each one is situated on a double bond and the individual assignments will be clarified later with help of the NOESY and HMBC spectra.

62 63 Supporting information Chlorophyll a

81 31,2 142 P -P COSY Spectrum 141 131 5 16 181 P 21 3a 91 92t 92c P 4 P 3 212 31 32 P δH / ppm 2 1 4

P11a,7a,16

P -P 5 15 P , 142 2 3a P4, 3 31,2, 31

131, 81 181, 212 2 14 , 3 4, P1

P2 92c 21, 92t

δH / ppm

Fig. S2.5-7 DQF-COSYspectrum of chlorophyll a

2 H2O 14 P16 P5 ------P14 P7a P3a P11a 1 1,2 2 3 3 3 P4 P15 δH / ppm

P11a P7a P16

P6 - P14

P5, P15 P3a 142

P4 32

31,2 H2O 31

δH / ppm Fig. S2.5-8 Expansion of the COSY spectrum of chlorophyll a

A multiplet of a proton at δH = 2.57 according to the COSY and HSQC spec-

tra belongs together with a multiplet at δH = 2.39 to the same C-atom at δC = 30.7 and is allocated to the protons H-31. Correspondingly, the multiplet at

δH = 2.42 and that at δH = 2.05 with the related C-atom at δC = 31.2 belongs 2 to H-3 . The COSY spectrum has the typical appearance of a –CH2-CH2-

64 Supporting information Chlorophyll a

coupling pattern. The multiplet at δH = 1.90/ δC = 40.6 is attributed to the al-

lylic methylene group P4. The COSY spectrum indicates that the methylene

group P5 is at δH = 1.36.

APT 13C NMR Spectrum 91

9 211 22 7 17, 10 13 19

20 5 2 12 15 14 P3 8 18 33

δC / ppm

Fig. S2.5-9 Expansion of the 13C NMR spectrum of chlorophyll a between 190 and 120 ppm

P2 16 11 6

2 9 1 P1

δC / ppm

Fig. S2.5-10 Expansion of the 13C NMR spectrum of chlorophyll a between 120 and 60 ppm

64 65 Supporting information Chlorophyll a

P11

3 P7 P15 P15a P7a 2 1 2 1 1 21 4 4 P16 P11a 14 18 , 13 1 P3a 8

32 * 141 31

P4 P6 P13 P14 P8,P10 P12

δC / ppm

Fig. S2.5-11 Expansion of the 13C NMR spectrum of chlorophyll a between 55 and 10 ppm, * = impurity

2 H2O 14 P3a P5 P16, 7a, 11a

P6 ----- P14 1 1,2 2 3 3 3 P4 P15

δC / ppm

P3a

P , P 7a 11a

P15a, P16

P5, P9, P13

31, 32

P11, P7

P6 - P12

P4, P14

δH / ppm

Fig. S2.5-12 Expansion 2 of the HSQC spectrum of chlorophyll a

66 Supporting information Chlorophyll a

HMBC Spectrum

2 1 1 2 21 18 8 H2O 14 P5 P5 - P14 P16,7a,11a P1 1 1 1 1,2 2 4 3 14 13 3 3 3 P4 P3a P15

δC / ppm

2 9 , P2

91 19, 18, 13 8, 9

P3, 14 15, 17 10, 12 7, 2

22 5 211 31

δH / ppm

Fig. S2.5-13 Excerpt from the HMBC spectrum of chlorophyll a in the aliphatic proton region

In the pyrrole ring II the protons of both methyl groups H-131 and H-142

couple with the C-atom 14 at δC = 144.4. The C-atom 13 at δC = 134.3 couples with H-11 and H-141. The latter shows further correlations to C-14 and

C-15 at δC = 146.4. Therefore, the assignment of C-12 to the signal at δC = 151.7 remains, which is also seen by H-11 and H-131.

In the pyrrole ring I the methyl protons H-181 couple with the three C-atoms

at δC = 132.2, 134.4 and 148.2. H-16 does not couple with the signal at δC = 132.2 that is therefore attributed to C-19. Of the other two signals the

chemical shift of δC = 148.2 suggests the assignment to C-17. H-21 couples 1 with C-20 at δC = 189.6 as well as with C-21 at δC = 171.3, which is seen 2 by H-21 . Two further correlations of H-21 at δC = 161.7 and 106.5 are assigned to C-22 and C-1 on the basis of their chemical shifts.

In the pyrrole ring IV the protons of the methyl group H-41 couple with the 3 C-atom C-5 at δC = 167.7, which is also recognized by H-4. A J coupling 2 1 leads from H-4 to C-2 at δC = 156.0, as well as a J coupling to C-4 at δC =

24.0 and to C-3 at δC = 51.8.

66 67 Supporting information Chlorophyll a

1 1 1 2 14 18 8 H2O 14 P3a P5 - P14 P16,7a,11a P1 2 1 1 1,2 P2 4 3 21 13 3 3 P4 P15

δC / ppm 131, 181, 81

2 P3a, 14 , P7a 1 P11a, 14 1 P16, P15a, 4

P9, P13, P5 32, 31

P7, P11

P8, P10, P12, P6

P14, P4

4, 3, 212

δH / ppm

Fig. S2.5-14 Excerpt from the HMBC spectrum of chlorophyll a in the aliphatic carbon region

The assignment of the signals in the phytyl residue is particularly difficult, since those at the end of the chain strongly overlap. The protons of the 3 methyl group P1 couple with the carbonyl group C-3 and also with P2 at δC

= 119.7 and with P3 at δC = 142.5. The olefinic proton P2 couples with the

methyl group P3a at δC = 16.3 and the methylene group P4 at δC = 40.6, which

is also seen by P3a.

The signals of the diastereotopic isopropyl groups P15a/P16 at δH = 0.85 cou-

ple in the COSY spectrum with the methine proton P15 at δH = 1.51, which in

the HSQC spectrum is linked with the C-atom P15 at δC = 28.9. The HMBC

spectrum finds the autocorrelation within the isopropyl group at δC = 23.1

and a further cross peak to the methylene group P14 at δC = 40.3, the protons

of which can be identified via the HSQC spectrum atδ H = 1.14.

The most strongly shielded doublet of the methyl group at δH = 0.79 corre-

lates with the C-atom at δC = 20.1 and is assigned to P11a. From this group

of signals P11 and P10 are reached. The same applies to the doublet of the

methyl group at δH = 0.83 that correlates with the C-atom at δC = 20.0 and is

attributed to P7a. This group of signals leads to the assignment of P6, P7 and

P8 at δC = 38.4. P13 is tentatively assigned to δC = 25.2 under the signal from

the solvent. Only P9 from the phytyl chain remains, which is assigned to δC

= 25.2 and couples in the HMBC spectrum with the protons from P7 and P11.

68 Supporting information Chlorophyll a

NOESY Spectrum

2 H2O 14 P16 P3a P6 ------P14 P7a P11a 2 1,2 2 3 3 3 P4 P15 P5 δH / ppm

P11a,7a,16

P6 - P14

P5, P15

P3a

142

P4 32

1,2 3 , H2O 31

δH / ppm

Fig. S2.5-15 Expansion of the NOESY spectrum of chlorophyll a

Literature [S1] H. Fischer, A. Stern, "Die Chemie des Pyrrols", Vol. 2, Akademischer Verlag Leipzig 1940, S. 173. [S2] IUPAC, J. Amer. Chem. Soc. 1960, 82, 5575–5583, 5582. [S3] G. Sherman, E. Fujimori "Chlorophyll-dioxane interaction in the solid state" Arch. Biochem. Biophys. 1969, 130, 624–628. [S4] R. M. Leblanc, C. Chapados "Aggregation of chlorophylls in monolayers" Biophys. Chem. 1977, 6, 77–85. [S5] I. A. Sutherland, L. Brown, S. Forbes, G. Games, D. Hawes, K. Hostettmann, E. H. Mckerrell, A. Marston, D. Wheatley, P. Wood "Countercurrent Chromatography (CCC) and Its Versatile Application as an Industrial Purification and Production Process"J. Liq. Chromatogr. Relat. Technol. 1998, 21 279–298. [S6] Y. Ito "Golden Rules and Pitfalls in Selecting Optimum Conditions for High-Speed Counter-Current Chro- matography" J. Chromatogr. A 2005, 1065, 145–168. [S7] A. Berthod "Countercurrent Chromatography – The support-free liquid stationary phase" Elsevier Science B.V., Amsterdam 2002. [S8] I. Sutherland, P. Hewitson, S. Ignatova "New 18-l process-scale counter-current chromatography centrifuge" J. Chromatogr. A 2009, 1216, 4201–4205. [S9] I. A. Sutherland, J. Muytjens, M. Prins, P. Wood, "A new hypothesis on phase distribution in countercurrent chromatography" J. Liq. Chromatogr. Relat. Technol. 2000, 23, 2259–2276.

68 69 Supporting information Chlorophyll a

Questions and Answers A. Which vitamin is like chlorophyll a and b and the different haems an essential complex and which metal ion is its central ion?

3+ It is vitamin B12 (cyanocobalamin), Co is its central ion.

B. Which commonalities and which principle differences are there between covalent and coordinate covalent (dative covalent) bonds? Both are directional covalent bonds that are formed by an electron pair. In a normal covalent bond, such as in hydrogen chloride (H-Cl), each of the atoms involved in the bond contributes one electron to the electron pair of the bond. With a coordinate covalent bond the electron pair of the bond originates entirely from one partner, the ligand, that inserts the electron pair in an empty orbital of the central ion, such as Mg2+, Fe2+, Fe3+ or Co3+, or a central atom, such as Ni (0).

C. What is the driving force for the fixing of Mg2+-ions in chlorophyll a and Fe2+ in haem b, which occurs in haemoglobin, myoglobin and cytochrome P450? The driving force is the attempt of both ions to attain a complete noble gas electron shell. The central ion Mg2+

has the electron configuration of the noble gas neon (10Ne) and can, by the acceptance of 8 electrons from the 4 electron pairs of the tetradentate porphyrin ligand in chlorophyll, attain 18 electrons, which corresponds to

the electron configuration of the next noble gas argon 18( Ar). With the blood pigment, haem b in haemoglobin the central ion Fe2+ has 24 electrons. By the acceptance of 12 electrons from 6 electron pairs, which are donated by its three ligands, it attains 36 electrons, corresponds to

the electron configuration of the noble gas krypton 36( Kr). The situation with vitamin B12 is analogous. There a Co3+-ion, which like the Fe2+-ion has 24 electrons, accepts 12 electrons, leading to a stable electron configu-

ration of 36Kr.

D. Which class of coloured, synthetic pigments has a structure related to that of chlorophyll and types of haem and is extremely stable? It is the class of the phthalocyanines. Phthalocyanines are similar to the porphyrins, they are tetrabenzo- tetraazaporphyrins. They are also tetradentate ligands. Phthalocyanine itself sublimes without decomposition at 500°C (!), a temperature that is much too high for most organic compounds. Its metal complexes are also very stable. The deep blue copper phthalocyanine, which is used in paints, printing inks and plastics, is industrially the most important blue pigment. Phthalocyanine pigments are thermally very stable, lightfast and weatherproof. Their synthesis can start for example from phthalic acid and urea or phthalic acid dinitrile. Apart from Cu2+ other central ions are possible.

E. Which colour receptors does the human eye have? From which colour can we differentiate the most shades? The eye has RGB-receptors and is therefore sensitive to red, green and blue, the three primary colours for additive colour. Other colours, such as yellow, magenta or cyan blue and all of the other about 2 million distin- guishable shades of colour are to be regarded as being the calculated result of our brain in regard to the mixture of colour. We can differentiate the largest number of nuances for the colour green (think of a forest in spring) and correspondingly also very many complementary colours belonging to the red tones in the magenta region. The opinion exists, that the ability to recognize and differentiate green colours represented an evolutionary advantage in the search for nutrition, for those living organism that possess it.

70 Supporting information Chlorophyll a

F. Why must the absorption spectra of Chl a and Chl b be so different?

Because on C-13 instead of a CH3-group (for Chl a) Chl b has a formyl group. The latter contains a double bond, whereas the methyl group does not. This and the circumstance that it conjugates to the π-system, enlarg- es and extends the conjugated region. In the short wavelength part of the visible spectrum, radiation of lower energy is therefore sufficient for excitation. With our eyes, we experience this as two clearly different shades of green.

G. In the Assignment Table it is evident, that the 13C-data calculated with the aid of the program ChemBioDraw® are in good agreement with the experimental data. Only the values marked in red for C-5, C-2, C-15, C-16, C-11 and C-6 show unusually strong deviations. Explain why this is. The predictive methods based on empirical incremental systems typically analyse the substitution over 4 bonds but do not take electronic effects, such as delocalisation of charge by π-conjugation in conjugated mac- rocyclic ring systems into account that for example is illustrated by the mesomeric contributing structures. In addition, C-5, C-2 and C-15 are affected by the magnesium coordination of the nitrogen atoms, which is not taken into account by empirical incremental rules. Quantum chemical calculations of chemical shifts consider the electronic environment of the nuclei, which is the origin of NMR chemical shifts, thus leading to more reasonable results for conjugated systems and organometallic compounds.

1 H. In chlorophyll b the CH3 group C-13 conjugated systems is replaced by a formyl group. In the proton spectrum, the order of the signals from H-16 and H-11 is changed. Why? This is illustrated by the contributing resonance structures, which predict a partial positive charge on C-11. Therefore, H-11 and not longer H-16 is the most deshielded proton of the porphyrin ring.

I. Calculate with the help of the natural abundance of the carbon and magnesium isotopes the isotopic pattern for M+• of chlorophyll a. 12C (98.93), 13C (1.07%), 24Mg (78.99%), 25Mg (10.0%), 26Mg (11.01%). The computer-simulated isotope pattern for M+• of chlorophyll as shown in Fig. S2.5-16 b was obtained with a software that takes the contribution of all non-monoisotopic elements into account. Strictly speaking, all the elements found in chlorophyll are polyisotopic. However, the natural abundances of 15N (0.37%), 17O (0.04%) and 18O (0.2%) are very low. Chlorophyll only contains four N and five O-atoms, so that the statistical probability of the occurrence of these heavy isotopes only slightly affects the appearance of the isotopic pattern. Chlorophyll a contains 72 H-atoms. With a natural abundance of 0.011% for 2H the statistical probability, that one of the H-atoms is deuterium, is despite the large number of H-atoms only about 0.1%. Therefore, we ignore the contributions from 2H, 15N, 17O and 18O and take into account only the C-isotopes 12C (98.93%), 13C (1.07%) and the Mg-isotopes 24Mg (78.99%), 25Mg (10.0%) and 26Mg (11.01%). According to the laws of calculation of probability, the frequency of the individual combinations of isotopes can be calculated with a polynomial approach.

m n o (a1 + a2 + ...) × (b1 + b2 + ...) × (c1 + c2 + ...) × etc.

Where a1, a2,….., b1, b2….etc. are the natural abundances of the isotopes of the elements A, B,…. etc. and m, n,… is the number of each element in the molecule. Going stepwise, we first calculate the isotopic pattern for the 55 C-atoms of chlorophyll a. This is possible with the binomial approach.

70 71 Supporting information Chlorophyll a

(m−2) 2 (m−3) 3 m m(m − 1)a × a m(m − 1)(m − 2)a × a (a + a ) = am + ma(m−1) × a + 1 2 + 1 2 + .... am 1 2 1 1 2 2! 3! 2

with a1 = 0.9893, a2 = 0.0107 and m = 55 If we were to write out all the terms, we would come to the large number of m+1 = 56 terms. The probability, that all C-atoms are 12C or respectively 13C is given by:

12 55 55 C55 : a1 = 0,9893 = 55,34% 13 55 55 C55 : a2 = 0,0107 = 0% To keep the calculatory effort required for the isotopic combinations lying between these extremes within reasonable bounds, the series is curtailed after the 5th term. This leads to the following distribution:

m/z 868.55 869.55 870.56 871.56 872.56

12 13 C(55–x) Cx x = 0 x = 1 x = 2 x = 3 x = 4 Intensity 0.5534 0.3292 0.0961 0.0184 0.0026

12 13 Subsequently, the intensities of each C(55–x) Cx-species is combined with the natural abundance of the three Mg-isotopes 24Mg (78.99%), 25Mg (10.0%), 26Mg (11.01%).

12 13 24 12 13 25 12 13 26 The resulting distribution of intensities for C(55–x) Cx Mg, C(55–x) Cx Mg and C(55–x) Cx Mg is summarized in the following table:

C55Mg m/z Intensity 12 24 C55 Mg 892.54 0.4371 12 25 C55 Mg 893.54 0.0553 12 26 C55 Mg 894.54 0.0609 12 13 24 C54 C Mg 893.54 0.2600 12 13 25 C54 C Mg 894.54 0.0329 12 13 26 C54 C Mg 895.54 0.0362 12 13 24 C53 C2 Mg 894.54 0.0759 12 13 25 C53 C2 Mg 895.54 0.0096 12 13 26 C53 C2 Mg 896.54 0.0106 12 13 24 C52 C3 Mg 895.54 0.0145 12 13 25 C52 C3 Mg 896.54 0.0018 12 13 26 C52 C3 Mg 897.54 0.0020 12 13 24 C51 C4 Mg 896.54 0.0020 12 13 25 C51 C4 Mg 897.54 0.0003 12 13 26 C51 C4 Mg 898.54 0.0003

To compare the manually calculated distribution of intensities with the computer-simulated pattern of peaks, the results are finally ordered according to m/z. The isotopic pattern obtained in this way agrees well with the exact distribution that the computer software delivers, despite the discussed approximations.

72 Supporting information Chlorophyll a

m/z Species (Int.) Species (Int.) Species (Int.) Summe der Int. rel. Int.

12 24 892.54 C55 Mg (0.4371) 0.4371 100 12 25 12 13 24 893.54 C55 Mg (0.0553) C54 C Mg (0.2600) 0.3153 72.1 12 26 12 13 25 12 13 24 894.54 C55 Mg (0.0609) C54 C Mg (0.0329) C53 C2 Mg (0.0759) 0.1697 38.8 12 13 26 12 13 25 12 13 24 895.54 C54 C Mg (0.0362) C53 C2 Mg (0.0096) C52 C3 Mg (0.0145) 0.0603 13.8 12 13 26 12 13 25 12 13 24 896.54 C53 C2 Mg (0.0106) C52 C3 Mg (0.0018) C51 C4 Mg (0.0020) 0.0144 3.3 12 13 26 12 13 25 897.54 C52 C3 Mg (0.0020) C51 C4 Mg (0.0003) 0.0023 0.5 12 13 26 898.54 C51 C4 Mg (0003) 0.0003 < 0.1

Fig. S2.5-16 Comparison of the manually calculated (a) and computer simulated (b) isotopic distributions for M+• of chlorophyll a.

What the computer generated isotopic pattern does not show is the composition of the individual peaks. For ex- 12 13 24 ample the peak with m/z 894.94 with a relative intensity of about 39% is composed to 44.7% of C53 C2 Mg, 12 26 12 13 25 35.9% of C55 Mg and 19.4% of C54 C Mg. An experimentally recorded exact mass corresponds to the weighted mean of these contributions. If you have persevered in following the answer this far, you will have gained an impression of how arduous it could be, in the young years of the author, to evaluate a mass spectrum, armed only with a pocket calculator. All the more reason to rejoice over the possibilities of the present day. However, you should not allow yourself to run into the danger of knowing no longer, what you are doing.

J. What sort of reaction is the ring opening of the cyclopentenone ring under basic conditions? It is the retro-Claisen condensation.

72 73 Supporting information

Chapter 3 3.1 Rafﬁ nose

Fig. S3.1-1 Structure of rafﬁ nose

Fig. S3.1-2 400 MHz 1H NMR spectrum

in D2O of rafﬁ nose obtained by puriﬁ cation by analytical HPLC δH/ppm

IR Spectrum in KBr

100

%T 65

60 %T 55

4000 3500 3000 2500 2000 1500 1000 500 WavenumbersWavenumbers (cm-1)(cm–1) Fig S3.1-3 IR spectrum of rafﬁ nose

Rafﬁ nose in the absence of any chromophore has no UV-Vis spectrum. The IR spectrum is dominated by the great number of OH-groups and their OH-vibrations. The CH-vibrations and the other backbone vibrations are 74 Rafﬁ nose less intense, some strong bands are found in the region of O-C-single bonds around 1100 cm–1.

Selective TOCSY Spectra at 700 MHz in D2O

4' 5' 3' 2' 6'

300 ms

250 ms

90 ms

40 ms

15 ms

1H-NMR

δH/ppm

Fig. S3.1-4 Selective TOCSY of rafﬁ nose, excitation of H-1'

In Fig. S3.1-4 the TOCSY spectrum is repeated for the galactose unit. Re- markable is, that in the fourth trace from the bottom H-4' appears almost as a singlet, since H-4' has an e/a-arrangement (equatorial/axial) to both

H-3' and H-5'. The diastereotopic terminal CH2-group H-6' shows no pro- nounced shift difference.

3'' 4'' 5'' 6a'' 6b''

60 ms

40 ms

20 ms

1H-NMR

δH/ppm

Fig. S3.1-5 Selective TOCSY of rafﬁ nose, excitation of H-3''

Similarly the assignment of the fructose unit is unproblematic. As to be expected, the singlet from H-1'' is not reached by the spin lock. For the last appearing signal of the CH2-group H-6'' it is noticeable, that only one of the two protons couples to H-5'', making a structural prediction possible (Fig. S3.1-5).

75 Supporting information

COSY Spectrum

δH / ppm

4, 2

1'' 6b 3, 6' 6b'' 6a'', 2 3', 5'' 5' 4' 6a, 5, 4''

3''

δH / ppm

Fig. S3.1-6 Expansion of the COSY spectrum of rafﬁ nose

NOESY Spectrum

δH / ppm

4, 2

1'' 6b 3, 6' 6b'' 6a'', 2' 3', 5'' 5' 4' 6a, 5, 4''

3''

δH / ppm

Fig. S3.1-7 Expansion of the NOESY spectrum of rafﬁ nose

76 Rafﬁ nose

HSQC Spectrum 3 6' 1'' 3'' 5 6a 3' 6a'' 4'' 4' 5' 5'' 2' 6b'' 6b 2 4

δC / ppm

6', 1'' 6''

2', 4' 4, 3' 2, 5', 5 3 4''

3''

5''

δH / ppm

Fig. S3.1-8 Expansion of the HSQC spectrum of rafﬁ nose

ESI-(+) MS² and MS³ Fragmentation

Fig. S3.1-9 Decay of the [M+Na]+- ion of rafﬁ nose with the elimination of the galactose residue (formation of C) or respectively of the fructose residue (formation of D) under MS² conditions.

77 Supporting information

As described in the book, in the ESI-(+) mass spectrum of raffi nose [M+Na]+-ions can be observed. This is caused by the solvation of Na+ ions by the uncharged raffi nose. The MS² spectrum of the [M+Na]+-ion (m/z 527) (not shown) shows as almost the only fragment an ion with m/z 365, that is the [M+Na]+-ion eliminates a neutral entity with the composition − C6H10O5 (Δm = 162), as does the [M−H] -ion (cf. main text). Apparently, the electron donating effect of an OH-group, which naturally is lower than that of O−, is suffi cient to trigger the 1,2 hydride migration, so that ulti- mately the elimination of a hexose residue occurs. However, a further proton must migrate, so that the hexose residue can be eliminated as a neutral entity. This can be described in a single step (with a favourable fi ve membered transition state for the H+-migration) or via an intermediary ion pair, in which the H+-transfer occurs. In Fig. S3.1-9 both possibilities are shown for the elimination of the galactose residue with the formation of C. It is evident, that in an analogous way the fructose residue could also be eliminated, leading to an isomeric disaccharide fragment D. There is no reason to believe, that both paths cannot take place simultaneously.

+ [M+Na–C6H10O5]

Fig. S3.1-10 MS3-spectrum of the [M+Na+ + –C6H10O5] -ion of rafﬁ nose after isolation of m/z 365

After the elimination of one of the terminal hexoses from the [M+Na]+-ion of rafﬁ nose the excitation energy is exhausted and consequently no further + fragmentation is observed. This can only occur, when the [M+Na-C6H10O5] - ions are again subjected to impact activation (MS³) (Fig. S3.1-10). Starting

from m/z 365 differences in mass of Δm = 60 (C2H4O2), 90 (C3H6O3) and

120 (2×C2H4O2), which all fulﬁ l the molecular formula CnH2nOn, appear.

78 Rafﬁ nose

Fig. S3.1-11 Formation of the ions m/z 245, 275 and 305 in the MS³ spectrum of the [M+- + Na-C6H10O5] -ion (m/z 365) Again there are several possibilities, how the degradation process of the galactose, glucose or fructose residues in the disaccharide entities C or D can take place. This will be exempliﬁ ed by the fragmentation of the glucose unit in C (Fig. S3.1-11). A ring opening between the anomeric C-atom and the O-atom of the ring fulﬁ ls the condition for the elimination of dihydroxy- acetone (formation of m/z 275), glycolaldehyde (formation of m/z 305) and glycolaldehyde + ethendiol (formation of m/z 245). The fragment with m/z

203 corresponds to the loss of C6H10O5 from m/z 365. This corresponds to the elimination of one of the hexose residues from C or D as an anhydroke- tose according to the scheme shown in Fig S3.1-9 for the fragmentation of the [M+Na]+-ion. Finally complexes of the three involved hexoses with Na+ (m/z 203), which can be partially dehydrated (m/z 185), remain.

Questions and Answers

A. Describe the ring chain tautomerism of D-glucose and explain, what is meant by mutarotation.

Whoever buys dextrose in a chemist's shop, buys pure, crystalline α-D-glucose. Immediately after it has been dissolved in water, the hemiacetal function reacts with water. A ring opening occurs and the open chain hydroxy-aldehyde is formed (central structural formula as Fischer projection). This is the structural representation, in which aldohexoses are usually shown, when the structural relationship to each other is to be depicted. In solution only about 0.002% is present in this form. The open chain form now has four possibilities to react.

79 Supporting information

The OH-group on C-5 can react with the planar (!) aldehyde group (–CHO) from either above or below the plane. This results in two anomeric pyranoses (six membered rings), namely α-D-glucose or the β-D-glucose. There is, however, the energetically less favourable possibility, to form furanoses as fi ve membered rings. For this the 4-OH group must react with the –CHO- group, whereby again two possibilities exist. Each pair of pyranoses or furanoses are diastereoisomers. The formation of the fi ve membered ring is considerably less favourable. Under each formula the amount in percent in solution after reaching equilibrium is shown. To reach equilibrium takes about 8 hours at room temperature! That the β-D-glucose prevails against the α-anomer, demonstrates the effect on the energy content of the molecule, when all substituents are positioned equatorially, that is with the greatest separation to each other (compare the discussion to Question B). Of course, the equilibrium can also be reached, starting from β-D-glucose. The degree of equilibration, also referred to as mutarotation, (from Latin mutare = change) can be determined, by measuring the rotation of linear polarised light with a polarimeter. Pure α-D-glucose has a specifi c rotation of +112°, pure β-D-glucose of +19°. That can only be determined by extrapolation and not by measurement, since the mutarotation starts instantaneously with dissolution. Equilibrium is reached, when a specifi c rotation of +52.7° is reached. N.B. The percentages are valid in aqueous solution at 31°C and are quoted from the book: T. K. Lindhorst "Essentials of Carbohydrate Chemistry and Biochemistry", Wiley-VCH, 2000, p.12.

B. Why, amongst the aldohexoses are D -glucose, D -mannose and D-galactose the monosaccharides that most frequently occur naturally? Explain this in terms of their structure. In comparison to other aldohexoses these sugars are molecules with a lower energy content. This can be demonstrated structurally. All arrangements, in which two substituents are arranged with the least steric hindrance are thermodynamically favoured than those that are in close proximity. For the pyra- nose ring the e/e-arrangement (equatorial/equatorial) of two neighbouring substituents, which is referred to as a trans-arrangement, is energetically more favourable than an a/e-arrangement (axial/equatorial), which is a cis-arrangement. The following diagram is to be understood as follows: in β-D-glucose there are four trans-arrangements and therefore it has an all trans-conﬁ guration. That is the most favoured case. In all other compounds there are three trans-arrangements and only one cis-arrangement, because in D-mannose the 2-OH and in D-galactose the 4-OH are arranged axially (all three sugars are related to each other as diastereomers). With all other monosaccharides this is not the case. They therefore have a higher energy and were rejected during evolution. The situation at L-glucose is the same like at D-glucose, however it does not occur in nature.

80 Rafﬁ nose

If two substituents that are separated by three C-atoms are cis to each other (i.e. both are in an axial position), then this is energetically particularly unfavourable. This is illustrated by the structure at the bottom right. It also shows β-D-glucose in an inverted conformation with three 1,3-diaxial interactions. This is the energetically most unfavourable conformation and does not occur in nature. This conformer has 13.4 kJ×mol–1 more energy than the conformer of β-D-glucose that actually occurs and is shown at the top left.

N.B. In all six of the most frequently occurring D-hexoses shown here, there is not a single 1,3-diaxial interaction.

C. Why from these is D-glucose the sugar that is produced by photosynthesis? Explain this also structurally.

β-D-glucose is the only simple sugar, in which all ﬁ ve substituents can be arranged equatorially. Amongst thousands of glucosides that exist as natural products, there is a marked predominance of the β-glucosides over the α-glucosides. The reasons are the same as those in the answer of Question B.

Paradoxically, photosynthesis is nature's process for storing energy and with D-glucose produces a substance that is rich in energy, to act as a supplier of energy for the entire food chain, although in comparison to its related diastereomers D-glucose has the least energy.

D. What is the difference between a reducing and a non-reducing sugar? What is used to prove the presence of a reducing sugar? A reducing sugar contains as structural element a cyclic hemiacetal. In solution it can open, to form an open chain aldehyde (cf. scheme in Question A). The general name is an aldose. A property of the aldehyde group is, that it is easily oxidized to a carboxyl group, whereby the oxidising agent is reduced. For this reaction Tollens’ reagent, an ammoniacal silver nitrate solution, which contains a soluble silver(I) + + complex, the ion [Ag(NH3)2] , can be used. The oxidative potential is the tendency of Ag ions to be converted 0 to metallic silver (Ag ). Everybody, who has had a drop of AgNO3 solution on his skin, has experienced this. In an instance the skin turns black from colloidal silver. Another reagent uses the reagent potential of a metal, namely the Cu2+ ion of copper. Fehling's solution is a

mixture of two solutions, i.e. a dilute copper sulphate (CuSO4) solution and an alkaline solution of potassium sodium tartrate. On mixing an ink blue copper tartrate complex is formed, in which the complexed Cu2+ is the

oxidising agent. In the reaction the red Cu2O, copper(I) oxide, with a lower oxidation number of copper, is formed.

E. What is the reason, that in a selective TOCSY spectrum only exactly one sugar unit is displayed? The transfer of magnetization by a spin-lock requires an uninterrupted spin coupling chain. Between the anomeric protons of sugars and the next proton of the subsequent sugar unit are, however, four chemical bonds and thus, a vanishing spin coupling. In the case of fructose, there is even a quaternary center.

F. The sequence of the chemical shifts of the anomeric C-atoms 1 and 1' is the opposite to that of the respective protons 1 and 1'. Why? A simple incremental estimation for the C-atoms without any corrections gives for C-1 a greater shielding, because in comparison to C-1' this C-atom sees a γ-C-atom and a γ-O-atom more than C-1'. This ﬁ nding is in agreement with experiment. An incremental calculation (ChemBioDraw) for both anomeric protons gives ex- 13 actly the same value of δH = 5.40, which contradicts experiment. Whereas for the C-atom the inﬂ uences from the closer vicinity are usually well documented, this is less certain for protons. H-1 as in sucrose is strongly

deshielded at δH = 5.42; the greater shielding of the H-1'of galactose supposedly has conformational grounds, which have less inﬂ uence on the C-atoms.

81 Supporting information

G. To record the ESI minus spectrum a solution of rafﬁ nose in methanol/water was used. How is it possible, that under neutral conditions [M-H]–-ions are formed by deprotonation? During measurements in the negative mode the capillary of the ESI ion source has a high negative potential. The aim is, that by coulombic repulsion of the anions and attraction of the cations more negative than positive ions accumulate in the drop that detaches itself from the end of the capillary. Consequently, after desolvation of the drop the excess anions are released into the gas phase. The negatively charged capillary can also act as a cathode and reduce a proton of water to hydrogen, whereby simultaneously OH–-ions are formed. These deprotonate the acid groups of the analyte:

− − − − H2O+e → ½ H2 + OH ; X-OH + OH → H2O + X-O

H. Why are the neutral molecules with the formula C6H10O5 that are eliminated during the fragmentation of the quasi-molecular ions of rafﬁ nose referred to as anhydroketoses?

Molecules with the molecular formula C6H10O5 and structure 2 or 4 can be formally derived from the open chain ketohexoses 1 or 3 by the elimination of water from two alcoholic OH-groups.

82 Carbohydrates and Glycosides Fraxin

3.2 Fraxin

Fig. S3.2-1 Structure of fraxin

Isolation Older methods such as the precipitation of phenols with poisonous lead acetate [S1] can be avoided, as can methods with instrumental requirements such as preparative HPLC [S2, S3] or even HSCCC (High-speed countercurrent chromatography) [S4]. The work up of the fractions from the column chromatography requires their evaluation by thin layer chromatography (TLC). It was found, that the eluent n-butanol/glacial acetic acid/water (4:1:2; v/v/v) [S5] described for both glucosides, worked well for the separation but required an extremely long development time. It is advantageous to replace it by a mixture of ethyl acetate/methanol/water (15:3:2; v/v/v) which has a similar separating power but the separation is much faster. The

Rf-values in this mixture are: aesculin 0.57, fraxin 0.50. Both substances show a fluorescence quenching at 254 nm but each has a different intrinsic fluorescence when illuminated at 366 nm (see below). Extraction 1 with distilled water: The ground bark of the ash (50 g) is placed in a glass beaker (1 L) and stirred using a magnetic stirrer at room temperature with distilled water (750 mg) for 4 hours. A yellow-brown extract forms. The suspension is filtered and the bark discarded. The filtrate is lyophilized, to give a red-brown solid (7.2 1 g). An H NMR spectrum in CD3OD shows a ratio of fraxin to aesculin of 1:2.7 (or 27:73 %) as well as further substances in the extract. Extraction 2 with methanol: The ground bark of the ash (50 g) is placed in a 1L round bottomed flask and stirred with methanol (500 mL) for 24 h. The extract changes its colour from yellow to orange brown and the bark ist nearly completely decolor- ised. The suspension is filtered and the bark is saved. Methanol is removed from the filtrate in vacuo. A brown solid is obtained (5.2 g).1 H NMR shows a fraxin to aesculin ratio of 25:75% and further compounds. The bark obtained by filtration of the cold, methanolic extract is added to methanol (600 ml) and heated for one hour under reflux on a water bath. The bark is removed by filtration, the methanol is removed by distillation under reduced pressure and the residue dried in vacuum. A further amount (2.7 g) of a brown solid is obtained. The 1H NMR analysis shows a ratio of fraxin to aesculin of 1:2.3 (or 30:70 %). Both processes can be combined without difficulty, also a single hot extraction is possible. 83 83 Carbohydrates and Glycosides Fraxin

An Extraction with Ethyl Acetate as a Variant, to Remove Aglucons from the Raw Extract: The TLC (for the eluent see above) on aluminium plates shows both glucosides as spots, which on illumination with longwave UV (366 nm) fluoresce with an intensive colour that is different for each glucoside. Fraxin fluoresces turquoise and aesculin sky-blue. Both quench the fluorescence of the fluorescence indicator of the TLC plate, when it is illuminated with shortwave UV-light (254 nm). On inspection of the TLC plate, spots from

clearly more hydrophobic substances with higher Rf-values are seen. These come from the aglucons fraxetin and aesculetin. If it is desired, to separate these substances, which would be advisable before a column chromatography over silica gel, the following process can be used. The solid residue from the extraction (obtained with water or methanol) is boiled with ethyl acetate. Ethyl acetate has the remarkable ability, moderately but selectively to dissolve distinctly hydrophilic substances of medium polarity (e.g. shikimic acid or arbutin). This also applies to glycosides. In this case, it initially dissolves the glycoside and aglycon from the raw extract to the same extent, but on concentrating the solution, the glycoside crystallizes first, whereas the more hydrophobic aglycon remains in solution. In this way, the proportion of aglycon can be successfully reduced, if desired. If the separation is to be performed on Sephadex, this step is unnecessary, since the effectivity of the separation is very good. The procedure is as follows: The raw extract (6 g) is heated under reflux with ethyl acetate (125 mL) in a round bottom flask (250 mL) for 15 minutes and filtered while hot. The procedure is repeated twice. The light beige extracts are united and reduced in volume (to about 50 mL) under vacuum at 40°C, whereby the solution becomes turbid. The suspension is filtered to obtain a beige solid (400 mg) that is enriched in fraxin and aesculin, which is suitable to isolate aesculin by column chromatography over silica gel.

A Short Description of the Structure and Mode of Operation of Sepha- dex LH-20 Sephadex is a tradename of the Swedish firm Pharmacia and stands for "Separation Pharmacia dextran" and is a chemically cross-linked dextran gel. Dextrans are strongly branched polymer carbohydrates of high molecular mass on the basis of α-D-glucose alone, which are formed by bacteria to store energy. The molecular mass can be from 10 kDa to 50,000 kDa (!). The subsequent chemical cross-linking of the glucose chains is achieved with glycerine ether bonds that are introduced with the aid of epichlorohydrin in alkaline solution. Through the cross-linking the dextrans lose their original solubility in water. The grade of cross-linking depends on the amount of epichlorohydrin, whereby the swelling power of the gel decreas- es with increasing cross-linkage. Through a special manufacturing process, the gel is obtained in the form of small spheres that have a characteristic internal pore structure and are therefore suitable for gel permeation chromatography (size exclusion chromatography, SEC). Decisive for the separating effect on passage through such a column is not a mechanical "filtration effect" according to the size of the molecule but the difference in the hydrodynamic volume of macromolecules of different sizes (typically used for

84 85 Carbohydrates and Glycosides Fraxin

biopolymers such as carbohydrates, proteins, DNA, RNA). It follows, that small molecules can penetrate the pores but large ones cannot. Small molecules are therefore eluted after the large ones that can quickly pass through the free volume between the Sephadex spheres, without being retained by permeation into the pores. Because of the three hydroxyl groups per glucose unit, Sephadex gels can only be used with aqueous solutions. To make the gels useable with polar to medium polar organic solvents the free hydroxyl groups are hydroxypro- pylated. The number of hydroxyl groups is thereby not diminished but because of the three propyl residues per glucose unit the lipophilic properties are increased and a quasi ambiphilic gel with amazing properties is obtained – also for the separation of very similar compounds of low molecular mass (Fig. S3.2-2). The separation of fraxin and aesculin is an example of this. A cross-linked and hydrophobized dextran gel can not only still be swelled by and used as a stationary phase with water but also with numerous organic solvents such as acetone, dichloromethane, ethyl acetate and chloroform. The monograph cited below describes hundreds of applications for this amazing material. Often it is possible to use a very high loading, which is not typical for silica gel. Mixtures as eluent and variation of the eluent, which causes a change in swelling, should with Sephadex be avoided as should air bubbles in the column, the latter applies to all chromatographic columns. Recommended literature: H. Henke, "Präparative Gelchromatographie an Sephadex LH-20", Monograph, Hüthig Verlag, Heidelberg, 1994. The combined principle of separation, which is based not entirely on the molecular sieve effect but apparently includes very selective adsorption effects and partition chromatography, is useful for the separation of Fig. S3.2-2 An excerpt from the structural formula two such glycosides of low molecular mass. Both glycosides interact with of Sephadex LH20 the gel matrix by the formation of differently strong hydrogen bonds and van der Waals forces that in contrast to silica gel allows the isolation of both substances in the pure form.

84 85 Carbohydrates and Glycosides Fraxin

Column Chromatography of the Ethyl Acetate Extract on Silica Gel for the Isolation of Aesculin Specification: length of column: 250 mm; diameter: 30 mm; stationary phase: silica gel 60, Merck, 0.035-0.070 mm; mobile phase: ethyl acetate → ethyl acetate: ethanol 4:1 (v/v) A portion of the above mentioned ethyl acetate extract of the raw extract (depleted in aglycon) is purified by column chromatography over silica gel 60. The solid ethyl acetate extract (350 mg) is dissolved in methanol (25 mL) and mixed with a spatula of silica gel 60. The solvent is removed under reduced pressure at 40°C, the residue dried under high vacuum and applied to the column. The column is eluted with ethyl acetate and fractions collected (15 mL each). The fractions collected are investigated with TLC (EtOAc/

MeOH/H2O 75:15:10 v/v/v; Rf-value for aesculin 0.57, for fraxin 0.50) and united to collective fractions. Collective fraction 1 (the first 300 mL eluate) contains apart from aesculin aglycon and is discarded. The collective fraction 2 (the next 2100 mL eluate!) contains only aesculin. After removal of the solvent in vacuum pure aesculin (165 mg) is isolated, mp 195-197°C with decomposition (Litera- ture value [7] 204-206°C, vide supra) Comment to the attempt, to obtain pure fraxin in this way. A mixture of aesculin and fraxin was eluted. The eluate became increasing- ly rich in fraxin but never contained fraxin alone. Using a more polar eluent (ethyl acetate:ethanol 4:1 (v/v)) did not change this. The separation was hindered by the extreme tailing of aesculin, which the mixture contained in excess. The two substances proved to be separable and obtainable in a pure form only by the use of Sephadex. The remaining fractions were therefore discarded.

HO 6'

4' 5' O 5 4 HO O HO 6 4a 3 2' 3' 1' OH 7 2 8a HO O O 8

δH / ppm

1 Fig. S3.2-3 H NMR spectrum of aesculin in CD3OD at 700 MHz 86 87 Carbohydrates and Glycosides Fraxin

23 –1 Specific rotation of aesculin: [αD ] = –67.5° (c=11.0 mg.mL , methanol) 1 Rf-values, IR and H NMR spectra of both aesculin samples proved to be identical. It is not intended, to deliver a comprehensive discussion on the spectroscopy of aesculin here. Only the two one-dimensional NMR spectra (see Fig. S3.2-3 and S3.2-4), which are a proof of the purity of the substance, are shown.

13 Fig. S3.2-4 APT C NMR spectrum of aesculin at 700 MHz in CD3OD

Additional spectra of fraxin IR Spectrum in KBr

100 95 90 85 80 75 70 65 60 55 %T

%T 50 45 40 35 30 25 20 15 10 4000 3000 2000 1500 1000 500 WavenumbersWavenumbers (cm(cm–1)-1)

Fig. S3.2-5 IR-spectrum of fraxin

86 87 Carbohydrates and Glycosides Fraxin

Additional NMR Spectra at 700 MHz in CD3OD

4 5 3

δH / ppm

δH / ppm Fig. S3.2-6 Excerpt of the DQF-COSY-spectrum of fraxin

4 5 3

δH / ppm

δH / ppm Fig. S3.2-7 Excerpt of the NOESY spectrum of fraxin in the aromatic region

88 89 Carbohydrates and Glycosides Fraxin

EI Mass Spectrum of Fraxetin 100 208 In the molecular ion of fraxetin, which delivers the base peak, the 80 cleavage of a methyl radical (m/z 193) competes with a decarbony- 60 lation (m/z 180). By a subsequent

elimination of CO or CH3• respec- 44 tively, both fragments can lead to a 40 % Intensity% common ion at m/z 165. From m/z 60 81 137 193 165 three successive CO-cleavages 20 73 109 165 180 take place that end at m/z 81 (Fig. 3.2-12). In the course of fragmen- 0 tation, four of the five O-atoms of 40 60 80 100 120 140 160 180 200 220 fraxetin are eliminated as CO. m/z

Fig. S3.2-9 EI Mass spectrum of fraxetin, that is formed in the ion source from fraxin by an initial thermal degradation

Fig. S3.2-10 Fragmentation scheme of fraxetin

• Fig. S3.2-11 CH3 and CO elimination from the molecular ion of fraxetin

Since CO can be eliminated from both the lactone and phenolic moieties of fraxetin, it is not easy, to give a mechanistic rationalization of the fragmentation scheme. In the fragmentation scheme shown in Fig. S3.2-11, it is assumed, that the first CO molecule lost comes from the lactone group.

88 89 Carbohydrates and Glycosides Fraxin

+ The C5H5O ion (m/z 81) that is found at the end of the degradation sequence is a heterocyclic analogue of the benzyl cation. It is therefore plausible, that corresponding to the equilibrium between the benzyl and tropylium cations, it is in equilibrium with the 6π-aromatic pyrylium ion.

Fig. S3.2-12 Scheme for the further fragmentation of fraxetin

Literature

[S1] Fürst zu Salm-Horstmar, Chemisches Zentralblatt, 1857, 1, 452. [S2] G. Stanic, B. Jurisic, D. Brkic, Croat. Chem. Acta, 1999, 72, 827–834. [S3] L. Zhou, J. Kang, L. Fan, X.-C. Ma, H.-Y. Zhao, J. Han, B.-R. Wang, D.-A. Guo, Pharm. Biomed. Anal., 2008, 47, 39–46. [S4] R. Liu, Q. Sun, A. Sun, J. Cui, J. Chrom. A, 2005, 1072, 195–199. [S5] CHROMATOGRAPHIC DATA, 1961, 5, D22.

Questions and Answers

A. What is to be understood by the term glycoside? What types of glycosides are there? How do they differ?

Glycosides are derivatives of monosaccharides that are formed by a condensation reaction between the anomeric (glycosidic) OH-group with other, for this purpose, suitable functional groups, such as alcohols, phenols, thiols and amines. Thereby, O-glycosides, S-glycosides and N-glycosides are formed. Glycosides of this type are relatively labile, because they can be hydrolysed, i.e. split into their component parts, by dilute acids, heating in water or enzymatically. In contrast C-glycosides are a special case, because in these an anomeric O-atom has been exchanged against a C-atom, making them considerably more stable (cf. carminic acid in chapter 2.3). If two sugars react together a holoside, if one sugar reacts with a non-sugar a heteroside is formed. Disaccharides such as sucrose, lactose, maltose, cellobiose or trehalose are holosides, as are oligosac- charides (e.g. raffinose) or polysaccharides, which are formed by further reaction according to this scheme. Heterosides are all glycosides that are formed with a non-saccharidic aglycon. Thousands of natural products are of this type, such as the O-glycosides amygdalin, fraxin, aesculin, hesperidin and the digitalis glycosides.

90 91 Carbohydrates and Glycosides Fraxin

B. What role, from a physical perspective, does a molecule that fluoresces play? Fluorescence is the effect, by which a material or molecule shortly after excitation through radiation itself spontaneously emits radiation. Such a molecule functions sequentially as receiver and sender. The emitted light is always of lower energy, i.e. of longer wavelength, than the absorbed light.

C. Fraxin, aesculin, skimmin and daphnin all contain umbelliferone as a substructure. What is the structure? What properties does it determine? How can umbelliferone be used? Umbelliferone is 7-hydroxycoumarin. It is the aglycon of the glycoside skimmin.

Daphnetin and daphnin are hydroxyl derivatives of these substances. Solutions of umbelliferone fluoresce on radiation with longwave UV (366 nm) as do the solutions of all four named glycosides that contain umbelliferone as a substructure. This shows, that as expected the fluorescence emanates from the aglycon. Typical families of plants, in which umbelliferone and skimmin occur, are umbel- lifer (Apiaceae) such as lovage, rue (Rutaceae) such as skimmia, and Thymelaeaceae such as daphne. Umbelliferone is a phenolic compound, weakly acidic with a yellow colour. Its anion, because of a bathochromic shift, is of a darker colour, so that umbelliferone can be used as a pH indicator with an indicator range of 6.5 to 8.9. However, of more interest is the analytical use of its structurally seldom fluorescence. A highly diluted aqueous solution of umbelliferone can be used as a fluorescence indicator for the detection of calcium and copper ions. As a non-hydrophilic substance, the aglycon umbelliferone can be added to sunscreens or used as an optical brightener for textiles.

D. Explain with the aid of an energy diagram, why fluorescence spectra are always at a longer wavelength than the corresponding UV spectrum.

The UV excitation at room temperature always occurs from the lowest vibrational state v0 of the electronic

ground state S0 to the excited state S1, whereby higher vibrational states can be occupied. However, these

relax immediately to the vibrational ground state v0 of the electronic state S1 (internal return). From here, the

fluorescent emission takes place, which can end in a higher vibrational level of the electronic ground state 0S . A comparison of the UV and fluorescence arrows shows, that the latter, with the exception of the 0-0 transition are always shorter and therefore of longer wavelength than the UV-arrows.

90 91 Carbohydrates and Glycosides Fraxin

E. What are excimers? Why does their emission have no structure? Why is the excimer fluorescence shifted to longer wavelengths? Excimers are dimers that only exist in an excited electronic state. The cohesion between the two molecules is based on excitation resonance and charge resonance. Since the emission occurs to a dissociated state without an energy minimum and therefore without discrete vibrational levels, the fluorescence from excimers appears as a structureless continuum. The transition from the excited electronic state is in comparison to that of mon- omers shifted to long wavelengths, because it connects a stabilized excimer with a destabilized, repulsive ground state.

Energy

Potential energy as a function of the intermolecular dis-

tance for the ground state and the first excited state (S1) of an excimer. (From "Physikalisch-Chemisches Fortgeschrit- tenen-Praktikum", K. Müller, H. Dilger, University of Stutt- Intermolecular distance gart, 2008)

F. In both the 1H and 13C NMR spectrum the resonances from the position 4 are distinctly more strongly deshielded than those from the position 3, although the latter are nearer to the electron attracting C=O group. Explain this finding.

This is an α/β-unsaturated carbonyl system, in which the mesomeric effect predominates over the inductive withdrawal of electrons by the carbonyl group. The mesomeric resonance structure illustrates the finding of the NMR.

G. In the supporting information the EI induced fragmentation of the aglycon fraxetin is discussed, starting with the elimination of CO from the pyranon ring. Make a suggestion for the cleavage of the first molecule of CO from the phenolic substructure of fraxetin.

If the first loss of CO is from the phenolic ring, or as shown in the Fig. above from the lactone group, or if both processes occur simultaneously, could only be clarified by an elaborate labelling experiment.

92 92 Supporting information Stevioside

3.3 Stevioside

Fig. S3.3-1 Structure of stevioside

Final Purification of Rebaudioside A From the 1H NMR spectrum it is evident, that a small amount of stevioside is still to be found in rebaudioside A. For the final purification, rebaudioside A (60 mg) from the first column chromatography is dissolved in eluent given below (2 mL) and placed on the column. During elution, fractions

(10 mL) are collected and investigated by TLC (eluent MeOH/CHCl3/H2O (25:65:4 v/v/v)). As for stevioside it is detected by treating with the See- bach's oxidative phosphomolybdic acid reagent. The fractions 15 to 20 are determined to be pure and are united and the solvent is removed under reduced pressure. Rebaudioside A (27.3 mg) is obtained as an amorphous, pale yellow solid with mp 196 – 200°C (literature value [S4]: 235 – 237°C).

Rf: 0.17 (MeOH/CHCl3/H2O (25:65:4 v/v/v)). HR-ESI(-)MS: m/z =965.42 [M-H]+, cf. calculated value 966.43 for neutral rebaudioside A.

Some Spectroscopic Data for Rebaudioside A Rebaudioside A is regarded to be the steviolglyco- side with the highest sensorial value, i.e. it is the most sweet and the least bitter one. Therefore, for those who are interested, we describe its data here.

UV Spectrum Fig. S3.3-2 Structure of rebaudioside A 1

0,9

0,8

0,7

0,6

0,5

0,4 Absorption 0,3

0,2

0,1 Fig. S3.3-3 UV spectrum of rebaudiosi- 0 de A in ethanol (0.7 mg substance in 10 200 300 400 500 600 700 800 900 1000 mL solvent, path length of the cuvette 10 λ/ nm Wellenlänge nm mm) 93 93 Supporting information Stevioside

This molecule also contains only two chromophores that are not conjugated, a C=C and a C=O double bond. Therefore the UV/Vis spectrum is un- spectacular and the substance colourless. The maximum of the absorption lies at 206 nm with a value of 0.588. This corresponds to an extinction coefficient of 8122.7 cm2×mmol–1. As with stevioside this could be the π→π* transition of the C=C bond.

IR Spectrum 100 *Rebaudiosid A 95 _

%T 60 % T 55

25 Fig. S3.3-4 IR spectrum of rebaudioside 4000 3500 3000 2500 2000 1500 1000 500 –1 A in KBr WavenumbersWavenumbers ( cm(cm-1) ) The broad absorption band between 3100 and 3600 cm–1 can be assigned to the stretching vibration of the OH-groups of the glucose residues. Between 2800 and 3000 cm–1 the stretching vibration frequencies of the C-H bonds

can be found. The C-H stretching vibration of the terminal C=CH2 group is probably covered by the intense OH absorptions. At about 2350 cm–1

the absorption of CO2 is detected. This band can be caused for example by breathing into the IR spectrometer. The stretching vibration of the ester function is found at 1730 cm–1. The C=C stretching vibration at about 1650 –1 –1 cm can be assigned to the C=CH2 group. Between 1300 and 1500 cm the deformation vibration frequencies of the CH bonds of the aglycone can be seen. At 1230 cm–1 the deformation vibration of the hydroxyl groups of the glucose residues are to be found. The C-O stretching vibrations of the glucose residues follow at 1080 and 1040 cm–1. The small band at 900 cm–1

can be assigned to a deformation vibration of the C=CH2 group. These data agree with those found in the literature [S2]. NMR Spectra of Rebaudioside A at 400 MHz A 1H and a 13C NMR spectrum of rebaudioside A were recorded, to check its purity. The chemical shift of the anomeric protons, which are characteristic for rebaudioside A, agree with those reported in the literature [S5]. A comparison of the glucose protons was not carried out, because these, as with stevioside, overlap each other in the region 3.75 ppm to 4.60 ppm. Spectra recorded at different temperatures show, that in solution rebaudioside A undergoes a dynamic process. Because of this, broad signals are 94 95 Supporting information Stevioside

found in the spectrum at ambient temperature (26°C). At higher temperatures (70°C or 90°C), these signals sharpen. Because of the dynamic of the molecule and the broad signals that it causes, a signal of the methylidene proton on C-17 is hidden. The signals in the 13C spectrum show only very small deviations from values in the literature. It can be concluded, that the substance isolated has a very high purity. This is in agreement with the finding, that in the TLC only one spot is found.

1 Fig. S3.3-5 H NMR spectrum of rebaudioside A measured at 26°C in pyridine-d5

1 Fig. S3.3-6 H NMR spectrum of rebaudioside A measured at 70°C in pyridine-d5 94 95 Supporting information Stevioside

1 Fig. S3.3-7 H NMR spectrum of rebaudioside A measured at 90°C in pyridine-d5

13 Fig. S3.3-8 C NMR spectrum of rebaudioside A in pyridine-d5

96 97 Supporting information Stevioside

1H NMR Assignment Table for Rebaudioside A Comparison of the chemical shifts of the protons of the isolated rebaudioside A with the values found in the literature [S5]

Number of Atom δH, Lit. δH 1 0.78; 1.77 0.76; 1.76 2 2.22; 1.45 2.22; 1.45 3 1.03; 2.35 1.03; 2.35 4 / / 5 1.05 1.05 6 2.45; 1.92 2.46; 1.92 7 1.30 1.29 8 / / 9 0.88 0.88 10 / / 11 1.68 1.68 12 2.25; 2.00 2.23; 2.01 13 / / 14 2.66; 1.81 2.65; 1.81 15 2.05 2.04 Fig. S3.3-9 From the pharmacy – a mix- 16 / / ture of steviolglycosides 17 5.01; 5.64 hidden; 5.64 18 / / 19 1.25 1.24 20 1.32 1.31 1' 6.12 6.12 1'' 5.08 5.06 1''' 5.58 5.56 1'''' 5.33 5.33

Exercise for the practical application of the knowledge acquired. Inves- tigation of powdered stevioside extract from the pharmacy To obtain information about the solubility of stevioside and rebaudioside A and the NMR spectroscopic data the following extract of stevia were investigated: VitaNatura.de, Stevioside Extract Powder from Stevia Rebaudi- ana, 50 g, Order Nr. 24-001-0004 (purchased from the "Europa-Apotheke", Leipzig, Reudnitz) (Fig. S3.3-9).

The investigation of this beige coloured powder by TLC (eluent: MeOH/

CHCl3/H2O (25:65:4, v/v/v)) showed two very intense blue spots after treatment with Seebach's reagent (Fig. S3.3-10). From the TLC plate, it is immediately clear, that the polar substance with

the lowest Rf has the highest concentration in the extract (higher than that Fig. S3.3-10 TLC of Stevia Extract from of the next but one substance above). Stevioside and rebaudioside A are VitaNatura.de

96 97 Supporting information Stevioside

the steviol glycosides with the highest concentrations in Stevia rebaudiana and we can therefore assign these substances to these spots. However, what belongs to which spot? Since rebaudioside A with the greater number of hydroxyl groups (four glucose residues) is more polar than stevioside (three glucose residues), it runs slower on silica gel and is to be found less eluted on the TLC plate. The most intensive spot can therefore be assigned to rebaudioside A and the second most intensive is from stevioside. What does this mean for this particular batch? The list of contents on the label of the stevia extract, which states "a content of reb-A of about 25%", cannot be correct. The content of rebaudioside A in the mixture of steviol glycosides is at least in this case much higher. This finding is confirmed by1 H NMR spectroscopy (Fig. S3.3-11).

Fig. S3.3-11 400 MHz 1H NMR spectrum of stevioside extract powder from

the pharmacy in pyridine-d5 (only the region of the anomeric protons is shown)

The ratio of stevioside to rebaudioside A can be calculated using the integrals of the characteristic signals for the anomeric protons. This shows, that if only stevioside and rebaudioside A were contained in stevia extract, the content of rebaudioside A would be about 61% and the content of stevioside about 39%. It seems likely, that since the substance is an extraction of naturally occurring material, the composition will be different from batch to batch. An NMR measurement on the powder is a quick way to clear up the situation.

Query to CAS-Helpdesk (Problem of nomenclature regarding ent-Kau- rene oder Kaurene) Dear helpdesk colleague, I would like to request your help in the following matter: Nomenclature of stevioside (CAS RN 57817-89-7), i.e. especially the name of the agylcone in this glycoside. What is correct?

Your name for this compound is: Kaur-16-en-18-oic acid, 13-[(2-O-β-D-glucopyranosyl-β-Dglucopyranosyl)oxy]-, β-D-glucopyranosyl ester, (4α)- (taken from Scifinder ®) In several papers and also in the book STEVIA (ISBN 0-415-26830-3) I have found the description that "steviosid is a diterpene glucoside having

98 99 Supporting information Stevioside

the ent-kaurene diterpene aglycone, steviol, ..." Furthermore, several papers set the number for the carboxylic acid group to 19 and not to 18 as Chemical Abstracts does. I guess, that 19 just wrong, or isn't it? But my main problem is, kaurene or ent-kaurene? I do completely trust in your work and I would like to know what is the (possible) reason for the continuous enantio-assignation in such a lot of papers? Is'nt that simply wrong? I do very much appreciate your comments on this, thank you in advance. ……….. (D.S.) This was directly answered: I asked our Production and Content Operations [PCO, formerly Editorial Operations (EO)] to field your inquiry; the following is PCO's response: The name for stevioside (CAS RN 57817-89-7) is correct as presented: Kaur-16-en- 18-oic acid, 13-[(2-O-β-D-glucopyranosyl-β-D-glucopyranosyl)oxy]-, β-D-glucopyranosyl ester, (4α)-. CAS has assumed that the predominant naturally-occurring enantiomeric form of Kaurane would be the CAS parent for this diterpene stereoparent. The literature calls this ent-kaurane, but CAS calls this both names. (See RN 1573-40-6). Names for kauranes on our database only cite stereochemistry which differs from the assumed ("ent") parent (1573-40-6), as opposed to having to cite all stereocenters present based on the literature structure "Kaurane". Inherent in this nomenclature is the assumption that for any stereoparent which has equivalent unsubstituted nodes (e.g. 18 and 19 in this parent), any substitution implicitly is on the lowernumbered node, in this case 18-. After this locant is assigned, the stereo at the resulting ring chiral center is stated based on CIP rank of C-18 vs. C-19; the higher ranking node will be assigned either α or β in an absolute sense based on the "standard orientation" of Kaurane. In the case of Stevia derivs., this will be 4α. Thank you for your inquiry; PCO appreciates the opportunity to clarify. … (CAS Customer Center)

Literature [S1] V. Jaitak, B. Bikram Singh, V. K. Kaul "An efficient microwave-assistant extraction process of Stevioside and Rebaudioside A from Stevia rebaudiana (Bertoni)" Phytochem. Anal. 2009, 20, 240–245. [S2] G. I. Kovylyaeva, G. A. Bakaleinik, I. Y. Strobykina, V. I. Gubskaya, R. R. Sharipova, V. A. Al'fonsov, V. E. Kataev, A. G. Tolstikov "Glycosides from Stevia rebaudiana" Chem. Nat. Compd. 2007, 43, 81–85. [S3] H. B. Wood, Jr., R. Allerton, H. W. Diehl, H. G. Fletcher, Jr. "Stevioside. I. The structure of the Glucose moieties" J. Org. Chem. 1955, 20, 875–883. [S4] M. Kobayashi, S. Horikawa, I. H. Degrandi, J. Ueno, H. Mitsuhashi "Dulcosides A and B, new diterpene glycosides from Stevia rebaudiana" Phytochemistry 1977, 16, 1405–1408. [S5] W.E. Steinmetz, A. Lin "NMR studies of the conformation of the natural sweetener rebaudioside A" Carbo- hydr. Res. 2009, 344, 2533–2538.

98 99 Supporting information Stevioside

Additional Spectra for stevioside, NMR spectra at 700 MHz in pyridine-d5 IR Spectrum in KBr

100 Steviosid 98

88 %T –1 % T Between 3100 and 3600 cm the 86 stretching frequencies of the OH- 84 groups of the glucose residues can 82 be found followed by the stretching 80 vibration of the sp3-hybridized C-H 78

76 bonds in the region 2800 to 3000 4000 3500 3000 2500 2000 1500 1000 500 cm–1. The C-H stretching vibration WavenumbersWavenumbers (cm-1) (cm–1) of the terminal C=CH2 group can be identified as a small shoulder at Fig. S3.3-12 IR spectrum of stevioside 3000 cm–1. At about 1750 cm–1 the C-O stretching vibration of the ester group is found. The absorption band at about 1640 cm–1 can be attributed to the C=C stretch-

ing vibration of the terminal C=CH2 group.

COSY Spectrum

17b 1' 17a 1''' 1''

δH / ppm

17x 1'' 1'''

17x

δH / ppm

Fig. S3.3-13 DQF COSY spectrum of stevioside in the sugar region

100 101 Supporting information Stevioside

Selective TOCSY Spectra

6'''a 4''' 3''' 2''' 5''' 6'''b

200 ms

80 ms

60 ms

40 ms

20 ms

normal 1H-NMR

δH / ppm

Fig. S3.3-14 700 MHz sel-TOCSY spectrum at the frequency of H-1'''

3'' 2'' 4'' 5'' 6''a

150 ms

80 ms

55 ms

40 ms

20 ms

normal 1H-NMR

δH / ppm

Fig. S3.3-15 700 MHz sel-TOCSY spectrum at the frequency of H-1''

100 101 Supporting information Stevioside

2''' HSQC Spectrum 3' 6' 4''' 3'' 6'' 2'' 5''' 6'' 6''' 6''' 6' 4' 3''' 2' 4'' 5' 5''

δC / ppm

6' 6''' 6''

4',4''' 4'' 2'

2''' 3' 5'' 3''' 3'' 5''' 5'

2'' 13

δH / ppm

Fig. S3.3-16 HSQC spectrum of stevioside in the sugar region

15 20 19 11 7 5 3 14 6 3 12 2 12 6 14 2 9 1 1 δC / ppm

2 11

12, 3 1, 7 14 15

9 5

δH / ppm

Fig. S3.3-17 HSQC spectrum of stevioside in the aliphatic region

102 103 Supporting information Stevioside

ROESY Spectrum Because of the relatively high molecular weight, the polarity of both stevioside and the solvent and the resulting viscosity, the NOESY spectra have little relevance. Therefore, the stereochemistry of the proton spin system was investigated with ROESY, because with this method there is no zero-crossing of the cross-relaxation signal. The cross-relaxation signals in the sugar region will not be discussed further here. Fig. S3.3-18 shows the spectrum in the aliphatic region that will be discussed in detail.

15 20 19 7 11 5 3 14 3 12 2 12 6 14 1 9 1 6 2 δH / ppm

1 9 3, 5 19 7, 20 2 11 1, 14 6, 12 15 2, 12 3 6

δH / ppm

Fig. S3.3-18 ROESY spectrum of stevioside in the aliphatic region

We begin again with ring A of the aglycone steviol, which is shown as a fragment in its chair conformation. The methyl group H-19 exhibits a dis-

tinct ROE signal to the signal of the protons H-3 at δH = 2.353. Therefore

this proton is equatorial and its geminal partner at δH = 1.025 axial. The methyl group H-20 shows a clear 1,3-diaxial interaction with the proton H-2 at

δH = 2.210 but not to its geminal partner at δH = 1.439. Therefore, the former is in an axial position, which is confirmed by a crosspeak between H-3e and

H-2e. A crosspeak between the axial H-9 and the signal from H-1 at δH = 0.756 determines its position as an axial proton and that of its geminal part-

ner at δH = 1.736 as equatorial. The 1,3-diaxial interaction between the three protons 1, 3 and 5 is also confirmed in the ROESY spectrum.

In ring B we only have to determine the stereochemistry of the two diastereotopic methylene groups H-6 and H-7. Whereas the signals of the methylene group H-7 are to a large extent overlapped by those of the methylene

group H-20, making the determination impossible, H-6 at δH = 2.487 shows a clear correlation to H-20, while its geminal partner correlates to H-19. The former is therefore determined to be an axial proton.

102 103 Supporting information Stevioside

The methylene protons H-11 show only an overlapping signal, making a further analysis impossible. The protons H-12 also do not allow a definite stereochemical assignment. In contrast, the methylene protons H-14 show a clear diastereotopism. Apart from the trivial, mutual ROE signal, both show

a ROE correlation to H-7, whereby the respective signal from H-14 at δH = 2.724 is distinctly more intensive. We therefore tentatively assign the latter to the axial position.

The protons 15 also appear as a overlapping signal, making a further assignment impossible. The ROESY spectrum shows (without illustration) a

distinct crosspeak between the proton signal at δH = 5.061 and the protons H-15, determining this to be H-17b. It is interesting, that the anomeric proton H-1''of sophorose exhibits ROE interactions with H-12 and H-14a.

EI Mass Spectra of Stevioside and Steviol

Fig. S3.3-19 Thermal decomposition of stevioside in the EI ion source

Because of its polar sugar residues stevioside does not volatilize without decomposition, therefore an EI spectrum of the intact molecule cannot be obtained. On heating to about 300°C thermal decomposition occurs, whereby the aglycone steviol is liberated. The spectrum of the products of thermolysis (Fig. S3.3-19) is strongly "contaminated" by the decomposition of the sugar residues, however, the molecular ion of steviol and the fragment ions that are to be found in its database spectrum (Fig. S3.3-20) are clearly recognizable.

104 105 Supporting information Stevioside

Fig. S3.3-20 Database EI spectrum of steviol

The fragment with m/z 300 can only

occur by the elimination of H2O from M+. After ionisation, alcoholic OH-groups can intramolecularly abstract an H-atom and then leave

as H2O. The H-abstraction occurs preferentially via 5 and 6 membered transition states. In the case of steviol this process would produce an unfavourable carbenium ion that, however, can be avoided by a simul- taneous opening of the C-8-C-14 bond, resulting in the loss of the bi- Fig. S3.3-21 Elimination of water from the molecular ion of steviol to m/z 300 and

cyclic partial structure. the subsequent elimination of CH3 (m/z 285) or H2O + CO (m/z 254)

If by a preceding H-shift the unpaired electron is brought to a position suit-

able for an α-cleavage, the ion with m/z 300 can eliminate a CH3-radical. After a further radical initiated H-shift an isomeric structure forms, which

by loss of H2O and CO fragments to the ion with m/z 254 (Fig. S3.3-21).

104 105 Supporting information Stevioside

Fig. S3.3-22 Elimination of 2-methylacrylic acid from the radical cation A (Fig. S3.3-21) to m/z 214 Between m/z 214 and m/z 300 is a difference in mass of Δm = 86 that can be achieved by two sequential α-cleavages in ion A of Fig. S3.3-21 forming 2-methylacrylic acid (Fig. S3.3-22).

Fig. S3.3-23 Formation of a + C14H17 -ion (m/z 185) from m/z 254 (Fig. S3.3-21) The initially suggested procedure leads from the radical cation with m/z

254 as the precursor, which must lose a C5H9 radical, to the cation with m/z 185. Fig. S3.3-23 illustrates a path with three H-shifts and two α-cleavages.

Fig. S3.3-24 Elimination of acetone from the molecular ion of steviol to m/z 260 and further fragmentation to m/z 146 106 107 Supporting information Stevioside

The formation of the ionic fragment with m/z 260 tests the skill of inter- + +• pretation, although as precursor only the M and [M-H2O] ions come in question. The difference in mass to the M+ ion (Δm = 58) seems more likely +• than that to [M-H2O] (Δm = 40). A possibility is the elimination of acetone or one of its isomers.

The cleavage of a C3H6O unit requires the involvement of the OH-group on C-13 and probably it is necessary, that it shifts from the quaternary C-atom to the neighbouring C-16 position. Starting with a molecular ion that is ionized on the exocyclic double bond, this can be achieved via an epoxide intermediate. The opening of the protonated epoxide with cleavage of the C-13-O bond is only possible, if the formation of a bridgehead carbenium ion is avoided by a hydride shift from C-12 to C-13. In the thus formed radical cation, a charge induced cleavage of the C-12—C-13 bond can occur that followed by a radical induced H-shift produces an intermediate, in which the enol of acetone can be eliminated (Fig. S3.3-24). The ion with m/z 260 that is formed in this way is the precursor of the ion with m/z 146. For this, the part of the ring A defined by the C-atoms 1-4 must be lost. This can be realized by two H-shifts, each followed by an α-cleavage and the elimination of 2-propylacrylic acid (Fig. S3.3-24).

The base peak is an ion with an even number of electrons at m/z 121. To find an explanation, how this ion is formed, is perhaps the most difficult task in the interpretation of the EI mass spectrum of steviol. After attempts, to derive the ion with m/z 121 from one of the fragments with a higher mass, proved to be unsuccessful, the only remaining possibility is, that despite its relatively low mass it is Fig. S3.3-25 Formation of the base peak with m/z 121 by three α-cleavages and two intermediary H-shifts formed directly from the molecular ion. The for- +• mation of the [M-H2O] ion (m/z 300), which can further fragment to the ions with m/z 217, 254 and 285, has already been explained by a molecular ion ionized on an OH-group (Fig. 3.3-47). In the following, it will be shown, that this molecular ion can also be the starting point for the formation of the ion with m/z 121, if an α-cleavage a is taken into account. If subsequently, H-shifts and further α-cleavages take place in the order shown in Fig. S3.3- + 25, then we come to a C8H9O ion (m/z 121) with an optimal stabilization of the positive charge.

106 107 Supporting information Stevioside

Questions and Answers A. What structural element is responsible for the sweet taste? Base your answer on a comparison of the structural formulae of a typical sugar, a sugar substitute and a sweetener. There is no such structural element! One could be of the opinion, that polyols are such an element, but that is not the case. The bitter substance in grapefruit, naringin, is a counter-example as are the structures of the synthetic sweeteners saccharin, cyclamate, acesulfame K and aspartame.

B. Think of a sweet tasting O-glycoside such as stevioside. In respect to a cleavage of the molecule, which structural element is the most sensitive? What conditions would be necessary to bring this cleavage about? O-glycosides are sensitive to acid hydrolysis. Heating in an aqueous, acid medium can lead to the degradation of such glycosides.

C. Bears withstand the attacks of bees to steal honeycombs, hummingbirds suck nectar from flowers and humans love confectionary. What, do you suppose, is the reason why "sweetness" is so popular amongst living beings? The sweetness of naturally occurring products such as fruit, honey or even maple syrup originates from "sugar", i.e. carbohydrates such as fructose, glucose and sucrose, and represents for living beings the source of energy! Today, that may not be of prime importance but in the course of evolution, it was crucial and sweetness was an indication of energy-rich nutrition.

D. Why do the NMR signals of the residual protons in deuterated pyridine appear as sharp singlets and not with a coupling pattern as in normal pyridine? In >99% deuterated pyridine there are no neighbouring protons and also no different protons in the same molecule. The residual signals come from isolated protons and appear therefore as singlets.

E. In the context of the formation of doubly charged ions in a mass spectrum, the term "half masses" is often used. What do you think of this? This term has no physical meaning, because it is not the mass that is halved but the charge that is doubled.

108 108 Linalool

Chapter 4 4.1 Linalool

Fig. S4.1-1 Structure of linalool

The Isolation of Lavender Oil with (R)-(–)-Linalool as the Main Com- ponent by Steam Distillation of Lavender Flowers Lavender flowers have served since time immemorial as a source of the, because of its scent, much loved lavender oil, the main components of which are (R)-(–)-linalool and its acetic acid ester linalyl acetate. In the work reported here, flowers that were frozen when fresh and thawed after three months were used. Preferentially, fresh flowers should be used. Lavender flowers (330 g), which have been frozen when fresh, are thawed, shredded and subjected to a steam distillation. The standard apparatus described and pictured in chapter 2.1 which includes an electrical steam generator, is used. The emulsion (1500 mL), which forms on shaking, is halved and each half extracted twice with methyl tert.-butyl ether (150 mL). The united ether phases are dried over

MgSO4 and the solvent removed under vacuum. A yield of lavender oil 25 –1 (5.4 g) with a specific rotation of [α]D = –7.1° (c. 0.03 g×mL , CH3OH) is obtained. The negative value of the specific rotation of the lavender oil isolated indicates, that only or predominantly the (R)-enantiomers of linalool and linalyl acetate are present in the essential oil, as described in Wikipedia (see https:// de.wikipedia.org/wiki/Lavendel%C3%B6l). Using NMR (see Fig. S4.1-3) the ratio of both main components, (R)-(–)-linalool and (R)-linalyl acetate, in the lavender oil obtained as described above was determined. To do this, suitable signals that do not overlap with other signals were chosen, so that the integrals could be compared. The SDBS Data Base (Spectral Data Base for Organic Compounds, see http://sdbs. db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi) delivers reference spectra for linalool and linalyl acetate, with which the actual spectrum of lavender oil can be compared and suitable signals selected. Their chemical shifts are shown in Fig. S4.1-2.

Fig. S4.1-2: 1H NMR chemical shifts for linalool and linalyl acetate from the SDBS data base

MF 109 Supporting information Linalool

Suitable peaks are for linalool the methyl group on the quaternary centre and for linalyl acetate the methyl group of the ester, i.e. the peaks at 1.27 and 1.99 ppm respectively (see Fig. S4.1-3).

Fig. S4.1-3: 1H NMR overview spectrum The signals in the spectrum between 1.1 and 2.2 ppm and between 5.0 and of lavender oil in CDCl 3 5.3 ppm as well as the signals from 5.8 to 6.0 ppm agree with the NMR data from the SDBS Data Base for linalool and linalyl acetate, which are the main components of lavender oil.

To determine the ratio of linalool to linalyl acetate the two signals given above were integrated. These signals are well suited for comparison, as they are characteristic and hardly over-

0 8 lap with other signals. Fig. S4.1-4 0 2 . . 2 1

450 shows the excerpt of the spectrum used. The chosen signals were inte- 400 grated and the sum of the integrals

350 normalised to 100. The ratio of linalool to linalyl acetate was deter- 300 mined to be approximately 90:10.

250

200

150

100

50 Fig. S4.1-4 Excerpt of the spectrum used 0 to determine the ratio of linalool to lina- 0 5 0 0 . . 1 9 lyl acetate in lavender oil by the integra-

2.15 2.10 2.05 2.00 1.95 1.90 1.85 1.80 1.75 1.70 1.65 1.60 1.55 1.50 1.45 1.40 1.35 1.30 1.25 1.20 1.15 1 f1 (ppm) tion of characteristic H NMR signals 110 Linalool

IR Spectrum as Film

58 56 54 52 50 48 46 44 42

%T 40 38 % T 36 34 32 30 28 26 24 22 4000 3000 2000 1500 1000 500 WavenumbersWavenumbers (cm-1)–1) Fig. S4.1-6 IR Spectrum of linalool The IR spectrum (Fig. S4.1-6) is dominated by the strong OH stretching vibration at 3250 cm–1. Noteworthy is the sharp peak of low intensity at 3100 cm–1 from the C-H vibrations at the sp2-hybridized C-atoms.

HMBC Spectrum at 600 MHz in CDCl3

8 10 9

dC / ppm 10 5 8, 9

1 6 7 2

dH / ppm

Fig. S4.1-7 Excerpt of the HMBC spectrum in th methyl group region

The expansion of the HMBC spectrum in the methyl group region (Fig. 4.1-17) shows for H-9 the expected correlations to C-4, C-3 and C-2. Both methyl groups H-8 and H-10 show correlations to C-6 and C-7 and to each other.

110 111 Supporting information Linalool

Further Aspects of the EI-Mass Spectrum

As mentioned, the mass spectra of myrcene, β-ocimene, α- and β-phel- landrenen are helpful for the interpretation of the electron impact induced fragmentation of linalool. Figures S4.1-8 to S4.1-11 show the relevant spectra from the NIST Data Base [21].

Fig. S4.1-8: EI spectrum of myrcene

Fig. S4.1-9 EI spectrum of β-ocimene

Fig. S4.1-10 EI spectrum of α-phelland- rene

112 Linalool

Fig. S4.1-11: EI spectrum of β-phelland- rene

+ At m/z 121 the EI spectrum of linalool contains a C9H13 -ion of medium • +• intensity. It corresponds to the loss of H2O and CH3 from the M -ion. Its +• formation can easily be explained, starting with the [M-H2O] -ion A3 (Fig. S4.1-12). By reversing the sequence of the elimination steps, it should be + obtainable from the [M-CH3] -ion (m/z 139). Analogous paths lead to the + C8H13 -ion (m/z 109) (Fig. S4.1-13).

+ +• + Fig. S4.1-12 Formation of C9H13 (m/z 121) from [M-H2O] or [M-CH3]

+ +• + Fig. S4.1-13: Formation of C8H13 (m/z 109) from [M-H2O] or [M-C2H3]

112 113 Supporting information Linalool

Fig. S4.1-14: The series of ions m/z 109, 107 and 105

+ Below m/z 109 follow each with a difference of 2 amu the ions C8H11 and + C8H9 . Their formation is understandable, if a six electron 1,6-sigmatropic + + H-migration to an isomeric C8H13 , in which an H2-elimination to C8H11 (m/z 107) can occur, is assumed. The ion m/z 107 can isomerise by electro-

cyclization to a methyl cycloheptadienylium ion that by a further H2-elimination gives the methyl tropylium ion (m/z 105) ( Fig. S4.1-14). Along this path all the radical cations involved are stabilised by resonance. Their relatively weak intensity (relative intensity ≤10%) is caused by the + low probability of the formation of the C8H13 -ion, because both paths that • could lead to it include the formation of an unfavourable vinyl radical C2H3 .

• Fig. S4.1-15: Cleavage of C3H7 from M+• after preceding rearrangement

The fragment found 2 amu above m/z 109 has nothing to do with the ion • series 109 – 105. The ion at m/z 111 corresponds to the cleavage of C3H9 from the molecular ion. Also in this case, rearrangement processes must create the requirements for the formation of an isopropyl residue as a leaving group and that after the α-cleavage the fragment remaining is stabilized by resonance (Fig. S4.1-15).

114 Linalool

+ Fig. S4.1-16 Formation of C4H7 and + C6H9 On careful inspection of the mass spectrum of linalool the presence of rep- + + resentatives of both homologous series [CnH2n–1] and [CnH2n–3] are found.

+ + + + CnH2n–1: C3H5 (m/z 41), C4H7 (m/z 55), C5H9 (m/z 69), C6H11 (m/z 83)

+ + + + CnH2n–3: C3H3 (m/z 39), C4H5 (m/z 53), C5H7 (m/z 67), C6H9 (m/z 81), + C8H13 (m/z 109)

+ + The formation of C5H9 and C6H11 from the CnH2n–1-series has already been + + discussed. The ions C4H7 and C3H5 , which are present in the spectrum +• with considerable intensity, remain. From the [M-H2O] -structures A1-A3 + • a stable C4H7 -ion and C6H6 -radical can first be formed after a preceding

H-migration, as exemplified byA 2 (Fig. S4.1-16).

+ With this sequence the formation of C6H9 (m/z 81), an ion from the CnH2n–3- + series, also becomes understandable. Allyl cations C3H5 (m/z 41) are almost always present in the mass spectrum of aliphatics. With the saturated hydro- + carbons they are formed from the C3H7 -ion by elimination of H2. Since in the spectrum of linalool an intense signal for the isopropyl cation is present, + a source for C3H5 seems certain. However, its high intensity suggests, that + there may be other predecessors. For example the C5H9 -ion (m/z 69) could + eliminate ethene to become C3H5 , however, a massive isomerisation is first

required. Typical for allyl cations is their further reaction with the loss of H2 + + to cyclopropenylium ions C3H3 (m/z 39). In a similar way C4H5 (m/z 53) + as a methyl cyclopropenylium ion can be derived from C4H7 (m/z 55). The + + other representatives of the CnH2n–3-series, C5H7 (m/z 67) and C8H13 (m/z 109) have already been discussed. In the mass spectrum of linalool some ions are to be found that differ from those that have been dealt with so far, because of their even numbered mass. CHO-ions with an even numbered mass must fundamentally be radical cations. Such ions can as a rule not be created from ions with an even number of electrons (for CHO-species this means with an uneven numbered mass).

114 115 Supporting information Linalool

Such a process would require the elimination of a radical and therefore in the overall balance the energetically unfavourable splitting of an electron pair. In other words, radical cations can only be formed from other radical cations by the elimination of neutral molecules.

Fig. S4.1-17 Suggestion for the +• cleavage of C4H10 from M with the +• formation of C6H8O (m/z 96)

+• Only the molecular ion and the [M-H2O] -ion come into consideration as predecessors for the radical cations with m/z 96, 92, 80 and 68. Only then are reasonable differences in mass to be found that can be attributed to neutral molecules. The ion with m/z 96 has a difference in mass (Δm) to the +• M -ion of 58, corresponding to C4H10. Assuming, that C4H10 contains the C-atoms 6,7,8 and 10 of the original structure, then the M+•-ion must be rearranged, that these C-atoms form an isobutyl group, which finally abstracts an H-atom to leave as isobutane. In Fig. S4.1-17 a suggestion is shown, how this process can take place, beginning with the cyclisation of the M+•-

ion. Subsequently a pinacol type rearrangement of the CH3-group occurs that stabilises the positive charge. Two intramolecular radical H-abstractions over a five or six membered transition state generate a radical cation, in which the bond between C-5 and C-6 can be broken by α-cleavage. The aim, to build an isobutyl group, can now be attained by another intramolecular H-abstraction of the new radical centre. In the newly formed radical cation the tert. CH-bond is weakened by the influence of the neighbouring radical centre. The detaching H-atom cleaves the σ-bond to the isobutyl residue homolytically with the elimination of isobutane. In the remaining +• radical cation C6H8 the radical and positive charge are well stabilised.

+• +• Fig. S4.1-18: Formation of C7H8 (m/z 92) by Cleavage of Propane from the [M-H2O] -Ion A2

116 Linalool

Of interest is the ion with m/z 92, which is reminiscent of the molecular ion + of toluene, particularly as it is accompanied by C7H7 (m/z 91). For this pro- +• pane must be eliminated from the [M-H2O] -ion. In Fig. S4.1-18 is shown, +• how this could occur. After cyclisation of the [M-H2O] -ion A2 and two successive 1,2 hydrogen shifts a structure is obtained, in which a charge induced separation of an isopropyl cation is possible. While still being formed the carbenium ion abstracts a hydride from the neighbouring position with the generation of an ionized aromatic compound.

Fig. S4.1-19: Formation of the +• C6H8 -Ion (m/z 80) from A2

+• The ion with m/z 80 has a difference in mass Δm = 56 to the [M-H2O] -

ion. This corresponds to C4H8, i.e. a butene. In this case it can again be assumed, that the C-atoms 6,7,8 and 10 form the backbone for the eliminated

C4H8. Our suggestion begins with an intramolecular attack of the electron

rich C6-C7 double bond on one of the protons of the CH3(9)-group of the +• [M-H2O] -ion A2. Out of the resulting radical ion with the shifted positive charge isobutene can be lost by a charge induced elimination (Fig. S4.1-19) The accompanying ion with m/z 79 would then be the result of a cyclisation and sequential H-elimination.

Fig. Fig. S4.1-20 Generation +• of the C5H8 -ion (m/z 68)

from A3

+• The ion with m/z 68 corresponds to exactly half of the mass of the [M-H2O] -

ion and can be generated in two steps from the cation A3 (Fig. S4.1-20).

Questions and Answers A. Assuming, that an olfactory receptor can detect the substance, on which physical parameter does it depend, if a compound can be smelled or not? The parameter is the vapour pressure, which at a given temperature may be build up or not. It is dependent on the molecular mass of a substance, which is related to the size of the molecule. It is further dependent on weak interactive forces, with which the molecules of the substance are held together. Are these e.g. relatively strong intermolecular interactions such as hydrogen bonds, which cause crystalline glucose to be nonvolatile, or e.g. extremely weak van der Waals forces, which allow hydrocarbons such as benzene to evaporate quickly.

116 117 Supporting information Linalool

B. Formally terpenes can be regarded as being composed of a number of isoprene molecules. From these O-func- tionalised derivatives can be formed, these are referred to as terpenoids. How are the terpenes classified according to the number of isoprene units? Monoterpenes: 2 isoprene units Sesquiterpenes: 3 isoprene units Diterpenes: 4 isoprene units Triterpenes: 6 isoprene units Tetraterpenes: 8 isoprene units Compounds derived only from one isoprene are depicted as hemiterpenes, eg. prenol (3-methyl-2-buten-1-ol = 3,3-dimethylallylalcohol).

C. Which naturally occurring polyisoprenes are there? From what can they be isolated? These polyisoprenes are natural rubber and gutta-percha. Natural rubber is won industrially from the latex of the rubber tree (Hevea brasiliensis) and processed to rubber by vulcanisation. Structurally it is cis-1,4-polyisoprene with a very high molecular mass of 2×106 Da crosslinked by Sulfur bridges. Guttapercha is a speciality, which comes from the latex of the gutta-percha tree (Palaquium gutta). Structur- ally it is trans-1,4-polyisoprene with a considerably lower molecular mass. Gutta-percha is harder and less elastic but durable. It was used as isolation for the first submarine telegraph cables. It is still used by dentists for provisional fillings.

D. In the CD spectrum (+)-linalool shows a positive, (–)-linalool a negative Cotton effect. Must this be so, or could it be the other way round?

The signs + and – are determined by polarimetry at the wavelength of the D-line of sodium (589 nm), whereas the Cotton effect is measured at the absorption maximum of the substance, i.e. at totally different wavelengths. The two parameters are not correlated to each other.

E. In the 1H NMR spectrum the signals of the terminal vinyl group show a distribution of the intensity. What is this effect called? Roof effect, it occurs, when the difference in chemical shift measured in Hz (Δδ) is smaller than the tenfold of the coupling constant J.

F. Even if, as discussed, it is not relevant for linalool, in the EI mass spectrometry the formation of [M+H]+-ions is sometimes observed. Explain this phenomenon. The proportion of the molecules that are converted by impact with kinetic electrons to molecular ions is very small. Therefore, there is a certain probability of a collision between molecular ions and neutral molecules, by which the transfer of an atom or group of atoms from the neutral molecule to the molecular ion can occur. The intensity of the ions formed is dependent on the square of the pressure of the sample, because it is a bimolecu- lar process. It is sometimes observed for the transfer of an H-atom and has been reported for e.g. ethers, esters, amines and nitriles.

The same onium-ions are formed as are generated by chemical ionisation, proton transfer MS etc.. However, a difference is, that the radical cation abstracts an H-atom, whereas in CI etc. a proton is transferred to the neutral molecule.

118 Linalool

+ G. How is it possible, that the α-cleavage that is explained in Fig. 4.1-23 (formation of the oxonium ion C4H7O • + (m/z = 71) by elimination of C6H11 from the molecular ion) can also lead to a charge migration to C6H11 (m/z 83)? At first glance this really seems difficult to understand, because the splitting of the electron pair of the σ-bond and transfer of an electron to the half-filled n-orbital of the O-atom is difficult to imagine. If the bond cleavage is carried out heterolytically, then the electron pair is localized on the OH-substituted C-atom of structure b.

This structure that is simultaneously a zwitter ion and a radical, is the resonance structure that is used in the Valence Bond Theory to explain the stabilization of C-radicals by donor substituents such as OH-groups [1]. Additionally there is the no bond resonance structure c and the allylic boundary formula d. The comforting conclusion is, that the fragmentation is a normal charge induced heterolysis. [1] R. Brückner: Reaktionsmechanismen, 3. Edition, p.9 Spektrum-Verlag, Heidelberg 2007

118 119 Supporting information Camphor

4.2 Camphor

Fig. S4.2-1 Structure of D-camphor

Sulfonic acid derivatives of camphor

(−)-camphor (1S,4S) or L-camphor is the enantiomer of (+)-camphor (1R,4R) or D-camphor.

In a mixture of concentrated sulphuric acid and acetic anhydride (−)-camphor reacts in 40% yield to (−)-camphor -10- sulfonic acid (Reychler's acid) [7-9], which has attained great importance for the separation of racemic bases. Frerejaques [10] obtained as a byproduct (5%) a sultone, the structure of which could only be elucidated much later. It is derived from an intermediate step of Reychler's reaction. Camphor and chlorosulfonic acid react together, to give an isomeric sulfonic acid (Kipping's acid). For a long time it was unclear if this was camphor-8- or camphor-9-sulfonic acid. Starting with (−)-camphor the acid is formed as a racemate. The definitive proof of structure showed it to be camphor-9-sulfonic acid. A direct sulphonation of the methyl group 9 or 10 is unthinkable for chemists with a mechanistic training. Actually the occupation with both reactions lasted many decades and delivered a complicated cascade of reactions, which explain, why Reychler's acid during its formation retains the optical activity, whereas Kipping's acid undergoes racemization [11-13]. To aid understanding, in the following mechanistic explanation carbenium ions will be formulated as discrete intermediates, although our present view suggests that σ-delocalised carbocations (carbonium ions) are intermediary. In this case the carbenium ions formulated are limiting resonance structures of the delocalised non-classical ions.

In the mixture of sulphuric acid and acetic anhydride, which is used in Re- ychler's reaction, an exothermic reaction takes place leading to an equilibrium with acetyl sulphuric acid and acetic acid: Fig. S4.2-2 Formation of cam-

H2SO4 + Ac2O AcO-SO3H + AcOH phor-10-sulfonic acid ⇌ 120 Camphor

The camphor added goes into solution as a carboxonium ion. The decreased electron density on the carbonyl C-atom effects a Wagner-Meerwein rearrangement, which leads to a low equilibrium concentration of the carbenium ion A (Fig. S4.2-2). Since the subsequent reactions are irreversible, the protonated camphor finally undergoes the rearrangement reaction completely. The deprotonation of the carbenium ion A leads to an alken- ic intermediate, which serves as a suitable substrate for the electrophilic sulphonation. Thereby a further carbenium ion is first formed, the sextet centre of which initiates a semipinacol rearrangement, leading back to the camphor framework. The byproduct, isolated by Frerejaque, can be derived from the carbenium ion A in the first rearrangement step (Fig. 4.2-3). In competition to the deprotonation some of A reacts with migration of the less sterically hindered exo-methyl group 8 to the carbenium ion B. The deprotonation on methyl group 9 now follows, accompanied by the acetylation of the tert.-OH group. An electrophilic attack of the sulphonation reagent on the exocyclic double bond leads to a new tert. carbenium ion C. In the next step the proposed Wagner-Meerwein rearrangement to D seems at first somewhat problematic, since a secondary carbenium ion is formed from a tertiary carbenium ion. It must, however, be remembered that the tert. ion C is destabilised by the sulfomethyl substituent. The subsequent internal

addition of the SO3H group on the carbenium ion delivers finally 4-acetoxy-2-exo-hydroxybornane-10-sulfonic acid sultone.

Fig. S4.2-3 Formation of Frerejaque's sultone in sulphuric acid/acetic anhydride

How can it be possible that by changing the sulphonation reagent to chlorosulfonic acid a totally different result occurs? In Fig. S4.2-2 the proposed deprotonation of carbenium ion A to an alkene requires as a conjugated base the presence of a proton acceptor. For the sulphonation according to Reychler acetic anhydride or the acetic acid from the equilibrium reaction are possibilities. However, in Kipping's reaction only chlorosulfonic acid, which will only very unwillingly accept a proton, is present. The deprotonation is hindered to such an extent, that the migration of the exo-methyl group 8, which is of importance for the formation of Frerejaque's sultone, predominates. The retarded deprotonation first occurs at the stage of the carbenium ion B (Fig. S4.2-4). The alkene formed reacts to carbenium ion E, which apart from the missing acetylation of the OH group corresponds to the carbenium ion C in Fig. 4.2-3. The subsequent methyl-migration leads to the more stable carbenium ion F, in which a semipinacol rearrangement is possible. After deprotonation and aqueous work-up the result is camphor-9-sulfonic acid, however, this sequence does not contain a racemization of the camphor structure.

120 121 Supporting information Camphor

The carbenium ion E can, apart from the exo-CH3-migration to F, also un- dergo a Wagner-Meerwein rearrangement, to produce the carbenium ion G. If this ion undergoes a hydride shift, then it can equilibrate with the enantiomeric ion G' and this finally with E'. From here the path is free to enantiomeric camphor-9-sulfonic acid. The total racemisation observed requires, that the equilibration

has entirely occurred before the exo-methyl-migration to F takes place.

Fig. S4.2-4 Formation of racemic camphor-9-sulfonic acid in chlorosulfonic acid

Questions and Answers A. Apply the Octant Rule (W. Mofitt, R. B. Woodward, A. Moscowitz, W. Klyne, C. Djerassi "Structure and the optically rotatory dispersion of saturated ketones" J. Amer. Chem. Soc. 1961, 83, 4013–4018.) to predict the sign of the Cotton effect in the CD spectrum. One places the molecule in a coordinate system with the origin in the center of the C=O double bond. The oxygen atom is placed in +z-direction whereas the carbonyl carbon atom is behind the paper plane in –z-direction. The C-atoms 1 and 3 have the x-coordinate zero. As the diagram shows, all methyl groups are in a positive octant, whereby the contributions of C-5 and C-6 neutralize each other. Therefore a positive Δε is to be expected.

122 Camphor

B. Why can cross signals appear in a two dimensional long range COSY NMR spectrum, although in the one dimensional 1H-NMR spectrum no coupling is observed? In a 1D-NMR spectrum the spin coupling appears as a splitting of the peak, provided that the peak width is smaller than the coupling constant. In a 2D-COSY spectrum the intensity of the cross peak results from the 4 spin coupling (in this case JHH) and is independent of the peak width of the signal.

C. Propose a process that leads to the loss of identity of the methyl groups 8, 9 and 10, so that these are all in- + volved statistically in the formation of the M–CH3 -ion (m/z 137).

+ D. The C8H12 -ion (m/z 108) is formed from a by elimination of C2H4O. Suggest a mechanism for the fragmentation.

E. In the structural formula of camphor label the C-atoms with the letters a and b, to show how the two isoprene units are joined together in this monoterpene.

122 123 Supporting information Cantharidin

4.3 Cantharidin

Fig. S4.3-1 Structure of cantharidin

Questions and Answers A. Why is cantharidin optically inactive?

The compound contains a plane of symmetry and belongs to the point group Cs. This may remind us of a but- terfly, the wings of which are mirror images of each other. Molecules with this element of symmetry have a mirror image that is identical to the object itself, i.e. object and mirror image are superposable. They are not chiral but achiral. Therefore enantiomerism does not occur, which would be the prerequisit for optical activity. A textbook example for this is the already mentioned meso-tartaric acid.

B. Why can cantharidin and camphor be sublimed? What are the conditions, for a substance to be volatile? Both are spherical molecules without functional groups such as carboxyl or hydroxyl groups, which can produce a strong intermolecular association by hydrogen bonding. The elemental formula and molecular mass –1 –1 have similar dimensions, C10H16O (MW 152.2 g×mol ) for camphor and C10H12O4 (MW 196.2 g×mol ) for cantharidin, giving a similar vapour pressure, which is a condition for sublimation. It is not without reason that Spanish flies are to be kept in well sealed containers. In general the volatility of a substance, i.e. its vapour pressure, is dependent upon its molecular mass and the strength of the intermolecular interactions. During sublimation a direct transition from the solid to the gaseous state occurs.

C. What are the conditions, for a substance to have an odour? Two conditions must be fulfilled before a substance can be smelt. It must be volatile and it must interact with the smell receptors in the nose, which is not to be taken for granted. A high vapour pressure alone is not sufficient. While we immediately experience the odour of ammonia as pungent, the putrid smelling hydrogen sulphide after a short time numbs our noses. On the other hand we do not smell the highly poisonous carbon monoxide in the same way as we do not smell oxygen or nitrogen in the air.

D. In view of the structure of cantharidin which reactions can be expected? With a cyclic carboxylic acid anhydride nucleophiles such as thiols, amines and alcohols should react with ring opening and formation of a covalent bond to give thioesters, acid amides and esters respectively.

E. Why does the carbonyl band in the IR spectrum of cantharidin appear as a double peak? This is a characteristic of carboxylic acid anhydrides (see W. G. Dauben, W. W. Epstein "Infrared Spectra of Some Cyclic Anhydrides" J. Org. Chem., 1959, 24, 1595–1596). With cyclic anhydrides the band from the symmetrical carbonyl stretching is less intense than the anti-symmetrical carbonyl stretching.

124 Cantharidin

F. Explain the formation of the ion m/z 100.

G. Explain the formation of the ion m/z 109. After the loss of the anhydride bridge firstly a semipinacol-type rearrangement occurs in competition to the

retro-Diels-Alder cleavage. The direct elimination of a CH3-group from this fragment would lead to an unfavourable anti-Bredt structure. This can be avoided by an alternative α-cleavage that opens the bicyclic structure.

124 125 Supporting information

4.4 Artemisinin

Fig. S4.4-1 Structure of artemisinin

EI mass spectrum

100 151

166 80

60 55 137 192

40 123 % Intensity 107 81 178

208 222 20 236

250 264 282 0 50 70 90 110 130 150 170 190 210 230 250 270 290 Fig. S4.4-2 EI mass spectrum of artemi- m/z sinin In the following, we give an interpretation of the EI spectrum, recorded at the lowest temperature possible on a quadrupole mass spectrometer (Fig. S4.4-2), that allows the measurement of masses at high resolution and does not require the assumption of thermal processes before ionisation. The radical cation at m/z 250 is formed by the loss of the peroxide

bridge as O2. This process can be plausibly formulated starting with the M+•-ion of artemisinin (Fig. S4.4-3). In the literature, a thermal

or catalytic loss of O2 before ionisation is considered. In a McLafferty +• type process, the [M-O2] -ion forms the stable radical cation m/z 192 and

in competition to this the [M-O2- +• H2O] -ion with m/z 232.

Fig. S4.4-3 Elimination of O2 from the molecular ion of artemisinin and subse-

quent loss of H2O and C3H6O

126 Artemisinin

The ion at m/z 236 can arise by the electron impact induced elimination of formic acid from artemisinin (Fig. S4.4-4).

Fig. S4.4-4 Elimination of formic acid from the molecular ion of artemisinin

In Fig. S4.4-5 the formation of m/z 222 from the artemisinin radical cation by the loss of acetic acid is postulated. The elimination of two +• CO leads to a C11H18O -ion (m/z 166), from which ﬁ nally by the loss • • • of HCO , CH3 and H the ions with m/z 137, 151 and 165 are formed.

Fig. S4.4-5 Electron impact induced elimination of acetic acid from artemisinin and formation of further fragment ions 127 Supporting information

The acetyl cation at m/z 43 is the base peak. It can be formed from the +• [M-O2] -ion (m/z 250, Fig. S4.4-3) as well as from the open ring structure with m/z 282 (Fig. S4.4-4) (Fig. S4.4-6).

Fig S4.4-6 Formation of the base peak + with m/z 43 (CH3CO )

GC/MS Investigation of the Thermolysis of Artemisinin

Fig. S4.4-7 Artemisinin 1 and the three main products of thermolysis 2, 3 and 4

Considering its peroxide structure, artemisinin shows an amazing kinetic stability. It melts without decomposition, it sublimes, survives three minutes, when it is heated to 50°C above its melting point [S1] and does not decompose in boiling toluene [S2]. From this one might expect, that it possesses the requirements for the recording of an EI mass spectrum. However, if it is heated in a molten form to 190°C for 5-10 minutes, or if it is heated under reﬂ ux in xylene it decomposes completely. By a preparative work-up of the products of thermolysis, the compounds 2-4 were isolated and structurally elucidated [S2, S3]. The three products of thermolysis were each obtained in preparative yields of about 15%.

128 Artemisinin

Fig. S4.4-8 Gas chromatogram of the products of thermolysis of artemisinin.

Column: 30 m×0.25 mm (ID), df = 0.25 μm DB-5MS; injector: 280°C, temp. program: 100°C 3 min isotherm, 10°C. min−1 to 320°C, ﬂ ow rate: 1.2 mL He.

To establish, if thermal processes before the ionisation might possibly infl u- ence the appearance of the mass spectrum, we heated 1 on a microscale at 190°C for 10 minutes and investigated the products of thermolysis by GC/ MS. In the gas chromatogram (Fig. S4.4-8) apart from at least 12 products in negligible concentration three main products in the ratio 1:4.5:1.5 can be recognized, which from their mass spectra can be identifi ed as 2, 3 and 4. The mass spectrum of 2 shows the M+•-peak at m/z 236 and signifi cant peaks m/z 192, 166 and 151. The mass spectrum of 3 is characterized by an M+•-peak of low intensity and intense fragments at m/z 166, 165, 151, 138 and 137. In the mass spectrum, compound 4 also shows a weak molecular peak (m/z 282) and its dominant +• fragment is the [M-CH3CO2H] -ion 151 at m/z 222. The m/z-values found in 100 the spectra of 2-4 actually do occur in the mass spectrum of artemisinin 80 1. However, there are distinct differ- 166 ences. In the higher region of mass 60 in the mass spectrum of 1, ions oc- +• cur with m/z 250 ([M-O2] ) and 232 55 +• 40 ([M-O2-H2O] ), which are missing 107 122 in the spectra of 2-4. The most in- % Intensity 93 69 81 236 tense ion in the spectrum of 1 with 20 135 192 180 m/z 192 is considerably less intense 207 0 in the spectrum of 2 and almost ab- 50 70 90 110 130 150 170 190 210 230 sent in the spectrum of 3 and 4. This also applies to the base peak in the m/z spectrum of artemisinin at m/z 43 (CH CO+). We interpret this as an Fig S4.4-9 EI mass spectrum of 2 3 indication, that the mass spectrum recorded for artemisinin originates from an intact molecular structure.

129 Supporting information

100 166 165 137 80 151

40 % Intensity

55 96 20 69 81 109 123 211 193 282 0 50 70 90 110 130 150 170 190 210 230 250 270 290 m/z Fig S4.4-10 EI mass spectrum of 3

100 222

80 A comparison of the mass spectra of 2, 3 and 4 (Fig. S4.4-8 - S4.4-11) 150 60 55 shows many common m/z-values 93 122 137 that, however, considering the dif- 40 107 166 81 ferent initial structures can hardly 207 % Intensity 67 178 194 be identical. This is exempliﬁ ed by 20 the peak with m/z 166. In the case 281 252 267 of 3 this is accompanied by a peak 0 at m/z 165 with almost the same 50 70 90 110 130 150 170 190 210 230 250 270 290 intensity. In contrast, the loss of an m/z H-atom from m/z 166 is of much less importance for the compounds 2 and 4. Fig S4.4-11 EI mass spectrum of 4

Fig. S4.4-12 Electron impact induced fragmentation of the product of thermolysis 2

130 Artemisinin

Our suggestions for the mechanism of the electron impact induced fragmentation of the three compounds are shown in Fig. S4.4-12 - S4.4-14. A comparison of our measurement of the EI spectrum of artemisinin with published data [24, 25] disclos- es a further discrepancy.

Fig. S4.4-13 Electron impact induced fragmentation of the isomer of artemisinin 3

The determination of the exact masses with a two sector spectrometer in Tübingen and Leipzig gave for the ions with m/z 166, 165, 151 and 137 the elemental composition +• + + C11H18O , C11H17O , C10H15O and + C10H17 . In contrast, according to the literature these correspond to isobaric ions with the composition +• + + C10H14O2 , C10H13O2 , C9H11O2 and + C9H13O . The fragment ions suggested for 2-4 in the fragmentation schemes (Fig. S4.4-7 - S4.4-9) with m/z 166, 165, 151 and 137 have the elemental composition found by us. This forces us to the assumption, that during the measurements of Fales et al. [24] and Madhusudanan et al. [25] a thermal rearrangement preceded the ionisation.

Fig. S4.4-14 Electron impact induced fragmentation of the isomer of artemisinin 4

131 Supporting information

Mechanism of Thermolysis

Fig. S4.4-15 Formation of 2 via an initial thermal homolysis of the peroxide bond according to [26]

In the literature [26] a homolysis of the peroxide bond is suggested as the initial step of the thermal rearrangement process leading to the products 2, 3 and 4. The mecha- nisms are summarized in Fig. S4.4- 15/16.

Fig. S4.4-16 Formation of 3 and 4 via an initial thermal homolysis of the peroxide bond according to [26]

Fig. S4.4-17 Formation of 2 by a radical chain mechanism

132 Artemisinin

It has been reported, that on heating to 190°C in the ﬁ rst minutes no reaction occurs but then the transformation of artemisinin 1 takes place to completion within a few minutes. From this, one can conclude, that the transformation of 1 into 2, 3 and 4 requires in incubation time, such as is typical for radical chain processes. A corresponding mechanism is shown in Fig. S4.4-17 and S4.4- 18.

Fig. S4.4-18 Formation of 3 and 4 by a Fig. S4.4-18 Formation of 3 and 4 by a radical chain mechanism radical chain mechanism

ESI(+)-MS/MS Spectrum

Fig. S4.4-19 Collision activated spectrum of [M+Na]+ of artemisinin (ESI(+)- MS/MS) 133 Supporting information

Fig.S4.4-20 Decay sequence with forma- + tion of C10H15O (m/z 151)

Literature [S1] D. L. Klayman, "Qinghaosu (Artemisinin): An antimalarial drug from China", Science 1985, 228, 1049−1055. [S2] X. D. Luo, H. J. C. Yeh, A. Brossi, J.-L. Flippen-Andersen, R. Giardi "The chemistry of drugs VI. Thermal decomposition of Qinghaosu" Heterocycles 1985, 23, 881−887. [S3] A. J. Liu, D. L. Klayman, J. M. Hoch, J. V. Silverton, C. F. George "Thermal rearrangement and decomposition products of Artemisinin" J. Org. Chem. 1985, 50, 4504−4508.

Questions and Answers A. What is to be understood by the term chiral pool? Chiral pool is an allegorical term for chiral, mostly enantiomerically pure natural products. This includes e.g. alkaloids, amino acids, carbohydrates or terpenes. An advantage is, that the arduous, stereoselective synthesis of such chiral compounds can be avoided. This work has already been undertaken by "nature", in that the chiral (!), enzymatic synthesis apparatus of living organisms performs stereoselective reactions, leading to enantiomerically pure compounds. Chiral pool substances are valuable. They are used as reactants and as chiral auxiliaries that as adjuvants in syntheses transfer stereochemical information.

B. Organic peroxides and hydroperoxides are derivatives of hydrogen peroxide. What chemical redox properties

does H2O2 have? Hydrogen peroxide can act as an oxidizing agent or as a reducing agent. What happens in a reaction depends on the reaction partner. In most cases, it reacts as a strong oxidizing agent, also in its industrial applications.

Only even stronger oxidizing agents such as KMnO4 oxidize it to oxygen. The formal oxidation number of O

in H2O2 is −1, it disproportionates to water and oxygen:

2 H2O2 → 2 H2O + O2

The oxidation number of O in H2O is −2 and that of O2 is 0.

134 Artemisinin

C. Which natural product from the in Asia growing star anise, the total synthesis of which would be ineffi cient, about ten years ago, caused an enormous demand for the fruits of the star anise that drove their price to an extreme height? Which drug against which illness was produced by partial synthesis and sold in enormous amounts? It is shikimic acid ((3R,4S,5R)-3,4,5-trihydroxycyclohex-1-ene-1-carboxylic acid: CAS RN 138-59-0), from which the active ingredient oseltamivir (INN), the antiviral agent of the infl uenza medication Tamifl u®, is synthesised. The occurrence of the virus type Infl uenza A/H5N1, as happened in the pandemic in 2006, is particularly feared. The panic that this caused led to an enormous increase in the demand for the chiral pool compound shikimic acid, as the starting material for a nine step partial synthesis of the active ingredient. (see also S. Berger, D. Sicker, "Classics in Spectroscopy" Chapter 6.1 Shikimic Acid, ISBN 978-3-527-32617-4, WILEY VCH 2009)

D. Discuss, how one can distinguish the 1H assignment of the methyl protons H-15 and H-16.

In the COSY spectrum the methyl group signal at δH = 1.211 couples with the proton H-9 at δH = 3.4, therefore

this methyl group is assigned to H-16. The methyl group signal at δH = 1.00 couples with the proton H-6 at δH= 1.423, therefore this methyl group is assigned to H-15.

+ + E. The CI spectrum that is discussed here was obtained with CH5 as the reactant ion. How is CH5 formed in the CI-source? What sort of species is it?

+• In a CI ion source with CH4 as the reagent gas CH4 -ions are ﬁ rst produced by electron impact ionisation. Because of the relatively high pressure of methane (≈102 Pa) subsequently reactions occur between ions and molecules. The most important reactions are:

+ CH5 is an extremely strong Brønsted acid, which protonates analyte molecules with n or π electrons.

+ CH5 is a hypercoordinated carbonium ion with a bond with two electrons and three centres. With the VB-method the σ-delocalisation can be described with the following resonance structures:

+ The fi ve H-atoms of CH5 are subject to a permutation process, by which all become equivalent. Each one of the fi ve H-atoms can in- terchangeably together with another H-atom be involved in the three centre bond. This gives 5!=5*4*3*2=120 identical structures of minimum energy that continuously interchange (fl uctuating structure, see fi gure on the right). Latest publication: K. M. T. Yamada, S. Brünken, A. Potapov, S. Schlemmer, Science, 2015, 347, 1346−1349 and the literature cited therein.

+ It is worthy of note, that J. J. Thomson in 1911 already found a comparable species, the H3 -ion, with his par- abolic mass spectrometer (J. J. Thomson Phil. Mag. 1911, 21, 225, ibid. 1912, 24, 209).

135 Supporting information Diosgenin

4.5 Diosgenin

Fig. S4.5-1 Structure of diosgenin

CD Spectrum

1 ) 1 - mmol 2 / (cm ε ∆ 0

-1 200 225 250 275 300 λ / nm Fig. S4.5-2 CD spectrum of diosgenin IR Spectrum in KBr

100 95 90 85 80 75 70 65 60 55 %T

%T 50 45 40 35 30 25 20 15 10 Fig. S4.5-3 IR spectrum 4000 3000 2000 1500 1000 500 WavenumbersWavenumbers (cm(cm-–11)) of diosgenin in KBr As already described for the UV spectrum, there are few functional groups in diosgenin. Therefore, the IR spectrum [S1] in Fig. S4.5-3 is without characteristic features. [S1] R. Schwartz, A. Juhasz, P. Coltea "Comparative study of the IR spectra of some B-nor and natural steroids" Rev. Roumaine de Chimie 1986, 31, 481–488. 136 137 Supporting information Diosgenin

Additional NMR Spectra at 700 MHz in CDCl3

COSY Spectrum 24 19 21 18 15 2 1 25 11 27 7 20 17 12 8 7 2 24 1 23 15 12 14 9 δH / ppm

18, 27

9, 21 19 1, 14 12 15

24, 2, 11, 7 24, 25, 8, 23 12, 17 1, 2, 20

15, 7

δH / ppm

Fig. S4.5-4 Excerpt from the DQF COSY spectrum of diosgenin in the aliphatic region

HSQC Spectrum 6 16 26 3 4

δC / ppm 4

26 3

δH / ppm

Fig. S4.5-5 Excerpt 1 from the HSQC spectrum of diosgenin

The CH-edited HSQC spectrum in Fig. S4.5-5 and S4.5-6 is particularly helpful, to recognize the C-atoms of the diastereotopic methylene protons. 136 137 Supporting information Diosgenin

19 21 27 18 15 2 1 17 8 25 24 7 11 2 9 7 20 12 23 15 12 14 1

δC / ppm

9, 21 18, 27 19 11

24 25, 8 7, 2, 15, 23 1 12 20

δH / ppm

Fig. S4.5-6 Excerpt 2 from the HSQC spectrum of diosgenin

NOESY Spectrum 19 21 27 18 20 1 17 23 25 15 2 12 8 24 7 11 1 9 4 7 2 24 15 12 14

δH / ppm

18, 27

9, 21 19 1, 14, 12

24, 2, 11, 7 24, 25, 8, 23 12, 17 1, 2, 20

15, 7

δH / ppm

Fig. S4.5-7 Excerpt from the NOESY spectrum of diosgenin

138 139 Supporting information Diosgenin

ESI (+) Mass Spectrum The ESI(+) mass spectrum of diosgenin, recorded in a solution with solvent acetone/methanol/formic acid, shows an [M+H]+-ion (m/z 415) as well as a [M+Na]+-ion (m/z 437) and [M+K]+-ion (m/z 453). Impact activation of the [M+H]+-ion leads to a MS/MS spectrum that interestingly contains the fragment ions with m/z 271 and 253, both of which are found in the EI spectrum.

Fig. S4.5-8 ESI(+)-MS/MS spectrum of diosgenin

In [S2] a method for the measurement of the concentration of diosgenin in blood plasma that uses the LC-ESI-MS/MS technique is described. The quantification is carried out by "multiple reaction monitoring" of the transition of m/z 415 ([M+H]+) to m/z 271. Without a detailed explanation for the formation of m/z 271, the structure A is suggested in [S2] for this ion, although it must have a mass that is two a.m.u. higher. Actually, it is the allylic cation that was already described in the discussion of the EI spectrum. For its formation in the ESI(+)-MS/MS spectrum (Fig. S4.5-8) we suggest the sequence shown in Fig. S4.5-9. [S2] L. Xu, Y. Liu, T. Wang, Y. Qi, X. Han, Y. Xu, J. Peng, X. Tang "Devel- opment and validation of a sensitive and rapid non-aqueous LC-ESI-MS/ MS method for measurement of diosgenin in the plasma of normal and hyperlipedemic rats" J. Chromatogr. B 2009, 877, 1530−1536.

138 139 Supporting information Diosgenin

Fig. S4.5-9 Cleavage of the rings E and F in the [M+H]+-ion of diosgenin

+ An ion with the relative intensity of 5% and the formula C21H31 is detected at m/z 283. Despite its low intensity, this ion awakes our mechanistic interest, because for its formation the elimination of the OH-group of ring A as

H2O and the elimination of ring F as C6H10O2 (presumably as 4-methylvalerolactone), i.e. two neutral fragments that have the maximum separation

in the diosgenin molecule, are necessary. For the cleavage of C6H10O2, a protonation of one of the acetal O atoms would be a hindrance, therefore we assume, that in a small fraction of the [M+H]+-ion the added proton is

taken up by the OH-group on C-3. The then possible elimination of H2O could initiate a cascade of six hydride shifts along the rings A to D, ending with an oxonium ion (m/z 397) (Fig. S4.5-10) that can isomerize to a further

oxonium ion. The latter has all the requirements for the cleavage of C6H10O2 to form an allylic cation at m/z 283.

+ Fig. S4.5-10 Cleavage of H2O and 4-methylvalerolactone from the [M+H] -ion of diosgenin. Questions and Answers A. What is the definition of the term hormone? Is there a common structural element for all hormones? Hormones are endocrine signalling substances that are produced in the body, usually in specialised organs, the glands, and distributed in the blood stream to transport information and to initiate specific effects in the target organ. They are a component of the biochemical regulatory circuit. Specific hormone receptors convey

140 141 Supporting information Diosgenin

the hormonal effect. Very low concentrations of hormones influence metabolism, growth and reproduction. There is no structural element common to all hormones. This is evident, when one compares the structure of adrenalin (a catecholamine), insulin (a peptide) and testosterone (a steroid).

B. Saponins occur in many autochthonic or introduced plants. What are the saponins of the following plants called: horse chestnut, ivy, foxglove, potatoes, soapwort, licorice? The seed of the horse chestnut contains the mixture of saponins aescin, with 5 components, of which β-aescin is the best known. Ivy contains hederin, which accounts for the poisonousness of ivy. Foxglove contains digi- tonin, which has a haemolytic effect (destroys erythrocytes), when present in the blood stream. Digitoxin and digoxin are further pharmaceutically useful cardiac glycosides, for which an exact dosage is very important, to avoid poisoning. α-Solanine, which is also poisonous, is found in the green parts of potatoes or in its shoots. Therefore, the water, in which unpeeled potatoes are cooked, should not be used further. Unripe tomatoes also contain α-solanine. Soapwort contains various saponins, so many in fact that it was used to make soap. To describe this in modern language, it was used as a mild non-ionic surfactant that is biologically degradable, which ecologically makes sense. Licorice contains glycyrrhizic acid as a natural sweetener, giving it its sweet and characteristic taste. This saponin is also edible. However, it should not be consumed in vast quantities, since if a maximum daily amount (100 mg) is exceeded, unpleasant side effects such as hypertension can be encountered. This is due to the physiological activity of the aglycone.

C. What is the principle physico-chemical difference between saponins and sapogenins? Saponins are amphiphilic. The molecule has a hydrophilic moiety, the saccharide part, and a large hydrophobic moiety, the aglycone as a terpenoid. Saponins are therefore naturally occurring surfactants and have the corresponding properties. They reduce the surface tension of water, above a certain concentration form micelles and as an emulsifying agent have a stabilizing effect on emulsions. Sapogenins are not amphiphilic; instead as aglycones, they are hydrophobic. As they do not foam as saponins do, their extraction is easier, as is the case for diosgenin.

D. Not used in this book, but potentially of use in cases, in which HMBC and NOESY do not help further, is an about 15 year old new NMR method that evaluates residual dipolar couplings. Which reagent is required for this method? So-called alignment media that to a small extent align the molecules of the analyte in respect to the magnetic field are required.

E. In which class of compounds have C-atoms with the chemical shift of C-22 a particular importance? The signal of the anomeric C-atom of all sugars appear in this region, since it is bound to two O-atoms by single bonds.

F. What could be the reason, that the EI spectrum of diosgenin shows no retro-Diels-Alder fragmentation, although a suitably situated double bond (C-4/C-5) is available? The ionisation of the double bond is a prerequisite for an RDA-cleavage. However, this apparently does not compete with the ionisation of the O-atom of the acetal function.

140 141 Supporting information Diosgenin

G. As mentioned, the fragments m/z 300, 285 and 271 eliminate water, whereas the molecular ion and the ions with m/z 355 and 342 do not? Try to explain this difference. In the ions with m/z 300, 285 and 271 the O-atoms of the acetal function of the rings E and F have been eliminated. The ion with m/z 300 corresponds to an ionized olefin. The ions with m/z 285 and 271 are substituted allylic cations. In the case of the radical cation with m/z 300, a radical H-shift and three 1.2-hydride shifts lead to an isomeric structure, in which the positive charge is conjugated to the double bond C-4/C-5. With the allylic cations with m/z 285 or 271 the same result can be reached by three hydride shifts. From here it is possible, +• to initiate a charge-induced elimination of H2O. The other discussed ions (M , m/z 355 and 342) are oxonium ions or in the case of the M+•-ion an ion with localisation of the charge on an O-atom of the acetal function. For these structures an isomerisation, in which the charge is moved in the direction of the double bond C-4/C-5 is not possible.

142 142 Friedelin

4.6 Friedelin

Fig. S4.6-1 Structure of friedelin

IR Spectrum in KBr

50 %T

%T 45

4000 3000 2000 1500 1000 500 Fig. S4.6-2 IR spectrum of WavenumbersWavenumbers (cm (cm-1)–1) friedelin

As already mentioned for the UV spectrum, friedelin has almost no signif- icant functional groups, consequently the IR spectrum (Fig. 4.6-2) lacks a characteristic proﬁ le. Naturally, a strong CH-stretching band for sp3 hybridized C-atoms is to be found. The strong C=O stretching band at 1720 cm–1 is very sharp. The broad band at 1460 cm–1 can be attributed to a CH-deformation.

143 Supporting information

Additional NMR Spectra at 600 MHz in CDCl3 HMBC Spectrum

2β 2α 4 1α 6β 1β 19α

δC / ppm 23

24 7 25 27 26 1

12 30 28 29 19 11 16 22 6 2 15 21 18

8 4 10

δH / ppm

Fig. S4.6-3 Excerpt from the HMBC spectrum of friedelin in the aliphatic region

23 2β 2α 4

δC / ppm

δH / ppm

Fig. S4.6-4 Excerpt from the HMBC spectrum of friedelin in the carbonyl region

144 Friedelin

Mass Spectrometry

Fig. S4.6-5 Further fragment ions that can be derived from m/z = 274 (Fig. 4.6-22)

Not unexpectedly for a triterpene, the greatest part of the total ion current is concentrated in the lower range of mass. The ion with m/z = 274 (Fig. 4.6-22) seems to play a central role in the fragmentation, because with the cleavage of ethene or butadiene the ions with m/z = 246 and 220 respectively and further following ions can be explained (Fig. S4.6-5).

Fig. S4.6-6 Formation of the fragments m/z = 193, 192, 191 and 189.

The fragment m/z 192 could also arise directly from M+• by cleavage of the bonds C8-C14 and C9-C11 and lead further to m/z 191 and 189. With two intermediary, radical induced H-shifts the formation of m/z 193 becomes understandable (Fig. S4.6-6).

145 Supporting information

+ Fig. S4.6-7 Formation of the ion m/z = 205 (C15H25 ) from the rings D and E

Fig. S4.6-7 outlines a branch of the fragmentation, by which the ring D and + E can be eliminated as neutral entities. To form the ion C15H25 (m/z 205) the charge must be transferred to this part of the molecule and in addition in the total balance an H-atom shifted to the radical that is to be eliminated.

The sequence starts with the ionisation of the C8-C14 bond. By three intramolecular, radical induced H-shifts, in each case via a ﬁ ve membered transition state, the radical centre reaches a position that facilitates an α-cleav- + age, leading to the allyl cation C15H25 .

Fig. S4.6-8 Cleavage of the rings A and E with the formation of m/z 179 A plausible explanation for the ion m/z 179 is difﬁ cult. It can be assumed, that this ion originates from ring C and contains in addition two C-atoms from each of the rings B and D. Theoretically if the rings A and E are cut off along the wavy lines shown (Fig. S4.6-8), a fragment with the formu-

la C13H22 remains. The real ion with m/z 179 corresponds to the formula + C13H23 and therefore requires an additional H-atom. The elimination of the ring E as an oleﬁ n has already been discussed in regard to the formation of m/z 302 (Fig. 4.6-22). A way must now be found, in which the bonds C-5-C-6 and C-9-C-10 are sequentially broken. For this, the radical centre must initially reside on C-4 and then on C-11, to facilitate the corresponding α-cleavages (Fig. S4.6-8). The H-atom which according to this consideration has to be captured by the charged fragment, originates from C-4 of the ring A. This ring is ﬁ nally lost as a radical.

146 Friedelin

Fig. S4.6-9 Stepwise dehydrogenation of m/z 179 The m/z 179 which cation is accompanied by ions with m/z 177 and 175, which could be formed by successive loss of H2. After a previous hydride and alkyl shift in the ion with m/z 179 a benzyl cation can be formulated as the ﬁ nal product of the decay path (Fig. S4.6-9)

Fig. S4.6-10 Breaking of the C13-C18 bond and the formation of m/z 125 (a) and m/z 124-119 (b)

The ions m/z 125 -119 emanate from the ring E. The cleavages of C-C bonds that are necessary correspond to those that lead to the ions with m/z 274 and 273 (Fig. 4.6-22). The charge remains in ring E and not ring C. To reach m/z 125 two intramolecular H-abstractions that lead to a radical cation with an ionized exo-methylene double bond on C13 are necessary. From here the m/z 125 is formed by a charge induced fragmentation (Fig. S4.6-10 (a)). If this already occurs in the initially formed molecular ion, m/z 124 is produced (Fig. S4.6-10 (b)), from which by a combination of elimination • of H and H2 the ions m/z 123 – 119 emanate. For the last step to m/z 119 a preceding methyl shift is necessary.

147 Supporting information

Fig. S4.6-11 Remnants of ring E The intensive group of peaks around m/z 109 and m/z 95 seem at ﬁ rst glance to be puzzling. It can be assumed, that the starting point is the radical cation • with m/z 124, from which by elimination of CH3 or C2H4 the corresponding C-8 or respectively C-7 fragmentation is reached. These ions are then also remnants of the ring E (Fig. S4.6-11).

Fig. S4.6-12 Formation of the ions with m/z 109, 107, 105 (a) and m/z 96, 95, 93, 91 (b) from m/z 124.

Fig. S4.6-12 summarises the mechanism of the two above decay sequenc- es. The formation of the ion with m/z 96 requires fi rstly an isomerisation of a double bond and then a retro-Diels-Alder cleavage (Fig. S4.6-12 (b)). Finally an elimination of H• delivers m/z 95 as the most intensive ion. The +• further decay as far as m/z 91 can best be understood, if the C7H11 undergoes a cyclisation. Questions and Answers A. Investigate the origin of the term "terpene". Who were the two scientists, who received Nobel prizes for their work on terpenes? This term was coined by August Kekulé in 1863 in his "Lehrbuch der organischen Chemie" (Textbook of Or- ganic Chemistry). The suggestion came from the French chemist M. Berthelot. It was inspired by turpentine, the resinous effl uence of conifers such as pine trees (not to be confused with oil of turpentine that is distilled from turpentine), that consists of resin acids and terpenes. The originally rather diffuse term was more precise- ly defi ned with growing knowledge of the structure and building principles of terpenes.

148 Friedelin

Nobel prizes for chemistry were awarded to Otto Wallach (1910), an assistant of Kekulé, and Leopold Ruzicka (1939), who is cited in this article. Wallach recognized that isoprene (or its synthetic equivalent) is the building block of the terpenes. In 1887 Wallach discovered the so-called biogenetic isoprene rule that was formulated under this name by Ruzicka in 1922.

B. According to the literature [20] the reduction of friedelin results in two stereoisomeric alcohols depending

if NaBH4 or metallic sodium is used. Carry out these experiments and determine the stereochemistry of the alcohols obtained with NMR. What is the reason for the difference in the reactions? This is a very good experiment for the ground course in practical organic chemistry and has been successfully performed many times by us. The steric shielding of the carbonylgroup C-3 by the methyl group on C-4

leads by the reduction with NaBH4 mainly to the less stable axial alcohol epifriedelinol. In contrast, by radical reduction with sodium metal predominantly the more stable equatorial alcohol friedelinol is formed. The corresponding NMR spectra differ drastically, because of the different coupling constants, which can be analysed with the aid of the Karplus equation.

C. Use the key words in the text to inform yourself about the reasons for a wine being “corked”. Discuss the use of other sealing systems for expensive wines. When wine is "corked", it is not the fault of the natural cork. "Cork taint" denotes the presence of 2,4,6- trichloroanisole (TCA). The compound is formed by microbial methylation of 2,4,6-trichlorophenol (TCP), which for example could as a component of a plant protection agent gain access to the bark of the cork-oak. Other anisole compounds have also been identiﬁ ed as causing cork taint and can be introduced into wine in the winery from chlorine containing cleaning and disinfecting agents or from wooden pallets or packing materials. The previously used pentachlorophenol has been replaced by 2,4,6-tribromophenol (TBP), a compound with fungicide and ﬂ ame-retarding properties, which is therefore used as an additive in cardboard packing, plastics and paints. It is known, that micro-organisms metabolise TBP to 2,4,6-tribromoanisole (2,4,6-TBA), a compound, which has been described as possessing the sensory attributes musty, earthy, chemical or smelling of solvents. There are stoppers made of plastic or glass, the latter with a thin plastic seal, and metallic screw caps also with a plastic seal. People opposed to these systems claim, that one can taste the plastic and that wines must breathe, which is only possible with a classical cork. Until the appropriate investigations have been carried out as double-blind tests and with a large number of experts, one can remain sceptical about the validity of such criticism.

D. Which acyclic triterpene C30H50 is regarded as being the precursor of all cyclic terpenoids? Give an example, of where this substance occurs. Why are the manufactures of cosmetics interested in this compound? The compound is called squalene and is an acyclic, unsaturated hydrocarbon without functional groups other than the C=C double bond. Squalene is colourless and oily. It was ﬁ rst isolated in 1906 from shark liver oil. Squalene is also found in humans (skin, blood serum) and in plants (e.g. up to 0.7% in olive oil). Squalene hinders the growth of dermatophytes (ﬁ lamentous fungi that infect the skin). Therefore, jojoba oil and avocado oil, which contain squalene, are contained in the emulsions used as cosmetic creams. The formula shows squalene in its stretched, zig-zag form and in another, folded conformation, which illustrates, the required conformation for enzymatic catalyzed cationic cyclisation which leads to steroids.

149 Supporting information

E. Why is the C=O group in the HMBC spectrum (Fig. S4.6-4 in supporting information) seen very well from H-2α and H-4 but less well from H-2β?

2 All three spin couplings are JCH couplings that show no clear stereochemical dependency. For C-atoms that are connected to a heteroatom with a free electron pair a Karplus-like dependency to this electron pair is often seen.

F. Explain the negative Cotton effect with the octant rule. Following the instructions of Djerassi (C. Djerassi, W. Klyne J. Amer. Chem. Soc. 1957, 79, 1506–1507) friedelin is placed in the octant coordinate system with the center of the C=O double bond at the origin. The methyl groups 23 and 24 are then in a negative octant and therefore a negative Cotton effect is to be expected.

G. Make a plausible suggestion, how starting with the ionisation on the keto group of friedelin eventually the +• ring E, at the other end of the molecule, is eliminated and a C21H34O -ion formed. If ring A is allowed to adopt a boat conformation, the unpaired electrons on the O-atom can abstract an H-atom from C-6 via a six membered transition state. In two further steps involving intramolecular H-abstractions via a six membered or respectively a ﬁ ve membered transition state the radical centre is ﬁ nally transfered to on C-12, just in the right position for the elimination of 1,4,4-trimethylcyclohexene following two α-cleavages. The radical cation formed has the same C-backbone as the corresponding ion (m/z 302) in Fig. 4.6-25. The charge is localised on an oxonium ion instead of a tert-carbenium ion and ring C has lost an H-atom that is now bound to the O-atom. With the aid of the thus formed ion with m/z 302 the existence of the ions with m/z 274 and 246 can be explained.

150 Friedelin

+ + H. The base peak in the EI mass spectrum of friedelin is C5H9 (m/z 69) that is accompanied by C5H7 (m/z 67). Make a suggestion for the formation of both ions.

For m/z 67 the dimethylcyclopropylium ion is a possibility.

151 Supporting information

4.7 3-O-Acetyl-11-keto-β-boswellic acid (β-AKBA)

Fig. S4.7-1: Structure of 3-O-Acetyl-11-keto-β-boswellic acid (β-AKBA)

Isolation The process (Method 1) consists of three steps. Initially all hydrophobic and amphiphilic substances are extracted from frankincense with diethyl ether using a Soxhlet extractor. Subsequently all acidic compounds contained in the ether extract are transferred as anions to the aqueous phase by an alkaline extraction. This is a critical point in the isolation, because the anions of boswellic acids and related substances act as surfactants (resin soaps). Consequently, they promote the formation of emulsions and hinder the phase separation. Therefore, patience is required. After acidifying the aqueous phase and extracting with ether a raw product is obtained, in which amongst other compounds the boswellic acids are enriched. The third step consists of the separation of 3-O-acetyl-11-keto-β-boswellic acid from the raw product with preparative RP-HPLC. The process described here is the result of a review of a copious literature on the separation of the components of frankincense. The authors are aware, that it will be possible to modify and improve this procedure. However, it was the aim of our work to isolate an amount of 3-O-acetyl-11-keto-β- boswellic acid that was sufficient for spectroscopic measurements and not to optimize the procedure for isolation. While this article was being written, the idea was born, to develop an alternative method for the preparation of the raw extract containing the resin acids, in which a formation of an emulsion does not occur (Method 2). This variant, as is the Method 1, is suited to obtaining material for the HPLC separation.

Material Granules of frankincense of about the size of peas can be obtained from vendors of natural products such as OMIKRON (see www.omikron-online. de/) or from specialised dealers of spices and natural products. Preparation of the Raw Extract Method 1 Frankincense (70 g) was ground to a fine powder in a mortar, then extracted for 4 h with diethyl ether (750 mL) in a Soxhlet extractor. This resulted

152 3-O-Acetyl-11-keto-β-boswellic acid (β-AKBA)

in the dissolution of a portion the frankincense (56 g) a residue (14 g) remained in the extraction thimble. A yellowish extract was obtained that was concentrated by evaporation to half its volume. Caution: during evaporation large persistent bubbles are formed. It is advisable to use the largest flask possible, which to avoid foaming over should be filled maximally to a quarter. The acidic components are extracted by alkaline extraction from the concentrated raw extract. The ether extract is shaken in a separating funnel three times with aqueous NaOH (each time 200 mL 0.1 M). The separation of the phases is difficult, because the emulsion that is formed separates very slowly. It helps to saturate the NaOH solution (0.1 M) with NaCl to increase its density and thus to increase the difference in density between ether and the NaOH solution, this aids the phase separation. Despite this, patience lasting in the range of hours or even overnight may be necessary. The aqueous alkaline phases are pooled and washed with a small amount of diethyl ether (2×150 mL). Between the aqueous and the ether phase a deep orange coloured oily phase can be observed, its origin is unknown. The NaOH extract and the oily phase are pooled and while stirring acidified with conc. hydrochloric acid to a pH-value of 2. The emulsion or yellow precipitate produced is extracted with diethyl ether (2×200 mL) to transfer the carboxylic acids (i.e. all resin acids) that are formed to the organic phase. The organic phase is washed once with distilled water

(100 mL), dried over MgSO4, filtered and the ether removed to dryness by evaporation. A light yellow amorphous powder (16 g) is obtained as a raw product.

Additional NMR Spectra at 700 MHz in CDCl3

Fig. S4.7-2 Expansion of the 1H-NMR spectrum in the aliphatic region

152 153 Supporting information

Fig. S4.7-3 DQF-COSY spectrum of 3-O-acetyl-11-keto-β-boswellic acid

27 24 26 25 30 18 29

δC / ppm 2, 24 15,16 28 21, 7 17,1, 10 19,20 22, 14 8,4 5

18, 9

δH / ppm Fig. S4.7-4 Excerpt 3 from the HMBC spectrum 30 CH3 29 20 H C 19 3 21 28 12 18 CH3 O 11 22 25 26 13 17 CH3 9 CH3 14 H 16 15 1 10 O 2 H 8 CH 3 5 7 3 4 6 27 C O 31 H H3C HOOC CH3 32 23 24

154 3-O-Acetyl-11-keto-β-boswellic acid (β-AKBA)

Fig. S4.7-5 Excerpt 4 from the HMBC spectrum

Fig. S4.7-6 NOESY-spectrum of 3-O-acetyl-11-keto-β-boswellic acid

Questions and Answers A. Where in the triterpenoid basic module of 3-O-acetyl-11-keto-β-boswellic acid, i.e. in β-boswellic acid, can the 6 isoprene units be found? Draw the structural formula of β-boswellic acid and mark the C-atoms of each isoprene unit with letters.

154 155 Supporting information

B. The alkali metal salts of α- and β-boswellic acid and their derivatives are resin soaps, i.e. they are anionic surfactants, in the same way that the anion of cholic acid in bile is a surfactant, which can act as a dispersant (intestines) or to remove stains (oxgall soap). Sodium acetate, also a carboxylate, is not a surfactant. Why is this so? What are the principle structural requirements for an anionic surfactant? A surfactant is an amphiphilic substance, in which a hydrophilic as well as a hydrophobic part of the molecule exists. Typical for the hydrophilic part are ionic groups such as carboxylate or sulfonate or carbohydrate residues as in saponins. The hydrophobic moiety must have a certain minimal number of C-atoms, because surfactant effects, such as a decrease of the surface tension of an aqueous solution and the formation of micelles only occur above a given number of C-atoms in the hydrophobic moiety. One methyl group, as in sodium

acetate, is not sufficient. Carboxylate soaps require C11 or more in the alkyl residue for good surfactant properties. This condition is naturally fulfilled for all resin acids and bile acids such as the examples shown below:

C. It is not always possible to see clearly through the air that surrounds us. Smoke or fog can make the view hazy. What in the physicochemical sense are the similarities and differences between fog and smoke. How does the effect of haze occur? Fog and smoke are aerosols, i.e. finely divided liquid droplets or solid particles dispersed in a gas such as air. From the colloid chemical point of view they are dispersions of a solid or liquid in a gas. The particle size of these colloids is in the range between 1 nm and 1 μm. They can interact with visible light and scatter it (Tyndall effect). All dispersions appear turbid and this also applies to emulsions, gels, soles and colloidal soap solutions that contain micelles.

D. The structural formula of the substituted β-boswellic acid described here is reminiscent of a steroid such as testosterone. Why is it not justifiable, to attribute these resin acids to the steroids? What is the reason that the structural formulas appear to be similar? Steroids are derived from the structure of the parent hydrocarbon sterane (cyclopentanoperhydrophenan- threne) with four fused rings (three six membered rings and one five membered ring). A sub-class are for

156 3-O-Acetyl-11-keto-β-boswellic acid (β-AKBA)

example the hormones of the gonane-type such as oestrogens, androgens, gestagens and the corticosteroids. All bile acids have the sterane structure (see above). With the boswellic acids three six membered rings are linked to each other, but a five membered ring is missing. The similarity to the steroids lies in the fact, that structurally both are built up from the basic unit isoprene.

E. Why does the EI mass spectrum of β-boswellic acid has so few fragment ions? The removal of an electron during the electron impact ionisation can in principle occur from an n-, π- or σ-electron pair, whereby the ionisation energy increases in that order. Therefore, the loss of an n-electron from the O-atom of the keto group is favoured. The assumption, that in the molecular ion formed the charge and radical character are localised, has proved to be of value as a working hypothesis for the interpretation of mass spectra. In the present case this means, that both degradation routes that are initiated by the unpaired electron on the O-atom of the keto group dominate the spectrum and other possibilities of degradation are to a greater extent excluded. With saturated hydrocarbons the ionisation can only occur on σ-bonds. Since the ionisation energy of the available C-C-bonds does not differ greatly many bonds can be cleaved and accordingly the spectra are richer in fragments.

F. Explain the ions of low intensity in the upper region of mass with m/z 497, 452 and 408.

+ The [M-CH3] ion (m/z 497) could be formed in an intermediate step of the fragmentation "A" (Fig. 4.7-27) or "B" (Fig. 4.7-28), if in competition to the degradation described a methyl radical is eliminated by α-cleavage. The low intensity can be explained by the fact, that in the case of competing α-cleavages it is always the loss of the larger neutral particle that will be favoured.

+ + A prerequisite for the formation of the [M-CH3COOH] (m/z 452) and [M-CH3COOH-CO2] ions (m/z 408) is the ionisation of the acetoxy group. The m/z 452 ion is formed by a McLafferty rearrangement with a charge localisation on the olefinic fragment. Finally by H-transfer over a six membered transition state a decarboxylation can occur. The main competition for this degradation route is the formation of the acetyl cation (m/z 43) by α-cleavage.

156 157 Supporting information

Chapter 5 5.1 Sinensetin

Fig S5.1-1 Structure of sinensetin

Questions and Answers A. Explain, why a considerable amount of sinensetin dissolves in water, although the molecule contains no hydrophilic substituents such as OH or COOH, for which one would expect this to apply. Sinensetin is not entirely hydrophobic, such as a condensed aromatic hydrocarbon is (e.g. naphthalene or pyrene), since it has a total of 7 oxygen atoms that can act as acceptors for hydrogen bonding to the solvent water. This is the structural reason for a certain solubility in water.

B. Explain, why, limonene can be so selectively separated from orange peel by steam distillation, as described. To be steam volatile a compound must fulfil two conditions. It must be hydrophobic and it must have a certain vapour pressure, i.e. it must not have a too low volatility, which is dependent on the molecular mass. As a rule the vapour pressure at 100 °C must be at least 10 mbar. Limonene fulfils both conditions. As a purely

monoterpene hydrocarbon C10H16 it is very hydrophobic. With a bp of 177 °C at normal pressure the vapour pressure at 100 °C is well above the value given above.

C. Why is a UV but no CD spectrum shown in this article? For a CD spectrum substances must possess a chromophore and be chiral. Sinensetin is achiral.

D. Suggest a reason, why the 13C NMR signals from C-9 and C-10 are slightly separated from those from C-7', C-8' and C-11, in contrast to the corresponding 1H NMR signals. C-7', C-8' and C-11 are methoxy groups that have to one side a methoxy group and to the other side an aromatic H-atom as neighbours. In contrast C-9 and C-10 have substituents as neighbours on both sides. This affects their chemical shift. Since the protons are situated further outside, on the edge of the molecule, they do not exhibit this differentiation.

E. What reasoning can be used, to differentiate between the 1H-NMR signals from H-3 and H-8? H-8 is an aromatic proton with two inductively electron withdrawing but mesomerically electron releasing oxygen atoms in the β-position. In contrast H-3 is an olefinic proton in the α-position of an α/β-unsaturated carbonyl system.

158 Sinensetin

F. Explain the formation of the peak with m/z 69 in the mass spectrum. Its mass was attributed to an ion with the + composition C3HO2 by high resolution MS.

G. The fragments c and d are found in the mass spectrum of many flavones.

Ar is the phenyl ring attached to C-2. The corresponding ions, albeit with a low intensity, can be found in the spectrum of sinensetin (m/z 162, 165). Explain their formation.

158 159 Supporting information

5.2 Rosmarinic acid

Fig. S5.2-1 Structure of rosmarinic acid

Biosynthesis The biogenesis of rosmarinic acid, which is an ester of caffeic acid (3,4-dihydroxycinnamic acid) with its product of hydration, 3-(3,4-di- hydroxyphenyl)lactic acid, proceeds via the shikimate pathway, one of the two important routes to aromatic compounds in plants. In the fi rst cycle erythrose-4-phosphate is joined to the heptose 3-de- oxy-D-arabino-heptulosonic acid 7-phosphate to give 3-hydroquinic acid. Further important and increas- ingly unsaturated intermediates are shikimic acid [9], chorismic acid and prephenic acid. The latter stands as the fi nal non-aromatic precursor at a junction from which the biosynthetic pathway divides in the direction of the aromatic amino acids L-phenylalanine or L-tyrosine [10]. From these rosmarinic acid is synthesized by 8 enzymes (see Fig. S5.2-2). Amongst others M. Peters- en et al. have made particular contributions to the elucidation of this process [7]. It should be noted, that rosmarinic acid is chiral, because the enantioselectively produced Fig. S5.2-2 Biosynthesis of rosmarinic acid in cell cultures of the painted nettle ac- (R)-(4-hydroxyphenyl)lactic acid is cording to [7]. The enzymes are: PAL=phenylalanine ammonia lyase, CAH=cinnamic esterifi ed at the aliphatic hydrox- acid 4-hydroxylase, 4CL=4-coumarate-CoA ligase, TAT=tyrosine aminotransferase, yl-group with 4-coumaroyl-CoA by HPPR=hydroxyphenylpyruvate reductase, RAS=rosmarinic acid synthase, 3-H and rosmarinic acid synthase to (R)-4- 3'-H=hydroxycinnamoylhydroxyphenyllactate 3- and 3'-hydroxylase coumaroyl-4'-hydroxyphenyllactic acid that then in a fi nal oxidative step is converted to rosmarinic acid.

160 Rosmarinic acid

CD spectrum in Ethanol

2 )

-1 1 mmol

2 0 / (cm ε

∆ -1

-2 200 300 400 λ / nm

Fig. S5.2-3 CD spectrum of rosmarinic acid The CD spectrum is difﬁ cult to interpret and shows at least one change of sign attributable to the Cotton effect.

IR spectrum in KBr

Fig. S5.2-4 IR spectrum of rosmarinic acid As might be expected the absorption bands of the OH groups dominate the IR spectrum. Additionally a carboxyl band at 1700 cm−1 and a band due to a C=C double bond at 1600 cm−1 can be identiﬁ ed.

161 Supporting information

NOESY-Spectrum at 700 MHZ in DMSO-d6 3 2' 5' 2''' 5''' 2 6' 6'''

δH / ppm

6''' 5''' 2''' 5'''

6' 2'

δH / ppm

Fig. S5.2-5 Expansion of the NOESY-spectrum in the aromatic region Because of the relatively high viscosity of DMSO the NOESY-spectrum (Fig. S5.2-5) only shows signals of the same phase, however, it conﬁ rms the assignments made from the analysis of the HMBC- spectrum. H-3 shows a cross-relaxation to H-2' and H-6', and this conﬁ rms that the aryl protons of the C-3 bonded aromatic ring are deshielded relative to those of the C-3'' bonded aromatic ring.

Mass Spectrometry

The dissociation of the [M-H]–-ion by collision activation (ESI(−)-MS/MS) leads to 3 fragments with m/z 197, 179 and 161 (Fig. S5.2-6). Interestingly the mechanistic interpretation of these decay paths shows analogies to the ion formation in the EI(+) mass spectrum. The formation of the ions at m/z 197 and 179 is summarized in Fig. S5.2-7. The ion at m/z 161 corresponds to the neutral ketene fragment that is lost during the formation of the m/z 197 minus a proton. How this proton belonging to the side of the molecule with the C1'-C6'-labeled ring can be transferred into the other side, so that this can be ejected as an uncharged fragment, is shown in Fig. S5.2-8.

Fig. S5.2-6 MS/MS spectrum of the [M−H]−-Ion

162 Rosmarinic acid

Fig. S5.2-7 Fragmentation path of the [M−H]–-Ion (m/z 359) from rosmarinic acid after collisional activation

Fig S5.2-8 Formation of m/z 161

Fig. S5.2-9 shows the GC/EI-MS of fully silylated rosmarinic acid. The McLafferty rearrangement of rosmarinic acid described in the main text (Fig. 5.2-24) leads to two radical cations with different degrees of silylation, depending upon which fragment takes the charge and radical (m/z 324 and 396), (Fig. S5.2-10). The fragment with m/z 198 found for rosmarinic acid (Fig. 5.2-25 in the main text) must be shifted to m/z 412 for the silylated derivative. The fact, that no ion is detected at m/z 412, supports the mechanistic proposition explaining the ion with m/z 198 in the EI-spectrum of rosmarinic acid, for which the presence of a phenolic H-atom is required.

163 Supporting information

Fig. S5.2-9 GC/MS of silylated romarinic acid from: S. Moldoveanu, "The utilization of gas chromatography/mass spectrometry in the proﬁ ling of several antioxidants in botanicals" Advances in Gas Chromatography, 2014, 103–132

Fig. S5.2-10 McLafferty rearrangement of silylated rosmarinic acid Questions and Answers A. The acidity of aliphatic carboxylic acids depends upon the substituents in the chain. Arrange the following carboxylic acids into the order of their increasing acidity and give reasons for your choice: lactic acid (regardless if D- or L-form), propionic acid, pyruvic acid. Which physico-chemical parameter indicates the order? The degree of acidity is expressed as the pKa-value, i.e. the negative decadic logarithm of the acid constant, which indicates the extent of dissociation of an acid in water. The weakest acid is propionic acid, noticeably more acidic is lactic acid (2-hydroxypropionic acid) and most acidic is pyruvic acid (2-oxopropionic acid). From the logarithmic pKa-values it can be seen that the acidity increases along this series by a factor of 10. This is caused by the withdrawal of electrons by the O-atom of the substituent in the 2-position, which sta- bilizes the carboxylate ion, thus weakening the O-H-bond of the carboxylic acid group and facilitating the dissociation in H+ and R-COO–. In the case of lactic acid the OH-group exerts a negative inductive effect (–I-effect), whereas the carbonyl group of pyruvic acid has in addition a negative mesomeric effect (–M-effect). The pKa-value for rosmarinic acid is 2.8 (H.-Y. Cheung, Q.-F. Zhang J. Chromatography A 2008, 1213, 231–238.) propinoic acid pKa=4.87 < 2-hydroxypropionic acid pKa=3.90 < 2-oxopropionic acid pKa=2.49

164 Rosmarinic acid

B. Viewed from a biochemical stand point, it is obvious that rosmarinic acid can only be a natural product from a plant. Why is this so? During the course of evolution living entities have divided into autotrophic and heterotrophic organisms. Autotrophic organisms such as plants are able to produce all necessary proteinogenic amino acids including the aromatic ones themselves. However, this is not the case for heterotrophic organisms such as animals, fungi and bacteria. These have lost the ability to biosynthesize aromatic rings and rely on obtaining these in their nutrition. Accordingly humans acquire for example the amino acid L-phenylalanine, which is essential because it cannot be produced in the human body, via their food. It is, however, possible that this amino acid can be oxidized (hydroxylation in the para position of the phenyl ring) to form L-tyrosine, which is also a proteinogenic amino acid. As always there are exceptions, such as in the biosynthesis of estrogens, in which in estrone, estradiol and estriol the A-ring is aromatic.

C. Rosmarinic acid shows three different protonic spin systems. One of these is an aliphatic ABX-system. For which class of substances is this spin system of particular importance? For many amino acids.

D. Give detailed reasons for the assignment of the 13C-signals to the aromatic ring with the numbers having a single prime. C-1': HMBC 3J to H-2 and 3J to H-5' C-2': HSQC 1J to H-2', HMBC 3J to H-3 and 3J to H-6' C-3': HMBC 2J to H-2' and 3J H-5' C-4': HMBC 3J to H-2', 2J to H-5' and 3J to H-6' C-5': HSQC 1J to H-5' C-6': HSQC 1J to H-6', HMBC 3J to H-2', 3J to H-3

Translation of Hortus Medicus on p. 393 In this country [Germany] 'wreath rosemary'1 requires particular care and attention2, whereas in the South of France3 it grows wild in such abundance, that its wood4 is even used for heating5. It fl owers twice per year, in spring and autumn and favours in particular dry soil, less so rich or moist earth. Plinius did not know, that it bares seeds, however, in contrast we have observed, that it produces them in copious amounts. They are self dispersing and if they are not gathered in time, a great number of them grow to young plants, however, these seldom survive the fi rst year. In milder regions if the stalk is wrapped with juniper straw to protect it from the cold, they may survive the winter even outdoors6. I can remember having seen it in this fashion in the Elector's7 Garden in Aschaffenburg. Since an- tiquity wreaths of honour8 have been displayed9 for the household deities. Horace wrote, "Bind a wreath only from rosemary and myrtle twigs for the minor gods." Its oil10, refi ned by alchemistic means11, helps well against forgetfulness12. Obtained from its fl owers during the sun's heat of the Dog Days13, it can be used advantageously to heal wounds and against deteriorating eyesight. Translation: Dr Walter Fischer (Allmendingen/Ehingen) 1 masculine form as Rosmarinus coronarius, now Rosmarinus offi cinalis. Cf.10 2 'mit fl eißiger Wartung erzogen und geheget' 3 Gallia Narbonensis: approximately Provence – Languedoc 4 It is well known that the humanists mocked the theological scholars for their poor Latin, however, here stands 'ligni utantur', from which we might deduce, that ligni is the masculine, genitive, plural from lignus instead of respectively lignum – ligna since uti takes the ablative it must be lignis.

165 Supporting information

5 According to Zedlers Universallexikon (1731–1754, Bd. 32, 499 ff.) used instead of ﬁ rewood and to heat the baking oven. 6 Probably reference is made to this simple protection for winter and not (Zedlers Camerarius-Interpretation), that a rosemary twig that has been grafted to a juniper stem will withstand the cold of winter. 7 Archbishop and Elector of Mainz 8 from rosemary, myrtle and laurel 9 See 4. Correct would be 'coronae suspendebantur' 10'externally applied' 11 circulatio 12 frigidius: neutr. from frigidior to frigidus – meaning run down, weak rather than freezing, although the oil should help in cases of 'weakness caused by coldness of the brain'. 13 Dies caniculares after canis maior, the dog star = Sirius

166

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