sign in sign up
<<
Home , Methoxy group, Narceine

4.

Definition: the term “” (alkali‐like) is commonly used to designate basic heterocyclic nitrogenous compounds of plant origin that are physiologically active. Classification: • True (Typical) alkaloids that are derived from amino acids and have in a heterocyclic ring. e.g Atropine • Protoalkaloids that are derived from amino acids and do not have nitrogen in a heterocyclic ring. e.g Ephedrine • Pseudo alkaloids that are not derived from amino acids but have nitrogen in a heterocyclic ring. e.g Caffeine • False alkaloids are non alkaloids give false positive reaction with alkaloidal reagents. Nomenclature:

Trivial names should end by "ine". These names may refer to: • The genus of the plant, such as Atropine from Atropa belladona. • The plant species, such as Cocaine from Erythroxylon coca. • The common name of the drug, such as Ergotamine from ergot. • The name of the discoverer, such as Pelletierine that was discovered by Pelletier. • The physiological action, such as Emetine that acts as emetic, acts as . • A prominent physical character, such as Hygrine that is hygroscopic. Prefixes and suffixes:

Prefixes: • "Nor‐" designates N‐demethylation or N‐demethoxylation, e.g. norpseudoephedrine and nornicotine. • "Apo‐" designates dehydration e.g. apomorphine. • "Iso‐, pseudo‐,neo‐, and epi‐" indicate different types of isomers. Suffixes: • "‐dine" designates isomerism as quinidine and cinchonidine. • "‐ine" indicates, in case of ergot alkaloids, a lower pharmacological activity e.g. ergotaminine is less potent than ergotamine. Physical Properties: I‐ Condition: • Most alkaloids are crystalline solids. • Few alkaloids are amorphous solids e.g. emetine. • Some are liquids that are either: Volatile e.g. nicotine and coniine, or Non‐volatile e.g. pilocarpine and hyoscine. II‐ Color: The majority of alkaloids are colorless but some are colored e.g.: • Colchicine and are yellow. • is orange. • The salts of are copper‐red. III- Isomerization: • Optically active isomers may show different physiological activities. • l‐ephedrine is 3.5 times more active than d‐ephedrine. • l‐ergotamine is 3‐4 times more active than d‐ergotamine. • d‐ Tubocurarine is more active than the corresponding l‐ form. • Quinine (l‐form) is antimalarial and its d‐ isomer quinidine is antiarrythmic. • The racemic (optically inactive) dl‐atropine is physiologically active. Chemical Properties:

I‐ Nitrogen:

• Primary R‐NH2 e.g. Norephedrine

• Secondary amines R2‐NH e.g. Ephedrine

• Tertiary amines R3‐N e.g. Atropine

• Quaternary salts R4‐N e.g d‐ Tubocurarine

II‐ Basicity:

• R2‐NH > R‐NH2 > R3‐N • Saturated hexacyclic amines is more basic than aromatic amines. According to basicity Alkaloids are classified into:

• Weak bases e.g. Caffeine • Strong bases e.g. Atropine • Amphoteric * Phenolic Alkaloids e.g. Morphine *Alkaloids with Carboxylic groups e.g. Narceine • Neutral alkaloids e.g. Colchicine III‐ :

• Most alkaloids contain Oxygen and are solid in nature e.g. Atropine.

• Some alkaloids are free from Oxygen and are mostly liquids e.g. Nicotine, Coniine. Extraction, Purification and Isolation of Alkaloids from Powdered plants

• Extraction and purification Method I: The powder is treated with alkalis to liberates the free bases that can then be extracted with immiscible organic solvents.

Method II: The powdered material is extracted with water or aqueous containing dilute acid. Alkaloids are extracted as their salts together with accompanying soluble impurities.

Method III: The powder is extracted with water soluble organic solvents such as MeOH or EtOH which are good solvents for both salts and free bases. • Liberation of the free bases: Alkalis are used to liberate free bases. Alkalis must be strong enough to liberate free bases. However, choice of strong alkalis must be avoidedinsomecases:

1‐ Alkaloids e.g. Solanaceous Alkaloids

2‐ Alkaloids e.g. Colchicine

3‐ Phenolic Alkaloids e.g. Morphine

4‐ Lactone Alkaloids e.g. Pilocarpine

5‐ Fatty Drugs due to saponification and emulsion formation. ¾Separation of Alkaloidal Mixtures:

• Fractional Crystallization:

Ephedrine & Pseudoephedrine Oxalates

Crystallization from water

Ephedrine Oxalate Pseudoephedrine Oxalate Crystals Solution Atropine & Hyoscyamineine Oxalates

Crystallization from /

Atropine Oxalate Hyoscyamine Oxalate Crystals Solution • Preparation of Derivatives: Separation of Primary, Secondary and Tertiary Alkaloids.

O Mixture + p-toluenesulphonyl chloride Cl S Add HCl and filter O

Filtrate Precipitate tertiary alkaloids as salt 1ry alk derivative 2ry alk derivative (no reaction with reagent O O R-N S R-HN S keto form R O O insoluble in alkalis acidic

OH enol form R-N S soluble in alkalis O

NaOH, filter

Filtrate Precipitate 1ry alk derivative 2ry alk derivative • Fractional Liberation:

Atropine & Hyoscyamine & Hyoscine the form of HCl salts

1- Alkalinize by NaHCO3 pH 7.5 2- Extract with Ether

Ether Aqueous layer Hyoscine free base Atropine & Hyoscyamine HCl (pKa = 6.2) (pKa = 9.3) • Fractional Distillation: e.g. Separation of Nicotine and Anabasine

• Chromatographic Separation. • Phenylalkylamines:

e.g. Ephedrine CH2 CH CH3

NH2

• and piperidine e.g. lobeline, nicotine NN H

• Tropane e.g. Atropine. NCH3 OH • Quinoline e.g.quinine and quinidine

N • e.g. N

• Phenantheren e.g. Morphine • Indole e.g.ergometrine N H

• Imidazole N e.g. pilocarpine

N

• Purine H 6 7 e.g. caffeine 5 N 1 N 8 2 4 N N 9 3 Purine • Steroidal e.g. Solanum and Veratrum alkaloids

• Terpenoid e.g. Taxol

Introduction:-

¾ As the molecular structure of alkaloids is quite complex , very little progress was achieved in the elucidation of their structures during 19th century .

¾ But now the new methods for the identification of unknown substances are known.

¾ The following pattern of procedure is adopted to establish the molecular structure of an alkaloid: 1) Molecular Formula determination: ¾ After a pure specimen has been obtained , its elemental composition , and hence the empirical formula , is found by combustion analysis. ¾ Then , its molecular weight is determined by the Rast procedure( depression of the freezing point) to establish its molecular formula. ¾ Its calculation is based upon the simple fact that introduction of a double bond or cyclisation of the chain decreases the molecular formula by two hydrogen atoms relative to the corresponding saturated aliphatic hydrocarbon. ¾ For example, the difference between hexene (C6H12) from hexane (C6H14) is two hydrogen's and this difference is called a double bond equivalent.

¾ Similarly, the difference between (C6H6) and hexane (C6H14) is eight hydrogen’s which will correspond to 8/2 or 4 double bond equivalents (accommodated by the three double bonds and one ring). ¾ The above procedure is valid for simpler compounds only. However, for complex formulae, where elements other than hydrogen and are present, the simpler method is that for any formula CaHbNcOd the number of double bond equivalents is given by the following expression: a – 1/2b + 1/ + 1 ¾ The above method for the calculation of double bond equivalents is useful to calculate the number of rings in a given compound. For example, hygrine has the molecular formula, C8H15NO which corresponds to 8 – 15/2 + ½ + 1 = 2 double bond equivalents. However, chemical tests reveal that hygrine contains only one (one double bond equivalent) and does not show other form of unsaturation. Thus hygrine must be monocyclic to account for the other double bond equivalent. ¾ The presence of unsaturation in an alkaloid may also be ascertained by treating the alkaloid with bromine or halogen acid or alkaline potassium permanganate when a glycol is obtained 2) Analysis: ¾ Application of classical techniques of organic analysis (especially if the alkaloid is available in appreciable amounts) and/or infra‐red examination (especially if the alkaloid is available only in small amounts) can reveal the nature of the functional groups present.

¾ This will also reveal the aromatic or aliphatic nature of the alkaloid and the unsaturation, if present. 3) Functional Nature Of Oxygen: ¾ If an alkaloid contains oxygen, it may be present as

–OH (phenolic or alcoholic), methoxy (–OCH3 ), acetoxy (– OCOCH3 ), benzoxyl (–OCOC6H5 ), carboxylic (–COOH), carboxylate (–COOK) or carbonyl (=C=O).

¾ Occasionally, lactone ring systems have also been encountered (e.g., narcotine, hydrastine). These functions have been detected by the usual methods of organic analysis including infra‐red examination.

¾ Various oxygen functional groups can be characterised according to the following characteristics: a) Hydroxyl group: 9 Its presence in an alkaloid can be ascertained by the formation of acetate, on treatment with acetic anhydride or acetyl chloride or by the formation of benzoate on treatment with benzoyl chloride in the presence of .

9 However, the above test for oxygen should be applied carefully because primary amines if present in an alkaloid also yield acetyl and benzoyl derivatives. ¾ Then the number of hydroxyl groups is determined by acetylation or Zerewitnoff’s method. In the former method, the number of hydroxyl groups is determined by acetylating the alkaloid and hydrolysing the acetyl derivative with a known volume of IN NaOH.

¾ The excess of alkali is estimated by titration with a standard solution of HCl. The number of acetyl groups or hydroxyl groups can be calculated from the volume of alkali used for hydolysis. b) Carboxylic group: 9 The solubility of an alkaloid in aqueous sodium carbonate or reveals the presence of carboxylic group. The formation of ester on treatment with an alcohol also reveals the presence of carboxylic group.

9The number of carboxylic groups may be determined by volumetrically by titration against a standard barium hydroxide solution using phenolphthalein as an indicator or gravimetrically by silver salt method. c) Oxo group: 9 The presence of this group is ascertained by the reaction of an alkaloid with hydroxylamine, semicarbazide or phenylhydrazine when the corresponding , semicarbazone or phenylhydrazone are formed.

9 Distinction between an and a can be made on the basis of reduction and oxidation reactions. d) : 9 The detection of this group and its number may be determined by the Zeisel determination, analogous to the Herzeg‐Meyer method for N‐methyl groups. 9 In this method, a known weight of alkaloid is heated with hydriodic acid at its boiling point (126 °C) when the methoxyl groups are thereby converted into methyl iodide which is then absorbed by ethanolic silver and the precipitated silver iodide is filtered, dried and weighed. From the weight of silver iodide, the number of methoxyl groups may be calculated. 9 For example, papavarine, C20H21O4N, when treated with hydrogen iodide, consumes 4 moles of hydrogen iodide, producing 4 moles of silver iodide and thus confirming the presence of four –OCH3 groups.

e)Ester and amide groups: 9These groups can be detected and estimated by observing The products of their alkali or acid hydrolysis. 4) Nature of Nitrogen: ¾ All alkaloids contain nitrogen . But in the majority of alkaloids it is present as a part of a heterocyclic system. Therefore, it must be either a secondary (=NH) or tertiary(=N–CH3 or =N–). ¾ However, there are phenylalkyl type of alkaloids (adrenaline, ephedrine, etc) which do not contain nitrogen as a part of a heterocyclic ring but in the form of a primary amino (–NH2) group. a) The general reactions of the alkaloid with acetic anhydride, methyl iodide and nitrous acid often show the nature of the nitrogen. 9If the alkaloid reacts with one mole of methyl iodide to form an N‐methyl derivative, it means that a secondary nitrogen atom is present. For example, coniine, C8H17N reacts with one mole of methyl iodide to form an N‐ methyl derivative, indicating that coniine must contain secondary nitrogen atom. 9If an alkaloid reacts additively with one mole of methyl iodide to form crystalline quaternary salt, this indicates that nitrogen atom present in this alkaloid is tertiary. For example, nicotine reacts additively with two moles of methyl iodide, indicating that it contains both nitrogen atoms as tertiary. 9 One can detect the tertiary nitrogen atom in an alkaloid by treating it with 30 % when tertiary nitrogen is oxidised to amine oxide.

b) Herzig‐Meyer’s method is used to detect by distillation of alkaloid with soda‐lime when methyl amine is obtained. For example, nicotine on heating with soda‐lime yields indicating that it must contain a N‐. 9 NMR may also be utilized for the rapid detection of N‐ methyl and N‐ethyl groups in alkaloids.

5) Estimation of C‐Methyl groups: ¾ C‐methyl groups are quantitatively estimated by the Kuhn‐Roth oxidation, the formed being distilled off and distillate titrated against standard base. 6) Degradation Of Alkaloids: ¾ The reactions used in degradation of alkaloids are as follows: (a) Hofmann exhaustive method (b) Emde’s degradation (c) Reductive degradation and zinc dust distillation (d) Alkali fusion (e) Oxidation (f) Dehydrogenation a) Hofmann’s Exhaustive Methylation Method: ¾ The principle of this method is that compounds, which + ‐ contain the structural unit =CH=C–N R3OH , eliminate a trialkylamine on pyrolysis at 200 °C or above to yield an olefin. ¾ If the nitrogen atoms forms a part of a cyclic structure, two or three such cycles are essential to liberate the nitrogen and expose the carbon skeleton. ¾However , this method is applicable only to reduced ring systems such as piperidine and actually fails with analogous unsaturated compounded such as pyridine and therefore the latter should be first of all converted into the former ¾ When a of water is eliminated from quaternary ammonium hydroxide, hydrogen atom is always eliminated from the β‐position, if this hydrogen is not available, the reaction fails ¾ The hofmann’s degradation method can be applied to hordenine methyl ether which yields p‐methoxy styrene. b) Emde’s degradation: ¾ If the alkaloid does not contain a β‐hydrogen atom, the Hofmann’s exhaustive methylation method fails. In such cases, Emde’s method may be employed. ¾ In this method, the final step involves reductive cleavage of quaternary ammonium salts either with sodium amalgam or sodium in liquid ammonia or by catalytic hydrogenation: ¾ Emde’s method can be demonstrated by considering the case of isoquinoline: c) Reductive Degradation and Zinc Dust Distillation: ¾ In some cases the ring may be opened by heating with hydiodic acid at 300 °C, e.g.,

HI 300 C ¾ Zinc dust distillation produces simple fragments from which one can draw the conclusion about the carbon framework of the alkaloid molecule. ¾ Zinc dust also brings about dehydrogenation or removal of oxygen if present. For example,

¾ As conyrine is formed by loss of six hydrogen atoms, it means that coniine must contain a piperidine ring d) Alkali fusion: ¾ This is very drastic method which is often employed to break down the complex alkaloid molecule into simpler fragments, the nature of which will give information on the type of nuclei present in the alkaloid molecule. For example, adrenaline when fused with solid potassium hydroxide yields protocateochic acid, indicating that adrenaline is a catechol deravative. e) Oxidation: ¾ This method gives useful information about the structure of alkaloid. By varying the strength of the oxidising agents, it is possible to obtain a variety of oxidation products. For example, (i) In order to carry out mild oxidation, hydrogen peroxide, iodine in ethanolic solution, or alkaline potassium ferricyanide are usually used. (ii) In order to carry out moderate oxidation, acid or alkaline potassium permanganate or chromium trioxide in acetic acid are generally used. (iii) For carrying out vigorous oxidation, potassium dichromate‐sulphuric acid, chromium trioxide‐sulphuric acid, concentrated nitric acid or manganese dioxide‐ sulphuric acid are used. These reagents usually break up an alkaloid into smaller fragments whose structures are either already known or can be readily ascertained. For example, ¾ From this reaction, it can be concluded that nicotene contains a pyridine ring having a side chain in β‐position. ¾ This classification of oxidising agents is not rigid because the ‘strength’ of an oxidising agent depends to some extent on the nature of the alkaloid which is being oxidised.

(f) Dehydrogenation: ¾ When an alkaloid is distilled with a catalyst such as sulphur, selenium or palladium, dehydrogenation takes place to form relatively simple and easy recognisable products which provide a clue to the gross skeleton of the alkaloid ¾ During dehydrogenation, there occurs the elimination of peripheral groups such as hydroxyl and C‐methyl.

(7) Synthesis: ¾ The structure of the alkaloid arrived at by the exclusive analytical evidence based on the foregoing methods is only tentative. The final confirmation of the structure must be done by the unambiguous synthesis. (8) Physical Methods: ¾ In alkaloid chemistry, the most important instrumental methods are as follows: (a) Ultraviolet spectroscopy (b) Infra‐red spectroscopy (c) Mass spectroscopy (d) Optically rotatory dispersion and circular dichroism (e) Conformational analysis, and (f) X‐ray diffraction (a) Ultraviolet Spectroscopy: ¾ This is mainly used to establish the class and/or structural type to which the alkaloid being investigated belongs. Such assignments are made because ultraviolet spectrum of a compound is not a characteristic of the whole molecule but only of the chromophoric system(s) present. ¾ The usual practice is to record the ultravoilet spectra of a very large number of different types of alkaloids. Then, the data are analysed and categorised with respect to structure correlation. ¾ Each group of alkaloids having a particular chromophoric system benzene, pyridine, indole, quinoline, etc . yields characteristic absorption maxima and extinction coefficients. ¾ Therefore, the comparison of these data with those observed for a new alkaloid may allow the identification of the exact nature of the aromatic or heterocyclic system in the new compound.

(b) Infra‐red spectroscopy: ¾ In alkaloid chemistry, it is mainly used to ascertain the presence and sometimes the absence of particular functional group. ¾ The presence of aldehyde, ketone, , , ester, amide, lactone, , carbonyl grroups and primary and secondary amines can rapidly be identified and distinguished by comparison of the observed frequencies with those reported for structural related compounds.

¾ One can also ascertain the presence of O‐methyl, N‐ methyl and aromatic groups from the infra‐red spectrum of an alkaloid but the quantitative analysis of such groups is best accomplished by NMR spectroscopy. (c) Mass spectroscopy: ¾ This technique is quite useful because it gives quite useful information about the alkaloid like. (i) The molecular weight. (ii) The empirical formula by accurate mass measurement of the molecular ion, and (iii) Knowledge of the molecular structure by comparison of the fragmentation pattern with those of analogous system. ¾ Most of the success has been achieved in the case of polycyclic indole alkaloids because the indole nucleus of these substances gives rise to an abundant, stable molecular ion which subsquently undergoes decomposition by highly specific bond fusion involving the acyclic portion of the molecule containing the other nitrogen atom(s). d) Optically rotatory dispersion and circular dichroism: ¾ These are only instrumental methods which are mainly used for elucidation of the stereochemistry of alkaloids but their application is restricted to those compounds which are optically active, i.e., to those in which a rotation‐ reflection symmetry axis is absent.

¾ Due to this reason, few alkaloids of the yohimbine, aprophine, morphine and benzlisoquinoline series have been examined so far by these techniques. e) Conformational Analysis: ¾ The principles of conformational analysis have been widely used to establish the stereochemistry as well as physical properties and chemical reactivity of alkaloids.

¾ The approach is mainly experimental which involves determination, correlation and interpretation of the kinetics and product ratios obtained from simple chemical transformations such as reduction of double bonds and carbonyl groups, hydrolysis and esterfication, oxidation of alcohols and quaternization of amines, epimerization, etc f) X‐ray Diffraction: ¾ This technique is widely used to study alkaloids because it gives the exact structure of he molecule, including bond angles and bond lengths; it also gives the information about the relative stereochemistry, including information on overcrowding twisted bonds, etc.

¾ X‐ray diffraction method is also useful to reveal the absolute configuration of the molecule. “ Structural elucidation of RESERPINE ” Content:‐ • Introduction • Constitution of Reserpine • Structure of Yobyrine & reserpic acid confirmation • Structure of reserpine confirmation by synthesis Introduction:‐ • Reserpine is the main constituent of Rauwolfia species,perticularly R.serpentina & R.vomitoria. • It is mainly used for the treatment of hypertension, headache, tension, asthma & dermatological disorders. Constitution of reserpine:‐

1. Molecular formula:‐ C33H40N2O9

2. Presence of five methoxy groups:‐ By zeisel method i.e. when treated with HI yields five of methyl iodide indicating the presence of five methoxy groups. 3. Nature of ‘N’ atom:‐ a) Secondary ‘N’ – Formation of monoacetyl derivative with acetic anhydride indicates secondary ‘N’ .

b) Tertiary ‘N’‐ Reserpine forms quaternary ammonium salts with CH3I that indicating one of ‘N’ is tertiary. 4. Hydrolysis:‐ Hydrolysis of reserpine yields mixture of

i) Methyl alcohol ii) 3,4,5‐trimethoxy benzoic acid iii) Acid (A) of composition

C22H28N2O5

COOH C H N O NaOH C H N O 33 40 2 9 + 2 H2 O CH3OH + 22 28 2 5 + Hydrolysis MeO OMe OMe • As reserpine does not contain ‐COOH & ‐OH groups, hence its hydrolysis product reveal that reserpine is a di‐ester. • The ester linkage in reserpine has been further confirmed by its reduction with LiAlH4 5. Reduction:‐

CH2OH

LiAlH C H N O 4 33 40 2 9 Reserpic Alcohol + C22H30N2O4 MeO OMe OMe #Structure of Reserpic acid:‐ a) Molecular formula:‐ C 22H28N2O5 b) Presence of one carboxyl group:‐ By usual tests e.g. silver salt method c) Presence one –OH group:‐ Reserpic acid on oxidation yields a ketone that means it has secondary alcoholic group. d) Nature of two methoxy groups:‐ By zeisels method. e) Nature of two ‘N’ atom:‐ It shown that it contains two ‘N’ atoms in heterocyclic ring in the form of‐ i) Secondary ‘N’ ii) Tertiary amino group f) Reduction of reserpic acid:‐ On reduction with yields reserpic alcohol. LiAlH g) Oxidation of reserpic 4acid:‐ On oxidation with it gives 4‐methoxy N‐oxalyl anthranilic acid.

KmnO4

COOH

KmnO4 C 22H28N2O5 oxidation MeO NHCOCOOH ThusReserpic one of acid the methoxyl group is present in meta position to –NH group 4-methoxy N-oxalyl Anthranilic acid h) Fusion with KOH :‐ When reserpic acid is fused with potash it yields 5‐hydroxy isophthalic acid. One of the acidic groups of isophthalic acid must be

COOH KOH C22H28N2O5 Fusion HOOC OH present in m‐ position to each other5-h ythisdroxy confirm isophthalic acibyd the fact that reserpic acid when heated with acetic anhydride yields a gamma lactone.

C 22H 28N 2 O 5 AC2 O OC O v-lactone i) Dehydrogenation:‐

When methyl reserpate is dehydrogenated with selenium it yields a hydrocarbon of molecular formula ‐ C 19H 16N 2

This hydrocarbon is also obtained by dehydrogenation of Yohimbine with selenium & was therefore named as Yobyrine.

Se Methyl reserpate C H N Dehydrogenation 19 16 2 Yobyrine

Se Dehydrogenation

Yohimbine Structure of Yobyrine:_

1. Molecular formula:‐ C H N 2. Zinc Distillation:‐ 19 16 2

H N Zn dust C H N C2H5 19 16 2 + 3. Oxidation:Distillation‐ N Isoquinoline When yobyrine is oxidized with permangnate3-ehtyl indole it yields phthalic acid

KmnO 4 COOH C 19H 16N 2 COOH cr o 2 3 Pthalic acid

CH3 COOH

o-Toulic acid 4. Condensation with :‐ Yobyrine gives condensation products suggesting the presence of pyridine ring with a substitution adjacent to the nitrogen. On the basis of above facts following structure has been postulated for yobyrine.

N N CH H 3

Yobyrine Synthesis of Yobyrine:‐

CH3 ClOC CH NH 2 N 2 + H 2-(3-indolyl)ethyl amine O-tolyl acetyl chloride

1) -HCl 2)Reduction

N N CH3 H POCl 3 NH CH CH2 N 3 H CO

CH2 - 2H dehydrogenation

N N CH H 3

Yobyrine j) As Yobyrine is formed from reserpic acid it means that reserpic acid may possesses the following types of skeleton structures.

N N H

SKELETON STRUCTURE k) From the fact 5(g) it follows that one of the methoxyl group is present in m‐ position to the NH group of indole i.e. on C‐11 i.e. Ome is at C‐11. • Reserpic acid when dehydrogenated yields 11‐hydroxy‐16‐methyl yobyrine,this may be only formed if –COOH group is present on C‐16

N OH N Se H C H N O CH2 22 28 2 5 dehydrogenation

3CH CH3

• From the step (h) it follows that –COOH11-hydroxy-16-methyl & ‐OH group are yobyri inn m‐position to each other but –COOH group is present at C‐16 therefore –OH group must be at C‐18. • From purely biogenic reasons, the 2nd methoxyl group has been assigned position C‐17. • On the basis above mentioned facts the structure of reserpic acid may be

N MeO N H

HOOC OH OMe Reserpic acid Structure of Reserpine:‐ • As reserpine is di‐ester of reserpic acid, it means that,

COOH

N MeO N CH OH + H + 3 -2 H2O MeO OMe OMe HOOC OH OMe Reserpic acid

N MeO N H OMe

O CO OMe H3COOC OMe OMe Reserpine Synthesis of Reserpine:‐ • The structure of reserpine has been confirmed by its synthesis given by Woodward et.al.

O O H2C CH Diels Alder 2 C CH + reaction C CH CH O CH O COOH p-Benzoquinone COOH Adduct

Reduction NaBH4 of less hindered ( =CO) OH OH

C H COOH O 6 5 phenoxy benzoic Ac2O acid O COOH O COOH Synthesis continue…..

O O O O CO -H2O Aluminium - isopropoxide O OH O CO O CO

O O CH3ONa Br CH3OH NBS- H2SO4

OH OMe OMe O CO O CO

Cro3 Synthesis continue…..

O

Br OH Zn / CH3COOH i) CH2N2 O OMe Ac O O OMe ii) 2 Pyridine COOH iii) OSO -H O O CO 4 2

COOCH3 COOCH3 COOCH3 CH MeO 2 ii) MeO O CH2N2 6-methoxy OAc CHO i) HIO OAc N NH tryptamine 4 OH MeO H 2 OH Synthesis continue…..

N H MeO N C H N i) NaBH4 H COOC MeO N O 3 CH2 H ii) CH OH H COOC 3 3 OAc OMe H COOC 3 OAc OMe

i) POCl3

ii) NaBH4 Synthesis continue…..

N i)Resolve OMe N ii)NaOH N H MeO N H iii)Hcl iv)DCC

H COOC 3 OAc OMe OMe OC O Synthesis continue…..

Isomerisation (CH3)3CCOOH

N MeO N H OMe CH OH N 3 MeO N H OCO OMe H3COOC OMe OMe OMe HOOC OMe OMe OC O Reserpine OMe Reserpic acid Lactone References :‐

1)“ Organic Organic Chemistry of Natural Products ”‐ by Gurudeep Chatwal ,Vol‐I.

2)“ Organic Chemistry ”by I.L.Finar , Vol‐II .

3)“Pharmacognosy” by C.K.Kokate , A.P.Purohit , S.B.Gokhale ‘O.D’