Natural Products

For Third Year Botany Students Major(2hrs./W.)2nd.Term

Prepared By Prof.Fathy Muhammad El-Taweel

ﻧﻤﻮذج رﻗﻢ ( ١٢ ) ﺟﺎﻣﻌﺔ: دﻣﯿﺎط ﻛﻠﯿﺔ: اﻟﻌﻠﻮم ﻗﺴﻢ : اﻟﻜﯿﻤﯿﺎء ﺗﻮﺻﯿﻒ ﻣﻘﺮر دراﺳﻰ ١ـ ﺑﯾﺎﻧﺎت اﻟﻣﻘرر اﻟرﻣز اﻟﻛودى : ٤٠٤ك اﺳم اﻟﻣﻘرر : ﻛﯿﻤﯿﺎء ﻋﻀﻮﯾﺔ (وﻣﻨﺘﺠﺎت طﺒﯿﻌﯿﺔ ) اﻟﻔرﻗﺔ/ اﻟﻣﺳﺗوى : اﻟﺛﺎﻟﺛﺔ اﻟﺗﺧﺻص : اﻟﻧﺑﺎت ﻋدد اﻟوﺣدات اﻟدراﺳﯾﺔ : ﻧظرى ٢ ﻋﻣﻠﻰ ٤

2ـ ھدف اﻟﻣﻘرر : Practice the students on chemistry of chromatographic analysis to qualify the .1 students to design and carry out experiments. 2. Habilitate the students to employ nautral products facts and theories to formulate, analyze, interpret practical and theoretical data, and solve problems. 3. Identify the role of of natural products and the development of society to the students, and develop chemical approaches that meet community needs taking into account economic, environmental, social, ethical, and safety requirements. 4. Enhance students self – learning using information technology, risk management, organization of time, working in a team, and reviewing of a quality control processes. ٣- اﻟﻣﺧرﺟﺎت اﻟﺗﻌﻠﯾﻣﯾﺔ اﻟﻣﺳﺗﮭدﻓﺔ ﻣن اﻟﻣﻘرر : By the end of this course, the student will be able to: أ ـ اﻟﻣﻌﻠوﻣﺎت واﻟﻣﻔﺎھﯾم : a1. Describe the processes and mechanisms supporting the structure and function of natural products . a2. Explain the related terminology, nomenclature and classification of natural products . a3. Identify the relation between dyes and colour and the environment in natural products. a4. Define chemical concepts, nomenclature, formulae and units in natural products . a5. Explain the major types of chemical reactions, their characteristics and mechanisms in natural products . a6. Describe the constitution and properties of the different chemical compounds, including the main synthetic pathways in natural products . ب ـ اﻟﻣﮭﺎرات اﻟذھﻧﯾﺔ : .b1. Differentiate between the different types of neutral products b2. Analyze chemical data to identify and confirm chemical structures in natural products . b3. Propose and conclude mechanisms for for the formation of in natural products . ﺟـ ـ اﻟﻣﮭﺎرات اﻟﻣﮭﻧﯾﺔ اﻟﺧﺎﺻﺔ . c1. Solve problems using a range of formats and approaches in natural products ﺑﺎﻟﻣﻘرر : c2. Assess risk in laboratory work taking into consideration the specific hazards associated with the use of chemical materials as well as the safe and proper operation of the laboratory techniques in in natural products . c3. Conduct standard laboratory procedures involved in analytical and synthetic work in in natural products . د ـ اﻟﻣﮭﺎرات اﻟﻌﺎﻣﺔ : . d1. Think independently, set tasks and solve problems on in natural products ٤ـ ﻣﺣﺗوى اﻟﻣﻘرر : :(I: Theoretical Part (3h/w Natural Products (2h/w) 1- Introduction to natural products. 2-Terpenoids:

2 a-Monoterpenes b-Diterpenes c-Sesquiterpenes 3-Steriodes; 4-Sterols e.,g. Cholesterol, ergosterol, stigmasterol II: Practical Part (4h/w) ٥ـ أﺳﺎﻟﯾب اﻟﺗﻌﻠﯾم واﻟﺗﻌﻠم .Lectures - - Practical. - Exercises. - Data show - White board - Models. ٦ـ أﺳﺎﻟﯾب اﻟﺗﻌﻠﯾم واﻟﺗﻌﻠم ﻟﻠطﻼب ذوى اﻟﻘدرات اﻟﻣﺣدودة ٧ـ ﺗﻘوﯾم اﻟطﻼب : أـ اﻷﺳﺎﻟﯾب اﻟﻣﺳﺗﺧدﻣﺔ Oral. a1, a2, a3, a4, a5, a6, b1, b2, b3, d1 1- 2- Written exam. a1, a2, a3, a4, a5, a6, b1, b2, b3, d1 3- Practical exam. c1, c2, c3, d1 ب ـ اﻟﺗوﻗﯾت ١- written exam أﺳﺒﻮع 15 ٢- oral exam أﺳﺒﻮع 15 ٣- Practical exam أﺳﺒﻮع 14 ﺟـ ـ ﺗوزﯾﻊ اﻟدرﺟﺎت ١- اﻣﺘﺤﺎن آﺧﺮ اﻟﺘﺮم ٦٠ % ٢- اﻻﻣﺘﺤﺎن اﻟﺸﻔﻮى ١٠ % ٣- اﻻﻣﺘﺤﺎن اﻟﻌﻤﻠﻰ ٣٠%

٨ـ ﻗﺎﺋﻣﺔ اﻟﻛﺗب اﻟدراﺳﯾﺔ واﻟﻣراﺟﻊ : أـ ﻣذﻛرات ب ـ ﻛﺗب ﻣﻘﺗرﺣﺔ (Organic (Organic Chemistry (7th Edition) by L. G. Wade (Feb 1, 2009 Organic Chemistry by Francis A. (Chemistry by John McMurry (Jan 16, 2007) Natural Product, (Chemistry of Natural Products by Sujata V. Bhat Jul 15, 2005) Carey (Jan 8, 2010) Chemistry at a Glance by Stephen P. Stanforth Aug 21, The Way of Synthesis: Evolution of Design and Methods for Natural (2006) (Products by Tomas Hudlicky (Sep 25, 2007) د ـ دورﯾﺎت ﻋﻠﻣﯾﺔ أو ﻧﺷرات ... اﻟﺦ

ﻣﺼﻔﻮﻓﺔ اﻟﻤﻌﺎرف واﻟﻤﮭﺎرات اﻟﻤﺴﺘﮭﺪﻓﺔ ﻣﻦ اﻟﻤﻘﺮر اﻟﺪراﺳﻰ I: Theoretical Part (3h/w) Natural Products (2h/w)

No أﺳﺒﻮع ﻣﮭﺎرات ﻣﮭﺎرات ﻣﮭﺎرات اﻟﻤﺤﺘﻮﯾﺎت ﻟﻠﻤﻘﺮر اﻟﻤﻌﺎرف اﻟﺮﺋﯿﺴﯿﺔ اﻟﺪراﺳﺔ ذھﻨﯿﺔ ﻣﮭﻨﯿﺔ ﻋﺎﻣﺔ ١ .d1 b1 a1 1-2 1- Introduction to natural products

٢ :d1 b1 , b2, a1, a2, a3, 3-9 2-Terpenoids b3 a4, a5, a6 a-Monoterpenes b-Diterpenes c-Sesquiterpenes

٣ d1 b1 , b2, a1, a2, a3, 10-14 3

b3 a4, a5, a6 :3-Steriodes: Sterols e.,g. Cholesterol, ergosterol, stigmasterol,steroid hormones

Chapter 1

Introduction: Naturaly occurring compounds are classified into three types according to their structures and their sources into : 1-Terpenes 2-Steroids 3-Alkaloids The work in this field classified into : a-Isolation of these compounds from their natural sources b-Structural elucidation by chemical reactions based on their function groups , for example, a-Oxo compounds e.g.RCOOH by esterification ,RCHO and RCOR by condensation with NH2NH2 or NH2OH ROH by esterification or oxidation

ArOH by FeCl3 or diazotization c-Compounds containing double bonds , may be with conjugated or separated double bonds Conjugated double bonds can be detected by Diels Alder Reaction ( D.A.R. ) by forming an adducts with maleic anhydride ,each two double bonds react with one molecule of maleic .

Separated double bonds ( no D.A.R.) and can be detected by H2 / Ni ,halogenations or by Each one double bond absorb one molecule of hydrogen and one molecule of halogen, thus, the number of double bonds can be determined . Each one double bond absorb one molecule of hydrogen during catalyitic hydrogenation and one molecule of halogen during halogenation, thus, the number of double bonds and the shape of the molecule can be determined .

Also, compounds with M.F. CnH2n+2 for alkane (acyclic compounds ) ;

M.F. CnH2n for alkene and monocyclic compounds ;

M.F.CnH2n-2 for alkyne and bicyclic compounds ;

M.F.CnH2n-4 for tricyclic compounds ;

M.F. CnH2n-6 for tetracyclic compounds .

Degradative oxidation : Using oxidizing agents such as O3,CrO3, NaOBr (Br2 / NaOH) ,

KMnO4 and total synthesis can also be used for structure elucidation of the naturally occuring compounds.

Chapter 2 Terpenes a-Compounds contains C,H and may be oxygen. b-Most of them isolated from plant source . c-All terpenes have M.F. ( C5H8 )n , because , the thermal degradation of terpenes yielded compound with M.F. C5H8 , called isoprene . These isoprene units joined together head to tail .

4 tail head 1 3 No.of carbons Class

5 (n =1) C5H8 isoprene

10 (n =2) C10H16 monoterpenoids

15 (n =3) C15H24 sesquiterpenoids

20 (n = 4) C20H32 diterpenoids

30 (n = 6) C30H48 triterpenoids

40 (n = 8) C40H64 tetraterpenoids (Carotenoids)

>40(n >8) polyterpenoids

A-Acyclic monoterpenes

1-Myrecene C10H16

This means that myrecene reacted with three molecules of hydrogen ,and three molecules of bromine ,thus, myrecene contains three double bonds .

The M.F. of decane is CnH2n+2 , means that myrecene is acyclic compound. Myrecene reacted with one molecule of maleic anhydride to give an adduct , this means that myrecene contain two conjugated double bonds , and the third is separated. Degradative oxidation is used to indicate position of the third double bond .

Thus , myrecene has the above structure .

2- Citral M.F. C10H16O Citral forming tetrabromo derivative on reacting with bromine , thus , citral contains two double bonds . Br

Br Br2 / CCl4

CHO Br CHO

citral tetrabromo derivative These two double bonds are separated , because citral did not react with maleic anhydride to form an adduct ( no D.A.R. ) . The oxo group of citral may be an aldehyde or ketone since it was condensed with hydrazine or hydroxyl amine to give the hydrazone or an oxime respectively . Since citral reduced to geraniol ( primary alcohol ) , then oxidized to geranic acid ( mono carboxylic acid ) , without loss of carbons during the oxidation process , thus , citral is an aldehyde not ketone .

The above equation indicates that , citral give p-cymene on heating with KHSO4,this means that citral consists of two isoprene units joined together head to tail .Also,indicates position of methyl group with respect to isopropyl group. Thus ,positions of the formyl group and the double bonds can be indicated by degradative oxidation .

Reaction of citral with aqueous potassium to give 6-methylhept-5-en-2-one means that citral is α,β-unsaturated aldehyde . There are two types of citral ;they are citral-a and citral-b as shown below :

CHO H

H CHO

citral-a (geranial;trans; E) citral-b (neral;cis; Z)

Ionones : Prepared by condensation of citral with acetone in presence of alkali,followed by cyclization with acid.

CHO O CHO acetone / H =

Ba(OH)2

O -ionone citral

O O O

-ionone -ionone -ionone The preparation of α-ionone and β-ionone varies with the nature of cyclizing agent used

e.g.with H2SO4 , β-ionone is the main product;with H3PO4 α-ionone is the main product.

3- Citronellal M.F. C10H18O It is an optically active compound occurs in citronella oil. It contains one double bonds from bromination and hydrogenation and the parent

hydrocarbon is acyclic compound with M.F.C10H22 = CnH2n+2 It is an aldehyde or a ketone since it condensed with hydrazine and hydroxyl amine to give hydrazone and oxime respectively. It is an aldehyde from the following reactions:

Degradative oxidation to indicate positions of the double bond and the formyl group.

H2 / Pt Na(Hg)

CHO CHO CH2OH

citral citronellal citronellol

4- Citronellol M.F. C10H20O Occurs in rose and geranium oils. Its structure was determined by the following: and by degradative oxidation:

7- Linalool M.F. C10H18O Occurs as(-) in rose oil and as (+) in orange oil. Its hydroxyl group is a tertiary alcohol because it resists oxidation ,esterification and easily dehydrated.

It contains double bonds ,since it adds two molecules of H2 and two molecules of Br2.

The M.F.of the product 3,7-dimethyloctane-1-ol is C10H22O = CnH2n+2 ; this means that linalool is acyclic compound.These two bonds are separated since, this compound does not react with maleic anhydride.

Degradative oxidation to indicate positions of the double bonds and the hydroxyl group.

KMnO4 O - acetone laevulic acid - oxalic acid CO2H

(linalool) b-Monocyclic monoterpenes :

Their parent hydrocarbon is p-menthane ,with M.F. C10H20

1- Limonene C10H16 Occurs in limonene and orange oils ,in pepperimt oils and in turpentine oils It contains two unconjugated double bonds , because it adds two bromine molecules to give tetrabromide and adds two hydrogen molecules to give p-menthane with M.F.CnH2n , thus , limonene is a monocyclic compound . The two double bonds are unconjugated since the copmpound did not react with maleic anhydride .

Br Br

Br2

Br Br

Limonene tetrabromide derivative

H2 / Ni

limonene p-menthane

To proof that there is one double bond at C1 using the following reactions , Also , the carbon skeleton of limonene will known .

To proof that there is one double bond at C-8 , Since , the structure of carvoxime is known , the structure of limonene must be has one

double bond at C-8 .

2- Menthol C10H20O Occurs in pepperiment oil It is a saturated compound since , it did not add hydrogen or bromine . The oxygen atom is an alcoholic , as shown by its reactions : Easily forming an ester and oxidized to menthone , therefore , menthol is a secondry alcohol

H2 / Ni or Br2 no reaction

menthol

CrO3

O OH

menthol menthone Since reduction of menthol with hydrogen iodide , gives p-menthane , thus , menthol most probably contains this carbon skeleton i.e.it is a monocyclic monoterpene .

p-menthane menthol Finally , since pulegone gives menthol on reduction , and since structure of pulegone is known , it therefore follows that menthol must be ,

3- Menthone C10H18O occurs in pepperiment oils It behaves as a ketone ,that it can be condensed with hydrazine and hydroxyl amine to give the hydrazone and oxime derivative respectively . It is a satutated compound since it did not react with bromine . When heated with hydrogen iodide / red phosphorous , it is reduced to p- menthane , thus , it a monocyclic compound .

Oxidation processes to indicate position of the carbonyl group .

KMnO4 H3COCH3 + O CO2H CO2H

3-methyladipic acid menthone 5- Pulegone C10H16O

It contains one double bond ,since it adds one H2 ,one Br2 It behaves as a ketone by condensation with hydrazine and hydroxyl amine. It is a monocyclic ,has p-mebthane structurewith one double bond and a carbonyl ketone at C-3 as shown:

To confirm that pulegone is α,β-unsaturated ketone i.e.to indicate the position of the carbonyl group and the double bond.this is can be done by the following reactions;

KMnO 4 + acetone

O CO2H CO2H 3-methyladipic acid

pulegone

4- Piperitone C10H16O Occurs in eucalyptus and is a valuable source of menthone and thymol

It contains one double bond ,since it adds one Br2 and one H2.

In quantitative yield for menthone.

On oxidation with FeCl3 ,thymol is obtained.

These reactions shows that piperitone is p-menyh-3-one,but do not show the position of the double bond.

This is had been shown on oxidation with KMnO4.

O O KMnO4 / OH [O] CO2H

CO2H O -CO2 O

piperitone 4-acetylisobutyric acid

NaOBr -CHBr3

CO2H

2-isopropylglutyric acid 2 CO2H

5- Carvone C10H14O Occurs in caraway oils.

It behaves as a ketone from its reactions,by forming an oxime with NH2OH and hydrazone with NH2NH2.

Bromination indicates that ,it adds two molecules of Br2,thus,it contains two double bonds, and its parent hydrocarbon with M.F.C10H20 ,i.e.CnH2n,means p-menthane structure,thus,it is monocyclic compound. Position of the carbonyl group can be indicated by the following reaction:

Thus ,it has p-cymene structure,and the keto group is in the ring ,in the ortho position to the methyl group. Degradative oxidation to indicate positions of the double bonds. To indicate that there is one double bond in the 8-position.

To indicate that there is one double bond in the 6-position

Structure of elucidation of carvone based on its synthesis from α-terpineol.

Cl Cl NO NOH NOCl H

OH OH OH

terpineol -HCl EtONa

O NOH dil.H2SO4 / warm

carvone Thus,carvone has the same carbon skeletone of α-terpineol .Also the above reactions indicate the position of the ketonic group in carvone.

2-Bicyclic monoterpenes :

c-Bicyclic monoterpenes

6 +5 6 +3 6 +4 10 pinane 1 thujane carane 6 7 2 9 5 8 3 4 camphane

10 10

2 1 1 3 8 6 7 2 9 7 9 6 4 5 8 3 5 4 camphane pinane

a-6+3 group

Carane group : e.g. Carone C10H16O It behaves as a ketone ,that it can be condensed with hydrazine and hydroxyl amine to give the hydrazone and oxime derivative respectively . It is a satutated compound since it did not react with bromine.

The parent hydrocarbon has M.F. CnH2n-2 ,called ,carane , it is bicyclic

The following reactions to indicate position of the carbonyl group , the presence of three membered ring and six membered ring .

Thujane group : Umbellulone C10H14O It is found in leaves of Califorina laurel It behaves as a ketone ,that it can be condensed with hydrazine and hydroxyl amine to give the hydrazone and oxime derivative respectively . It contains one double bond , because it adds one bromine molecules to give dibromide ,thus

, the parent hydrocarbon of umbellulone is thujane with M.F.CnH2n-2 , thus , it is a bicyclic compound. It is α,β-unsaturated ketone from the following reaction ,

Structure of it was established by degradative oxidation ,

25 b) 6+4 group, Pinane group : There are three isomers for pinenes, ,α-pinene , β-pinene and δ-pinene

Bredt's rule: States that a double bond cannot be formed by a carbon atom occupying the bridge-head (of a bicyclic system).

α-pinene C10H16 This compound reacted with one molecule of hydrogen and one molecule of bromine ,thus, α-pinene contains one double bond .

Since M.F. of pinane is C10H18 , = CnH2n -2 thus , α-pinene is bicyclic This compound contains six membered ring , since it can be converted into α-terpineol , position of the double bond in α-pinene as in α-terpineol. 26

Degradative oxidation for the six membered ring( to indicate the presence of the four membered ring), the two compounds have the same carbon skeleton); i.e.α-pinene has the same carbon skeleton of α-terpineol and thus, α-pinene contains a six membered ring and ring opening occurs at C-6.

The second ring is a four membered ring by degradative oxidation of the six membered ring OH 1 OH

6 2 1% alkaline KMnO4 pinene glycol

1,2-diol warm alk.

KMnO4 COOH O COOH COOH NaOBr

- CHBr3

pinic acid pinonic acid

Br2 COOH COOH COOH COOH CrO COOH Ba(OH) 3 2 COOH

Br OH

hydroxypinic acid cis - norpinic acid bromopinic acid Formation of norpinic acid(2,2-dimethylcyclobutane-1,3-dicarboxylic acid),(its parent hydrocarbon has M.F.CnH2n ; monocyclic indicates that cyclobutane ring is present. 27 c)Camphane group : Camphor C10H16O Obtained from camphor laurel trees in china and japan It behaves as a ketone ,that it can be condensed with hydrazine and hydroxyl amine to give the hydrazone and oxime derivative respectively . It is a satutated compound since it did not react with bromine . Position of the carbonyl group at C-2 from this reaction

It is a bicyclic compound from the following reactions

This compound contains six membered ring , It is a ketone not an aldehyde ,also contains five membered ring as follows :

Since camphoric acid is a dicarboxylic acid and has the same number of carbons as camphor

, camphoric acid is a monocyclic compound ,has the M.F.C10H16O4 and its parent hydrocarbon is a saturated hydrocarbon by neglecting two carboxylic groups and three methyl groups ,the parent hydrocarbon of camphoric acid is has the M.F.C5H10 equivalent to

CnH2n .It is a monocyclic and a cyclopentane ring and this is confirmed by synthesis of camphor.

Sesquiterpenes or sesquiterpenoids ( C15),forming the higher boiling fractions of the essential oils.The general M.F.is (C5H8)n ; where is n = 3

Classification of sesquiterpenoids Number of double bonds

Acyclic 4

Monocyclic 3

Bicyclic 2

Tricyclic 1

a)Acyclic sesquiterpenes 1-Farnesene

M.F. is C15H22 ; it has two isomers ,are α-farnesene and β- farnesene.

α-Farnesene is a liquid and not occurs naturally ; β- Farnesene is a liquid and occurs naturally in oil of hops. Both of α-farnesene and β- farnesene are catalytically reduced to a compound with

M.F,C15H32 corresponding to CnH2n+2 ;thus,both of them is acyclic compound. Also ,catalytic hydrogenation indicates that both of contains four double bonds. Both of them contains two conjugated double bonds ,since it reacts with one molecule of maleic anhydride,thus,the other two double bonds separated.

2-Farnesol M.F.C15H26O

This is occurs in the oil of amberette seeds . The oxo function is primary alcohol since farnesol oxidized with chromic acid to farnesal( an aldehyde).

Farnesol is acyclic compound with three double bonds;from catalytic hydrogenation.

The M.F.of C15H31OH corresponds to CnH2n+1OH , corresponding to acyclic compound. Position of the double bonds can be indicated as follows: Since farnesol does not react with maleic anhydride (No D.A.R.);the three double bonds are separated.

The farnesal is an α,β-unsaturated aldehyde by this reaction

The structure of farnesol has been confirmed by its synthesis from synthetic nerolidol.

Diterpenoids

Monocyclic Diterpenoids Phytol

o Phytol C20H40O ,liquid ;b.p.145 C. The reactions of phytol showed that it is a primary alcohol. An acyclic diterpenoids; it is produced from hydrolysis of chlorophyll.

It also forms part of the molecules of vitamins E and K. It was composed of four isoprene units joined together head to tail.

Since on catalytic reduction phytol, C20H40O forms dihydrophytol, C20H42O,it therefore follows that phytol contains one double bond. Degradative oxidation to indicate the position of the doubleas follows:

20 19 18 17 13 9 8 5 1 12 10 6 4 16 15 11 7 3 OH 14 2 phytol

i)O3 + ii)H3O 20 19 18 17 13 9 8 5 12 10 6 4 16 15 11 7 O 14 + CO H HO 2

glycolic acid

Its structure was established based on its synthesis from farnesol as follows:

15 14 13 9 8 5 1 12 10 6 4 11 7 3 OH 2 farnesol

H2 / Pd 15 14 13 9 8 5 1 12 10 6 4 11 7 3 OH 2 hexahydrofarnesol PBr 15 14 3 13 9 8 5 1 12 10 6 4 11 7 3 Br 2 hexahydrofarnesyl bromide sodio-acetoacetic ester 15 14 13 O 9 8 5 1 12 10 6 4 2 11 7 3 i)KOH dil. ii)HCl CO2Et 15 14 13 O 9 8 5 1 12 10 6 4 2 11 7 3 + H H,NaNH2 then H 15 14 13 9 8 5 1 12 10 6 4 2 11 7 3 H / Pd OH 15 14 2 9 8 5 1 10 6 4 2 11 7 3 Ac2O OH 20 19 18 17 13 9 8 5 1 12 10 6 4 16 15 11 7 3 OH 14 2 phytol

Ferrugenol : C20H30O

It contains phenolic hydroxyl group ,since it gives a positive test with FeCl3. It is tricyclic compound with three double bonds from these reactions.

OH OH

H2 / Ni

hexahdroferrugenol ferrugenol C20H36O = CnH2n-4 C H O 20 30 tetracyclic

S / heat

7 5 4 8 3

2 1

retene

Position of the hydroxyl group can be indicated by this reaction:

Chapter 2 Steroids

Includes : a-Sterols b-Steriod hormones c-Vitamines d-Bile acids ; e-The adrenal cortex hormones ; f-Some carcinogenic hydrocarbons g-……..etc Steriods are compounds most of them isolated from animal sources and some of them isolated from plant source. Steroids produces Diels hydrocarbon and small amount of chrysene on distillation with Se/ at 3600C If distillation carried out at 4200C , chrysene. Is the mainly product with small amoumt os picene .

All steroids containing perhydrocyclopentenophenanthrene with M.F. CnH2n-6

Preparation of Diels hydrocarbon ( 3'-methyl-1,2-cyclopentenophenanthrene )

MgBr O

+ H3O +

b-naphthyl MgBr 2,5-dimethylcyclopentanone

o P2O5 /140 C - H2O H - H2

H3PO4

Se / 3600C

- CH4

3 3- 4 2 - 5 2 6 1 1- 3,-methyl-1,2-dimethylcyclopentenophenanthrene 7 10 8 9 Some nomenclature of steroids:

For substituents : Dotted line refers to α-configuration; Solid line refers to β-configuration; Wavy line refers to unknown configuration

A)Sterols

They are present in animal and plant in oils and fats , three types : a-Zoosterols : They are sterols of animals , b-Phytosterols :They are sterols of plants c-Mycosterols :They are sterols of yeast and fungi

All sterols are containing Diel,s hydrocarbon nucleus or

perhydrocyclopentenophenanthrene with M.F.CnH2n-2 = C17H28

1- Cholesterol C27H46O

It is present in the animal cell free or as fatty esters especially in the brain and spinal cord

Color reactions of cholesterol:

A solution of cholesterol in chloroform gives a red color with c.H2SO4 and gives greenish color with c.H2SO4 / (CH3CO)2O.

a-Structure of the ring system (Tetracyclic form of cholesterol):

Formation of cholestanol from cholesterol indicates the presence of one double bond in cholesterol. Reduction of cholestanone to cholestane indicates the presence of a secondary hydroxyl group in cholesterol. Consequently , cholesterol is a tetracyclic ring compound with one double bond and secondary hydroxyl group . Distillation of cholesterol with selenium give Diels hydrocarbon , indicates the presence of this nucleus in it .

b- Position of the hydroxyl group(at C-3) :

Thus, the hydroxyl group in cholesterol at C-3. c-Position of the double bond can be indicated by the following reactions:

Consequently ,the double bond at C5---C6 in the ring . d- The nature and position of the side chain in cholesterol : It has been found that :

The side chain is methylisohexylketone ( C8 ) and the nucleus is ( C19)

Barbier-Wieland degradation ( B.W. )

Degradative oxidation for the side chain in coprostane

Formation of acetone means that the side chain of cholesterol ends with isopropyl group .

COOH

17 COOH B.W. 13 B.W. cholanic acid

nor-cholanic acid bis-nor-cholanic acid B.W.

O O COOH

B.W. NaOBr D

ketone O HNO3

COOH Ac O 17 2 O COOH 13

O dicarboxylic acid cyclic anhydride

Se / 360oC 13 2

The last reactions indicates that the side chain consists of eight carbons attached with the ring D which is a five membered ring . 46

Also , formation of Diel,s hydrocarbon on heating cholesterol with Selenium ,indicate that the side chain attached to the nucleus at C-17. Formation of 1,2-dimethylphenanthrene indicate that there is an angular methyl at C-13. e-Presence of an angular methyl at C-10 :

The carboxylic group at C-10 resists esterification and resists decarboxylation , this is due to the presence of an angular methyl group and the carbon is tertiary carbon . f- Presence of an angular methyl group at C-13: cholesterol + H2 / Pt → cholestanol cholestanol + CrO3 → cholestanone cholestanone + Na ( Hg ) / HCl → cholestane

2-Ergosterol C28H44O ; Occurs in yeast

The oxo function is one hydroxyl group from its reactions such as acetylation , benzoylation , or esterification .or formation of monoacetyl or monobenzoyl derivatives. 22 23

10 8

3 7 HO 5 6 ergosterol

acetic anhydride

O O

H3C ergostanyl acetate

22 23

10 8

3 7 HO 5 6 ergosterol 22 H /Ni 2 23

10 8

3 7 HO 5 C H O 6 ergostanol 28 50

Thus, ergosterol contains three double bonds , because it absorbs three hydrogen molecules . There are two conjugated double bonds ,since it reacts with one molecule of maleic anhydride to give an adduct and the third is isolated .

There is one double bond at C22-C23 in the side chain from this reaction .

Thus , the other two double bonds must be in the nucleus ,

Since , one double bond transefered from C5-C6 to C4-C5 ,means that the another double

bond shoulde be at C7-C8 . Route A :

ergosterol ergostanol ergostanyl acetate

CO2H

CrO3

AcO H H OAc -

ergostanyl acetate 3- -acetatenor-5a-cholanic acid Route B :

Cholesterol + H2 / Pt → cholestanol + Ac2O →cholestanyl acetate

This means that both of ergosterol and cholesterol have the same nuclei . Also have the same position of both hydroxyl groups , two angular methyl groups, and the side chain.

3-Stigmasterol

C29H48O ; Isolated from soya bean

The oxo function is hydroxyl group from its reactions such as acetylation , benzoylation , or esterification.

HO stigmasterol

H3C O stigmastanyl acetate This compound contains two bonds since it reacts with two molecules of hydrogen to give stigmastanol with M.F.C29H52O= CnH2n-6 and reacts with two bromine molecules to give tetrabromide derivative ,these two bonds are isolated ,because there is no D.A.R . Reaction of stimasterol with ozone methylisopropylacetaldehyde and give other products , this give an evidence for the presence of one double bond in the side chain at C22-C23 . 53

Thus , the another double bond must be in the nucleus . The following reactions indicates the presence of in the nucleus at C5-C6

H2O2 / CH3CO2H

HO OH stigmasterol triol OH OH

CrO3

reduction - H2O O

OH O O O

CrO3 HO2C

HO2C O CO2H CO H O 2 1,4-diketone tetracarboxylic acid Route A:

22 23

H2 / Ni 3 stigmastanol 5 Ac O 6 2 OH stigmasterol

AcO stigmastanyl acetate H CrO3

COOH O

OAc ethylisopropylketone H 3 acetate-nor-5 -cholanic acid 3β-acetate –nor-5α-cholanic acid also prepared from cholesterol as in ergosterol Route B