1Dha. Ilangeswaran

Introduction

A secondary alcohols with 27 to 29 carbon atoms ofanimal or plant origin.Crystalline solids with m.p. range of 100 – 200oC.The terms ‘stero’ – means solid & ‘ol’ – alcohol.Number of compounds such as vitamin D, bile acids, sexhormones, adrenal cortex hormones are belonging tosteroids.The structure of steroids are based on 1,2-cyclopentenophenantherene skeleton.Sterols occur in animal and plant oils & fats. They occuras free or an ester of the higher fatty acids.They are isolated from unsaponifiable portion of oils &fats.

2Dha. Ilangeswaran

1

23

4

5

6

7

8 9

10

1'

2'

3'

1,2-Cyclopentenophenanthrene

Any compound which gives Diels hydrocarbon with Se

distillation is defined as steroid. When steroids are distilled at

420oC yields mainly chrysene and a little picene.

CH3

Diels Hydrocarbon Chrysene Picene

1

1

2

2

3

3

4

4

5

5

6

6

778

8

9

9

10

11

12

1314

11

10

12

3Dha. Ilangeswaran

ClassificationSterols

Zoosterols

(Animal Sources)

Phytosterols

(Plant Sources)

Mycosterols

(Yeast & Fungi Sources)

CH3

CH3

CH3

OH

CH3

CH3

CH3

H

CH3

CH3

CH3CH3

CH3

OH

1

2

34

56

7

89

10

1112

13

1415

16

17

18

19

2021 22

23

24 25

26

27

CH3

CH3

CH3

OH

CH3

CH3

CH3

20

21 22

23

24

25 26

27

28

29

Cholesterol (zoosterol)

C27

H46

O m.p. 149oC

Ergosterol (mycosterol)

C28

H44

O m.p. 165oC

Stigmasterol (phytosterol)

C29

H48

O

4Dha. Ilangeswaran

Constitution of Cholesterol

1. Molecular formula: C27H46O

2. Presence of double bond and hydroxyl group:

a. The conversion of cholesterol into cholestanol showsthe presence of double bond.

b. The oxidation of cholestanol with chromic acid intocholestanone shows the presence of a secondaryalcoholic group in cholesterol.

C27H48OH2 - Pt

C27H46O

Cholesterol Cholestanol

C27H48O C27H46OCrO3

Cholestanol Cholestanone

c. The Clemmenson’s reduction of cholestanone

yields a saturated hydrocarbon called cholestane.

3. Presence of a ring system: The molecular formula

of cholestane corresponds to the general formula

CnH2n-6, of a tetracyclic system. On distillation with

Se at 360oC, cholesterol yields Diel’s hydrocarbon.

Zn/Hg - HClC27H48C27H46O

Cholestanone Cholestane

C27H46O

Cholesterol

Se

360 °C

CH3

1

2

1'

2'

3'

3'-Methy-1,2-cyclopentenophenanthrene(Diel's hydrocarbon)

+

Chrysene

6Dha. Ilangeswaran

The above reaction shows the presence of a

cyclopentenophenanthrene nucleus in cholesterol and thus

cholesterol is a steroid.

The various rings in the cholesterol opened in

different conditions to give various dicarboxylic acids.

The relative positions of two carboxylic groups with

respect to each other were determined with the help of

Blanc’s rule.

Blanc’s rule: It states that on heating with acetic

anhydride, 1,5-dicarboxylic acids form cyclic anhydrides

and 1,6-dicarboxylic acids form cyclopentanone with the

elimination of carbon dioxide.

7Dha. Ilangeswaran

Ring A: Cholesterol and cholic acid (which is structurally

related to the former) were converted into the dicarboxylic

acid A which on heating with acetic anhydride gave a

cyclopentanone. Therefore the ring A in cholesterol may be

six membered on the basis of Blanc’s rule.

HOOC

HOOC

R

B

C D (CH3CO)2O

O

R

B

C D

Dicarboxyllic acid A

8Dha. Ilangeswaran

Ring B: The tricarboxylic acid derived from cholesterol on

heating with acetic anhydride gave cyclopentanone derivative

and carbon dioxide. Therefore on the basis of Blanc’s rule the

ring B in cholesterol may be a six membered.

HOOC

COOH

R

COOH

C D

O

R

HOOC

C D

(CH3CO)2O

Tricarboxyllic acid (B)

9Dha. Ilangeswaran

Ring C: Deoxycholic acid (structurally related to cholesterol &

cholic acid) was converted into a dicarboxylic acid which gave a

cyclic anhydride when subjected to Blanc’s reaction condition.

Hence the ring C was assumed to be a five membered. On this basis

Windaus and Wieland proposed the following structure to

cholesterol.

OH

CH3

C2H5CH - (CH2)3 - CH (CH3)2

CH3

A

B

C

D

10Dha. Ilangeswaran

Later in 1930, Wieland et al proved that there was no ethyl

group at C10 and proposed that steroids contain chrysene

nucleus and cyclopentenophenanthrene structure. Here Blanc’s

rule failed. If we use the correct structure of cholesterol, the

cyclisation reaction results in the formation of seven

membered cyclic anhydride. Thus ring C also six membered.

HOOC

HOOC

R

D

A B

(CH3CO)2O

R

O O

O

A B

D

Dicarboxyllic acid (C)

11Dha. Ilangeswaran

Ring D: Cholestane was converted into etiobilianic acid, D

which gave a cyclic anhydride. Hence according to Blanc’s

rule the ring D may be a five membered one.

(CH3CO)2O

COOH

COOH

O

O

OA B

C

A B

C

Etiobilianic acid (D) Anhydride

12Dha. Ilangeswaran

Position of the Hydroxyl Group & Double Bond

The positions of hydroxyl group and double bond were fixedwith the help of following reaction.

Here the dicarboxylic acid and cholestanone contain the samenumber of carbon atoms. Hence cholestanone must contain theketo group in the ring. Further pyrolysis of dicarboxylic acidgave a ketone, c. According to Blanc’s rule compound b may be1,6 or 1,7-dicarboxylic acid. Already it was established thatcholesterol contains 3 six membered & 1 five membered ring.Here the dicarboxylic acid may be produced by the opening ofring A or B or C. Hence –OH group may be present any one ofthe above three rings.

C27H46OHNO3

C27H46O4

300 °CC26H44O

Cholestanone Dicarboxyllic acid Ketone

a b c

When cholestanone is oxidized, we get two isomers of

dicarboxylic acids. This will be possible only if keto group in

cholestanone is flanked on either side with methylene group

(-CH2COCH2-). This arrangement is possible only in ring A.

Hence the hydroxyl group in cholesterol must be present in ring

A.

Now consider the following set of reactions.

C27H46OH2O2

CH3COOHC27H48O3

CrO3C27H44O3

(i) - H2O

(ii) Zn - CH3COOHC27H44O2

CrO3

C27H44O8

Cholesterol I

Cholestanetriol II

Hydroxycholestanedione III

Cholestanedione IV

Tetracarboxylic acid V

14Dha. Ilangeswaran

In the above reactions, the conversion of cholestane triol(II)

to dione(III) showed that in compound II, two –OH groups are 2o and

the third one which resists oxidation may be 3o.

The formation of IV from III without loss of carbon atoms

showed that the two keto groups may be present in different rings.

Hence in cholesterol the –OH group and the double bond may be

present in different rings.

Compound IV, cholestane dione forms a pyridazine derivative

with hydrazine. This is a characteristic property of -diketone. Hence

compound, IV must be a -diketone.

The position of hydroxyl group is already fixed with ring A.

All the above reactions can be explained well if we place –OH

group at C3 position of ring A and the double bond between C5 and C6

of ring B.

15Dha. Ilangeswaran

OH

A B

Cholesterol I

H2O2

CH3COOHOH

OHOH

A B

Cholestanetriol II

CrO3

O

OOH

A B

Hydroxycholestanedione III

(i) -

H 2O

(ii) Z

n - CH 3

COOH

O

OH

A B

Cholestanedione IV

CrO3

COOH

HOOC

COOHHOOC

Tetracarboxylic acid V

NH2NH2

HN

NPyridazine derivative

16Dha. Ilangeswaran

The above fact is further supported by the following reactions

Cholesterol on heating with copper oxide at 290oC gave

cholestenone, which on oxidation with permanganate yielded

keto acid with a loss of carbon atom. The formation of keto acid

revealed the presence of keto group and a = bond in the same

ring of cholestenone. These results can be explained if it is

assumed that the double bond in cholesterol migrates during the

formation of cholestenone.

CuO

290 °COH

Cholesterol I

OCholestenone VII

KMnO4

O

O

OH+ CO2

Keto acid VIII

17Dha. Ilangeswaran

The position of hydroxyl group at C3 is finally proved

by Kon et al. The formation of 3’,7-dimethyl

cyclopentenophenanthrene from cholesterol by the following

steps is possible only if –OH group is considered at position

C3.

Thus we concluded that cholesterol contains –OH group

at position C3 and a double bond between C5 to C6.

OH

Cholesterol I

H2 / Pt

OHH

CrO3

OH

CH3MgBr

OHH

CH3

Se

350 °C

CH3

CH3

3'

7

3',7-Dimethylcyclopentenophenanthrene

18Dha. Ilangeswaran

Nature of the Side Chain

The cholesteryl acetate derived from cholesterol

on oxidation with CrO3 forms a ketone (steam volatile) &

acetate of hydroxy acetone (non-steam volatile). The ketone was

found to be isohexyl methyl ketone, which may be the side

chain of cholesterol. The point of attachment of side chain may

be at the carbon of the keto group, i.e. at C17 – position.

O

AcO

+CH3 CH3

CH3O

Isohexylmethyl ketone

The nature of side chain and the linkage have been

studied by Barbier – Wieland. The B-W degradation offers a

method of getting a lower acid with one carbon atom less as

described below.

RCH2COOHCH3OH

HClRCH2COOCH3

2C6H5MgBrRCH2C(OH)(C6H5)2

- H2O

RCH=C(C6H5)2

CrO3O=C(C6H5)2+RCOOH

20Dha. Ilangeswaran

Cholesterol was first converted into 5 -cholestane (a

stereoisomer of cholestane). The nucleus of 5 -cholestane is

represented as Ar and side chain as Cn, then the degreadations

can be expressed as follows.

5 -Cholestane (or coprostane)

Ar-Cn

CrO3 CH3COCH3 + Cholanic acid

Ar-Cn-3

B - W

(C6H5)2CO + Norcholanic acid

Ar-Cn-4

B - W(C6H5)2CO + Bisnorcholanic acid

Ar-Cn-5

B - W

(C6H5)2CO + Etiocholylmethyl ketone

Ar-Cn-6

CrO3 Etianic acid

Ar-Cn-7

21Dha. Ilangeswaran

The formation of acetone from 5 -cholestane shows

that side chain terminates in an isopropyl group. The

conversion of bisnorcholanic acid into a ketone reveals the

presence of an alkyl group on the -carbon in bisnorcholanic

acid. As etiocholyl methyl ketone is oxidised to etianic acid

with a loss of one carbon atom, the ketone may be a methyl

ketone, i.e., there is a methyl group in -carbon of bisn

orcholanic acid.

When etianic acid is subjected to one more B – W

degradations, a ketone (etiocholanone) is obtained which on

oxidation using nitric acid yields dicarboxylic acid

(etiobilianic acid) without loss of carbon atoms. Hence

etiocholanone must be a cyclic ketone.

These reactions shows the presence of eight C atoms

in side chain.

22Dha. Ilangeswaran

The above degradations can only be explained if the

side chain has the following structure.

Position of Side Chain: Etiobilianic acid derived frometiocholanone forms an anhydride on treatment with aceticanhydride. Acc. to Blanc’s rule etiocholanone may be a 5membered ketone. So the side chain may be attached to the 5membered ring D. The formation of Diel’s hydrocarbon fromcholesterol suggests that the side chain is at C17 asdehydrogenation degrades a side chain to methyl group.

Ar - CH - CH2 - CH2 - CH2 - CH

CH3 CH3

CH3

Nucleus Side chain

23Dha. Ilangeswaran

Further 5 -cholanic acid is formed from 5 -

cholestane on oxidation. Deoxycholic acid also on oxidation

followed by Clemmenson’s reduction yields 5 -cholanic acid.

the side chain in cholesterol & deoxycholic acid may be

present in the same position. Now the nature & position of

side chain is known the conversion of 5 -cholestane to

etiobilianic acid can be expressed as follows.

24Dha. Ilangeswaran

CH3

CH3

CH3

5 -Cholestane

CrO3

CH3

COOH

CH3

CH3

O

5 -Cholanic acid

+

B - W

CH3 COOH

Nor-5 -Cholanic acidB - W

CH3

COOH

Bisnor-5 -Cholanic acid

B - W

CH3O

Etiocholylmethyl ketoneCrO3

COOH

Etianic acid

B - W

O

Etiocholanone

HNO3

CH3 COOHCOOH

Etiobilianic acid

25Dha. Ilangeswaran

Position of Two Angular Methyl GroupsWhen anhydride of etiobilianic acid is distilled with Se,

1,2-dimethylphenanthrene is obtained. It shows the presence ofphenanthrene nucleus and an angular methyl group at C13 ofcholesterol.

The cyclopentenophenanthrene nucleus of cholesterolaccounts for 17 carbon atoms & the side chain for 8 C atoms.This accounts for 25 C atoms and the remaining 2 are assumedto be angular methyl groups.

When the position of –OH group & = bond aredetermined one of the compounds formed was keto acid, VIII.This on subjected to Clemmenson’s reduction followed by twoB-W degradations, we get an acid which is difficult to estrifyand gives CO on warming. This clearly shows that –COOHgroup must be linked to a tertiary C atom & side chain must beof the following type.

26Dha. Ilangeswaran

This shows an alkyl group at C10-position.

To determine the position of 2nd methyl group, consider

the Se dehydrogenation of cholesterol to yield chrysene & Diel’s

hydrocarbon. The formation of chrysene can be explained if

there is an angular methyl group at C13 of cholesterol, which

enter into 5 membered ring during Se dehydrogenation to give

4th six membered ring of chrysene.

C

C - C - C - C - COOH

C

2 (B - W)C - C - COOH

C

C

The reactions are shown below

COOH

O

Keto acid VIII

Zn - Hg

HClCOOH

2 (B - W)

HOOC

Acid group with

3o carbon atom

27Dha. Ilangeswaran

The positions of 2 angular methyl groups at C10 &

C13 are supported by the following reactions.

OH

OH

COOH

CrO3

OH

O

COOH

HNO3

HOOCCOOH

O

COOH

Deoxycholic acid Dehydrodeoxycholic acid Deoxybillianic acid

O

O

COOHKMnO4

OCOOH

O

COOH

HNO3

COOH COOH

COOH

+

HOOC

COOH

HOOC

COOH COOHO

HOOCH

HNO3

COOHHOOC

HOOCH

Diketo-dicarboxylic acidA

B C D

28Dha. Ilangeswaran

The compound A was found to be butane-2,2,4-tricarboxylic

acid. This shows an angular methyl group at C10. Compound B is a

tetracarboxylic acid with a cyclopentane ring and a side chain having

–C-Me group. D is a tricarboxylic acid where –COOH is linked to

tertiary carbon. This proves an angular methyl group at C13.

The conversion of etiobilianic acid into 1,2-dimethylphen

anthrene shows angular methyl group at C13. If it is at C14 the

product should be 1-methylphenanthrene.

COOH

COOHCH3

CH3

CH3

O

O

OCH3

Se

CH3

CH3

Etiobilianic acid Anhydride 1,2-Dimethylphenanthrene

29Dha. Ilangeswaran

Androsterone

Molecular formula: C19H30O2

Melting point: 183oC, [ ]D = +94o

It was first isolated by Butenandt et al. in 1931. Androsterone

behaves as a saturated compound.

As it forms mono-ester, one oxygen atom is present as a

hydroxyl group. The 2nd O atom was shown to be oxo, since

androsterone forms an oxime.

The parent hydrocarbon of androsterone is C19H32, and since

this corresponds to the general formula CnH2n-6, it may be

tetracyclic.

D.B.E. of C19H30O2 = (a+1) - (b-c)/2 = (19+1) – (30-2)/2 = 5;

1 double bond due to C=O, & 4 rings.

30Dha. Ilangeswaran

The above facts suggest that androsterone may conatain

a steroid nucleus. Butenandt proposed a structure which was

proved to be correct by Ruzicka as follows.

Ruzicka oxidized 5 -cholestanyl-3 -acetate with

chromic oxide in acetic acid to epiandrosterone, which is a

hydroxyketone.

CH3

CH3

CH3

CH3

CH3H

H

H

AcO

(i) CrO3

(ii) Hydrolysis

CH3

CH3

O

H

H

H

OH

5 -Cholestanyl-3 -acetate Epiandrosterone

31Dha. Ilangeswaran

But the oxidation of 5 -cholestanyl-3 -acetate only

yields androsterone.

Thus the configuration of the –OH group at C3 is and

not as Butenandt suggested. Epiandrosterone, m.p. 174oC,

[ ]D + 88o, has about 1/8th of the activity of androsterone. In

1955 Sondheimer et al. converted epiandrosterone into

androsterone as follows.

CH3

CH3

CH3

CH3

CH3H

H

H

AcO

(i) CrO3

(ii) Hydrolysis

CH3

CH3

O

H

H

H

OH

5 -Cholestanyl-3 -acetate Androsterone

32Dha. Ilangeswaran

CH3

CH3

O

H

H

H

TsO

AcONa

CH3

CH3

O

H

H

H+

CH3

CH3

O

H

H

H

AcO

Epiandrosterone-p-toluenesulphonate54%

AcOH-Ac2O

H2O2

CH3

CH3

O

H

H

H

O

39%

NaOH

CH3

CH3

O

H

H

H

OH

PhCO3H

CH3

CH3

O

H

H

H

O

O

LiAlH4

CH3

CH3

O

H

H

H

O

OH

Aq. AcOH

Androsterone

33Dha. Ilangeswaran

But a conventional preparation of androsterone starts

from dehydroepiandrosterone.

CH3

CH3

O

H

H

OH

H

Dehydroepiandrosterone

H2O2

TsOH / PhH

CH3

CH3

H

H

OH

H

O

O

Oppenauer

oxidation

CH3

CH3

H

H

O

H

O

O

(i) B2H6

(ii) Ac2O

CH3

CH3

H

HH

O

H

H2O2

TsOH / PhH

CH3

CH3

H

HH

O

H

O (i) B2H6

(ii) H2O2/OH

(iii) acid

CH3

CH3

O

H

H

H

OH

H

Androsterone

34Dha. Ilangeswaran

A total synthesis of androsterone has been carried out

by Woodward et al. using the following ester, which is used

in the synthesis of cholesterol.

Soon after the discovery of androsterone, Butenandt et

al. isolated the following hormones from male urine.

CH3

CH3

COOMe

H

H

H

O

H

CH3

CH3

O

H

H

H

OH

H

5 -androsterone

m.p. 151oC, [ ]D = +105o

CH3

CH3

OH

H

H

O

H

Testosterone

35Dha. Ilangeswaran