Cholesterol Prepared by

Dr. N.GOPINATHANASSISTANT PROFESSOR

DEPARTMENT OF PHARMACEUTICAL CHEMISTRYFACULTY OF PHARMACY

SRI RAMACHANDRA UNIVERSITYCHENNAI-116TAMILNADU

It is present in all mammalian tissues either in free state or esterified with fatty acid. Microsomal and cytosolfraction of cell is responsible for cholesterol synthesis.Large quantities present in brain and nerve tissue.It is optically active levo rotatory in nature.

• It is an animal sterol occurs as free and fatty ester.

• Main source are brain, spinal cord gallstone and fish liver oil.

• It was first isolated from human gallstone deposited in the bile duct. It is known as cholesterol.

• Human body not only synthesis it but can absorb it from food through the intestine into the blood stream

• Too high concentration of cholesterol in the blood can lead to its precipitation in the circulatory vessel results in high Blood pressure and arteriosclerosis.

• It is white crystal and optically active solid with melting point 149◦ C

Structure

CH3

CH3

OH

CH3

CH3

CH3

Identification test

• Libermann Burchard test• Sample is treated with chloroform,

acetic anhydride and sulphuric acid is added along the sides the chloroform layer becomes green in color.

• Dehydrogenated with selenium at 360 degree celsius yield Diels hydrocarbon

Salkowski Reaction To 2ml of the extract

Add 2ml of chloroform

2ml conc. Sulphuric acid

Shake well

Chemistry• On acetylation it forms mono

acetate indicates the presence of OH group.

• It takes up two bromine atoms suggest that the presence of double bond.

I-Cholesterol reduction

II-cholestanol cro3 oxidation

III-choletanone Zn/Hg HCl

IV-cholestane

I-II proves presence of double bondII-III proves presence of secondary alcoholIII-IV saturated parent hydrocarbon cholestane C27H48 which corresponds to the formulae CnH2n-6 tetracyclic nature hence cholesterol has tetracyclic ring system

Cholestanone on oxidation with nitric acid gives dicarboxylic acid which on pyrolysis yield ketone

Conclusion of OH in sterol

• Oxidation of cholestanone to acid reveals that ketonic group is present inside the ring because the acid formed are less carbon number.

• Conversion of dicarboxylic acid yield ketone it is either 1,6 or 1,7 dicarboxylic acid

Conclusion of OH in sterol• Blanc Rule:• Dicarboxylic acid upto 1,5 gives

anhydride on heating or pyrolysis • But 1,6 or 1,7 dicarboxylic acid yield

ketone with loss of one carbon atom, on similar treatment.

• OH is not in D because it forms 1,5 dicarboxylic acid .

• OH might be in A,B or C.

The formation of two isomeric dicarboxylic acid Suggest that keto group is between two methylene group

CH3

CH3

CH3

CH3

CH3

OOH

O

OHCH3

CH3

O

CH3

CH3

CH3

H N O 3

It is possible if OH is present in Ring A but confusion is in position whether it is in 2 or 3

Cholestanone on treatment with methyl magnesium iodide followed by selenium dehydrogenation yield 3,7 dimethyl cyclo pentano perhydro phenanthrene proved by its synthesis

CH3

CH3

O

CH3

CH3

CH3

CH3

CH3

OH

CH3

CH3

CH3

CH3CH 3M gI

CH3

CH3

CH3

CH3

CH3

CH3

OH is at 3rd position

Position of double bond

I to II represent hydroxylation of doublebond

II on oxidation yield diketone indicating that II possess two secondary OH and Third OH is tertiary because resistant to oxidation

IV is oxidised to V tetra carboxylic acid without any loss of carbon atom suggest that two keto groups are in different ring. IF they are in same ring, the carbon atom would have been lost during oxidation

Double bond and OH are in different ring. Already OH is in Ring A therefore double bond must be in B , C or D

Cholestanedione IV forms pyridazine derivative. Two keto groups must be in gamma position. Thus double bond must be in C5 andC6.UV spectra of cholestenone gives λ max at 240 nm which shows that keto group and double bond are in conjugation.

Nature and position of side chain

Iso hexyl methyl ketone forms side chain and it is attached to nucleus through the carbon atom oxidise to carbonyl group

Position of Angular methyl group

• Ring and side accounts for 17 & 8 carbon atom thus the 25 out of 27 atoms are accounted rest found to be as angular methyl group.

• Keto acid on clemmenson reduction followed by twice barbier wieland degradation gives tertiary acid so one of the angular methyl group is at C10

Barbier wieland degradation- stepping down an acid by one carbon less.

On selenium dehydrogenation it yield diel’s hydrocarbon and chrysene.it is explained by the fact that angular methyl group might be at C13 or C14 which enter into 5 member ring D to form six member ring Chrysene

Indicates that methyl group in position 13 if it is at C14 then mono methyl phenanthrene would have been formed

• The structure of cholesterol is further confirmed by its synthesis

Stereo chemistry of cholesterol

• The fusion of ring A to B, B/C, C/D may be either in CIS or Trans manner but the steroid molecule is essentially flat so the ring B and C always fused in trans.

• Naturally occuring steroids except those of heart poison belongs either to the cholestane series or coprostane series

Coprostane series5αcholestane A/B-Trans B/C – Trans and C/D - Trans

5 β cholestane A/B-CIS B/C – Trans and C/D - Trans

• Both the angular methyl groups are CIS.• The compound derived from cholestane

are known as Allo compound whereas the compounds derived from coprostane are known as normal compound

If OH is present above the lane of ring i.e in CIS position with respect to methyl group at C10 then it is βon the other hand if hydroxyl group is present below the plane of ring is known as α or epi compound

Thank you