Cholesterol is present in all mammalian tissues, either in a free state or esterified with fatty acids. It is synthesized in the microsomal and cytosolic fractions of cells and is present in large quantities in brain and nerve tissue. Cholesterol has an optically active levo-rotatory structure and is an animal sterol that occurs as a free alcohol and in fatty acid esters. It was first isolated from human gallstones and deposits in the bile duct. The structure of cholesterol was determined through a series of chemical reactions and degradations.

1. Cholesterol
Prepared by
Dr. N.GOPINATHAN
ASSISTANT PROFESSOR
DEPARTMENT OF PHARMACEUTICAL CHEMISTRY
FACULTY OF PHARMACY
SRI RAMACHANDRA UNIVERSITY
CHENNAI-116
TAMILNADU

2. 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.

3. • 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.

4. • 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

5. Structure
CH3
CH3
OH
CH3
CH3
CH3

6. 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

7. Salkowski Reaction
To 2ml of the extract
Add 2ml of chloroform
2ml conc. Sulphuric acid
Shake well

8. 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.

9. I-Cholesterol
reduction
II-cholestanol
cro3 oxidation
III-choletanone
Zn/Hg HCl
IV-cholestane

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

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

12. 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

13. 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.

14. The formation of two isomeric dicarboxylic
acid Suggest that keto group is between
two methylene group
CH3
CH3
CH3
CH3
CH3
OOH
O
OH
CH3
CH3
O
CH3
CH3
CH3
HNO3
It is possible if OH is present in Ring A but confusion is in
position whether it is in 2 or 3

15. 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
CH3
CH3MgI
CH3
CH3
CH3
CH3
CH3
CH3
OH is at 3rd position

16. Position of double bond

17. 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

18. 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

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

20. 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.

21. 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

22. 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

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

25. 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

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

27. • The structure of
cholesterol is further
confirmed by its
synthesis

30. 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

31. Coprostane series
5αcholestane A/B-Trans B/C
– Trans and C/D - Trans
5 β cholestane A/B-CIS B/C –
Trans and C/D - Trans

32. • 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

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

34. Thank you