STRUCTURAL ELUCIDATION OF CHOLESTEROL
1. Structure of the nucleus
i. Analysis and molecular weight determination corresponds to the molecular formula C27H46O.
ii. On acetylation, it gives monoacetate, indicating the presences of one –OH group.
iii. It takes up two bromine atoms to give dibromocholesterol, indicating the presence of one double bond.
iv. On reduction, it gives cholestanol, which on oxidation gives cholestanone, cholestanone on reduction gives cholestane.
a) Conversion of I to II shows the presence of double bond.
b) Oxidation of II to III shows that cholesterol is secondary alcoholic group in nature.
c) Cholestane (IV) is a saturated hydrocarbon and corresponds to the general formula (C27H48) CnH2n-6 and it is tetracyclic.
v. The formation of Diel’s hydrocarbon shows that cholesterol (in general steroids) contains cyclopentenophenanthrene nucleus.
vi. The size of the rings is explained by Blanc’s Rule.
a) The 1, 5-dicarboxylic acids on heating with acetic anhydride forms a cyclic anhydride indicating the presence of a five membered ring.
b) The 1, 6-dicarboxylic acids form cyclophentanone with elimination of CO2 indicating the presence of six membered ring.
Ring A
Cholesterol & the cholic acids were converted into the dicarboxylic acid (A) which gave a cyclopentanone derivative on heating with acetic anhydride. Therefore the ring A is six membered.
Ring B
Cholestrol was converted into the tricarboxylic acid (B) which gave the cyclopentanone derivative. Hence, ring (B) is six membered.
Ring C
Deoxycholic acid was converted into dicarboxylic acid (C) which gave a cyclic anhydride.This reaction results in the form of seven membered cyclic anhydride. In this case, Blanc’s rule fails. But, it is six membered ring.
Ring D
5β-cholestane (coprostane) was converted into etiobilianic acid and this gave a cyclic anhydride, Hence, ring D is five membered.
2. Position of Double bond
Consider the following reaction for the position of double bond The above reaction leads to the following out come
a) Conversion of I to II indicates the double bond in I is hydroxylated.
b) Conversion of II to III indicates that (of the three –OH groups present in II) the two –OH groups are secondary and the 3rd–OH group is tertiary in nature.
3. Position of hydroxyl group (-OH) and double bond
Consider the following reaction, Oxidation of IV to V (without loss of C-atom) shows that the two keto groups in IV must be present in
different rings. Also it follows that the –OH group & the double bond in cholesterol must be in different rings.
(d.) Conversion of IV to VI indicates the two ketonic groups are in γ-position (γ-diketone) with respect to each other. This is possible only if the double bond is present in the position of V and VI the – OH group at position III.
4. Nature and position of the side chain
The formation of this voltaic ketone (isohexyl methyl ketone) in the above reaction shows that this ketone is the side chain being at the carbon of the keto group (non-volatile). The above reactions do not show where the side chain is attached to the nucleus. If we assume, that it must be attached to C-17 in the above reaction can be formulated.
5. Nature of side chain
Barbier-wieland Degradation
a) The formation of acetone shows that the side chain terminates in an isopropyl group.
b) The conversion of IV to V shows that there is an alkyl group on the α-carbon in the former compound IV.
c) Conversion of V to VI ketone is oxidized to 5β-etianic with the loss of one carbon atom. Therefore the ketone must be a methyl ketone. So the alkyl group on α-carbon atom in bis nor-5β-cholanic acid is a methyl group.
d) Now the carboxyl group in etianic acid is directly attached the nucleus. This is shown by the conversion of V to VI. The ketone
etiocholanone obtained by step VI. On oxidation with HNO3 gives a dicarboxylic acid etiobilianic acid without any loss of carbon atom. Thus etiocholanone must be a cyclic ketone. Hence, it follows that there are 8-carbon atoms in the side chain which is represented by the following structure.
6. Position of the side chain
The dicarboxylic acid of etiobilianic acid an anhydride when heated with Ac2O. Thus, the ketone etiocholanone is probably a five membered ring. And the side chain is attached to the five membered ring D.
7. Actual point of attachment
The formation of Diel’s hydrocarbon from cholesterol shows that the side chain is C-17. Since, Selenium-dehydrogenation the side chain to a methyl group. The position of C-17 is also supported by X-ray diffraction studies and surface film measurements. When the anhydride of etiobilianic acid is distilled with Se,1,2-dimethyl phenanthrene is obtained. This shows the presence of phenanthrene nucleus in cholesterol and also gives evidence for the position of the C-13 angular methyl group.
