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Question

A biologically active compound bombykol (C16H30O) is obtained from a natural source. The structure of the compound is determined by the following reactions:
(a) On hydrogenation, bombykol gives a compound (A), C16H34O, which reacts with acetic anhydride to give an ester.
(b) Bombykol also reacts with acetic anhydride to give another ester, which on oxidation ozonolysis (O3/H2O2) gives a mixture of butanoic acid, oxalic acid, and 10-hydroxy decanoic acid.
Determine the number of double bonds in bombykol. Write the structures of compound (A) and bombykol. How many geometrical isomers are possible of bombykol?

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Solution

(i) D.U in bomykol (X) (C16H30O)=(2nC+2)nH2=34302=2o
(ii) this suggests that (X) has either two (C=C) bonds or one (CC) bond
(iii) Reaction (X) with acetic anhyride to form an ester suggests that (X) contains (OH) group
Proceed reverse from oxidative ozonolysis (O3/H2O2) products of ester of (X)
(iv) Number of double bonds in (X)=2
(v) Number of geometrical isomers of (X)=22=4 isomers (since terminal groups are different)

1776288_1769164_ans_263b60a4707e4ad9988ca1722e46522e.png

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