A biologically active compound bombykol `(C_(16) H_(30)O)` 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), C_(16) H_(34) O`, which reacts with acetic anhydride to give an ester.
(b) Bombykol also reacts with acetic anhydride to give another ester, which on oxidation ozonolysis `(O_(3)//H_(2) O_(2))` gives a mixture of butanoic acid, oxalic acid, and `10-hydroxy` decanoic acid.
Determone the number of double bonds in bombykol. Write the strcutures of compounds `(A)` and bombykol. How many geometrical isomers are possible for bombykol ?

Explanation

(i) `D.U`. In bombykol `(X) (C_(16) H_(30) O)`
=`((2n_C +2)-n_H)/(2) = (34 -30)/(2) = 2^@`
(ii) This suggests that `(X)` has either two `(C=C)` bonds or one `(C -= C)` bond.
(iii) Reaction of `(X)` with acetic anhydride to form an ester suggests that `(X)` contains `(OH)` group.
Proceed reverse from oxidative ozonolysis `(O_(3)//H_(2) O_(2))` products of ester of `(X)`.
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(iv) Number of double bonds in `(X) = 2`
(v) Number of geometrical isomers of `(X) = 2^2 = 4` isomers (since terminal groups are different).