Question

A biologically active compound bombykol $\left(C_{16}{H}_{30}O\right)$ is obtained from a natural source. The structure of the compound is determined by the following reactions:
(a) On hydrogenation, bombykol gives a compound $\left(A\right)$, $C_{16}{H}_{34}O$, which reacts with acetic anhydride to give an ester.
(b) Bombykol also reacts with acetic anhydride to give another ester, which on oxidation ozonolysis $\left(O_3/H_2{O}_2\right)$ 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 $\left(A\right)$ and bombykol. How many geometrical isomers are possible of bombykol?

Answer 1

(i) D.U in bomykol $\left(X\right)$ $\left(C_{16}{H}_{30}O\right)=\frac{(2n_C+2)-n_H}{2}=\frac{34-30}{2}=2^o$

(ii) this suggests that $\left(X\right)$ has either two $(C=C)$ bonds or one $(C\equiv C)$ bond

(iii) Reaction $\left(X\right)$ with acetic anhyride to form an ester suggests that $\left(X\right)$ contains $\left(OH\right)$ group

Proceed reverse from oxidative ozonolysis $\left(O_3/H_2{O}_2\right)$ products of ester of $\left(X\right)$

(iv) Number of double bonds in $\left(X\right)=2$

(v) Number of geometrical isomers of $\left(X\right)=2^2=4$ isomers (since terminal groups are different)