Hydrocarbons - Result Question 71
57. A biologically active compound, Bombykol $\left(C _{16} H _{30} O\right)$ is obtained from a natural source. The structure of the compound is determine 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 oxidative ozonolysis $\left(O _3 / H _2 O _2\right)$ gives a mixture of butanoic acid, oxalic acid and $10 $-acetoxy decanoic acid.
Determine the number of double bonds in Bombykol. Write the structures of compound $A$ and Bombykol. How many geometrical isomers are possible for Bombykol?
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Solution:
- From oxidation products, structure of starting compound can be deduced as :
BETA
