1. Tardigrade
  2. Question
  3. Chemistry
  4. A biologically active compound, Bombykol ( C 16 H 30 O ) 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 ( O 3 / H 2 O 2) 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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Q. 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?

Hydrocarbons

Solution:

From oxidation products, structure of starting compound can be
deduced as :

Solution Image