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, C₁₆.H₃₄O, 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 for bombykol.?

Answer 1

Let us summarise the given facts

(i) Hydrogenation of bombykol (C₁₆H₃₀O) to C₁₆HH₃₄O (A) indicates the presence of two double bonds in bombykol.

(ii) Reaction of with acetic anhydride to form ester indicates the presence of an alcoholic group in A and hence, also in bornbykol.

(iii) Products of oxidative ozonolysis of bombykol ester suggests the structure of bombykol.

The structure of bombykol ester suggests that bombykol has the following structure :

Four geometrical isomers are possible for the above bombykol structure (as it has two double bonds)