Structure of Bombykol is CH 3 –CH 2 –CH 2 –CH=CH–CH=CH(CH 2 ) 8 CH 2 OH

Sol. Ozonolysis of acetate ester of Bombykol gives.

CH 3 –CH 2 –CH 2 –COOH + HOOC–COOH + HOOC–CH 2 ) 8 .CH 2 OCOCH 3

Thus acetate ester of Bombykol is

CH 3 –CH 2 –CH 2 –CH=CH–CH=CH.(CH 2 ) 8 –COCOCH 3

So the structure of Bombykol is

CH 3 –CH 2 –CH 2 –CH=CH–CH=CH(CH 2 ) 8 CH 2 OH

This Bombykol on hydrogenation to give compound ‘A’ (C 16 H 34 O)

CH 3 –CH 2 –CH 2 –CH 2 –CH 2 –CH 2 –CH 2 (CH 2 ) 8 CH 2 OH

This compound on reaction with acetic anhydride to give an ester

CH 3 –CH 2 –CH 2 –CH=CH–CH=CH(CH 2 ) 8 CH 2 OH

↓ (CH 3 CO) 2 O(–CH 3 COOH)

CH 3 –CH 2 –CH 2 –CH=CH–CH=CH(CH 2 ) 8 CH 2 OCOCH 3

Four geometrical isomers are possible in Bombykol because it has two C = C bonds, which exhibit the property of geometrical isomers.

So no. of geometrical isomers = 2 n = 2 2 = 4 which are cis-cis, cis-trans, trans-trans and trans-cis.