Hydrocarbons 2 Question 54
57. A biologically active compound, Bombykol $\left(\mathrm{C}{16} \mathrm{H}{30} \mathrm{O}\right)$ is obtained from a natural source. The structure of the compound is determine by the following reactions.
(a) On hydrogenation, Bombykol gives a compound $A, \mathrm{C}{16} \mathrm{H}{34} \mathrm{O}$, which reacts with acetic anhydride to give an ester.
(b) Bombykol also reacts with acetic anhydride to give another ester, which on oxidative ozonolysis $\left(\mathrm{O}{3} / \mathrm{H}{2} \mathrm{O}_{2}\right)$ 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 :
$$ \begin{aligned} \underset{\text { butanoic acid }}{\mathrm{C}{3} \mathrm{H}{8}-\mathrm{COOH}} & +\underset{\text { oxalic acid }}{\mathrm{HOOC}-\mathrm{COOH}} \ & +\mathrm{HOOC}-\left(\mathrm{CH}{2}\right){8}-\mathrm{CH}{2} \mathrm{O}-\mathrm{C}-\mathrm{CH}{3} \ \qquad & \underset{10 \text {-acetoxy decanoic acid }}{-} \mathrm{O}{3} / \mathrm{H}{2} \mathrm{O}{2} \ \mathrm{CH}{3} \mathrm{CH}{2} \mathrm{CH}{2} \mathrm{CH} & =\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\left(\mathrm{CH}{2}\right){8} \end{aligned} $$
Therefore, Bombykol is :
$$ \begin{gathered} \mathrm{CH}{3} \mathrm{CH}{2} \mathrm{CH}{2}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\left(\mathrm{CH}{2}\right){8}-\mathrm{CH}{2} \mathrm{OH} \ \text { Bombykol } \ \mathrm{CH}{3}-\left(\mathrm{CH}{2}\right){14} \stackrel{\downarrow}{\downarrow} \mathrm{CH}{2} \mathrm{OH} \end{gathered} $$
(A)
