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 :
$$ \begin{aligned} \underset{\text { butanoic acid }}{C _3 H _8-COOH} & +\underset{\text { oxalic acid }}{HOOC-COOH} \ & +HOOC-\left(CH _2\right) _8-CH _2 O-C-CH _3 \ \qquad & \underset{10 \text {-acetoxy decanoic acid }}{-} O _3 / H _2 O _2 \ CH _3 CH _2 CH _2 CH & =CH-CH=CH-\left(CH _2\right) _8 \end{aligned} $$
Therefore, Bombykol is :
$$ \begin{gathered} CH _3 CH _2 CH _2-CH=CH-CH=CH-\left(CH _2\right) _8-CH _2 OH \ \text { Bombykol } \ CH _3-\left(CH _2\right) _{14} \stackrel{\downarrow}{\downarrow} CH _2 OH \end{gathered} $$
(A)
