Question 5.SSE.16: A biologically active compound, Bombykol (C16H30O) is obtain......
A biologically active compound, Bombykol (C_{16}H_{30}O) is obtained from a natural source. The structure of the compound is determined by the following reactions.
(i) On hydrogenation, Bombykol gives a compound A, C_{16}H_{34}O , which reacts with acetic anhydride to give an ester;
(ii) Bombykol also reacts with acetic anhydride to give another ester, which on oxidation ozonolysis (O_2 / H_2O_2) gives a mixture of butanoic acid, oxalic acid and 10-acetoxydecanoic 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?
[IIT, 2002]
\begin{matrix} C _{16} H _{30} O & \overset{\left( CH _3 CO \right)_2 O}{\xrightarrow {\hspace{2 cm}} } & \text {Ester} & \overset{\text{Oxidative}}{\underset{\text{ozonolysis}}{\xrightarrow {\hspace{2 cm}} }} &CH _3 CH _2 CH _2 COOH &+&( COOH )_2 &+& AcOCH _2\left( CH _2\right)_8 CH _2 COOH \\ &&&& \text {Butanoic acid} && \text {Oxalic acid} && \text {10-Acetoxydecanoic acid}\end{matrix}
(i) Hydrogenation of bombykol \left( C _{16} H _{30} O \right) to C _{16} H _{34} O (A) shows the presence of two double bonds in bombykol.
(ii) Reaction of (A) with acetic anhydride to form ester shows the presence of an alcoholic group in (A) and also in bombykol.
(iii) Products of oxidative ozonolysis of bombykol ester suggests the structure of bombykol
\begin{matrix} CH _3 CH _2 CH _2 CH = CH – CH = CH .\left( CH _2\right)_8 CH _2 OAc \\ \text {Bombykol ester} \\ \hspace{90 pt} \downarrow \text {Oxidative ozonolysis}\end{matrix}\begin{matrix} CH _3 CH _2 CH _2 COOH &+&\text { HOOC.COOH } &+& HOOC \cdot\left( CH _2\right)_8 \cdot CH _2 OAc \\ \text {Butanoic acid} && \text {Oxalic acid} && \text {10-Acetoxydecanoic acid}\end{matrix}
The structure of Bombykol ester indicates that Bombykol has the following structure:
\begin{matrix} CH _3 CH _2 CH _2 CH = CH – CH = CH \left( CH _2\right)_8 CH _2 OH \\ \text {(Bombykol)} \end{matrix}and structure of (A) is
CH _3 CH _2 CH _2 CH _2- CH _2- CH _2- CH _2\left( CH _2\right)_8 \cdot CH _2 OH or C _{16} H _{13} OH
\begin{matrix} &&&&&& H &&&&&& \left( CH _2\right)_8 \stackrel{1}{ C } H _2 OH\\ &&&&&&& \diagdown &&&& \diagup \\ \eta &-&C_3H_7 &&&&&&\overset{11}{C} &=& \overset{10}{C} \\ &&& \diagdown &&&&\diagup &&&&\diagdown \\ &&&& \overset{13}{C} &=&\overset{12}{C} &&&&&& H \\ &&&\diagup &&&&\diagdown \\ &&H &&&&&&H \end{matrix}
10 E, 12 Z
\begin{matrix} &&&&&& H &&&&&& H\\ &&&&&&& \diagdown &&&& \diagup \\ \eta &-&C_3H_7 &&&&&&\overset{11}{C} &=& \overset{10}{C} \\ &&& \diagdown &&&&\diagup &&&&\diagdown \\ &&&& \overset{13}{C} &=&\overset{12}{C} &&&&&& \left( CH _2\right)_8 \stackrel{1}{ C } H _2 OH \\ &&&\diagup &&&&\diagdown \\ &&H &&&&&&H \end{matrix}
10 Z, 12 Z
\begin{matrix} &&&&&& H &&&&&& H\\ &&&&&&& \diagdown &&&& \diagup \\ &&H &&&&&&\overset{11}{C} &=& \overset{10}{C} \\ &&& \diagdown &&&&\diagup &&&&\diagdown \\ &&&& \overset{13}{C} &=&\overset{12}{C} &&&&&& \left( CH _2\right)_8 \stackrel{1}{ C } H _2 OH \\ &&&\diagup &&&&\diagdown \\ \eta &-&C_3H_7 &&&&&&H \end{matrix}
10 Z, 12 E
\begin{matrix} &&&&&& H &&&&&& H\\ &&&&&&& \diagdown &&&& \diagup \\ \eta &-&C_3H_7 &&&&&&\overset{11}{C} &=& C \\ &&& \diagdown &&&&\diagup &&&&\diagdown \\ &&&& \overset{13}{C} &=&\overset{12}{C} &&&&&& \left( CH _2\right)_8 \stackrel{1}{ C } H _2 OH \\ &&&\diagup &&&&\diagdown \\ &&H &&&&&&H \end{matrix}
10 E, 12E