Chapter 7, Problem 7.76P

Muscalure, the sex pheromone of the common housefly, can be prepared by a reaction sequence that uses two nucleophilic substitutions. Identify compounds A-D in the following synthesis of muscalure.

Interpretation Introduction

Interpretation: Compounds A, B, C and D are to be identified in the given synthesis of muscalure.

Concept introduction: The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. The electron rich chemical species that contains negative charge or lone pair of electrons are known as a nucleophile. In a nucleophilic substitution reaction, nucleophile takes the position of leaving group by attacking the electron deficient carbon atom.

Answer to Problem 7.76P

Compounds A, B, C and D are $\text{HC}\equiv {\text{C}}^{-}{\text{Na}}^{+}$, ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv \text{CH}$, ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ and ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv \text{C}-{\text{CH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{11}}{\text{CH}}_{\text{3}}$, respectively.

Explanation of Solution

Sodium hydride on reaction with $\text{HC}\equiv \text{CH}$ leads to the formation of $\text{HC}\equiv {\text{C}}^{-}{\text{Na}}^{+}$. Acetylene lose one of its proton on reaction with strong base like $\text{NaH}$.

$\text{HC}\equiv \text{CH}+\text{NaH}\to \text{HC}\equiv {\text{C}}^{-}{\text{Na}}^{+}+{\text{H}}_{\text{2}}$

Therefore, compound A is $\text{HC}\equiv {\text{C}}^{-}{\text{Na}}^{+}$.

The compound A acts like a nucleophile and attack on the electron deficient carbon atom of ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}\text{Br}$ at which $\text{Br}$ is present. It is expressed as,

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}\text{Br}+\text{HC}\equiv {\text{C}}^{-}{\text{Na}}^{+}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv \text{CH}$

Therefore, compound B is ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv \text{CH}$.

Compound B on reaction with strong base $\text{NaH}$ loss its acidic proton and becomes negatively charged.

${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv \text{CH}+\text{NaH}\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}+{\text{H}}_2$

Therefore, compound C is ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$.

The compound C acts like a nucleophile and attack on the electron deficient carbon atom of ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{11}}{\text{CH}}_{\text{2}}\text{Br}$ at which $\text{Br}$ is present. It is expressed as,

$\left(\begin{array}{l}{\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}\text{Br}+\\ {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{11}}{\text{CH}}_{\text{2}}-\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}\end{array}\right)\to {\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv \text{C}-{\text{CH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{11}}{\text{CH}}_{\text{3}}$

Therefore, compound D is ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv \text{C}-{\text{CH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{11}}{\text{CH}}_{\text{3}}$.

Conclusion

Compound A, B, C and D is $\text{HC}\equiv {\text{C}}^{-}{\text{Na}}^{+}$, ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv \text{CH}$, ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv {\text{C}}^{-}{\text{Na}}^{+}$ and ${\text{CH}}_{\text{3}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{7}}{\text{CH}}_{\text{2}}-\text{C}\equiv \text{C}-{\text{CH}}_{\text{2}}{\left({\text{CH}}_{\text{2}}\right)}_{\text{11}}{\text{CH}}_{\text{3}}$, respectively.