Question

Show the terminal alkyne and alkyl halide from which the following products can be obtained. If two routes look feasible, list both.

$\mathbf{a}\mathbf{.}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{C}\mathbf{\equiv }\mathbf{CH}$

Short answer

Answer

Method

1. $\mathrm{a}.{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{C}\equiv \mathrm{CH}+{\mathrm{CH}}_3\mathrm{Br}\to {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{C}\equiv {\mathrm{CCH}}_3$

2.$\mathrm{a}.{\mathrm{CH}}_3\mathrm{C}\equiv \mathrm{CH}+{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}\to {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{C}\equiv {\mathrm{CCH}}_3$

Step-by-step solution

Alkylation

Migration of the alkyl group from one molecule to another via carbanion or carbocation formation

Alkylation of acetylide ion

Acetylide ion is a strong nucleophile that reacts with the electrophile like alkyl halide that replaces the halide and forms alkyne. Hence new C-C bond is formed, and the old C-Br bond is broken.

Illustration of the second mechanism

The attack of alkyne acts as a nucleophile and leads to the formation of a transition state, and the removal of halogen takes place.