Products can be

Engineering

Chemistry

Nucleophilic Substitution

Preparation of Alkyl Halide

Free radical

Question

${\overset{14}{CH}}_{2} = CH - C H_{3} \underset{{CCl}_{4}, Δ}{\overset{NBS}{\to}} \underset{dry ether, Δ}{\overset{Na}{\to}} Products .$
Products can be

CH₂ = CH – CH₂–CH₂–CH = CH₂

${\overset{14}{CH}}_{2} = CH - {CH}_{2} - {CH}_{2} - CH = {\overset{14}{CH}}_{2}$

${CH}_{2} = CH - {\overset{14}{CH}}_{2} - {\overset{14}{CH}}_{2} - CH = {CH}_{2}$

${CH}_{2} = CH - {\overset{14}{CH}}_{2} - {CH}_{2} - CH = {\overset{14}{CH}}_{2}$

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This question involves the Wurtz reaction and allylic bromination. The reactant is propene (CH₃-CH=CH₂). NBS/Δ performs allylic bromination, giving 3-bromopropene (CH₂=CH-CH₂Br). This alkyl halide then undergoes the Wurtz reaction with Na/dry ether, where two molecules couple to form a symmetrical product.

The Wurtz reaction: $2 R - X \overset{Na, dry ether}{\to} R - R$

For 3-bromopropene (R = CH₂-CH=CH₂), the symmetrical coupled product is CH₂=CH-CH₂-CH₂-CH=CH₂ (1,5-hexadiene).

Final Answer: CH₂ = CH – CH₂–CH₂–CH = CH₂