Question

An alkyl halide $(RX)$ reacts with Na to form $4, 5$-diethyloctane. Compound $RX$ is

• A. $ C{{H}_{3}}{{(C{{H}_{2}})}_{3}}Br $
• B. $ C{{H}_{3}}{{(C{{H}_{2}})}_{2}}CH(Br)C{{H}_{2}}C{{H}_{3}} $
• C. $ C{{H}_{3}}{{(C{{H}_{2}})}_{3}}CH(Br)C{{H}_{3}} $
• D. $ C{{H}_{3}}{{(C{{H}_{2}})}_{5}}Br $
• E. $ C{{H}_{3}}Br $

Answer 1

Answer: (B)

Since, the alkyl halide RX gives 4, 5-diethyloctane, when reacts with Na, it must be
$ C{{H}_{3}}{{(C{{H}_{2}})}_{2}}CH(Br)C{{H}_{2}}C{{H}_{3}}. $ $ 2C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ C{{H}_{2}}C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-Br\xrightarrow[Dry\,ether]{Na} $ $ C{{H}_{3}}{{(C{{H}_{2}})}_{2}}\overset{\begin{smallmatrix} C{{H}_{2}}C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,-\overset{\begin{smallmatrix} \,\,\,\,\,\,C{{H}_{2}}C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{CH}}\,-{{(C{{H}_{2}})}_{2}}\,-C{{H}_{3}} $
This reaction is known as Wurtz reaction.