An alkyl bromide (A) forms a Grignard reagent | vedclass

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(C) $1$. The alkyl bromide $(A)$ forms $n$-hexane $(C_6H_{14})$ upon treatment with water after forming a Grignard reagent. This implies $(A)$ is a $6

Vedclass · Accepted Answer

$CH_3CH_2CH_2CH(Br)CH_2CH_3$

Vedclass · Answer

(C) $1$. The alkyl bromide $(A)$ forms $n$-hexane $(C_6H_{14})$ upon treatment with water after forming a Grignard reagent. This implies $(A)$ is a $6$-carbon chain alkyl bromide,$C_6H_{13}Br$. $2$. Wurtz reaction of $(A)$ with sodium in dry ether yields $4,5$-diethyloctane. The structure of $4,5$-diethyloctane is $CH_3CH_2CH_2CH(C_2H_5)CH(C_2H_5)CH_2CH_2CH_3$. $3$. In the Wurtz reaction,$2R-Br + 2Na \rightarrow R-R + 2NaBr$. The product $4,5$-diethyloctane is formed by coupling two $3$-hexyl radicals. $4$. Therefore,$(A)$ is $3$-bromohexane,which is $CH_3CH_2CH_2CH(Br)CH_2CH_3$.

An alkyl bromide $(A)$ forms a Grignard reagent which on treatment with water yields $n$-hexane. When $(A)$ is treated with sodium in dry ether,$4,5$-diethyloctane is formed. The structure of $(A)$ is:

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