The correct sequence of reagents for the conv | vedclass

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Question

(D) The conversion of $1-	ext{phenylethyl bromide}$ $(X)$ to $2-	ext{phenylacetic acid}$ $(Y)$ involves the following steps:$1$. Dehydrohalogenation

Vedclass · Accepted Answer

$(i) NaOEt \ (ii) B_{2}H_{6}/H_{2}O_{2} \ (iii) 	ext{Jones reagent}$

Vedclass · Answer

(D) The conversion of $1-	ext{phenylethyl bromide}$ $(X)$ to $2-	ext{phenylacetic acid}$ $(Y)$ involves the following steps:$1$. Dehydrohalogenation: Treatment of $1-	ext{phenylethyl bromide}$ with $NaOEt$ (a strong base) leads to the formation of $styrene$ $(Ph-CH=CH_{2})$ via an $E2$ elimination mechanism.$2$. Hydroboration-Oxidation: $Styrene$ reacts with $B_{2}H_{6}$ followed by $H_{2}O_{2}/OH^{-}$ to yield $2-	ext{phenylethanol}$ $(Ph-CH_{2}-CH_{2}-OH)$ via anti-Markovnikov addition of water.$3$. Oxidation: Finally, $2-	ext{phenylethanol}$ is oxidized to $2-	ext{phenylacetic acid}$ $(Ph-CH_{2}-COOH)$ using a strong oxidizing agent like Jones reagent $(CrO_{3}/H_{2}SO_{4})$.Thus, the correct sequence is $(i) NaOEt, (ii) B_{2}H_{6}/H_{2}O_{2}, (iii) 	ext{Jones reagent}$.

The correct sequence of reagents for the conversion of $X$ to $Y$ is :

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