Identify the reagents S and mu in the scheme | vedclass

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Question

(B) The reaction sequence is a Friedel-Crafts acylation followed by a nucleophilic substitution ($S_N2$ reaction). $1$. The starting material $R$ is b

Vedclass · Accepted Answer

$C_6H_6, 	ext{m-BrCH}_2C_6H_4COCl (AlCl_3), KCN$

Vedclass · Answer

(B) The reaction sequence is a Friedel-Crafts acylation followed by a nucleophilic substitution ($S_N2$ reaction). $1$. The starting material $R$ is benzene $(C_6H_6)$. $2$. The reagent $S$ is $m$-bromomethylbenzoyl chloride $(m-BrCH_2C_6H_4COCl)$ in the presence of $AlCl_3$ (a Lewis acid catalyst). This performs a Friedel-Crafts acylation to form the intermediate benzylic halide. $3$. The reagent $\mu$ is $KCN$,which acts as a nucleophile to replace the bromine atom in the benzylic position via an $S_N2$ mechanism to form the nitrile product. Therefore,$R = C_6H_6$,$S = m-BrCH_2C_6H_4COCl (AlCl_3)$,and $\mu = KCN$.

List-$I$ (Reagents)	List-$II$ (Reaction Name)
$A$. $NH_2-NH_2, KOH$	$I$. Tollen's Test
$B$. $[Ag(NH_3)_2]OH$	$II$. Clemmensen Reduction
$C$. Aq. $CuSO_4$, Sodium Potassium tartarate, $KOH$	$III$. Wolff-Kishner Reduction
$D$. $Zn-Hg, HCl$	$IV$. Fehling's Test

Identify the reagents $S$ and $\mu$ in the scheme below in which $R$ is converted to the nitrile via the benzylic halide. $R, S$ and $\mu$ respectively are:

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