The product (B) of the reaction sequence is:C | vedclass

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(A) Step $1$: The reaction of $4-chlorobenzyl bromide$ with $Mg$ in $dry$ $ether$ $(Et_2O)$ forms the Grignard reagent,$4-chlorobenzylmagnesium$ $brom

Vedclass · Accepted Answer

$Cl-C_6H_4-CH_2CH(OH)CH_3$

Vedclass · Answer

(A) Step $1$: The reaction of $4-chlorobenzyl bromide$ with $Mg$ in $dry$ $ether$ $(Et_2O)$ forms the Grignard reagent,$4-chlorobenzylmagnesium$ $bromide$ $(A)$,which is $Cl-C_6H_4-CH_2MgBr$.Step $2$: The Grignard reagent $(A)$ reacts with acetaldehyde $(CH_3CHO)$ followed by acid hydrolysis $(H_3O^+)$ to form a secondary alcohol.The nucleophilic $CH_2-C_6H_4-Cl$ group attacks the carbonyl carbon of $CH_3CHO$ to form $Cl-C_6H_4-CH_2-CH(O^-)-CH_3$,which upon protonation gives $Cl-C_6H_4-CH_2-CH(OH)-CH_3$ as product $(B)$.

The product $(B)$ of the reaction sequence is:
$Cl-C_6H_4-CH_2Br$ $\xrightarrow{Mg, Et_2O} A$ $\xrightarrow[2. H_3O^+]{1. CH_3CHO} B$

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The product $(B)$ of the reaction sequence is:$Cl-C_6H_4-CH_2Br$ $\xrightarrow{Mg, Et_2O} A$ $\xrightarrow[2. H_3O^+]{1. CH_3CHO} B$