The products A and B in the following reactio | vedclass

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Question

(B) The reaction sequence is as follows:$1$. Styrene $(Ph-CH=CH_2)$ reacts with $HBr$ to form $1-$bromo$-1-$phenylethane $(Ph-CH(Br)-CH_3)$ via Markov

Vedclass · Accepted Answer

$A$ = $2-$phenylpropanoic acid; $B$ = $N$-methyl$-2-$phenylpropanamide

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(B) The reaction sequence is as follows:$1$. Styrene $(Ph-CH=CH_2)$ reacts with $HBr$ to form $1-$bromo$-1-$phenylethane $(Ph-CH(Br)-CH_3)$ via Markovnikov addition.$2$. This alkyl bromide reacts with $Mg$ in dry ether to form the Grignard reagent $(Ph-CH(MgBr)-CH_3)$.$3$. The Grignard reagent reacts with $CO_2$ followed by acidic workup $(H_3O^+)$ to form $2-$phenylpropanoic acid $(Ph-CH(CH_3)-COOH)$,which is product $A$.$4$. $2-$phenylpropanoic acid $(A)$ reacts with $SOCl_2$ to form the corresponding acid chloride $(Ph-CH(CH_3)-COCl)$.$5$. This acid chloride then reacts with methylamine $(CH_3NH_2)$ to form the amide,$N$-methyl$-2-$phenylpropanamide $(Ph-CH(CH_3)-CONHCH_3)$,which is product $B$.

The products $A$ and $B$ in the following reaction sequence are:

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