Identify the final product formed on ammonoly | vedclass

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(D) Step $1$: Ammonolysis of benzyl chloride $(C_6 H_5 CH_2 Cl)$ with alcoholic $NH_3$ yields benzylamine $(C_6 H_5 CH_2 NH_2)$.$C_6 H_5 CH_2 Cl + NH_

Vedclass · Accepted Answer

$C_6 H_5 CH_2 N(CH_3)_2$

Vedclass · Answer

(D) Step $1$: Ammonolysis of benzyl chloride $(C_6 H_5 CH_2 Cl)$ with alcoholic $NH_3$ yields benzylamine $(C_6 H_5 CH_2 NH_2)$.$C_6 H_5 CH_2 Cl + NH_3 \rightarrow C_6 H_5 CH_2 NH_2 + HCl$Step $2$: Benzylamine reacts with two moles of methyl iodide $(CH_3 I)$ via nucleophilic substitution (alkylation) to form $N,N$-dimethylbenzylamine $(C_6 H_5 CH_2 N(CH_3)_2)$.$C_6 H_5 CH_2 NH_2 + 2CH_3 I \rightarrow C_6 H_5 CH_2 N(CH_3)_2 + 2HI$Thus,the final product is $C_6 H_5 CH_2 N(CH_3)_2$.

Identify the final product formed on ammonolysis of benzyl chloride followed by the reaction with two moles of $CH_3 I$.

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$CH_3-CH_2-NH_2 \xrightarrow{CHCl_3, KOH} \text{Product}$

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