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The product $D$ of the reaction
$\mathrm{CH}_3 \mathrm{Cl} \xrightarrow{\mathrm{KON}}(A) \xrightarrow{\mathrm{H}_2 \mathrm{O}}(B) \xrightarrow{\mathrm{NH}_3}(C) \xrightarrow{\Delta}(D)$ is
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$\mathrm{CH}_3 \mathrm{Cl} \xrightarrow{\mathrm{KON}}(A) \xrightarrow{\mathrm{H}_2 \mathrm{O}}(B) \xrightarrow{\mathrm{NH}_3}(C) \xrightarrow{\Delta}(D)$ is
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The correct answer is:
$\mathrm{CH}_3 \mathrm{CONH}_2$
$\mathrm{CH}_3 \mathrm{Cl} \xrightarrow[-\mathrm{KCl}]{\mathrm{KCN}} \mathrm{CH}_3 \mathrm{CN} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{NH}_3} \mathrm{CH}_3 \mathrm{COONH}_4 \xrightarrow{\Delta} \mathrm{CH}_3 \mathrm{CONH}_2$
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