CH3Cl Alc.AgCN (A) [ ]LiAlH4 (B) , what would be the product B ? a) Methanamine b) Ethanamine c) N methyl methanamine d) N methyl ethanamine

The reaction proceeds as follows: 1. First step (A): CH3Cl reacts with alcoholic silver cyanide ( Alc.AgCN ) to form methyl cyanide (acetonitrile) as the major product. This is because CN acts as a nucleophile, attacking the carbon atom of CH3Cl via an S N2 mechanism, forming CH3CN (A). CH3Cl + AgCN CH3CN + AgCl 2. Second step (B): CH3CN (A) is then reduced by LiAlH4 (a strong reducing agent) under heating ( ). The nitrile group ( CN ) is converted into an amine group ( CH2NH2 ), yielding ethanamine (B). CH3CN + 4[H] CH3CH2NH2 Thus, the product B is ethanamine , which corresponds to option (b) .

CH3Cl Alc.AgCN (A) [ ]LiAlH4 (B) , what would be the product B ? a) Methanamine b)… – Chemistry NEB 2081 (MCQs) Solution

The reaction proceeds as follows:

First step (A): $CH X_{3} Cl$ reacts with alcoholic silver cyanide ( $Alc \cdot AgCN$ ) to form methyl cyanide (acetonitrile) as the major product. This is because $CN X^{-}$ acts as a nucleophile, attacking the carbon atom of $CH X_{3} Cl$ via an $S_{N} 2$ mechanism, forming $CH X_{3} CN$ (A). $CH X_{3} Cl + AgCN CH X_{3} CN + AgCl$
Second step (B): $CH X_{3} CN$ (A) is then reduced by $LiAlH X_{4}$ (a strong reducing agent) under heating ( $Δ$ ). The nitrile group ( $- CN$ ) is converted into an amine group ( $- CH X_{2} NH X_{2}$ ), yielding ethanamine (B). $CH X_{3} CN + 4 [H] CH X_{3} CH X_{2} NH X_{2}$

44[H]LiAlH₄, Δethylamineethylamine

Reduction of acetonitrile (A) to ethanamine (B) using LiAlH4 under heating

Thus, the product B is ethanamine, which corresponds to option (b).

CH3Cl Alc.AgCN (A) [ ]LiAlH4 (B) , what would be the product B ? a) Methanamine b) Ethanamine c) N methyl methanamine d) N methyl ethanamine

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