Maximum number of molecules of CH_3I that can | vedclass

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(A) $CH_3NH_2$ reacts with $CH_3I$ through a process called exhaustive alkylation (Hofmann's alkylation). The reaction proceeds as follows: $1. CH_3NH

Vedclass · Accepted Answer

$3$

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(A) $CH_3NH_2$ reacts with $CH_3I$ through a process called exhaustive alkylation (Hofmann's alkylation). The reaction proceeds as follows: $1. CH_3NH_2 + CH_3I \rightarrow (CH_3)_2NH + HI$ (Secondary amine) $2. (CH_3)_2NH + CH_3I \rightarrow (CH_3)_3N + HI$ (Tertiary amine) $3. (CH_3)_3N + CH_3I \rightarrow (CH_3)_4N^+I^-$ (Quaternary ammonium salt) Thus,a total of $3$ molecules of $CH_3I$ react with one molecule of $CH_3NH_2$ to reach the final stable product.

Maximum number of molecules of $CH_3I$ that can react with a molecule of $CH_3NH_2$ are

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