When methyl iodide is heated with ammonia,the | vedclass

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Question

(D) The reaction of $methyl$ $iodide$ $(CH_3I)$ with $ammonia$ $(NH_3)$ is a nucleophilic substitution reaction known as $Hofmann$ $ammonolysis$.Since

Vedclass · Accepted Answer

$A$ mixture of the above three amines

Vedclass · Answer

(D) The reaction of $methyl$ $iodide$ $(CH_3I)$ with $ammonia$ $(NH_3)$ is a nucleophilic substitution reaction known as $Hofmann$ $ammonolysis$.Since the primary amine $(CH_3NH_2)$ formed is more nucleophilic than $ammonia$,it further reacts with $CH_3I$ to form secondary and tertiary amines.Therefore,the reaction proceeds as follows:$CH_3I + NH_3 \rightarrow CH_3NH_2 + HI$$CH_3NH_2 + CH_3I \rightarrow (CH_3)_2NH + HI$$(CH_3)_2NH + CH_3I \rightarrow (CH_3)_3N + HI$Thus,a mixture of primary,secondary,and tertiary amines is obtained.

List-$I$ (Amine)	List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline	$I$. $9.30$
$B$. Aniline	$II$. $8.92$
$C$. $N$-Ethylethanamine	$III$. $9.38$
$D$. $N$-Methylaniline	$IV$. $3.00$

When $methyl$ $iodide$ is heated with $ammonia$,the product obtained is

Similar Questions

Which of the following reactions does $NOT$ yield an amine?

The product $(A)$ of the given reaction is:

Match the following:
List-$I$ (Amine) List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline $I$. $9.30$
$B$. Aniline $II$. $8.92$
$C$. $N$-Ethylethanamine $III$. $9.38$
$D$. $N$-Methylaniline $IV$. $3.00$

The correct answer is

$9.3 \ g$ of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume $100\%$ yield/conversion) is . . . . . $g$. (nearest integer)

Identify the product $B$ in the following reaction:
$C_6H_5NH_2$ $\xrightarrow[0^\circ-5^\circ C]{NaNO_2/HCl} A$ $\xrightarrow{Ph-N(CH_3)_2} B$

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