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MHTCET 16 May 2024 Shift 2 Paper

Question : 83 of 150

Marks: +1, -0

Which of the following conversions is Hofmann Elimination reaction?

$\mathrm{RCONH}_{2} \xrightarrow{\mathrm{Br}_{2}, \mathrm{KOH}_{(\mathrm{aq})}} \mathrm{R}-\mathrm{NH}_{2}$
$\mathrm{R}-\mathrm{NH}_{2} \xrightarrow[\text { excess }]{\mathrm{R}-\mathrm{X}} \mathrm{R}_{4} \mathrm{NX}$
$\mathrm{R}_{4} \mathrm{NX} \xrightarrow[\Delta]{\text { mois } \mathrm{Ag}_{2} \mathrm{O}}$ Alkene $+\mathrm{R}_{3} \quad \mathrm{~N}$
$\mathrm{RCN} \xrightarrow{\mathrm{Na}_{2} / C_{2} \mathrm{H}_{5} \mathrm{OH}} \mathrm{RCH}_{2} \mathrm{NH}_{2}$

Solution:

Concept:

Hofmann elimination converts a quaternary ammonium salt into an alkene using moist silver oxide and heat.

Explanation:

Hofmann elimination is an E2 elimination reaction.

A quaternary ammonium salt,

\mathrm{R}_4\mathrm{NX}

, reacts with moist

\mathrm{Ag}_2\mathrm{O}

and heat.

The products are an alkene and a tertiary amine,

\mathrm{R}_3\mathrm{N}

This reaction favours the less substituted alkene, called the Hofmann product.

Option A is Hofmann bromamide degradation, which forms

\mathrm{RNH}_2

from an amide.

Option B is exhaustive methylation, forming a quaternary ammonium salt from a primary amine.

Option D is reduction of a nitrile to a primary amine using

\mathrm{Na}/\mathrm{C}_2\mathrm{H}_5\mathrm{OH}

Only option C matches the Hofmann elimination conditions and products.

Answer:

Option C:

\mathrm{R}_4\mathrm{NX} \xrightarrow[\Delta]{\text{moist }\mathrm{Ag}_2\mathrm{O}} \text{Alkene} + \mathrm{R}_3\mathrm{N}

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