Contrast the reduction of propanamide with against treating it with Bromine and aqueous (Hofmann bromamide degradation). What is the major organic product of this specific Hofmann reaction?

Question

Contrast the reduction of propanamide with $LiAlH_4$ against treating it with Bromine and aqueous $NaOH$ (Hofmann bromamide degradation). What is the major organic product of this specific Hofmann reaction?

Prashant Kotian · Accepted Answer

Correct answer: B. The Hofmann bromamide degradation is a classical step-down reaction. The reagent combination () actively excises the carbonyl carbon of the primary amide as a carbonate ion. Thus, the 3-carbon propanamide loses exactly one carbon atom, yielding the 2-carbon ethanamine.

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