Contrast the reduction of propanamide with against treating it with Bromine and aqueous . What are t — Amines Chemistry Question

💡 Solution & Explanation

$LiAlH_4$ directly converts the $-C=O$ group to $-CH_2-$, retaining all carbons (yields Propan-1-amine). $Br_2/NaOH$ is the Hofmann bromamide degradation, which excises the carbonyl carbon as carbonate, stepping down the chain (yields Ethanamine).