JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Compounds Containing Nitrogen

$Cl\text{-}(C{H}_2{)}_4\text{-}Cl$ treated with excess ammonia undergoes two successive nucleophilic substitutions; using ammonia in large excess suppresses intramolecular cyclisation and favours the intermolecular pathway, giving the diammonium salt $C{l}^{-}\stackrel{+}{N}{H}_3\text{-}(C{H}_2{)}_4\text{-}\stackrel{+}{N}{H}_{3}C{l}^{-}$ as intermediate A.

Treating A with NaOH deprotonates both ammonium centres, liberating the free diamine and the by-products $NaCl$ and $H_{2}O$: $A+2NaOH\to H_{2}N\text{-}(C{H}_2{)}_4\text{-}N{H}_2+2NaCl+2H_{2}O$.

So B is putrescine, $H_{2}N\text{-}(C{H}_2{)}_4\text{-}N{H}_2$.