Identify B formed in the reaction: Cl-(CH_2)_ | vedclass

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(B) When $1,4$-dichlorobutane reacts with excess ammonia,it undergoes nucleophilic substitution to form the ammonium salt $A$ $([H_3N^{+}-(CH_2)_4-NH_

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$H_2N-(CH_2)_4-NH_2$

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(B) When $1,4$-dichlorobutane reacts with excess ammonia,it undergoes nucleophilic substitution to form the ammonium salt $A$ $([H_3N^{+}-(CH_2)_4-NH_3^{+}] 2Cl^{-})$. Treatment of this salt with a strong base like $NaOH$ releases the free diamine $B$ $(H_2N-(CH_2)_4-NH_2)$,along with water and sodium chloride. The reaction is: $Cl-(CH_2)_4-Cl + 2NH_3 \rightarrow [H_3N^{+}-(CH_2)_4-NH_3^{+}] 2Cl^{-} (A)$ $[H_3N^{+}-(CH_2)_4-NH_3^{+}] 2Cl^{-} + 2NaOH \rightarrow H_2N-(CH_2)_4-NH_2 (B) + 2H_2O + 2NaCl$

Identify $B$ formed in the reaction: $Cl-(CH_2)_4-Cl$ $\xrightarrow{\text{excess } NH_3} A$ $\xrightarrow{NaOH} B + H_2O + NaCl$

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