Consider the following sequence of reactions and find B

The correct answer is H₂N− (CH₂)₄ −NH₂.

Let us break down the reaction sequence step-by-step:

Step 1: Nucleophilic Substitution with Excess Ammonia
In the first step, 1,4-dichlorobutane, ${\text{Cl−(CH}}_2{\text{)}}_4\text{−Cl}$, reacts with excess ammonia (${\text{NH}}_3$). Ammonia acts as a nucleophile and undergoes nucleophilic substitution (${\text{S}}_{\text{N}}2$) at both carbon atoms holding the chlorine leaving groups. Since ammonia is present in excess, the reaction proceeds to form the corresponding diammonium salt, which is intermediate A:
${\text{Cl−(CH}}_2{\text{)}}_4\text{−Cl}+2{\text{NH}}_3\to {\text{[H}}^3{\text{N}}^{+}{\text{−(CH}}_2{\text{)}}_4{\text{−NH}}^3{\text{]}}^{2+}\cdot 2{\text{Cl}}^{-}$

This intermediate A is the diammonium chloride salt shown in the attached image as ${\text{H}}_3{\text{N}}^{+}{\text{−(CH}}_2{\text{)}}_4{\text{−NH}}^3{}^{+}$.

Step 2: Neutralization with Sodium Hydroxide (NaOH)
In the second step, the acidic diammonium salt A is treated with a strong base, sodium hydroxide ($\text{NaOH}$). The hydroxide ions (${\text{OH}}^{-}$) deprotonate the ammonium groups ($-{\text{NH}}^3{}^{+}$) to regenerate the free amine groups ($-{\text{NH}}_2$), producing water (${\text{H}}_2\text{O}$) and sodium chloride ($\text{NaCl}$):
${\text{[H}}^3{\text{N}}^{+}{\text{−(CH}}_2{\text{)}}_4{\text{−NH}}^3{\text{]}}^{2+}\cdot 2{\text{Cl}}^{-}+2\text{NaOH}\to {\text{H}}_2{\text{N−(CH}}_2{\text{)}}_4{\text{−NH}}_2+2{\text{H}}_2\text{O}+2\text{NaCl}$

Thus, the final product B is the free diamine, 1,4-butanediamine (also known as putrescine), which corresponds to the formula:
${\text{H}}_2{\text{N−(CH}}_2{\text{)}}_4{\text{−NH}}_2$