Consider the following sequence of reactions and find B

Correct Answer: H₂N− (CH₂)₄ −NH₂ (1,4-diaminobutane / putrescine)

To determine the product B, we can analyze the reaction sequence step-by-step from the provided images:

1. Analysis of the Image Data:
- The first image displays the reaction sequence:
$\text{Cl}-({\text{CH}}_2{)}_4-\text{Cl}\stackrel{{\text{Excess NH}}_3}{\to }\text{A}\stackrel{\text{NaOH}}{\to }\text{B}+{\text{H}}_2\text{O}+\text{NaCl}$
- The second image shows the structure of the dication:
${{\text{H}}_3\text{N}}^{+}-({\text{CH}}_2{)}_4-{{\text{H}}_3\text{N}}^{+}$
which corresponds to the protonated intermediate salt A (along with two chloride counter-ions, $2{\text{Cl}}^{-}$).

2. Step-by-Step Reaction Mechanism:

Step 1: Nucleophilic Substitution with Excess Ammonia
The starting material is 1,4-dichlorobutane, $\text{Cl}-({\text{CH}}_2{)}_4-\text{Cl}$. When it is treated with excess ammonia (${\text{NH}}_3$), ammonia acts as a nucleophile and undergoes nucleophilic substitution (${\text{S}}_{\text{N}}2$ mechanism) at both chlorine-bearing carbon atoms. Since ammonia is basic, the hydrochloric acid ($\text{HCl}$) produced during substitution protonates the newly formed amine groups. This yields the dihydrochloride salt of 1,4-diaminobutane, which is compound A:
$[{{\text{H}}_3\text{N}}^{+}-({\text{CH}}_2{)}_4-{{\text{NH}}_3}^{+}]2{\text{Cl}}^{-}$

Step 2: Deprotonation with Strong Base (NaOH)
When intermediate salt A is treated with sodium hydroxide ($\text{NaOH}$), a strong base, it undergoes an acid-base neutralization reaction. The hydroxide ions (${\text{OH}}^{-}$) deprotonate the ammonium groups ($-{{\text{NH}}_3}^{+}$), converting them back to free neutral amine groups ($-{\text{NH}}_2$). This releases the free diamine, 1,4-diaminobutane (compound B), alongside water and sodium chloride:
$[{{\text{H}}_3\text{N}}^{+}-({\text{CH}}_2{)}_4-{{\text{NH}}_3}^{+}]2{\text{Cl}}^{-}+2\text{NaOH}\to {\text{H}}_2\text{N}-({\text{CH}}_2{)}_4-{\text{NH}}_2+2{\text{H}}_2\text{O}+2\text{NaCl}$

Thus, compound B is H₂N− (CH₂)₄ −NH₂.