To determine compound B formed in the reaction, we proceed as follows:
- Starting material is 1,4-dichlorobutane, represented as \(\text{Cl} - \text{(CH}_2\text{)}_4 - \text{Cl}\).
- Reaction of 1,4-dichlorobutane with excess ammonia (\(\text{NH}_3\)) yields compound A, 1,4-butanediamine (\(A: \text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2\)), via nucleophilic substitution where chlorine atoms are replaced by amine groups.
- Treatment of compound A with sodium hydroxide (\(\text{NaOH}\)) results in no structural change to the 1,4-butanediamine, as amine groups are unreactive under these conditions.
Therefore, compound B is 1,4-butanediamine (\(\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2\)). This aligns with the provided correct answer option.
Alternative options are incorrect as they represent different chemical species or incomplete reaction products, failing to account for the complete substitution of chlorine by amine groups. Thus, \(\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2\) is the correct structure for compound B.