Identify B formed in the reaction.
\[ \text{Cl} - \text{(CH}_2\text{)}_4 - \text{Cl} \xrightarrow{\text{excess NH}_3} A \xrightarrow{\text{NaOH}} B + \text{H}_2\text{O} + \text{NaCl} \]
\[ \text{Cl} - \text{(CH}_2\text{)}_4 - \text{Cl} \xrightarrow{\text{excess NH}_3} A \xrightarrow{\text{NaOH}} B + \text{H}_2\text{O} + \text{NaCl} \]
- \[ \text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2 \]
- \[ \text{ClNH}_3 - \text{(CH}_2\text{)}_4 - \text{NH}_3\text{Cl}^- \]
The Correct Option is B
Approach Solution - 1
To identify compound B formed in the given reaction, let's follow the step-by-step process:
- We start with the compound \(\text{Cl} - \text{(CH}_2\text{)}_4 - \text{Cl}\), which is 1,4-dichlorobutane.
- When 1,4-dichlorobutane is treated with excess ammonia (\(\text{NH}_3\)), a nucleophilic substitution reaction occurs. In this reaction, the chlorine atoms in 1,4-dichlorobutane are replaced by amine groups. As a result, we form a diamine compound \(A: \text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2\), which is 1,4-butanediamine.
- Next, the compound A is treated with sodium hydroxide (\(\text{NaOH}\)). However, as the amine groups don't undergo any further transformation in the presence of NaOH, the structure of 1,4-butanediamine remains unchanged in this step.
Thus, compound B is \(\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2\), which is 1,4-butanediamine. This is consistent with the initially provided correct answer option:
The other options can be ruled out because they represent different compounds or partial reactions that do not account for the complete process given in the question, particularly the full substitution of chlorine with amine groups. Hence, option \(\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2\) is the correct structure for compound B.
Approach Solution -2
The compound \( \text{Cl-(CH}_2\text{)}_4-\text{Cl} \) reacts with excess ammonia (\( \text{NH}_3 \)) to form an intermediate \( \text{A} \), which is \( \text{NH}_3^+-(\text{CH}_2)_4-\text{NH}_3^+ \cdot 2\text{Cl}^- \). This intermediate compound is a diammonium salt. Upon treatment with \( \text{NaOH} \), it undergoes deprotonation to yield compound \( \text{B} \), which is \( \text{H}_2\text{N}-(\text{CH}_2)_4-\text{NH}_2 \), also known as 1,4-diaminobutane or putrescine. Therefore, the correct answer is \( \text{H}_2\text{N}-(\text{CH}_2)_4-\text{NH}_2 \).
The Correct answer is: \(\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2\)
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