Question

Identify B formed in the reaction.
\[ \text{Cl} - \text{(CH}_2\text{)}_4 - \text{Cl} \xrightarrow{\text{excess NH}_3} A \xrightarrow{\text{NaOH}} B + \text{H}_2\text{O} + \text{NaCl} \]

• A.

  

• B. \[ \text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2 \]
• C. \[ \text{ClNH}_3 - \text{(CH}_2\text{)}_4 - \text{NH}_3\text{Cl}^- \]
• D.

  

Answer 1

The Correct Option is B

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: \(\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2\)