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