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Chemistry Question on Organic Chemistry

Identify B formed in the reaction.
Cl−(CH2)4−Cl→excess NH3A→NaOHB+H2O+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}

A

Fig 1

B

H2N−(CH2)4−NH2\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2

C

ClNH3−(CH2)4−NH3Cl−\text{ClNH}_3 - \text{(CH}_2\text{)}_4 - \text{NH}_3\text{Cl}^-

D

Fig 2

Answer

H2N−(CH2)4−NH2\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2

Explanation

Solution

The compound Cl-(CH2)4−Cl\text{Cl-(CH}_2\text{)}_4-\text{Cl} reacts with excess ammonia (NH3\text{NH}_3) to form an intermediate A\text{A}, which is NH3+−(CH2)4−NH3+⋅2Cl−\text{NH}_3^+-(\text{CH}_2)_4-\text{NH}_3^+ \cdot 2\text{Cl}^-. This intermediate compound is a diammonium salt. Upon treatment with NaOH\text{NaOH}, it undergoes deprotonation to yield compound B\text{B}, which is H2N−(CH2)4−NH2\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 H2N−(CH2)4−NH2\text{H}_2\text{N}-(\text{CH}_2)_4-\text{NH}_2.
The Correct answer is: H2N−(CH2)4−NH2\text{H}_2\text{N} - \text{(CH}_2\text{)}_4 - \text{NH}_2