Question

A compound is treated with $NaNH_2$ to give sodium salt. Identify the compound.

• A. $C_2 H_2$
• B. $C_6 H_6$
• C. $C_2 H_6$
• D. $C_2 H_4$

Answer 1

The Correct Option is A

 To identify the compound that forms a sodium salt when treated with \(NaNH_2\), we need to understand the behavior of the given compounds in the presence of \(NaNH_2\).

Concept: \(NaNH_2\), known as sodamide, is a strong base used to deprotonate acidic hydrogen atoms in organic chemistry.

1. Acidity of Hydrogen: To form a sodium salt, the compound must have an acidic hydrogen atom. Among the options provided, the terminal hydrogen in acetylene \((C_2H_2)\) is acidic enough to be removed by \(NaNH_2\).
2. Reaction: In the presence of \(NaNH_2\), acetylene \((C_2H_2)\) reacts to form sodium acetylide \((C_2HNa)\) by deprotonation of the terminal hydrogen:

\[C_2H_2 + NaNH_2 \rightarrow C_2HNa + NH_3\]

Acetylene is the only compound in the provided options that has hydrogen atoms acidic enough to be deprotonated by \(NaNH_2\). Thus, it reacts to form a sodium salt.

1. Elimination of Other Options:
   • \(C_6H_6\) (Benzene) does not have any acidic hydrogens. Its hydrogens are not deprotonated by \(NaNH_2\).
   • \(C_2H_6\) (Ethane) contains only saturated C-H bonds that are not acidic.
   • \(C_2H_4\) (Ethylene) also has non-acidic hydrogens.

Thus, the compound that forms a sodium salt when treated with \(NaNH_2\) is \(C_2H_2\).