Q17 / 160 · AP EAMCET 2019 (22 Apr Shift 1) · Chemistry
Wurtz reaction of bromoethane gives \(n\)-butane. Sodium salt of \(X\) on heating with sodalime also results in \(n\)-butane. Compound \(X\) is
- A\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}\)
- Correct answerB\(\mathrm{CH}_3\left(\mathrm{CH}_2\right)_3 \mathrm{COOH}\)
- C\(\mathrm{CH}_3\left(\mathrm{CH}_2\right)_4 \mathrm{COOH}\)
- D\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\)
Solution
In Wurtz reaction, two molecules of bromoethane react with 2 molecules of sodium to form \(n\)-butane.
\(\begin{aligned}
\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Br}+2 \mathrm{Na}+\mathrm{Br}-\mathrm{CH}_2-\mathrm{CH}_3 \\
-2 \mathrm{NaBr} \longrightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3
\end{aligned}\)
\(n\)-butane can also be obtain when sodium salt of pentanoic acid \((X)\) gets heated with sodalime. The reaction is given as under :
\(\underset{(X)}{\mathrm{CH}_3\left(\mathrm{CH}_2\right)_3 \mathrm{COOH}} \xrightarrow{\mathrm{NaOH}}\)
\(\underset{\text { Sodium salt }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COONa}} \xrightarrow{\mathrm{NaOH}+\mathrm{CaO} \text { (Sodalime) }} \underset{\text{n-butane}}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3}+\mathrm{Na}_2 \mathrm{CO}_3\)
Hence, \(X\) is \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}\).
Hence, the correct answer is option: 'B'.