Q4 / 155 · WBJEE 2019 · Chemistry
The compound, which evolves carbon dioxide on treatment with aqueous solution of sodium bicarbonate at $25^{\circ} \mathrm{C},$ is
- A$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
- Correct answerB$\mathrm{CH}_{3} \mathrm{COCl}$
- C$\mathrm{CH}_{3} \mathrm{CONH}_{2}$
- D$\mathrm{CH}_{3} \mathrm{OOOC}_{2} \mathrm{H}_{5}$
Solution
$\mathrm{CH}_{3}-\mathrm{C}O-\mathrm{Cl}$ hydrolyses to form $\mathrm{CH}_{3} \mathrm{COOH}$
even at $25^{\circ} \mathrm{C},$ which subsequently reacts with $NaHCO_{3}$ (sodium bicarbonate) present in the same medium to form carbon dioxide $\left(\mathrm{CO}_{2}\right)$.
The reaction involved is given below. $\mathrm{CH}_{3} \mathrm{COCl}+\mathrm{NaHCO}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{HCl}+\mathrm{CO}_{2}$
Thus, the option (b) is correct answer.
Hence, the correct answer is option: 'B'.