The compound, which evolves carbon dioxide on treatment with aqueous solution of sodium bicarbonate at…
- $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$
- $\mathrm{CH}_{3} \mathrm{COCl}$
- $\mathrm{CH}_{3} \mathrm{CONH}_{2}$
- $\mathrm{CH}_{3} \mathrm{OOOC}_{2} \mathrm{H}_{5}$
Solution
even at $25^{\circ} \mathrm{C},$ which subsequently reacts with $NaHCO_{3}$ (sodium bicarbonate) present in the same medium to form carbon dioxide $\left(\mathrm{CO}_{2}ight)$.
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. ,
Asked in: JEE-TOPICTESTS-CHEMISTRY
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