Consider the following sequence of reactions $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2} \xrightarrow[700…
$\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2} \xrightarrow[700 \mathrm{~K}]{\mathrm{Cl}_{2}} \mathrm{~A} \xrightarrow[420 \mathrm{~K}, 12 \mathrm{~atm}]{\mathrm{Na}_{2} \mathrm{CO}_{3}} \mathrm{B}$
Compound 'B' is
- $\mathrm{HOCH}_{2}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}$
Asked in: JEE-TOPICTESTS-CHEMISTRY