In the following reaction, A and B, respectively are $A\xrightarrow{{HBr}}{{\rm{C}}_2}{{\rm{H}}_5}{\rm{Br}}\xleftarrow{B}A$

${{\rm{C}}_2}{{\rm{H}}_4},{\rm{alc}}{\rm{.KOH/}}\Delta $

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{Cl}},{\rm{aq}}{\rm{.KOH/}}\Delta $

${\rm{C}}{{\rm{H}}_3}{\rm{OH}},{\rm{aq}}{\rm{.KOH/}}\Delta $

${{\rm{C}}_2}{{\rm{H}}_5},{\rm{PB}}{{\rm{r}}_3}$

NEET UG CHEMISTRY MCQ: HYDROCARBONS

1 Easy

1 of 1

Application +4 | -1

In the following reaction, A and B, respectively are

A\xrightarrow{{HBr}}{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{Br}\xleftarrow{B}A

View answer and explanation

Correct option: A

{\mathrm{C}_2}{\mathrm{H}_4},\mathrm{alc}\mathrm{.KOH/}\Delta

Option A (

C_2H_4

, alc.KOH/

\Delta

) is correct because ethene reacts with HBr to form ethyl bromide, and ethyl bromide reacts with alcoholic KOH to form ethene.