Consider the following reaction {\CH}_3- {\CH}= {\CH}_2 \stackrel{\text{ 1. }{──── ▸} {\HBr}} _{\text{ 2. aq. } {\KOH}} The major end product is

Correct Answer is : CH3−CH|OH−CH3

{\CH}_3- {\CH}_2- {\CH}_2- {\OH}

{\CH}_3- {\CH}_2- {\CH}_2- {\Br}

CBSE Class 12 Chemistry 2022 Term 1 Paper - Question 5

Question : 5 of 55

Marks: +1, -0

Consider the following reaction

{\CH}_3- {\CH}= {\CH}_2 {──── ▸}↖{\;\text " 1. "\; {\HBr}} ↙{\;\text " 2. aq. "\; {\KOH}}

The major end product is

Solution:

Explanation: Propene yields two products, however only one predominates as per Markovnikov's rule i.e., 2-bromopropane which on heating with aq.

{\KOH}

gives secondary alcohol. Aq.

{\KOH}

is alkaline in nature so it gives hydroxide ion which is a nucleophile. It replaces halide(bromide in this case) ion and form alcohols.

{{\CH}_3- {\CH}= {\CH}_2}↙{ \Propene}+ {\HBr}   {──── ▸}{{\CH}_3- {\CH}↖{|}↖{\Br}- {\CH}_3} ↙{2-\bromopropane}

{──── ▸}↖ { {\;\text " aq. "\;}{\KOH}} {\CH}_3- {\CH}↖{|}↖{\OH}- {\CH}_3

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