Hydrocarbons's Question 51 | Chemistry XI - Part 2 | NCERT Exemplar Questions

Physical Chemistry

Inorganic Chemistry

Hydrogen The s-Block Elements The p-Block Elements Environmental Chemistry

Organic Chemistry

Some Basic Principles and Techniques Hydrocarbons

NCERT

Physics XI - Part 1 Physics XI - Part 2 Chemistry XI - Part 1 Chemistry XI - Part 2 Mathematics XI Physics XII - Part 1 Physics XII - Part 2 Chemistry XII - Part 1 Chemistry XII - Part 2 Biology XI Mathematics XII - Part 1 Mathematics XII - Part 2

Subjective

In the presence of peroxide addition of HBr to propene takes place according to anti Markownikoff's rule but peroxide effect is not seen in the case of HCl and HI . Explain.

Explanation

$\underset{\substack{\text { propene }}}{\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2}+\mathrm{HBr} \xrightarrow{\text { Peroxide }} \underset{\substack{n \text {-propyl bromide }}}{\mathrm{CH}_3-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}}$

The mechanism of the reaction is

Peroxide effect is effective only in the case of HBr and not seen in the case of HCl and HI . This is due to the following reasons.

(i) $\mathrm{H}-\mathrm{Cl}$ bond $(103 \mathrm{kcal} / \mathrm{mol})$ is stronger than $\mathrm{H}-\mathrm{Br}$ bond $(87 \mathrm{kcal} / \mathrm{mol})$ $\mathrm{H}-\mathrm{Cl}$ bond is not decomposed by the peroxide free radical whereas the $\mathrm{H}-\mathrm{I}$ bond is weaker ( $71 \mathrm{kcal} / \mathrm{mol}$ ) form iodine free radicals.

(ii) lodine free radical $\left(\mathrm{I}^{\circ}\right)$ formed as H - I bond is weaker but iodine free radicals readily combine with each other to form iodine molecules rather attacking the double bond.