Hydrogen bromide is added to ${C H}_{3} C H = {C H}_{2}$ in presence of peroxides. The major compound formed is:

{C H}_{3} {C H B r C H}_{3}

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C_{2} H_{5} {C H}_{2} B r

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{C H}_{2} = {C H}_{2} {C H}_{2} B r

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{C H}_{3} {C H}_{2} {C H}_{3}

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Solution

The correct option is B $C_{2} H_{5} {C H}_{2} B r$
In presence of peroxide, anti markownikoff addition takes place due to the formation of bromine free radical and $C_{2} H_{5} {C H}_{2} B r$ is formed as the product.
The reaction is as follow:
$C H_{3} C H = C H_{2} + H B r p e r o x i d e - ---- \to C H_{3} C H_{2} - C H_{2} B r$
The reaction follows anti-Markownikoff's rule which states that "when the reaction between an unsymmetrical alkene and unsymmetrical reagent takes place, then negative part of the reagent gets attached to that C atom which has more number of H atom".

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