Question

${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{CH}} = {\text{C}}{{\text{H}}_2} + {\text{HBr}}\xrightarrow{{{\text{ROOR (peroxide)}}}}{\text{(X)Major}} + {\text{(Y)Minor}}$. X and Y respectively are:
A) ${\text{BrC}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{CH}} = {\text{C}}{{\text{H}}_2}$ and ${{\text{C}}_2}{{\text{H}}_5} - {\text{CH(Br)}} - {\text{C}}{{\text{H}}_3}$
B) ${{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{Br}}$ and ${\text{Br}} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2} - {\text{CH}} = {\text{C}}{{\text{H}}_2}$
C) ${{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{Br}}$ and ${{\text{C}}_2}{{\text{H}}_5} - {\text{CH(Br)}} - {\text{C}}{{\text{H}}_3}$
D) ${{\text{C}}_2}{{\text{H}}_5} - {\text{CH(Br)}} - {\text{C}}{{\text{H}}_3}$ and ${{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{Br}}$

Answer 1

Hint: Alkenes react with hydrogen bromide in the presence of peroxide occurs contrary to Markownikoff’s rule. This is known as peroxide effect.

Complete step by step answer:
In the given reaction, but-1-ene reacts with hydrogen bromide in presence of peroxide.
The reaction occurs in contrast to Markownikoff’s rule.
The Markownikoff’s rule states that when an unsymmetrical alkene reacts with an unsymmetrical reagent, the negative part of the reagent gets attached to that carbon atom which has less number of hydrogen atoms.
The anti – Markownikoff’s rule states that when an unsymmetrical alkene reacts with an unsymmetrical reagent, the negative part of the reagent gets attached to that carbon atom which has more number of hydrogen atoms.
Thus, the reaction of but-1-ene with hydrogen bromide in presence of peroxide will give two products. The product obtained by anti – Markownikoff’s rule will be the major product while the other is the minor product.
Thus, the reaction of but-1-ene with hydrogen bromide in presence of peroxide is as follows:
${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{CH}} = {\text{C}}{{\text{H}}_2} + {\text{HBr}}\xrightarrow{{{\text{ROOR (peroxide)}}}}{{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{Br (Major)}} + {{\text{C}}_2}{{\text{H}}_5} - {\text{CH(Br)}} - {\text{C}}{{\text{H}}_3}{\text{ (Minor)}}$
Thus, the major product obtained by the anti – Markownikoff’s addition is ${{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{Br}}$ and the minor product obtained is ${{\text{C}}_2}{{\text{H}}_5} - {\text{CH(Br)}} - {\text{C}}{{\text{H}}_3}$.
Thus, X is ${{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{Br}}$ and Y is ${{\text{C}}_2}{{\text{H}}_5} - {\text{CH(Br)}} - {\text{C}}{{\text{H}}_3}$.

Thus, the correct option is (C).

Note: The reaction is also known as Kharasch effect or abnormal addition or anti – Markownikoff’s addition. The products obtained by Markownikoff’s addition are reversed. The major product obtained will be ${{\text{C}}_2}{{\text{H}}_5} - {\text{CH(Br)}} - {\text{C}}{{\text{H}}_3}$ and the minor product obtained will be ${{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{Br}}$.