Question

3-phenylpropene on reaction with ${\text{HBr}}$ gives __________ as a major product.
(A)${{\text{C}}_6}{{\text{H}}_5}{\text{C}}{{\text{H}}_2}{\text{CH}}\left( {{\text{Br}}} \right){\text{C}}{{\text{H}}_3}$
(B)${{\text{C}}_6}{{\text{H}}_5}{\text{CH}}\left( {{\text{Br}}} \right){\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3}$
(C)${{\text{C}}_6}{{\text{H}}_5}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{Br}}$
(D)${{\text{C}}_6}{{\text{H}}_5}{\text{CH}}\left( {{\text{Br}}} \right){\text{CH}} = {\text{C}}{{\text{H}}_2}$

Answer 1

According to Markownikoff’s rule, when an unsymmetrical reagent is added to a carbon-carbon double bond of an alkene, the negative part of the reagent is added to more substituted doubly bonded carbon atom and the positive part of the reagent is added to less substituted doubly bonded carbon atom.

Complete step by step answer:
The chemical formula of 3-phenylpropene is ${{\text{C}}_6}{{\text{H}}_5}{\text{C}}{{\text{H}}_2}{\text{CH}} = {\text{C}}{{\text{H}}_2}$ . It has an aliphatic part containing 3 carbon atoms and an aromatic part containing 6 carbon atoms. The aromatic part is a benzene ring. The aliphatic part is a 3 carbon atom chain containing a carbon-carbon double bond.
Let us look at the reaction:
Proton from ${\text{HBr}}$ is donated to doubly bonded carbon atoms to form less stable secondary carbocation.
${{\text{C}}_6}{{\text{H}}_5}{\text{C}}{{\text{H}}_2}{\text{CH}} = {\text{C}}{{\text{H}}_2}{\text{ + }}{{\text{H}}^ + } \to {\text{ }}{{\text{C}}_6}{{\text{H}}_5}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}^ + } - {\text{C}}{{\text{H}}_3}$
Less stable, secondary carbocation rearranges to more stable benzylic carbocation via 1,2-hydride shift.
${{\text{C}}_6}{{\text{H}}_5}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}^ + } - {\text{C}}{{\text{H}}_3} \to {\text{ }}{{\text{C}}_6}{{\text{H}}_5}{\text{C}}{{\text{H}}^ + } - {\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3}$
More stable benzylic carbocation then accepts a bromide ion to form 1-bromo-1-phenylethane.
${{\text{C}}_6}{{\text{H}}_5}{\text{C}}{{\text{H}}^ + } - {\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3}{\text{ + B}}{{\text{r}}^ - } \to {\text{ }}{{\text{C}}_6}{{\text{H}}_5}{\text{CH}}\left( {{\text{Br}}} \right){\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3}$
3-phenylpropene on reaction with ${\text{HBr}}$ gives ${{\text{C}}_6}{{\text{H}}_5}{\text{CH}}\left( {{\text{Br}}} \right){\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3}$ as a major product. The IUPAC name of ${{\text{C}}_6}{{\text{H}}_5}{\text{CH}}\left( {{\text{Br}}} \right){\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3}$ is 1-bromo-1-phenylethane.

Hence, the option (B) is the correct answer.

Note: The order of stability of carbocations is benzylic > tertiary > secondary > primary.
But if peroxides are used, then 3-phenylpropene on reaction with ${\text{HBr}}$ will give ${{\text{C}}_6}{{\text{H}}_5}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{C}}{{\text{H}}_2}{\text{Br}}$ as the major product. In the presence of peroxides, anti Markownikoff’s rule is followed. According to peroxide effect, when an unsymmetrical reagent is added to a carbon-carbon double bond of an alkene, in presence of a peroxide, the negative part of the reagent is added to less substituted doubly bonded carbon atom and the positive part of the reagent is added to more substituted doubly bonded carbon atom.