Question

3-phenyl propene on reaction with HBr gives (major product).
(A) ${{C}_{6}}{{H}_{5}}C{{H}_{2}}CH\left( Br \right)C{{H}_{3}}$
(B) ${{C}_{6}}{{H}_{5}}CH\left( Br \right)C{{H}_{2}}C{{H}_{3}}$
(C) ${{C}_{6}}{{H}_{5}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br$
(D) ${{C}_{6}}{{H}_{5}}CH\left( Br \right)CH=C{{H}_{2}}$

Answer 1

Hint: 3-phenyl propene on reaction with HBr undergoes hydrohalogenation reaction to form the products. There would be formation of two products but the major product will follow the Markownikoff’s rule of addition.

Complete step by step solution:
Let us first know what we mean by hydrohalogenation reaction and the markownikoff’s rule of addition before actually looking into the reaction:
Hydrohalogenation- It is the reaction in which addition of hydrohalic acids (HCl or HBr) takes place to give corresponding haloalkanes. In this type of reaction, Markownikoff’s rule is applicable.
Markownikoff’s rule- This rule states that when two carbon atoms are attached with a double bond and are linked with different numbers of hydrogen atoms, then the halogen will get embedded on the carbon atom having a low number of hydrogen atoms attached. In other words, the electropositive part of the acid gets attached to the carbon with more hydrogen substituents and the electronegative part of the same gets attached to the carbon with more alkyl substituents. This rule also states that, the more stable carbocation will always be the carbon who is more substituted. If benzene or cyclohexane are attached to the carbon, it will be most stable carbocation as both benzene and cyclohexane have a high number of carbons i.e. 6 within them. Now, according to the topics discussed above, the illustration can be solved as, When 3-phenyl propene reacts with HBr two corresponding products are formed as,
${{C}_{6}}{{H}_{5}}C{{H}_{2}}CH=C{{H}_{2}}+HBr\to {{C}_{6}}{{H}_{5}}CH\left( Br \right)C{{H}_{2}}C{{H}_{3}}+{{C}_{6}}{{H}_{5}}C{{H}_{2}}CH\left( Br \right)C{{H}_{3}}$
But, by Markownikoff’s addition rule the major product will be 1-Bromo-1-phenylethane. The mechanism for the reaction is,

Therefore, option (B) ${{C}_{6}}{{H}_{5}}CH\left( Br \right)C{{H}_{2}}C{{H}_{3}}$ is correct.

Note: Do note that the formation of two products will give two choices in the option, so we need to choose the right one by applying Markownikoff’s rule. Option (B) will be the major product whereas, option (A) will be the minor product.