$3 -$ Phenylpropene on reaction with $H B r$ gives (as a major product)

C_{6} H_{5} C H_{2} C H (B r) C H_{3}

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C_{6} H_{5} C H (B r) C H_{2} C H_{3}

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C_{6} H_{5} C H_{2} C H_{2} C H_{2} B r

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C_{6} H_{5} C H (B r) C H = C H_{2}

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Solution

The correct option is B $C_{6} H_{5} C H (B r) C H_{2} C H_{3}$
The addition of halogen acid to the alkenes follow Markownikoff's rule. According to Markownikoff's rule, the negative part of the attacking reagent adds to the less hydrogenated carbon atom of the double bond. ( This rule can be explained by the reaction mechanism. In the mechanism after the addition of a proton to the double bond, a primary carbocation is formed. Then a hydride shift occurs to form further stable benzyl carbocation ).
So the correct option is B.

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H B r

The reaction of $C_{6} H_{5} C H = C H C H_{3}$ with HBr produces :

	Column I		Column II
A	$C H_{3} - C H B r - C D_{3}$ /on treatement with alc.KOH gives $C H_{2} = C H - C D_{3}$ as a major product	P	E1 reaction
B	$P h - C H B r - C H_{3}$ reacts faster than $P h - C H B r - C D_{3}$	Q	E2 reaction
C	$P h - C H_{2} - C H_{2} - F$ on treatement with $C_{2} H_{5} O D / C_{2} H_{5} O^{-}$ gives $P h - C D = C H_{2}$ as the major product	R	E1cB reaction
D	$P h - C H_{2} - C H_{2} - B r$ and $P h - C D_{2} - C H_{2} - B r$ react with same rate	S	First order reaction

C H_{3} - C H_{3} | C | O H - H | C = C H_{2} + H B r \to

C H_{2} = C H - C H = C H_{2} + H B r - 80^{o} C ⟶ A (M a j o r) + B (M i n o r)

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