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(B) The reaction is the electrophilic addition of $HBr$ to $1-phenylpropene$ $(Ph-CH=CH-CH_3)$.According to Markovnikov's rule,the electrophile $(H^+)

Vedclass · Accepted Answer

$Ph-CH(Br)-CH_2-CH_3$

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(B) The reaction is the electrophilic addition of $HBr$ to $1-phenylpropene$ $(Ph-CH=CH-CH_3)$.According to Markovnikov's rule,the electrophile $(H^+)$ adds to the carbon atom of the double bond that has more hydrogen atoms,resulting in the formation of the most stable carbocation.In $Ph-CH=CH-CH_3$,the carbocation formed at the benzylic position $(Ph-CH^+-CH_2-CH_3)$ is highly stabilized by resonance with the phenyl ring.Therefore,the bromide ion $(Br^-)$ attacks this benzylic carbocation to form $1-phenyl-1-bromopropane$ $(Ph-CH(Br)-CH_2-CH_3)$ as the major product.

The major product of the following reaction is:

Similar Questions

Which of the following potential energy $(PE)$ diagrams represents the $S_N^1$ reaction?

$CH_3Br + Nu^{-} \rightarrow CH_3-Nu + Br^{-}$
The decreasing order of the rate of the above reaction with nucleophiles $(Nu^{-})$ $A$ to $D$ is
$[Nu^{-} = (A) \, PhO^{-}, (B) \, AcO^{-}, (C) \, HO^{-}, (D) \, CH_3O^{-}]$

Arrange the following in order of ease of dehydrohalogenation:
$(i)$ Bromocyclohexane
(ii) $3-$Bromocyclohexene
(iii) $3-$Bromocyclohexa$-1,4-$diene
(iv) Bromobenzene

The major product of the reaction is:

In the reaction shown,the major product obtained is:

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