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NEET CHEMISTRYMedium

The reaction of C6H5CH=CHCH3C_6H_5CH=CHCH_3 with HBr produces:

A

C6H5CH(Br)CH2CH3C_6H_5CH(Br)CH_2CH_3

B

C6H5CH2CH(Br)CH3C_6H_5CH_2CH(Br)CH_3

C

C6H5CH2CH2CH2BrC_6H_5CH_2CH_2CH_2Br

D

Option 4

Step-by-Step Solution

  1. Reaction Type: The reaction is an electrophilic addition of hydrogen bromide (HBr) to an unsymmetrical alkene (C6H5CH=CHCH3C_6H_5CH=CHCH_3, 1-phenylprop-1-ene).
  2. Mechanism: The reaction proceeds via a carbocation intermediate.
  • Step 1 (Protonation): The proton (H+H^+) from HBr adds to the double bond to form the most stable carbocation. There are two possibilities:
  • Addition to C2C_2 forms a benzylic carbocation: C6H5−C+H−CH2CH3C_6H_5-C^+H-CH_2CH_3.
  • Addition to C1C_1 forms a secondary carbocation: C6H5−CH2−C+H−CH3C_6H_5-CH_2-C^+H-CH_3.
  1. Stability Analysis: The benzylic carbocation is significantly more stable than the normal secondary carbocation due to resonance delocalisation of the positive charge into the benzene ring [NCERT 11th, Ch 13, Sec 13.5.6; NCERT 12th, Ch 10, Sec 10.6]. Therefore, the reaction proceeds through the benzylic intermediate.
  2. Step 2 (Nucleophilic Attack): The bromide ion (Br−Br^-) attacks the positively charged benzylic carbon to form the final product. C6H5−C+H−CH2CH3+Br−→C6H5−CH(Br)−CH2CH3C_6H_5-C^+H-CH_2CH_3 + Br^- \rightarrow C_6H_5-CH(Br)-CH_2CH_3
  3. Product: The major product is 1-Bromo-1-phenylpropane.
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