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NEET CHEMISTRYHydrocarbonsMedium

Question

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.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Hydrocarbons. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYHydrocarbonsreactionchchchchproduces

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