The reaction of $C_6H_5CH=CHCH_3$ with HBr produces:

Question

Accepted Answer

1. **Reaction Type:** The reaction is an electrophilic addition of hydrogen bromide (HBr) to an unsymmetrical alkene ($C_6H_5CH=CHCH_3$, 1-phenylprop-1-ene).
2. **Mechanism:** The reaction proceeds via a carbocation intermediate.
   - **Step 1 (Protonation):** The proton ($H^+$) from HBr adds to the double bond to form the most stable carbocation. There are two possibilities:
     - Addition to $C_2$ forms a **benzylic carbocation**: $C_6H_5-C^+H-CH_2CH_3$.
     - Addition to $C_1$ forms a **secondary carbocation**: $C_6H_5-CH_2-C^+H-CH_3$.
3. **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.
4. **Step 2 (Nucleophilic Attack):** The bromide ion ($Br^-$) attacks the positively charged benzylic carbon to form the final product.
   $$ C_6H_5-C^+H-CH_2CH_3 + Br^- \rightarrow C_6H_5-CH(Br)-CH_2CH_3 $$
5. **Product:** The major product is 1-Bromo-1-phenylpropane.

Step-by-Step Solution

Practice All CHEMISTRY Questions