Question

Consider the reaction: \[ \text{Ph–CH=CH}_2 \xrightarrow[\text{peroxide}]{\text{HBr}} \text{Product} \] Which of the following statements are correct?
[A.] The reaction proceeds through a more stable radical intermediate.
[B.] The role of peroxide is to generate \(\mathrm{H^\bullet}\) radical.
[C.] During this reaction, benzene is formed as a byproduct.
[D.] \(1\)-Bromo-\(2\)-phenylethane is formed as a minor product.
[E.] The same reaction in absence of peroxide proceeds via a carbocation intermediate. Choose the correct answer.

• A. A, B & D only
• B. C, D & E only
• C. A, C & E only
• D. A & E only

Hint:

The peroxide effect operates only with HBr and switches the mechanism from ionic to radical.

Answer 1

The Correct Option is D

Concept: Addition of HBr to alkenes in presence of peroxides follows the {free radical (anti-Markovnikov) mechanism}, known as the peroxide effect or Kharasch effect.
Step 1: Nature of the intermediate (Statement A) In styrene (\(\text{Ph–CH=CH}_2\)), radical addition leads to a {benzylic radical}, which is highly stabilised due to resonance. \[ \Rightarrow \text{Statement A is true.} \]
Step 2: Role of peroxide (Statement B) Peroxides decompose to form \(\mathrm{RO^\bullet}\) radicals, which abstract hydrogen from HBr to generate \(\mathrm{Br^\bullet}\), not \(\mathrm{H^\bullet}\). \[ \Rightarrow \text{Statement B is false.} \]
Step 3: Formation of benzene (Statement C) No rearrangement or elimination produces benzene in this mechanism. \[ \Rightarrow \text{Statement C is false.} \]
Step 4: Product distribution (Statement D) Anti-Markovnikov addition gives \(1\)-bromo-\(2\)-phenylethane as the {major} product, not a minor one. \[ \Rightarrow \text{Statement D is false.} \]
Step 5: Absence of peroxide (Statement E) Without peroxide, HBr adds via the ionic (Markovnikov) pathway involving a carbocation intermediate. \[ \Rightarrow \text{Statement E is true.} \] Final Conclusion: \[ \boxed{\text{Statements A and E are correct}} \]