Question

Ph-CH=CH$_2 \xrightarrow[\text{(PhCOO)}_2]{\text{HBr}}$ Product.
Correct statement(s) regarding product :
(a) Ph-CH(Br)-CH$_3$ is minor product
(b) Benzene is also form a bi product
(c) Reaction follow free radical mechanism
(d) In absence of peroxide carbocation mechanism is followed

• A. b, c
• B. a, c, d
• C. c, d
• D. a, b, c

Hint:

Peroxide Effect applies only to HBr. Major product is determined by the stability of the intermediate free radical (Benzyl>$3^\circ>2^\circ>1^\circ$).

Answer 1

The Correct Option is C

The reaction is the addition of HBr to Styrene in the presence of Peroxide.
This condition leads to the Anti-Markovnikov addition via a free radical mechanism (Kharasch effect).
The bromine radical attacks the $\beta$-carbon to form a stable benzylic radical:
$\text{Ph}-\dot{\text{C}}\text{H}-\text{CH}_2\text{Br}$.
This radical abstracts H from HBr to form the major product: $\text{Ph}-\text{CH}_2-\text{CH}_2\text{Br}$ (1-Bromo-2-phenylethane).
Statement (a): The Markovnikov product ($\text{Ph-CH(Br)-CH}_3$) is indeed minor under these conditions. So (a) is technically correct.
Statement (c): Mechanism is free radical. Correct.
Statement (d): Without peroxide, electrophilic addition occurs via a stable carbocation ($\text{Ph}-\text{CH}^+-\text{CH}_3$), leading to the Markovnikov product. Correct.