The major product in the following reaction is

$C_6H_5CH=CH_2 + HI→ ?$

$C_6H_5CHICH_3$

The correct answer is Option (2) → $C_6H_5CHICH_3$

Core Concept

This is electrophilic addition of $\text{HX}$ to an alkene following Markovnikov’s rule.

Styrene $(\text{C}_6\text{H}_5\text{-CH=CH}_2)$ forms a benzylic carbocation, which is highly stabilized by resonance with the benzene ring.

Stepwise Mechanism

Step 1: Protonation of double bond

$H^+$ from $\text{HI}$ adds to the terminal carbon $(\text{CH}_2)$, forming the benzylic carbocation:

$\text{C}_6\text{H}_5\text{-}\overset{+}{\text{C}}\text{H - CH}_3$

This carbocation is resonance stabilized, so it forms preferentially.

Step 2: Nucleophilic attack

$I^-$ attacks the carbocation:

$\text{C}_6\text{H}_5\text{-CH(I)-CH}_3$

Why other options are wrong

Option 1: $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{I}$

Would require anti-Markovnikov addition. That occurs only with peroxide effect, and $\text{HI}$ does NOT show peroxide effect.

Option 3: $\text{IC}_6\text{H}_4\text{CH=CH}_2$

Would require substitution on the benzene ring. $\text{HI}$ does not cause electrophilic substitution here.

Option 4: $\text{C}_6\text{H}_5\text{CH}_2\text{CH}\text{I}_2$

Would require addition of two iodines, not possible with just $\text{HI}$.