Write the major product A in the following chemical reaction.

The correct answer is Option (2) → 1-bromo-3-phenylpropane

HBr adds to an unsymmetrical alkene in two different ways, depending on whether a peroxide is present:

• Without peroxide (ionic mechanism): \( \ce{H^+} \) adds first to give the more stable carbocation, so Br ends up on the more substituted carbon: Markovnikov addition.

• With peroxide (free radical mechanism, peroxide or Kharasch effect): the peroxide gives radicals that take H from HBr and form a Br• radical. Br• adds first, to the terminal \( \ce{CH2} \), to give the more stable (secondary) carbon radical. That radical then takes H from HBr, so Br ends up on the less substituted carbon: anti-Markovnikov addition.

Only HBr shows the peroxide effect; HCl and HI give the Markovnikov product even with peroxide.

The alkene here is 3-phenylprop-1-ene (allylbenzene), \( \ce{C6H5-CH2-CH=CH2} \). With peroxide the radical chain runs as follows:

    RO–OR → 2RO• (on heating)
    RO• + HBr → ROH + Br•
    C₆H₅–CH₂–CH=CH₂ + Br• → C₆H₅–CH₂–C•H–CH₂Br
    C₆H₅–CH₂–C•H–CH₂Br + HBr → C₆H₅–CH₂–CH₂–CH₂Br + Br•

The Br• formed in the last step starts the next cycle. The net reaction is:

    \( \displaystyle \ce{C6H5-CH2-CH=CH2 + HBr} \xrightarrow{\text{peroxide}} \ce{C6H5-CH2-CH2-CH2Br} \)

So the major product A is 1-bromo-3-phenylpropane.

Why the other options are incorrect

• Option (1) 2-bromo-1-phenylpropane — Incorrect. Here Br is on the middle carbon, the more substituted carbon of the double bond. This is the Markovnikov product, formed when HBr adds in the absence of peroxide.

• Option (3) 1-bromo-1-phenylpropane — Incorrect. Br is on the carbon next to the ring, which is not part of the double bond. Addition of HBr across C=C puts Br on one of the two double-bond carbons only.

• Option (4) 3-(2-bromophenyl)prop-1-ene — Incorrect. The double bond is untouched and Br is on the ring. Putting Br on the ring is an electrophilic substitution with \( \ce{Br2/FeBr3} \), not an addition of HBr.

Note: The same/similar question has been asked in different years by NTA as per the details given below: