Product formed is:

The correct answer is Option (3) → CH₃–C(OH)(Ph)–CH(CH₃)–CH₃

Acid-catalysed hydration of an alkene goes through a carbocation, and a carbocation rearranges if a more stable one can form. The net change is:

    \( \displaystyle \ce{CH2=CH-C(CH3)2Ph} \xrightarrow{\ce{H3O+}} \ce{(CH3)2CH-C(OH)(CH3)Ph} \)

First, \( \ce{H+} \) adds to the terminal \( \ce{CH2} \) (Markovnikov’s rule), giving a secondary carbocation next to the quaternary carbon:

    \( \displaystyle \ce{CH2=CH-C(CH3)2Ph + H+ -> CH3-C^+H-C(CH3)2Ph} \)

Next, a 1,2-methyl shift: a \( \ce{CH3} \) group moves with its bonding pair from the neighbouring carbon, giving a tertiary benzylic carbocation, stabilised further by resonance with the phenyl ring:

    \( \displaystyle \ce{CH3-C^+H-C(CH3)2Ph -> CH3-CH(CH3)-C^+(CH3)Ph} \)

Then water attacks and a proton is lost:

    \( \displaystyle \ce{CH3-CH(CH3)-C^+(CH3)Ph + H2O -> (CH3)2CH-C(OH)(CH3)Ph + H+} \)

The product is 3-methyl-2-phenylbutan-2-ol, the structure in Option (3).

Why the other options are incorrect

• Option (1) CH₃–C(CH₃)(Ph)–CH(OH)–CH₃ — Incorrect. This would come from water attacking the first, secondary carbocation; that cation rearranges to the far more stable tertiary benzylic cation before water attacks.

• Option (2) CH₃–C(CH₃)(OH)–CH(Ph)–CH₃ — Incorrect. This would need a phenyl shift, which gives a tertiary carbocation without benzylic resonance; the methyl shift gives the more stable cation.

• Option (4) CH₃–C(OH)(Ph)–CH₂CH₂CH₃ — Incorrect. This has an unbranched five-carbon chain; the carbon skeleton of the starting alkene cannot change in this way.

Note: The question is from beyond the NCERT textbook.

NCERT (2026-27 edition):
By Acid Catalysed Hydration – “Alkenes react with water in the presence of acid as catalyst to form alcohols. In case of unsymmetrical alkenes, the addition reaction takes place in accordance with Markovnikov’s rule.
Mechanism
The mechanism of the reaction involves the following three steps:
Step 1: Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺.
Step 2: Nucleophilic attack of water on carbocation.
Step 3: Deprotonation to form an alcohol.”