2-Methylpropene, \( \ce{(CH3)2C=CH2} \), reacts with water in the presence of an acid as catalyst. Which of the following gives the major product and the way the addition takes place?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2-Methylpropan-2-ol, as the addition follows Markovnikov’s rule
In acid-catalysed hydration, an unsymmetrical alkene adds water in accordance with Markovnikov’s rule: the hydrogen goes to the doubly bonded carbon that already has more hydrogen atoms (the CH2 end), and –OH goes to the other carbon.
Step 1: H3O+ protonates the alkene at the CH2 carbon, giving the carbocation \( \ce{(CH3)3C+} \).
Step 2: water attacks the carbocation.
Step 3: loss of a proton gives the alcohol.
\( \displaystyle \ce{(CH3)2C=CH2 + H2O ->[H+] (CH3)3C-OH} \)
The –OH ends up on the carbon bonded to three methyl groups, so the product is the tertiary alcohol 2-methylpropan-2-ol.
Why the other options are incorrect
Option (1) 2-Methylpropan-1-ol, as the addition follows Markovnikov’s rule — Incorrect. In 2-methylpropan-1-ol the –OH is on the CH2 carbon, which is the carbon that receives the hydrogen under Markovnikov’s rule.
Option (3) 2-Methylpropan-1-ol, as the addition is opposite to Markovnikov’s rule — Incorrect. An alcohol that looks like an anti-Markovnikov addition of water is obtained by hydroboration–oxidation, not by acid-catalysed hydration.
Option (4) 2-Methylpropan-2-ol, as the addition is opposite to Markovnikov’s rule — Incorrect. The product is right, but putting –OH on the carbon with fewer hydrogen atoms is Markovnikov addition, not the opposite.
NCERT (2026-27 edition):
7.4.1 Preparation of Alcohols – “(i) 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.”
“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.”