Question

What would be the major product when propene reacts with water in presence of acid catalyst?

• A. \(\mathrm{CH_3CH(OH)CH_3}\) (Propan-2-ol)
• B. \(\mathrm{CH_3CH_2CH_2OH}\) (Propan-1-ol)
• C. \(\mathrm{CH_3C(OH)_2CH_3}\)
• D. \(\mathrm{CH_3CH_2CH_2COOH}\)

Hint:

Hydration of alkenes in acidic medium follows Markovnikov addition and generally produces the more substituted alcohol.

Answer 1

The Correct Option is A

Concept: Addition of water to an alkene in the presence of an acid catalyst is called acid-catalysed hydration. The reaction follows Markovnikov's rule. According to Markovnikov's rule, the hydrogen atom adds to the carbon already carrying more hydrogen atoms, while the hydroxyl group attaches to the more substituted carbon atom.

Step 1: Write the structure of propene.
\[ CH_3-CH=CH_2 \]

Step 2: Apply Markovnikov addition.
During protonation, the more stable secondary carbocation is formed. \[ CH_3-CH=CH_2 \rightarrow CH_3-CH^+-CH_3 \] The hydroxyl group subsequently attacks this carbocation.

Step 3: Formation of alcohol.
\[ CH_3-CH^+-CH_3 \xrightarrow{H_2O} CH_3-CH(OH)-CH_3 \] The product formed is propan-2-ol.

Step 4: Final answer.
\[ \boxed{CH_3CH(OH)CH_3} \]