Read the passage carefully and answer the Questions.
Alcohols and phenols are obtained when a hydrogen atom in aliphatic and aromatic hydrocarbons respectively, is replaced by - OH group. These compounds find wide applications in industry as well as in our daily life. You may notice that ordinary spirit is used for polishing wooden furniture which is chiefly a compound containing hydroxyl group. An alcohol contains one or more hydroxyl (OH) group(s) directly attached to carbon atom(s), of an aliphatic system (\( CH_3OH \)) while a phenol contains -OH group(s) directly attached to carbon atom(s) of an aromatic system (\( C_6H_5OH \)). Alcohols are prepared by the hydration of alkenes, fermentation of glucose, and reduction of carbonyl compounds. Alcohols produce alkenes upon dehydration whereas alkenes on acid catalyzed hydration convert into alcohols.
The product of the following reaction is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → propan-2-ol (\( \ce{CH3-CH(OH)-CH3} \))
Alkenes react with water in the presence of an acid catalyst to form alcohols (acid-catalysed hydration). With an unsymmetrical alkene the addition follows Markovnikov’s rule: the H atom goes to the double-bond carbon that already has more hydrogen atoms, and the –OH group goes to the other carbon.
The reason is the carbocation. The alkene is first protonated by \( \ce{H3O+} \); adding the proton to the terminal \( \ce{CH2} \) of propene gives the more stable secondary carbocation. Water attacks this carbocation, and loss of a proton gives the alcohol:
\( \displaystyle \ce{CH3-CH=CH2 + H+ -> CH3-CH^{+}-CH3} \)
\( \displaystyle \ce{CH3-CH=CH2 + H2O} \xrightarrow{\ce{H+}} \ce{CH3-CH(OH)-CH3} \)
So the product is propan-2-ol.
Why the other options are incorrect
Option (1) prop-1-en-2-ol (\( \ce{CH3-C(OH)=CH2} \)) — Incorrect. It still has the C=C double bond; in hydration the double bond is used up when H and OH add across it.
Option (2) prop-1-en-1-ol (\( \ce{CH3-CH=CH-OH} \)) — Incorrect. It also keeps a double bond, so it cannot be an addition product of water.
Option (4) prop-2-en-1-ol (\( \ce{HO-CH2-CH=CH2} \)) — Incorrect. This is allyl alcohol; it keeps the double bond and has the –OH on a carbon that was not part of the double bond.
NCERT (2026-27 edition):
By Acid Catalysed Hydration
