Which of the following compound will form a cyclic product on reaction with ethylene glycol?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Acetone
The carbonyl carbon of aldehydes and ketones is attacked by alcohols in nucleophilic addition. Dry hydrogen chloride protonates the carbonyl oxygen and makes the carbon more electrophilic. A ketone with ethylene glycol, which has two –OH groups in one molecule, forms a cyclic ketal: both oxygens bond to the same carbonyl carbon and close a five-membered ring, and water is eliminated:
\( \displaystyle \ce{(CH3)2C=O + HOCH2CH2OH} \xrightarrow{\text{dry }\ce{HCl}} \ce{(CH3)2C(OCH2)2 + H2O} \)
In the product the two \( \ce{OCH2} \) groups are joined to each other, giving the ring \( \ce{C-O-CH2-CH2-O} \). Of the four compounds only acetone has a carbonyl group that can do this.
Why the other options are incorrect
Option (2) Ethylether — Incorrect. Diethyl ether has no carbonyl group and is unreactive towards alcohols; it is used as a solvent for that reason.
Option (3) Benzoic acid — Incorrect. A carboxylic acid reacts with an alcohol by esterification, replacing its –OH by an alkoxy group; it does not form a cyclic ketal.
Option (4) Benzyl alcohol — Incorrect. An alcohol has no carbonyl carbon for ethylene glycol to attack, so no ring is formed.
NCERT (2026-27 edition):
Chemical Reactions – “… (d) Addition of alcohols: Aldehydes react with one equivalent of monohydric alcohol in the presence of dry hydrogen chloride to yield alkoxyalcohol intermediate, known as hemiacetals, which further react with one more molecule of alcohol to give a gem-dialkoxy compound known as acetal as shown in the reaction. Ketones react with ethylene glycol under similar conditions to form cyclic products known as ethylene glycol ketals.”