Answer the question on basis of passage given below:
A ketone [A] which undergoes haloform reaction gives compound [B] on reduction. Compound [B] on heating with sulphuric acid gives compound [C] which forms mono-ozonide [D]. The compound [D] on hydrolysis in the presence of zinc dust gives compound [E]. Compound [E] on oxidation gives \(C_2H_4O_2\) [F].
The boiling point of [A] is observed exceptionally high because of:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Dipole-Dipole interaction
The passage identifies [A] as butan-2-one, \( \ce{CH3COCH2CH3} \) (see Q5). It is a ketone: it has a polar C=O group but no O–H or N–H bond, so it cannot form hydrogen bonds with its own molecules.
The Cδ+=Oδ− group makes the molecule a strong dipole. Neighbouring molecules line up and attract each other through dipole-dipole interactions. This weak association is why butan-2-one (b.p. 353 K) boils far above a hydrocarbon or ether of similar molar mass, though below an alcohol of similar molar mass.
Why the other options are incorrect
Option (1) Intermolecular H-bond — Incorrect. Needs an H attached to O, N or F; a ketone has none, so its molecules cannot hydrogen-bond with one another.
Option (2) Intramolecular H-bond — Incorrect. Also needs an O–H or N–H group within the molecule; and intramolecular bonding would lower, not raise, the boiling point.
Option (3) Like interaction — Incorrect. Not a recognised type of intermolecular force; it does not explain the higher boiling point.
NCERT (2026-27 edition):
Physical Properties – “… The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of comparable molecular masses. It is due to weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions. Also, their boiling points are lower than those of alcohols of similar molecular masses due to absence of intermolecular hydrogen bonding.”