Choose the correct sequence for the increasing boiling point of the following compounds.
(A) Pentan-1-ol
(B) Butan-2-ol
(C) Ethanol
(D) Propan-1-ol
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C), (D), (B), (A)
All four compounds are alcohols, so all of them form intermolecular hydrogen bonds; their boiling points differ in the van der Waals forces between the alkyl parts. Two rules decide the order:
The boiling point increases with the number of carbon atoms, since a bigger molecule has stronger van der Waals forces.
Among isomers, the boiling point decreases with branching, since a branched molecule has a smaller surface area.
The given alcohols, with their number of carbon atoms:
(C) Ethanol, \( \ce{CH3CH2OH} \): 2 C, b.p. 351 K
(D) Propan-1-ol, \( \ce{CH3CH2CH2OH} \): 3 C, b.p. 370 K
(B) Butan-2-ol, \( \ce{CH3CH(OH)CH2CH3} \): 4 C, a secondary alcohol, b.p. 373 K
(A) Pentan-1-ol, \( \ce{CH3(CH2)4OH} \): 5 C, b.p. 411 K
Butan-2-ol is branched at the carbon bearing –OH, so it boils lower than butan-1-ol (391 K), but its extra carbon still keeps it above propan-1-ol.
Shortcut: Ethanol has the fewest carbons and pentan-1-ol the most, so the order starts with (C) and ends with (A): Options (1) and (4). A three-carbon alcohol boils lower than a four-carbon one, so (D) comes before (B), which leaves Option (1).
NCERT (2026-27 edition):
Boiling Points
