Arrange the following compounds in increasing order of boiling points…

The correct answer is Option (2) → (A) < (C) < (B) < (D)

All four compounds are alcohols, so all of them form hydrogen bonds; the order is decided by the van der Waals forces between the alkyl parts. Two factors act:

• The boiling point rises with the number of carbon atoms, because larger molecules have stronger van der Waals forces.

• For the same number of carbon atoms, the boiling point falls with branching, because a branched molecule is more compact and has a smaller surface area for contact.

• (A) Propan-1-ol, \( \ce{CH3CH2CH2OH} \): three carbon atoms, the smallest; b.p. about 370 K.

• (C) Butan-2-ol, \( \ce{CH3CH2CH(OH)CH3} \): four carbon atoms, but branched at the carbon carrying –OH; b.p. about 373 K.

• (B) Butan-1-ol, \( \ce{CH3(CH2)3OH} \): four carbon atoms in a straight chain; b.p. about 390 K.

• (D) Pentan-1-ol, \( \ce{CH3(CH2)4OH} \): five carbon atoms, the largest; b.p. about 411 K.

In increasing order of boiling point:

    \( \displaystyle \text{propan-1-ol} \lt \text{butan-2-ol} \lt \text{butan-1-ol} \lt \text{pentan-1-ol} \)

Shortcut: Options (1) and (2) both start with (A) and end with (D). The branched butan-2-ol (C) boils below butan-1-ol (B), so only Option (2) fits.

NCERT (2026-27 edition):
Boiling Points

Note: The same/similar question has been asked in different years by NTA as per the details given below: