Arrange the following compounds in increasing order of their boiling…

The correct answer is Option (3) → (B), (D), (C), (A)

The boiling point of a compound depends on the strength of the forces between its molecules. For molecules of comparable size, the forces become stronger in this order:

• Alkanes: non-polar, held only by weak van der Waals forces.

• Ethers: the C–O–C group is only weakly polar, so the dipole-dipole attraction is weak, and ethers cannot form hydrogen bonds among themselves.

• Aldehydes and ketones: the C=O group is strongly polar, giving stronger dipole-dipole attraction, but still no hydrogen bonding.

• Alcohols: the O–H group forms intermolecular hydrogen bonds, the strongest attraction of the four.

So the boiling points increase as:

    alkane < ether < aldehyde < alcohol
    \( \ce{CH3CH2CH2CH3} \) < \( \ce{C2H5OC2H5} \) < \( \ce{CH3(CH2)3CHO} \) < \( \ce{CH3(CH2)4OH} \)

Matching the given compounds:

• (B) n-butane: alkane, b.p. about 273 K

• (D) ethoxyethane: ether, b.p. 308 K

• (C) pentanal: aldehyde, b.p. 376 K

• (A) Pentan-1-ol: alcohol, b.p. 411 K

Shortcut: n-Butane (B) has the lowest boiling point and the alcohol (A) the highest. Options (3) and (4) both start with (B) and end with (A). The ether boils lower than the aldehyde, so (D) comes before (C), and only Option (3) remains.

NCERT (2026-27 edition):
Physical Properties – …

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