Arrange the following compounds in decreasing order of their boiling…

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

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 decrease as:

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

Matching the given compounds:

• D. Pentan-1-ol: alcohol, b.p. 411 K

• C. Pentanal: aldehyde, b.p. 376 K

• A. Ethoxyethane: ether, b.p. 308 K

• B. n-Butane: alkane, b.p. about 273 K

Shortcut: The alcohol (D) has the highest boiling point and n-butane (B) the lowest. Only Option (2) starts with D and ends with B.

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: