Arrange the following compounds in the order of their increasing boil…

The correct answer is Option (1) → n-butane <ethoxy ethane < pentanal < pentan-1-ol

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:

• n-Butane: alkane, b.p. about 273 K

• Ethoxy ethane: ether, b.p. 308 K

• Pentanal: aldehyde, b.p. 376 K

• Pentan-1-ol: alcohol, b.p. 411 K

Shortcut: n-Butane has the lowest boiling point and the alcohol the highest. Options (1) and (2) both start with n-butane and end with pentan-1-ol. The ether boils lower than the aldehyde, so ethoxy ethane comes before pentanal, and only Option (1) 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: