Arrange the following compounds in order of increasing boiling points…

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

Haloalkanes are polar molecules held together by dipole-dipole and van der Waals forces. Their boiling points are decided by two factors:

• Nature of the halogen: for the same alkyl group, the boiling point rises with the size and mass of the halogen, because van der Waals forces become stronger: R–F < R–Cl < R–Br < R–I.

• Number of halogen atoms: each extra halogen atom raises the molecular mass and the attractive forces, so the boiling point rises: \( \ce{CH3X} \) < \( \ce{CH2X2} \) < \( \ce{CHX3} \).

Combining the two factors for the given compounds:

    \( \displaystyle \ce{CH3Cl} < \ce{CH3Br} < \ce{CH2Br2} < \ce{CHBr3} \)

    \( \displaystyle \text{249 K} < \text{277 K} < \text{370 K} < \text{423 K} \)

Matching the given compounds:

• (C) Chloromethane: \( \ce{CH3Cl} \) (one Cl, lightest)

• (A) Bromomethane: \( \ce{CH3Br} \) (one Br)

• (D) Dibromomethane: \( \ce{CH2Br2} \) (two Br)

• (B) Bromoform: \( \ce{CHBr3} \) (three Br, heaviest)

Shortcut: Chloromethane (C) has the lowest boiling point and bromoform (B) the highest. Only Option (2) starts with (C) and ends with (B).

NCERT (2026-27 edition):
Melting and Boiling Points – “… Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals) are stronger in the halogen derivatives. … The attractions get stronger as the molecules get bigger in size and have more electrons. … For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF. This is because with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.”

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