Arrange the following halocompounds in increasing order of their boil…

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

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:

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

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

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

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

Shortcut: Chloromethane (A) has the lowest boiling point and bromoform (C) the highest, which leaves Options (2) and (4). Bromomethane (B) boils below dibromomethane (D), so the answer is Option (2).

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: