Arrange the following haloalkanes in decreasing order of boiling point:
Chloromethane, Bromomethane, Dibromomethane, Bromoform
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Bromoform > Dibromomethane > Bromomethane > Chloromethane
Haloalkanes are held together by dipole–dipole and van der Waals forces. These forces grow as the molecule becomes bigger and heavier (more electrons), so the boiling point rises:
for the same alkyl group, with a heavier halogen: R–Br > R–Cl;
with more halogen atoms on the same carbon: \( \ce{CHBr3} \) > \( \ce{CH2Br2} \) > \( \ce{CH3Br} \).
The four compounds and their molar masses:
Bromoform, \( \ce{CHBr3} \): 253 g mol−1, b.p. about 422 K
Dibromomethane, \( \ce{CH2Br2} \): 174 g mol−1, b.p. about 370 K
Bromomethane, \( \ce{CH3Br} \): 95 g mol−1, b.p. about 277 K
Chloromethane, \( \ce{CH3Cl} \): 50.5 g mol−1, b.p. about 249 K
Shortcut: only Option (1) starts with bromoform (the heaviest) and ends with chloromethane (the lightest).
NCERT (2026-27 edition):
Melting and Boiling Points – “… 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.”