Arrange the following alkyl halides in increasing order of boiling points
(A) 1-Iodobutane
(B) 1-Bromobutane
(C) 1-Chlorobutane
(D) Butane
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (D), (C), (B), (A)
Boiling points depend on the strength of the intermolecular forces. Two factors decide this order.
Haloalkane or alkane. A polar C–X bond and a heavier molecule make a haloalkane boil much higher than butane.
Which halogen. For the same alkyl group, a larger halogen gives stronger van der Waals forces.
The four compounds with their boiling points:
(D) Butane, \( \ce{CH3CH2CH2CH3} \): about 273 K
(C) 1-Chlorobutane, \( \ce{CH3CH2CH2CH2Cl} \): about 351 K
(B) 1-Bromobutane, \( \ce{CH3CH2CH2CH2Br} \): about 375 K
(A) 1-Iodobutane, \( \ce{CH3CH2CH2CH2I} \): about 403 K
Shortcut: Butane, the only compound without a halogen, boils lowest, and Option (1) is the only order that starts with (D).
NCERT (2026-27 edition):
Melting and Boiling Points – “Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature. Higher members are liquids or solids. As we have already learnt, molecules of organic halogen compounds are generally polar. 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. That is why the boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass.
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.”