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NCERT Exemplar · Q8

Q.Arrange the following compounds in increasing order of their boiling points.

(a) (CH3)2CH−CH2Br\mathrm{(CH_3)_2CH-CH_2Br}
(b) CH3CH2CH2CH2Br\mathrm{CH_3CH_2CH_2CH_2Br}
(c) (CH3)3C−Br\mathrm{(CH_3)_3C-Br}
(i)
(b) <
(a) <
(c)
(ii)
(a) <
(b) <
(c)
(iii)
(c) <
(a) <
(b)
(iv)
(c) <
(b) < (a)
Mizoram MbseMCQ· 1mImportance★★★★★
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Boiling points of alkyl halides depend on molecular weight and branching. More branching lowers boiling point due to reduced surface area. The correct order is (c) < (a) < (b), which corresponds to option (iii).

Boiling point trends in organic compounds are governed by two main factors: molecular mass and intermolecular forces. For isomeric alkyl halides like these three C₄H₉Br compounds, the molecular weight is identical (137 g/mol), so the deciding factor is the strength of van der Waals forces — specifically, how well the molecules can pack together.

The key insight: more branching means a more compact, spherical molecule. A compact molecule has less surface area available for intermolecular contact, so the van der Waals forces are weaker, and the boiling point is lower. A straight-chain molecule, by contrast, is long and flexible, allowing many points of contact between neighbouring molecules.

Let’s examine each compound:

  1. Compound (c): 2-bromo-2-methylpropane — This is the most branched. The central carbon is bonded to three methyl groups and one bromine. The molecule is nearly spherical. Very little surface area for neighbouring molecules to “grip” each other. This will have the lowest boiling point.

  2. Compound (a): 1-bromo-2-methylpropane — Here the longest chain is only three carbons (a propane backbone) with a methyl branch on the middle carbon and the bromine on C-1: (CH3)2CHCH2Br\mathrm{(CH_3)_2CHCH_2Br}. The single branch makes the molecule less elongated than the straight chain, yet it is still not as compact as (c). Its surface area is intermediate, so its boiling point falls in the middle. …

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