organic chemistry - Why do branched chain compounds have lower boiling points than the corresponding straight chain isomers?

The branched chain compounds have lower boiling points than the corresponding straight chain isomers. For example,

• $\ce{CH_3CH_2CH_2CH_2CH_3}$ - No branching-Pentane (n-pentane) ($\mathrm{b.p.}=309~\mathrm{K}$)


• $\ce{CH_3CH(CH_3)CH_2CH_3}$ - One branching-2-Methylbutane (Iso-pentane) ($\mathrm{b.p.}=301~\mathrm{K}$)


• $\ce{C(CH_3)_4}$ - Two branches-2,2-Dimethylpropane (Neo-pentane) ($\mathrm{b.p.}=282.5\mathrm{K~}$)

This is due to the fact that branching of the chain makes the molecule more compact and thereby decreases the surface area. Therefore, the intermolecular attractive forces which depend upon the surface area, also become small in magnitude on account of branching. Consequently, the boiling points of the branched chain alkanes are less than the straight chain isomers.

The above extract from my book, mentions clearly that branching makes the molecule more compact and thereby decreases the surface area. Even for isomeric alkyl halides, the boiling points decrease with branching. The reason is said to be because of the decrease in surface area, same as explained before. How does decrease in the surface area make the intermolecular forces small in magnitude?