Answer:
In alkanes, the main attractive forces present in the molecules are weak van der Waals' forces which are linked with the surface area or the size of the molecule. The branching of the carbon atom chain brings the atoms closer, thereby decreasing the surface area and molecular size. Therefore, the branching of the carbon atom chain decreases the magnitude of the attractive forces. As a result, the branched chain alkanes have less boiling points as compared to straight chain isomers. For example,
\[\underset{\begin{smallmatrix} \,\,\,\,\, \\ \,\,\,\,\,\text{Pentane} \\ \text{(b}\text{.p}\text{.}\,\text{=}\,\text{309}\,\text{K)} \end{smallmatrix}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}}}\,\] \[\underset{\begin{smallmatrix} \\ \text{2-Methylbutane} \\ \,\,\,\,\text{(b}\text{.p}\text{.=}\,\text{301}\,\text{K)} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-C{{H}_{3}}}}\,\] \[\underset{\begin{smallmatrix} \\ \,\,\,\text{2,}\,\text{2-Dimethylpropane} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(b}\text{.p}\text{.}\,\text{=}\,\text{283}\ \text{K)} \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}}}\,\]
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