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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?

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The simple explanation is that weak intermolecular forces (the forces that make something condense to a liquid when things are cold enough) depend on the surface area (as well as many other things). But in the case of relatively similar non-polar isomers (where weak intermolecular forces are the dominant forces), the larger surface area will lead to the large force and hence the highest boilling point.

2,2 dimethyl propane is a compact, almost spherical molecule. Pentane is long and "floppier" so will experience more forces between molecules.

The underlying, but slightly simplified, explanation of this is that intermolecular forces (often called van der Waal's forces) depend on attractions caused by quantum fluctuations in the surface electrons of the molecule. These lead to short-lived dipole moments that can also induce dipoles in neighboring molecules to which the original dipole is attracted. The larger the surface area, the more opportunity for such dipoles to exist and therefore a stronger force.

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Straight chain molecules have more places along its length where they can be attracted to other molecules, so there are more chances of London Dispersion Forces to be developed. Hence, they have stronger intermolecular forces as compared to the branched chain molecules which have a compact shape, therefore fewer spaces where they can be attracted to other molecules.

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The London / van der Waals forces are weak forces but their effect depends both on the magnitude of the forces themselves and the distance over which they act. Linear molecules can align and lie parallel to each other, giving small distances of separation. This makes attraction produced by the same forces/moments greater.

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Straight chain compounds have large size and hence have large polarizability and have strong London dispersion forces hence high boiling points while branched compounds have compact structure and hence have low polarizability and have low boiling points.

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  • $\begingroup$ Could you elaborate on this? Currently your answers adds nothing of value to the question. $\endgroup$ – M.A.R. ಠ_ಠ Nov 9 '15 at 17:29
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Straight chain compounds have higher boiling point then branched chain because in straight chain, molecules are strongly entangled with each other (like noodles)and have more contact with other molecules so strong force is required to remove such molecules consequently straight chain compounds have higher boiling point than branched compounds

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protected by orthocresol Oct 14 '17 at 14:36

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