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The general explanation given is that the chains are further apart, so the intermolecular forces are less, and as they are further apart, they occupy more volume, and the density is lesser. But the length of each chain (and thus its mass) also increases, so how do we know that the density overall decreases?

As for the melting point, longer chains = more electrons = more Van Der Waals forces, so the melting point should increase.

Edit: I'm referring to branching in the context of polymerisation.

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I'm not sure that your statement about density is always correct. Looking at octane and its isomers the density changes very little, for example

octane $703~ \pu{kgm^{-3}}$

2-methylheptane $698~ \pu{kgm^{-3}}$

2,5-dimethylhexane $694~ \pu{kgm^{-3}}$

2,2,4-trimethylpentane $692~ \pu{kgm^{-3}}$

although the melting points do change considerably. In the same order as above they are $-57, -112,-93,-107$ so octane is the exception here; (data from Wikipedia pages).

This can be explained by the fact that melting can be considered to be governed by the repulsive part of the intermolecular potential this having a greater influence than the attractive part when the molecule has an odd shape. Rod like octane has to have deeper intermolecular potential than the odd shaped isomers as it has a higher melting point.

In the isomers the interaction should be dominated by the repulsive part of the potential as the odd shapes must try to push methyl groups into to one another. The potential itself has the same overall geometry, i.e. similar intermolecular minimum distance, as density is the same in all molecules above, but is shallower in nature due to the extra repulsion. Hence the isomers have significantly lower boiling points than octane.

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As a side note, whenever we talk about melting points, structural symmetry comes into play. More symmetrical molecules pack into the crystal lattice better thus allowing attractive tendencies of molecules to dominate (facilitated due to a closer approach between molecules)1. As a rule of thumb, as packing efficiency increases, melting point increases. A similar reasoning is given whenever we're to compare the melting points of trans and cis-2-butene.

For instance, let's say you're planning to go to a nice long vacation (ask me, I'd say to Hawaii). But you've decided to carry just one suitcase for the trip. Being a chemist, your love for chemicals is intense, and you've decided that you'll either carry butane or iso-butane for the trip. But which one? More chemicals means more fun, you say. Thus, a molecule with higher packing efficiency would be a clear winner in this scenario. Butane wins this game because it has a symetrical structure and you would be able to "stack" those molecules one over the other which accounts for its higher packing efficiency. Butane thus would have a higher melting point. Data supports our speculation; melting point of butane is $-140 ^\circ \pu{C}$, while for iso-butane it's $-159.6 ^\circ \pu{C}$.

Longer chains can also mean larger molecular masses, which in turn lead to an increased london force between the molecules, leading to an increased melting point.

Further, a branched molecule doesn't always mean a decreased melting point. Take for example the case of pentane and 2,2-dimethyl propane. The latter has a more "compact" spherical shape, which allows it to pack better into the crystal lattice, resulting in a significantly higher melting point.

$ \begin{array}{|c|c|} \hline \pu{Compound} & \pu{Melting point (in ^\circ C)} \\ \hline \pu{Pentane} & \pu{-130 ^\circ} \\ \hline \pu{2,2-dimethyl propane} & \pu{-71 ^\circ} \\ \hline \end{array} $


References:

  1. Organic Chemistry, Joseph M. Hornback
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  • $\begingroup$ Okay, so for the melting point, we have three different effects : 1) The molecule becomes more sphere-like, so the surface area decreases, and the strength of the Van Der Waals forces decrease. 2) The length of each chain becomes longer, so the Van Der Waals forces increase. 3)In general, branching decreases the symmetry of the structure, and thus the packing efficiency (and hence m.p ) decrease, but of course, there are exceptions. Also, I meant to ask this in the context of polymers. Does your answer still apply? Could you give a similar answer for the density? $\endgroup$ – Saad May 8 '17 at 9:10

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