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From a video on organic chemistry, I found this table that shows the boiling point temperatures of alkynes $\ce{C_{n}H_{2n-2}}$ with $\,n=2...10$.

$$\begin{array}{|c|c|c|c|} \hline n & t_\mathrm{boil}/^\circ\mathrm{C} & m & T_\mathrm{boil}/\mathrm{K}\\ \hline 2 & -83.6 & 26 & 189.6\\ \hline 3 & -23.2 & 40 & 250.0\\ \hline 4 & 8.1 & 54 & 281.3\\ \hline 5 & 39.7 & 68 & 312.9\\ \hline 6 & 71 & 82 & 344.2\\ \hline 7 & 99.7 & 96 & 372.9\\ \hline 8 & 125.2 & 110 & 398.4\\ \hline 9 & 150.8 & 124 & 424.0\\ \hline 10 & 174 & 138 & 447.2\\ \hline\end{array}$$

Here $\,m=14n-2\,$ is the molecular mass and $\,T_\mathrm{boil}\,$ is the boiling point temperature converted to kelvins. Surprisingly, when I plot $T_\mathrm{boil}$ v.s. $m$, I find a power law $T_\mathrm{boil}\propto \sqrt{m}$. The melting point temperatures were also shown in the video but were not so regular. Can someone explain why the boiling point satisfies such a simple power law?

Zhuoran He
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  • Which isomers are these? – Mithoron Jan 13 '18 at 22:55
  • related https://chemistry.stackexchange.com/questions/28576/alkane-alkene-alkyne-boiling-point – Mithoron Jan 13 '18 at 22:59
  • @Mithoron, based on the boiling point of $1$-butyne ($8.1,^\circ\mathrm{C}$) and $2$-butyne ($27,^\circ\mathrm{C}$), these should all be $1$-alkynes to exhibit such consistent pattern. – Zhuoran He Jan 13 '18 at 23:15
  • For the curious, I plotted the first ten boiling points, melting points, and their square roots in kelvin of alkanes, and there's no discernable pattern in them. Here's the graphs. Here's the sheet to fiddle with – Gaurang Tandon Jan 17 '18 at 03:06
  • @GaurangTandon, the melting points are not so regular as I observed for alkynes as well. For the boiling points, if you plot $T_{\mathrm{boil}}$ v.s. $\sqrt{m}$ with $m=14n+2$, you get a straight line. I'm not sure if there's a simple theory behind this empirical $\sqrt{m}$-law. – Zhuoran He Jan 17 '18 at 03:21
  • @ZhuoranHe oh, you're right. I should have plotted against molar mass instead (facepalm). Editing – Gaurang Tandon Jan 17 '18 at 09:50
  • @ZhuoranHe Ok, here's the new sheet. You're right. There's a straight line relation between $Tb$ and $\sqrt{m}$. – Gaurang Tandon Jan 17 '18 at 10:10
  • I did linear regression analysis (fancy name for data entry into calculator :P) and obtained $T_b=0.431287764+0.024704463\sqrt{m}$ for alkynes. Not a very good linear relation though :( Let's check if it fits higher alkynes. – Gaurang Tandon Jan 17 '18 at 10:18

1 Answers1

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I would like to try to give a more detailed answer, but this is all I can give right now. I am fairly confident this is what is going on.

The dominant intermolecular force which determines the boiling point is dispersion. Because dispersion is the result of instantaneous dipoles, the strength of the interaction is proportional to the "surface area" of the molecule. There is no well-defined measure of the surface area of a molecule, but the most common way to describe this would be by using a van der Waals' radius for the atoms/groups.

The important point, however, is that when we add another carbon with its hydrogens, to a hydrocarbon, the mass increases linearly, while the surface area increases proportional to the square of the van der Waals' radius.

So, if we believe that dispersion forces are truly proportional to surface area and hence truly proportional to the boiling temperature, then a change of mass proportional to $\sqrt m$ should make a linear change in the surface area of the molecule, which should make a linear change the boiling temperature.

Notice, however, that this argument should be exactly the same for a linear alkane, and not just an alkyne. I made this same plot for the first fifteen n-alkanes, and the trend is very linear.

sqrt(m) vs. T

As to why there isn't a similar trend for the melting points, melting temperature is much more sensitive to symmetry than the boiling point, so how the molecules stack will make quite a difference. I would expect that hydrocarbons do not form very good crystals, so the trends may be highly dependent on the specific molecule.

Also, notice that the more branched these molecules are the less this argument should apply to the boiling point because each groups added at branching points will add less surface area than for the straight alkane because some of the surface area will be "redundant" so to speak.

Great observation! I hope my answer is right. I'm sure this has been noticed before, so it would be interesting if someone could find references.

Another way of thinking about this which I like is this:

Imagine that we can add mass to the carbon chain continuously rather than in lumps of atoms. Because adding mass also adds length to the chain, adding mass adds surface area. From this perspective, the relationship between $\sqrt m$ and surface area is the same as the relationship between length and surface area. This follows because these molecules are shaped a lot like cylinders.

So, the only jump left is from surface area to boiling point, which is already understood to be related through dispersion forces.

Mathew Mahindaratne
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jheindel
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  • The empirical $\sqrt{m}$-law is found to work for small molecules. For example, if you extrapolate the alkane data to $n=0$, the boiling point of $\ce{H2}$ is reproduced with good accuracy. For very big $n$ I found a paper that uses a logarithmic function to fit the boiling point temperature (no explanation). To understand the $\sqrt{m}$-law, we would need a model for small molecules that allows to do calculations. – Zhuoran He Jan 19 '18 at 14:38
  • The geometrical explanation is very attractive, but the outer surface of a cylinder scales linearly with the cylinder length $S = 2\pi r * L$, so I would expect a linear scaling with the number of atoms rather than a square root ? – DSuchet Feb 05 '21 at 12:48