Which of the following has higher boiling points? Alkanes, alkenes, or alkynes? And why?
Answer
Disclaimer: All of this "jazz" will be about reaching a mere rule-of-thumb. You can't just compare whole families of organic compounds with each other. There are more factors to consider than below, mostly based on isomerism notions. However, as most of the A grade exams emphasize on the lighter aliphatic compounds, we can understand each other here. :)
Polarizability is the ability for a molecule to be polarized.
When determining (aka comparing) the boiling points of different molecular substances, intermolecular forces known as London Dispersion Forces are at work here. Which means, these are the forces that are overcome when the boiling occurs. (See here for example)
London forces get stronger with an increase in volume, and that's because the polarizability of the molecule increases. (See the answer to this recent question)
Alkanes vs. Alkenes
In their simplest form (where no substitution etc. has occurred) alkanes tend to have very close boiling points to alkenes.
The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids.
Boiling points of alkenes depends on more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules.
\begin{array}{|c|c|}\hline \text{Compound} & \text{Boiling point / }^\circ\mathrm{C} \\ \hline \text{Ethene} & -104 \\ \hline \text{Propene} & -47 \\ \hline \textit{trans}\text{-2-Butene} & 0.9 \\ \hline \textit{cis}\text{-2-Butene} & 3.7 \\ \hline \textit{trans}\text{-1,2-dichlorobutene} & 155 \\ \hline \textit{cis}\text{-1,2-dichlorobutene} & 152 \\ \hline \text{1-Pentene} & 30 \\ \hline \textit{trans}\text{-2-Pentene} & 36 \\ \hline \textit{cis}\text{-2-Pentene} & 37 \\ \hline \text{1-Heptene} & 115 \\ \hline \text{3-Octene} & 122 \\ \hline \text{3-Nonene} & 147 \\ \hline \text{5-Decene} & 170 \\ \hline \end{array} In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons.
Alkanes vs. Alkynes
As explained, since there is a bigger volume to an alkane than its corresponding alkyne (i.e. with the same number of carbons) the alkane should have a higher boiling point. However, there's something else in play here:
Alkynes, have a TRIPLE bond!
I currently can think of two things that happen as a result of this:
London Dispersion Forces are in relation with distance. Usually, this relation is $r^{-6}$. (See here) The triple bond allows two alkynes to get closer. The closer they are, the more the electron densities are polarised, and thus the stronger the forces are.
Electrons in $\pi$ bonds are more polarizable$^{10}$.
These two factors overcome the slight difference of volume here. As a result, you have higher boiling points for alkynes than alkanes, generally.
\begin{array}{|c|c|}\hline \text{Compound} & \text{Boiling point / }^\circ\mathrm{C} \\ \hline \text{Ethyne} & -84^{[1]} \\ \hline \text{Propyne} & -23.2^{[2]} \\ \hline \text{2-Butyne} & 27^{[3]} \\ \hline \text{1,4-Dichloro-2-butyne} & 165.5^{[4]} \\ \hline \text{1-Pentyne} & 40.2^{[5]} \\ \hline \text{2-Heptyne} & 112\text{–}113^{[6]} \\ \hline \text{3-Octyne} & 133^{[7]} \\ \hline \text{3-Nonyne} & 157.1^{[8]} \\ \hline \text{5-Decyne} & 177\text{–}178^{[9]} \\ \hline \end{array} 1: http://en.wikipedia.org/wiki/Acetylene
2: http://en.wikipedia.org/wiki/Propyne
3: http://en.wikipedia.org/wiki/2-Butyne
4: http://www.lookchem.com/1-4-Dichloro-2-butyne/
5: http://en.wikipedia.org/wiki/1-Pentyne
6: http://www.chemsynthesis.com/base/chemical-structure-17405.html
7: http://www.chemspider.com/Chemical-Structure.76541.html
8: http://www.thegoodscentscompany.com/data/rw1120961.html
9: http://www.chemsynthesis.com/base/chemical-structure-3310.html
10: https://chemistry.stackexchange.com/a/27531/5026
Conclusion: We can't fully determine the boiling points of the whole class of alkanes, alkenes and alkynes. However, for the lighter hydrocarbons, comparing the boiling points, you get: $$\text{Alkynes > Alkanes > Alkenes}$$
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