Intermolecular forces (IMFs) can be used to predict relative boiling points. The stronger the IMFs, the lower the vapor pressure of the substance and the higher the boiling point. Therefore, we can compare the relative strengths of the IMFs of the compounds to predict their relative boiling points.
H-bonding > dipole-dipole > London dispersion (van der Waals)
When comparing compounds with the same IMFs, we use size and shape as tie breakers since the London dispersion forces increase as the surface area increases. Since all compounds exhibit some level of London dispersion forces and compounds capable of H-bonding also exhibit dipole-dipole, we will use the phrase "dominant IMF" to communicate the IMF most responsible for the physical properties of the compound.
In the table below, we see examples of these relationships. When comparing the structural isomers of pentane (pentane, isopentane, and neopentane), they all have the same molecular formula C5H12. However, as the carbon chain is shortened to create the carbon branches found in isopentane and neopentane the overall surface area of the molecules decreases. The visual image of MO theory can be helpful in seeing each compound as a cloud of electrons in an all encompassing MO system. Branching creates more spherical shapes noting that the sphere allows the maximum volume with the least surface area. The structural isomers with the chemical formula C2H6O have different dominant IMFs. The H-bonding of ethanol results in a liquid for cocktails at room temperature, while the weaker dipole-dipole of the dimethylether results in a gas a room temperature. In the last example, we see the three IMFs compared directly to illustrate the relative strength IMFs to boiling points.
The observable melting and boiling points of different organic molecules provides an additional illustration of the effects of noncovalent interactions. The overarching principle involved is simple: the stronger the noncovalent interactions between molecules, the more energy that is required, in the form of heat, to break them apart. Higher melting and boiling points signify stronger noncovalent intermolecular forces.
Consider the boiling points of increasingly larger hydrocarbons. More carbons means a greater surface area possible for hydrophobic interaction, and thus higher boiling points.
As you would expect, the strength of intermolecular hydrogen bonding and dipole-dipole interactions is reflected in higher boiling points. Just look at the trend for hexane (nonpolar London dispersion interactions only ), 3-hexanone (dipole-dipole interactions), and 3-hexanol (hydrogen bonding).
Of particular interest to biologists (and pretty much anything else that is alive in the universe) is the effect of hydrogen bonding in water. Because it is able to form tight networks of intermolecular hydrogen bonds, water remains in the liquid phase at temperatures up to 100 OC, (slightly lower at high altitude). The world would obviously be a very different place if water boiled at 30 OC.
1. Based on their structures, rank phenol, benzene, benzaldehyde, and benzoic acid in terms of lowest to highest boiling point.
By thinking about noncovalent intermolecular interactions, we can also predict relative melting points. All of the same principles apply: stronger intermolecular interactions result in a higher melting point. Ionic compounds, as expected, usually have very high melting points due to the strength of ion-ion interactions (there are some ionic compounds, however, that are liquids at room temperature). The presence of polar and especially hydrogen-bonding groups on organic compounds generally leads to higher melting points. Molecular shape, and the ability of a molecule to pack tightly into a crystal lattice, has a very large effect on melting points. The flat shape of aromatic compounds such as napthalene and biphenyl allows them to stack together efficiently, and thus aromatics tend to have higher melting points compared to alkanes or alkenes with similar molecular weights.
Comparing the melting points of benzene and toluene, you can see that the extra methyl group on toluene disrupts the molecule's ability to stack, thus decreasing the cumulative strength of intermolecular London dispersion forces.
Note also that the boiling point for toluene is 111 oC, well above the boiling point of benzene (80 oC). The key factor for the boiling point trend in this case is size (toluene has one more carbon), whereas for the melting point trend, shape plays a much more important role. This makes sense when you consider that melting involves ‘unpacking’ the molecules from their ordered array, whereas boiling involves simply separating them from their already loose (liquid) association with each other.
If you are taking an organic lab course, you may have already learned that impurities in a crystalline substance will cause the observed melting point to be lower compared to a pure sample of the same substance. This is because impurities disrupt the ordered packing arrangement of the crystal, and make the cumulative intermolecular interactions weaker.
An interesting biological example of the relationship between molecular structure and melting point is provided by the observable physical difference between animal fats like butter or lard, which are solid at room temperature, and vegetable oils, which are liquid. Both solid fats and liquid oils are based on a ‘triacylglycerol’ structure, where three hydrophobic hydrocarbon chains of varying length are attached to a glycerol backbone through an ester functional group (compare this structure to that of the membrane lipids discussed in section 2.4B).
Interactive 3D image of a saturated triacylglycerol (BioTopics)
Saturated vs mono-unsaturated fatty acid (BioTopics)
In vegetable oils, the hydrophobic chains are unsaturated, meaning that they contain one or more double bonds. Solid animal fat, in contrast, contains saturated hydrocarbon chains, with no double bonds. The double bonds in vegetable oils cause those hydrocarbon chains to be more rigid, and ‘bent’ at an angle (remember that rotation is restricted around double bonds), with the result that they don’t pack together as closely, and thus can be broken apart (ie. melted) more readily. Shown in the figure above is a polyunsaturated fatty acid chain (two double bonds), and you can click on the link to see interactive images of a saturated fatty acid compared to a monounsaturated fatty acid (one double bond).
2. Arrange the following compounds in order of decreasing boiling point.
Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
Last updated: March 21st, 2021 |
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Figuring out the order of boiling points is all about understanding trends. The key thing to consider here is that boiling points reflect the strength of forces between molecules. The more they stick together, the more energy it will take to blast them into the atmosphere as gases.
There are 3 important trends to consider.
Let’s have a closer look:.
Table of Contents
1. Trend #1: The relative strength of the four intermolecular forces .
Compare the different butane alcohol derivatives shown below. Molecules of diethyl ether, C4H10 O, are held together by dipole-dipole interactions which arise due to the polarized C-O bonds. Compare its boiling point of (35 °C)with that of Its isomer butanol (117 °C). The greatly increased boiling point is due to the fact that butanol contains a hydroxyl group, which is capable of hydrogen bonding. Still, the attractive forces in butanol pale in comparison to those of the salt sodium butoxide, which melts at an extremely high temperature (well above 260 °C) and actually decomposes before it can turn into a liquid.
Then think about butane, C4H10, which contains no polar functional groups. The only attractive forces between individual butane molecules are the relatively weak Van der Waals dispersion forces. The result is that butane boils at the temperature at which water freezes (0° C), far below even that of diethyl ether.
Moral of the story: among molecules with roughly similar molecular weights, the boiling points will be determined by the functional groups present.
You could tell a similar tale for the similar amine and carboxylic acid isomers shown below.
For a previous discussion of the 4 intermolecular forces, see here. For the reference in Reusch’s textbook, see here.
2. Trend #2 – For molecules with a given functional group, boiling point increases with molecular weight.
Look at the dramatic increases in boiling points as you increase molecular weight in all of these series:
Here’s the question: How, exactly do intermolecular forces increase as molecular weight increases?
Well, the key force that is acting here are Van der Waals dispersion forces, which are proportional to surface area. So as you increase the length of the chain, you also increase the surface area, which means that you increase the ability of individual molecules to attract each other.
On an intuitive level, you could compare these long molecules to strands of spaghetti – the longer the noodles, the more work it takes to pull them apart. As the chain length increases, there will be regions where they can line up next to each other extremely well.
Individually, each interaction might not be worth very much, but when you add them all up over the length of a chain, Van der Waals dispersion forces can exert tremendous effects.
3. The Role Of Symmetry (or lack thereof) On Melting And Boiling Points
This is another byproduct of the surface-area dependence of Van der Waals dispersion forces – the more rod-like the molecules are, the better able they will be to line up and bond. To take another intuitive pasta example, what sticks together more: spaghetti or macaroni? The more spherelike the molecule, the lower its surface area will be and the fewer intermolecular Van der Waals interactions will operate. Compare the boiling points of pentane (36°C) and 2,2-dimethyl propane (9 °C).
It can also apply to hydrogen bonding molecules like alcohols – compare the boiling points of 1-pentanol to 2-pentanol and 3-pentanol, for instance. The hydroxyl group of 1-pentanol is more “exposed” than it is in 3-pentanol (which is flanked by two bulky alkyl groups), so it will be better able to hydrogen bond with its fellows.
In summary, there are three main factors you need to think about when confronted with a question about boiling points. 1) what intermolecular forces will be present in the molecules? 2) how do the molecular weights compare? 3) how do the symmetries compare?
One last quick question for the road (see comments for answer).
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