Kevin B. answered • 09/19/22
Enthusiastic Orgo Professor Can Help You Unlock Organic Chemistry
This is a great boiling point question. When answering questions about BP, it's always key to remember -- when a compound boils, you are NOT breaking the covalent bonds which hold the molecule together. It is the non-covalent intermolecular attractive forces which are being overcome (or "broken") in order to cause the compound to boil.
There are generally three types of non-covalent intermolecular forces to be considered when answering a question like this -- dipole-dipole interactions, hydrogen bonds, and London dispersion forces. Hydrogen bonds are the strongest kind, and occur when a "protic" (proton-like) hydrogen "sticks" to a large partial negative charge. This is almost always between a N-H, O-H or F-H hydrogen, and a lone pair on another N, O, or F atom. Interactions between other partially positive hydrogen atoms (like Cl-H, or S-H) do not "count" as H bonds, as they are qualitatively weaker -- same thing with interactions with other partially negative atoms with lone pairs (like C-Cl, or C-S). That said, for this pair of molecules, CCl4 obviously has no H-bonding, since it has no hydrogens at all. CH2Cl2 also has no H-bonding, because the H's are attached to C, and thus do not have a large partial positive (plus, there is no N or O or F for them to "stick" to). So hydrogen bonding is absent in both molecules, and thus cannot determine the relative boiling points.
The consideration of the dipole-dipole forces requires a little thinking, as there are potentially a few incorrect assumptions you could make. First, don't assume that CCl4 is more polar than CH2Cl2 because "there are more chlorine atoms -- electronegative atoms -- so the dipole must be larger". In CCl4, the chlorine atoms point to the corners of a tetrahedron. So while each individual C-Cl bond is withdrawn toward the -Cl, forming a partial negative charge on each -Cl atom, the net dipole is ZERO because all 4 individual bond dipoles cancel each other. It's a little harder to see for a tetrahedron, but consider BF3. Fluorine is super electronegative (and boron is even less electronegative than carbon) so each B-F bond is withdrawn toward the fluorine. This puts fairly large partial negative charges on each F, and a large partial positive on the B in the middle. But since those bond dipoles are vectors, and they point 120 degrees with respect to each other, those three bond dipoles completely cancel, and the overall molecular dipole of BF3 is zero. Same deal for CCl4.
On the other hand, CH2Cl2 IS polar (well, a little). Remember also that this is tetrahedral, and not square planar -- the bonds do not point "north-south-east-west" as we might be tempted to think. If that were the case, it would matter if the chlorine atoms were adjacent (like, north and east) or opposite (like north and south), since the bond dipoles would completely cancel in the second case. But this is not physically true -- it's a tetrahedron. And it does not matter which two "points" of the tetrahedron are -Cl, the molecular dipole will point half way between those two chlorine atoms. Make a molecular model and inspect it if you need.
So! CH2Cl2 is polar, and CCl4 is not. That should settle it -- except it predicts the wrong result! Based on the dipole interactions, one would expect CH2Cl2 to have a HIGHER boiling point, and it does not! That's because, despite the fact that the dipole interactions are larger for CH2Cl2, the London dispersion forces are enough larger for CCl4 that its boiling point ends up being higher. The dispersion forces are interactions between transient dipoles which temporarily form when two molecules are in proximity (look up a textbook section on this for a deeper explanation). The magnitude of the dispersion forces depends on several things, but most strongly on the total number of electrons in the molecules. This is because the more electrons you have to shift a little bit (to make the transient dipole), the larger a transient dipole which can be created. Smaller molecules tend to have lower boiling points primarily because they have smaller dispersion forces. Note that this effect is NOT an "inertia" effect (which is a tempting incorrect explanation). It would make sense that the heavier molecule would be harder to boil, because it takes more energy to get it moving fast enough to enter the gas phase. This is not physically untrue -- but it is NOT the primary reason for the increase in boiling point.
In this case, CCl4 has a much larger molecular weight, and many more electron pairs than CH2Cl2. As a result, its dispersion forces are MUCH larger than those of CH2Cl2, and the increase in BP due to the dispersion forces overwhelms the difference due to dipole interactions. The latter would have predicted CH2Cl2 would have a higher BP, but dispersion forces are apparently a larger determining force.
