Does CCl4 have a higher boiling point than CH2Cl2? I though London forces were weaker than dipole-dipole forces?

Apparently yes, but London dispersion forces ARE weaker than dipole-dipole forces.

It looks like the reason for the exception here in boiling point trends is that there is a greater increase in entropy due to boiling #"CH"_2"Cl"_2# than #"CCl"_4#, and it requires less thermal energy to boil #"CH"_2"Cl"_2# than #"CCl"_4#. (These are not competing data.)

#"CCl"_4# is completely symmetrical, and #"CH"_2"Cl"_2# is not. Therefore, the latter has dipole-dipole interactions, and the former has only London dispersion.

I went into this in more detail here.

As a result, #"CH"_2"Cl"_2# is anticipated to have the HIGHER boiling point; it is EXPECTED to interact with itself more strongly than #"CCl"_4# interacts with itself, and is EXPECTED to be harder to boil.

[However, #"CH"_2"Cl"_2# does indeed have a lower boiling point than #"CCl"_4#. Look these up on NIST.]

#T_b("CH"_2"Cl"_2) = ul"313 K"#, #DeltaH_"vap" ~~ ul"28.06 kJ/mol"#
#T_b("CCl"_4) = ul"349.8 K"#, #DeltaH_"vap" ~~ ul"29.82 kJ/mol"#

which is opposite to the trend we expected.

It may have to do with the idea that #"Cl"# is so much heavier than #"H"# when comparing #"CH"_2"Cl"_2# with #"CCl"_4#. I would suspect that the sheer size of #"CCl"_4# compared with #"CH"_2"Cl"_2#, as well as its RMS speed, is enough to push its boiling point higher from a borderline position.

Both the changes in enthalpy and entropy agree and do not compete (they are not conflicting trends).

Higher #DeltaH_"vap"# #-># harder to boil

For any phase change at constant temperature and pressure,

#DeltaH_"tr" = TDeltaS_"tr"#

Therefore,

#DeltaH_"vap" = T_bDeltaS_"vap"#

And so, the changes in entropy due to boiling are:

#DeltaS_"vap"("CH"_2"Cl"_2) = ("28.06 kJ/mol")/("313 K") xx "1000 J"/"kJ" = ul("89.6 J/mol"cdot"K")#

#DeltaS_"vap"("CCl"_4) = ("29.82 kJ/mol")/("349.8 K") xx "1000 J"/"kJ" = ul("85.2 J/mol"cdot"K")#

Lower #DeltaS_"vap"# #-># harder to boil

As it turns out, #"CH"_2"Cl"_2# boils at a lower temperature because its increase in entropy is larger, and #"CCl"_4# boils at a higher temperature because its increase in entropy is lower. But also, it requires less thermal energy to boil #"CH"_2"Cl"_2#.

[So, both enthalpy and entropy conspire to boil #"CH"_2"Cl"_2# at a lower temperature.]