I cannot figure out a question on my review sheet. The full question is:

Methane ($\ce{CH4}$) is a gas at room temperature but chloroform ($\ce{CH3Cl}$) is a liquid. Explain why changing just one $\ce{H}$ atom for a $\ce{Cl}$ atom changes the compound's state at room temp.

I thought about it, and I think it does have to do with $\ce{CH4}$ is non polar, so it does not tend to stick to each other (gas state), while $\ce{CH3Cl}$ is polar, so it does stick, such as like $\ce{H2O}$, which is liquid and is cohesive, so $\ce{CH3Cl}$ would be as well. Is anyone able to tell me if I am on the right path?

  • $\begingroup$ see chemistry.stackexchange.com/questions/6859/… $\endgroup$
    – Mithoron
    Nov 16, 2015 at 2:40
  • 1
    $\begingroup$ The question itself if just wrong: chloromethane is not chloroform and chloromethane is not a liquid but a gas at room temperature. That's a pretty bad epic fail for a chemistry question. $\endgroup$
    – matt_black
    Sep 17, 2019 at 10:29

2 Answers 2


Yes, I think that is the right answer expected for this question.

Three factors to consider.

(1) Most importantly there four intermolecular forces which effect BP

Ionic bonds > Ion-dipole interactions > H bonding > dipole-dipole interactions > Van der Waals dispersion forces.

(2) Increasing the number of carbon atoms (e.g. n-alkanes) increases the BP

(3) Branching on carbon chain lowers the BP.

The answer being given though is a bit of hand-waving that oversimplifies the situation. Dipole moment alone can't explain the overall trends. Van der Waals dispersion forces must be considered also.

For instance with increasing substitution $\ce{CH2Cl2}$, and $\ce{CHCl3}$ both have lower dipole moments than $\ce{CH3Cl}$ but higher BP's. $\ce{CCl4}$ has no dipole moment like methane, but has the highest BP of all.

                                          Bond Lengths
           Dipole     B.P.       Mol.
           Moment     (°C)       Wt.       C-H     C-X
 CH4         0        −161.49    16.04     108.7    ---
 CH3Cl       1.9       −23.8     50.49     111      178.3   
 CH2Cl2      1.6        39.6     84.93     106.8    177.2
 CHCl3       1.15       61.15    119.37    107.3    176.7  
 CCl4        0          76.72    153.81    -----    176.6

Also look at the halomethanes as a series. $\ce{CH3F}$ has a smaller dipole moment than $\ce{CH3Cl}$ because the bond length is shorter.

           Dipole   B.P      Mol.      Electro-  C-X     C-H
           Moment   (°C)     Wt.       Negat.    Length  Length
 CH3F       1.85    −78.4    34.03     4.0       138.5   109.5
 CH3Cl      1.87    −23.8    50.49     3.0       178.4   
 CH3Br      1.81      4.0    94.94     2.8       192.9
 CH3I       1.62      42     141.94    2.5       213.9 

The trend for molecular weight looks good, but the deuterated form of methane, $\ce{CD4}$, has the same (or at least nearly the same) boiling point as $\ce{CH4}$.


The answer to predict boiling points of the four substances would indeed be expected to be given as:

$\ce{CH3Cl}$ is a dipole. The dipole-dipole interactions are much stronger than the van der Waals interactions present in methane, so it’s boiling point is much higher.

When comparing $\ce{CH4}$ to $\ce{CCl4}$, the latter has a much higher boiling point due to its larger number of electrons meaning stronger van der Waals forces.

That said, the teacher that gave you the above information should be removed from chemistry classes immediately. Here’s why:

  • $\ce{CH3Cl}$ is not chloroform but methyl chloride or chloromethane. It is a gas at room temperature with a boiling point $\vartheta_\mathrm{b} = - 23.8~\mathrm{^\circ C}$.

  • Chloroform, which is indeed a liquid at room temperature ($\vartheta_\mathrm{b} = 61.2~\mathrm{^\circ C}$) is $\ce{CHCl3}$ or trichloromethane. You need to substitute three hydrogens with chlorine atoms to create chloroform.


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