Why does $\ce{CH3Cl}$, methyl chloride, have a larger dipole moment than $\ce{CHCl3}$, chloroform?

Let us consider $\ce{C-Cl}$ bond moment to be $x$ and that of $\ce{C-H}$ bond to be $y$.

In $\ce{CCl4}$, the dipole moment of any three $\ce{C-Cl}$ atoms is balanced by the dipole moment of the fourth $\ce{C-Cl}$ bond dipole moment. This means the resultant of dipole moment of three $\ce{C-Cl}$ bonds in tetrahedral structure is equal to $x$, which is equal and opposite to the dipole moment $x$ of the fourth $\ce{C-Cl}$ bond.

Now, when we replace one of the chlorines with a hydrogen, the net dipole moment equals $x+y$, (as they are in the same directions). Similarly if we carry out the same procedure for $\ce{CH3Cl}$ it can be proved that its dipole moment is also $x+y$.

enter image description here
Where have I gone wrong?

  • $\begingroup$ chemistry.stackexchange.com/questions/48066/… $\endgroup$
    – Mithoron
    Jan 9 '17 at 13:16
  • $\begingroup$ @Mithoron the discussion on the first answer is on the same topic but hasn't been concluded, and I am not able to get the answer from it...could you help? $\endgroup$
    – oshhh
    Jan 9 '17 at 14:36
  • $\begingroup$ Bond lengths aren't constant. $\endgroup$
    – Mithoron
    Jan 9 '17 at 15:20
  • 1
    $\begingroup$ @Mithoron At least the $\ce{C-Cl}$ bonds seem practically constant ($177~\mathrm{pm}$) – L. E. Sutton, L. O. Brockway, J. Am. Chem. Soc. 1935, 57, 473. $\endgroup$
    – Jan
    Jan 10 '17 at 2:25
  • $\begingroup$ tetrahedral structure is a 3D structure, not a 2D one $\endgroup$ Sep 6 '19 at 15:48

Your conclusion is built on the assumption that the bond lengths will be constant. A bond’s dipole moment varies with bond length and thus for the vector addition to give the identical product, all $\ce{C-Cl}$ and all $\ce{C-H}$ bonds would have to be identical. They’re probably not.

Sutton and Brockway studied the bond lengths in chloromethane, dichloromethane and chloroform back in 1935 by electron diffraction.[1] Their results indicate that the $\ce{C-Cl}$ bond lengths don’t vary between the compounds in question as indicated in the table below. (Note that all chlorine atoms in each molecule are homotopic and thus equal bond lengths are theoretically expected.)

$$\begin{array}{lccc}\hline \text{Compound} & \ce{CH3Cl} & \ce{CH2Cl2} & \ce{CHCl3}\\ \hline d(\ce{C-Cl})/\mathrm{pm} & 177 & 177\phantom{^\circ} & 178\phantom{^\circ} \\ \angle(\ce{Cl-C-Cl}) & \text{n. a.} & 111^\circ & 111^\circ \\ \hline\end{array}$$

These results point towards the chlorine side of things not having an impact. But there is still the hydrogen side of things. Hydrogen atoms are generally very hard to locate in diffraction type experiments. X ray diffraction is observed if photons interact with electrons in a certain manner; hydrogen atoms do not have many electrons around them and are thus hardly seen. One needs to observe diffraction at very large angles to be able to locate hydrogens. I assume, although I don’t know, that it’s similar in electron diffraction. Indeed, the paper notes:

[…] and $l_{\ce{C-H}}$ was taken as $1.06~\mathrm{\overset{\circ}{A}}$.

The two carbons in chloromethane and chloroform are, however, very much not identical. One is surrounded by three electronegative chlorine atoms which will strongly remove electron density from it, polarising the $\ce{C-Cl}$ bonds towards each of the three chlorines. The other will have only one such withdrawing partner. The different electron densities around chlorine are well represented by the corresponding $\ce{^13C}$ NMR shifts: $\delta(\ce{CH3Cl}) \approx 22~\mathrm{ppm}$; $\delta(\ce{CHCl3}) \approx 77~\mathrm{ppm}$.[2] The hydrogen shifts are also different in similar magnitudes, showing that hydrogen in chloroform is much more polarised than in chloromethane.

These different electronics must have an influence on the $\ce{C-H}$ bond length of the two molecules. I can’t imagine two electronically such different systems to have equal bond lengths. My assumption is that $\ce{CH3Cl}$ should have the shorter $\ce{C-H}$ bond lengths, since carbon is more negatively charged while hydrogen remains positively charged; this should give a slightly greater attractive force and hydrogen, being such a small element, immediately feels these. Therefore, instead of the final dipole vector $\vec d$ being equal in both cases ($\vec d = \vec x + \vec y$), we have one lesser and one stronger vector.

$$\begin{gather}\vec d_{\ce{CH3Cl}} = \vec x + \vec y + \delta\vec y\tag{1}\\ \vec d_{\ce{CHCl3}} = \vec x + \vec y - \delta\vec y\tag{2}\end{gather}$$


[1]: L. E. Sutton, L. O. Brockway, J. Am. Chem. Soc. 1935, 57, 473. DOI: 10.1021/ja01306a026.

[2]: Experimental spectra found for chloromethane/chloroform on SciFinder.


Because they are vectors and as such are not in the same direction within the molecule. from your example 4 x Cl atoms could be said to have dipole x - x = 0. You are effectively replacing 'x' with the dipole 'y' so the overall dipole for CCl3H would be 'y - x'. Similarly CH3Cl would be 'x - y' (so same magnitude but opposite direction).

What isn't taken into account with your model is steric deformation of the molecules. The three chlorine atoms missing their fourth to force a tetrahedral arrangement will adopt a much more planar arrangement; thus their contribution to the overall dipole would be reduced. The deformation and thus dipole in CH3Cl with the introduction of a single bulkier group would be less and therefore the overall dipole would be greater

  • $\begingroup$ Not opposite directions! $\endgroup$
    – oshhh
    Jan 9 '17 at 13:26
  • $\begingroup$ The deformation is just by 1 degree so why is there so much difference in dipole moment $\endgroup$
    – oshhh
    Jan 9 '17 at 13:26
  • $\begingroup$ If we consider deformation the dipole moment of $CH_3Cl$ is $x+1.05y$ and that of $CHCl_3$ is $1.05x+y$ which is very small and cannot account for the large difference in dipole moment. $\endgroup$
    – oshhh
    Jan 9 '17 at 13:29

It is a vector, parallel to the bond axis.This vector can be physically interpreted as the movement undergone by electrons when the two atoms are placed a distance d apart and allowed to interact, the electrons will move from their free state positions to be localised more around the more electronegative atom.

you can compare the dipole moment of both the molecules by these digrams. source:wikipedia. CHCl3 enter image description here

CH3Cl enter image description here

For polyatomic molecules there is more than one bond, and the total molecular dipole moment may be approximated as the vector sum of individual bond dipole moments. Often bond dipoles are obtained by the reverse process: a known total dipole of a molecule can be decomposed into bond dipoles. This is done to transfer bond dipole moments to molecules that have the same bonds, but for which the total dipole moment is not yet known. The vector sum of the transferred bond dipoles gives an estimate for the total (unknown) dipole of the molecule

  • $\begingroup$ you can reason like this: the electron cloud is distributed in case of CHCl3. $\endgroup$ Jan 9 '17 at 13:07
  • $\begingroup$ But what is wrong in the way I have calculated it? $\endgroup$
    – oshhh
    Jan 9 '17 at 13:11
  • $\begingroup$ the way you add dipole.... i didn't understand, that's not the way it is. vectors cannot be added algebraically. like the way you did. $\endgroup$ Jan 9 '17 at 13:27
  • $\begingroup$ okay, does the resultant of the dipole by 3 C-Cl point downwards? Is it parallel to the 4th C-H bond moment? Net dipole moment is the vector addition of resultant of 3C-Cl and bond moment of C-H. Both these vectors are parallel and in the same direction. So how else do you add them? $\endgroup$
    – oshhh
    Jan 9 '17 at 13:33
  • $\begingroup$ yes you are true the resultant is downward but please see the figure it is clear that they make a large angle measured from downward. so every components get added up and creates a single dipole but still it not more than the dipole moment of one C-Cl bond. i urge you please apply vector you will understand everything. $\endgroup$ Jan 10 '17 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.