# Why are DCM and chloroform so resistant towards nucleophilic substitution?

In the book Organic Chemistry by J. Clayden, N. Greeves, S. Warren, and P. Wothers I found the following reasoning:

You may have wondered why it is that, while methyl chloride (chloromethane) is a reactive electrophile that takes part readily in substitution reactions, dichloromethane (DCM) is so unreactive that it can be used as a solvent in which substitution reactions of,other alkyl halides take place. You may think that this is a steric effect: Indeed, $$\ce{Cl}$$ is bigger than $$\ce{H}$$. But $$\ce{CH2Cl2}$$ is much less reactive as an electrophile than ethyl chloride or propyl chloride: there must be more to its unreactivity. And there is: Dichloromethane benefits from a sort of 'permanent anomeric effect.' One lone pair of each chlorine is always anti-periplanar to the other $$\ce{C–Cl}$$ bond so that there is always stabilization from this effect.

So, in MO-terms the situation would look something like this.

The reasoning looks plausible to me. The interaction between the free electron pair on $$\ce{Cl}$$ and the $$\sigma^{*}$$ orbital of the neighboring $$\ce{C-Cl}$$ bond, which would be the LUMO of DCM, lowers the energy of the free-electron-pair-orbital, thus stabilizing the compound and it raises the energy of the LUMO, thus making DCM less reactive towards nucleophiles.

But how important is this anomeric effect actually for explaining the unreactiveness of DCM toward nucleophiles, especially compared to the steric effect? And how important is the steric effect actually: Is the steric hindrance exerted by the second $$\ce{Cl}$$ in $$\ce{CH2Cl2}$$ really that much larger than the steric hindrance exerted by the methyl group in $$\ce{CH3CH2Cl}$$ (which is a good electrophile for $$\mathrm{S_N2}$$ reactions)?

I'm a little skeptical because if this anomeric effect was very important I would have expected that the $$\ce{C-Cl}$$ bond length in DCM would be slightly higher than in methyl chloride because there should be some transfer of electron density from the free electron pair into the antibonding $$\sigma^{*}$$ orbital, which should weaken the $$\ce{C-Cl}$$ bond. But the actual bond lengths don't show this. They show rather the contrary:

$$\begin{array}{c|c} \hline \text{Species} & \text{Average }\ce{C-Cl}\text{ bond length / Å} \\ \hline \ce{CH3Cl} & 1.783 \\ \ce{CH2Cl2} & 1.772 \\ \ce{CHCl3} & 1.767 \\ \ce{CCl4} & 1.766 \\ \hline \end{array}$$ (source: Wikipedia)

Now, I know that the stronger polarization of the $$\ce{C}$$ atomic orbitals in di-, tri-, and tetrachlorinated methane as compared to methyl chloride (due to the electronegativity of $$\ce{Cl}$$) should lead to an overall strengthening of the $$\ce{C-Cl}$$ bonds in those compounds. But I would have expected a trend that would show only a slight decrease (or maybe even an increase) of bond length when going from methyl chloride to DCM and then a more pronounced decrease when going from DCM to chloroform followed by a similar decrease when going from chloroform to tetrachloromethane. But instead the polarization effect seems to only slightly increase on adding more chlorine atoms.

• Can you tell which chapter if it's not a problem? – RBW Aug 29 '14 at 20:43
• @Marko It's in the first edition, chapter 42, page 1133. – Philipp Aug 29 '14 at 20:48
• Never mind if one of the starting orbitals was an anti-bonding one. Then electrons aren't put in this anti-bonding orbital, but in the newly formed orbital which has the lowest energy in that system. The new anti-bonding orbital (new LUMO) is very high on the energy scale, but without any electrons in it, no destabilization in the system occurs. – RBW Aug 30 '14 at 12:51
• @Marko But the in-phase combination of the $\sigma^{*}(\ce{C-Cl})$ orbital and the free-electron-pair-orbital will have to some extent a $\ce{C-Cl}$-antibonding character. The amount of $\sigma^{*}(\ce{C-Cl})$-character in the newly formed orbital will of course depend on the energetic difference between the interacting orbitals and their orbital overlap. But if the interaction is strong enough to raise the LUMO enough to make DCM unreactive then I would definitely expect some bond-lengthening effect as well. – Philipp Aug 30 '14 at 12:58
• @Marko, I don't understand what your objection is. I think the claim is that in-phase combination of $\sigma^{*}\ce{(C-Cl)}$ with $n\ce{(e^{-})}$ is energy-lowering in the reactant ground state. The destabilization would occur in the transition state as the electrons of the nucleophile interact with the LUMO (which has been made more energetic by inclusion of the orbital combination described above). – Greg E. Aug 30 '14 at 19:45

# Introduction (and Abstract, TLDR)

In very short words you can say, that the anomeric effect is responsible for the lack of reactiveness. The electronic effect may very well be compensating for the the steric effect, that could come from the methyl moiety. In any way, most of the steric effects can often been seen as electronic effects in disguise.

# Analysis of Molecular Orbitals

I will analyse the bonding picture based on calculations at the density fitted density functional level of theory, with a fairly large basis set: DF-BP86/def2-TZVPP. As model compounds I have chosen chloromethane, dichloromethane, chloroform and chloroethane.

First of all let me state, that the bond lengths are a little larger at this level, however, the general trend for shortening can also be observed. In this sense, chloroethane behaves like chloromethane. An attempt to explain this will be given at the end of this article.

\begin{array}{lr}\hline \text{Compound} & \mathbf{d}(\ce{C-Cl})\\\hline \ce{ClCH3} & 1.797\\ \ce{Cl2CH2} & 1.786\\ \ce{Cl3CH} & 1.783\\\hline \ce{ClCH2CH3} & 1.797\\\hline \end{array}

In the canonical bonding picture, it is fairly obvious, that the electronic effects dominate and are responsible for the lack of reactivity. In other words, the lowest unoccupied molecular orbital is very well delocalised in the dichloromethane and chloroform case. This is effectively leaving no angle to attack the antibonding orbitals.

In the mono substituted cases there is a large coefficient at the carbon, where a nucleophile can readily attack.

One can also analyse the bonding situation in terms of localised orbitals. Here I make use of a Natural Bond Orbital (NBO) analysis, that transforms the canonical orbitals into hybrid orbitals, which all have an occupation of about two electrons. Due to the nature of the approach, it is no longer possible to speak of HOMO or LUMO, when analysing the orbitals. Due to the nature of the calculations, i.e. there are polarisation functions, the values do not necessarily add up to 100%. The deviation is so small, that it can be omitted.
The following table shows the composition (in $\%$)of the carbon chloro bond and anti bond. \begin{array}{lrr}\hline \text{Compound} &\sigma-\ce{C-Cl} & \sigma^*-\ce{C-Cl}\\\hline \ce{ClCH3} & 45\ce{C}(21s79p) 55\ce{Cl}(14s85p) & 55\ce{C}(21s79p) 45\ce{Cl}(14s85p)\\ \ce{Cl2CH2} & 46\ce{C}(22s77p) 54\ce{Cl}(14s85p) & 54\ce{C}(22s77p) 46\ce{Cl}(14s85p)\\ \ce{Cl3CH} & 48\ce{C}(24s76p) 52\ce{Cl}(14s86p) & 52\ce{C}(24s76p) 48\ce{Cl}(14s86p)\\\hline \ce{ClCH2CH3} & 44\ce{C}(19s81p) 56\ce{Cl}(14s85p) & 56\ce{C}(19s81p) 44\ce{Cl}(14s85p)\\\hline \end{array} As we go from mono- to di- to trisubstituted methane, the carbon contribution increases slightly, along with the percentage of $s$ character. More $s$ character usually means also a stronger bond, which often results in a shorter bond distance. Of course, delocalization will have a similar effect on its own.

The reason, why dichloromethane and chloroform are fairly unreactive versus nucleophiles, has already been pointed out in terms of localised bonding. But we can have a look at these orbitals as well.
In the case of chloromethane, the LUMO has more or less the same scope of the canonical orbital, with the highest contribution from the carbon. If we compare this antibonding orbital to an analogous orbital in dichloromethane or chloroform, we can expect the same form. We soon run into trouble, because of the localised $p$ lone pairs of chlorine. Not necessarily overlapping, but certainly in the way of the "backside" of the bonding orbital. In the case of chloroethane we can observe hyperconjugation. However, this effect is probably less strong, and from the canonical bonding picture we could also assume, that this increases the polarisation of the antibonding orbital in favour of carbon.
In the following pictures, occupied orbitals are coloured red and yellow, while virtual orbitals are coloured purple and orange.

(Note that in chloroform two lone pair orbitals are shown.)

# Conclusion

Even though this article does not use the Valence Bond Approach, one can clearly see the qualitative manifestation of Bent's Rule (compare also: Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?). A higher $s$ character means a shorter bond. The lack of reactivity towards nucleophiles can be explained electronically with a delocalised LUMO. In terms of localised bonding, the lone pairs of any additional chlorine atom would provide sufficient electron density, to shield the backside attack on the carbon.

• Awesome answer, thank you very much. But since I'm not very familiar with NBO could you maybe explain why you say that the localised $\mathrm{p}$ lone pairs of chlorine are "not necessarily overlapping" with the (anti-)bonding orbital of the $\ce{C-Cl}$ bond? Because looking at the form of the orbitals it look to me that they could overlap quite well, but then again, as I said, I don't know much about NBO. Also would you share your opinion on my own answer which is mostly concerned with solving the apparent contradiction concerning the the bond lengths? Do you think the reasoning is sound? – Philipp Sep 1 '14 at 17:06
• NBO transforms canonical orbitals into localised orbitals, i.e. it uses linear combinations to form orbitals that are concentrated to a certain spatial area. Symmetry constraints are lifted and these orbitals are no eigenfunctions of the Schrödinger equation. The canonical orbitals are already the optimal solution. With "not necessarily overlapping" I mean, that in the ground state you have two entirely separate orbitals. They do not form new orbitals. They could interact with each other when an external potential would necessitate this. – Martin - マーチン Sep 2 '14 at 3:42

But the main question remains: Is the anomeric effect the main cause for the unreactiveness of DCM towards nucleophiles or is it the steric hinderance exerted by the second Cl ?

Assessment of Steric Effects

Bonds, like humans, have a length and a girth. In some reactions, the reaction is most likely to occur when the attacking group strikes an atom or substituent "head on". In other reactions, like the $\ce{S_{N}2}$, the attacking group approaches from the back side and slips by the various substituents in a side-ways manner as it approaches the carbon under attack.

There is a nice way to measure this side-ways (girth of a bond, if you will) interference in organic systems. Axial substituents in a cyclohexane ring are subject to what is termed "A-strain" produced by the interactions depicted in the figure below. These interactions are very analogous to the gauche interactions

in butane and arise from side-ways interaction of the substituent and the two diaxial hydrogens located two carbons away. Equatorial substituents avoid these repulsive interactions. The smaller the substituent, the less repulsive the 1,3 interaction will be and, at equilibrium, more of the substituent should exist in the axial position. Conversely, the larger the substituent, more of it should reside in the equatorial position. By measuring the axial/equatorial ratio for various substituents we can meaningfully assess the side-ways steric bulk of the substituent.

For chlorocyclohexane the axial/equatorial ratio is ca. 31/69 ($\ce{\Delta G = 0$.$48 kcal/mol}$) at room temperature. For methylcyclohexane the axial/equatorial ratio is ca. 5/95 ($\ce{\Delta G = 0$.$1.7 kcal/mol}$) at room temperature. This analysis clearly suggests that the chloro substituent is "smaller" than a methyl substituent when we use this side-ways steric probe.

For the interested reader, here is a Table of axial/equatorial $\ce{\Delta G's}$ (in kcal/mol) for other substituents.

Since this analysis suggests that a chlorine substituent is smaller than a methyl substituent, in terms of the side-ways interaction expected in a back side $\ce{S_{N}2}$ attack, it would be difficult to use a steric argument to explain the reduced reactivity in the polyhalomethanes towards $\ce{S_{N}2}$ reaction. This leaves the anomeric effect as a reasonable explanation for the reduced reactivity of the polyhalomethanes. But...

Other Thoughts

• How unreactive are these compounds to $\ce{S_{N}2}$ reaction? Dichloromethane, chloroform and carbon tetrachloride will all undergo the $\ce{S_{N}2}$ reaction with various nucleophiles, under various conditions. Before I sign up for the anomeric effect explanation, it would be nice to know exactly what reaction Clayden, et. al. were discussing. Perhaps some other factor (hydrogen bonding, who knows what) suppressed the nucleophile's reactivity in Clayden's series of compounds.
• Bond strengths: $\ce{C-Cl}$ bond strengths decrease from chloromethane (80 kcal/mol) to carbon tetrachloride (70 kcal/mol). This would cause us to suspect the chloromethane might react the slowest and carbon tetrachloride the fastest. That this is not the case is another argument in support of the anomeric effect playing a key role.
• Thanks for bringing the A-strain into the discussion and explaining it so nicely. As for Clayden et al.: You can find the passage on p. 1133 of the book (first ed.) if you want to read it. They discuss this anomeric effect in the context of heterocycles and bring in some unusual examples such as that of DCM. The statement about the unreactiveness of DCM is made without reference to specific reactions. But they also write in a small box: "Dichloromethane will react as an electrophile, but it needs a very powerful nucleophile and long reaction times" and give the following example reaction... – Philipp Sep 1 '14 at 18:30
• ... $\ce{PhSNa ->[\substack{\ce{CH2Cl2}\\ \text{as solvent}} ][\text{several days}] (PhS)2CH2}$ – Philipp Sep 1 '14 at 18:30
• I really dislike the concept of strain. If you put a hydrogen and a chlorine atom in close proximity then this would be primarily an attractive interaction (dipole-dipole and electrostatics and orbitals). The same applies to other substituents. Only if you put them too close the nuclei start to repel each other (but this only starts when it is significantly shorter than the covalent radii). I would expect that hyperconjugation has a much larger effect in stabilising the equatorial position, than stabilising the axial position of the chlorine. – Martin - マーチン Sep 2 '14 at 6:31
• On a completely unrelated matter: Why do you put pictures in the middle of a sentence? I always have the feeling there is missing something... – Martin - マーチン Sep 2 '14 at 6:43
• @Martin As far as I know the origin of the strain is the interaction of filled orbitals on the two neighboring atoms. Only the axial bonds are in the right position to let the bond orbitals overlap appreciably. So, in an axial position of a cyclohexylchloride ring you have the $\sigma(\ce{C-Cl})$ orbital interacting with a neighboring axial $\sigma(\ce{C-H})$ orbital. As both orbitals are filled you get a 2-center-4-electron interaction which will be destabilizing (because the out-of-phase combination will more destabilized than the in-phase combination is stabilized) as it raises the energy. – Philipp Sep 2 '14 at 19:34

Thinking about it some more made me realize how the problem with the bond lenghts might be resolved. Let's look at the orbital picture again:

I still think, that the in-phase combination of the $\sigma^{∗}(\ce{C-Cl})$ orbital and the free-electron-pair-orbital $n(\ce{e-})$ will have to some extent - depending on the energetic difference between the interacting orbitals and their orbital overlap - an $\ce{C-Cl}$-antibonding character, thus weakening the blue $\ce{C-Cl}$ bond (a simplified view would be, that on interaction electron density flows from the free electron pair into antibonding $\sigma^{∗}(\ce{C-Cl})$ orbital and more electron density in an antibonding orbital mean a weakening of the corresponding bond). But my mistake was not to also consider the $\pi$-bonding interaction between the red $\ce{Cl}$ and $\ce{C}$ in the in-phase combination. This $\pi$-bonding interaction will of course strengthen the red $\ce{C-Cl}$ bond. So resulting from this one interaction pictured above there should be a bond-shortening of the red $\ce{C-Cl}$ bond and a bond-lengthening of the blue $\ce{C-Cl}$ bond.

But the anomeric effect works not only in one direction in DCM (or Chloroform). The blue $\ce{Cl}$ atom has free electron pairs, too. And there is also a $\sigma^{∗}(\ce{C-Cl})$ orbital for the red bond that can interact with one of those free electron pairs in exactly the same way described before. Now, this interaction will strengthen the blue $\ce{C-Cl}$ bond while weakening the red $\ce{C-Cl}$ bond. So, it counteracts the effects of the interaction presented above and the bond-shortening and bond-lengthening effects should cancel each other out to some extent. But the raising of the LUMO, which makes DCM so unreactive remains unchanged.

Now, the only thing that needs some thinking is: Do the $\pi$-bonding and the $\sigma$-antibonding in the in-phase combination have a comparable magnitude? This question I can't answer in any quantitative way. So I will only list some factors of influence here. The $\ce{C-Cl}$ bond is rather long. This means the $\pi$-overlap won't be optimal. And the $n(\ce{e-})$ orbital will lie somewhat lower in energy than the $\sigma^{∗}(\ce{C-Cl})$ orbital thus weakening the covalency of the $\pi$-interaction. On the other hand, the same factors will also lower the amount of $\sigma^{∗}(\ce{C-Cl})$-character contained in the in-phase combination MO. All this leaves me with the gut feeling that the $\pi$-bonding might indeed be comparable to or even a little stronger than the $\sigma$-antibonding effect. And this would mean that the observed bond lengths are in accord with the orbital situation and the anomeric effect might very well be quite important for explaining the unreactiveness of DCM towards nucleophiles.

But the main question remains: Is the anomeric effect the main cause for the unreactiveness of DCM towards nucleophiles or is it the steric hinderance exerted by the second $\ce{Cl}$?

Maybe one could go about it by employing the van der Waals radii of a chlorine atom and a methyl group. According to Wikipedia, the radius of a chlorine atom is 175 pm while the radius of a carbon atom (without any hydrogen atoms around) is 170 pm (so it should be safe to say that the radius of a methyl group is higher than that). Here is a picture of the van der Waals surface of methyl chloride created with Avogadro

which seems to support the view that the spacial requirements of a chlorine atom and a methyl group are rather similar (but I don't know how reliable the van der Waals surface calculations by Avogadro actually are). If those lengths mirror the real spacial requirements of the groups in the context of a nucleophilic substitution reaction then a chlorine atom wouldn't exert a larger steric hinderance than a methyl group and the anomeric effect would clearly be the determining factor when explaining the unreactiveness of DCM. But I'm unsure whether resorting to van der Waals radii is justified.

• Your qualitative analysis is sound. A (partial) $\pi$ contribution would strengthen the one bond while weaken the other. But you are interpreting only a concept here, something that is not there. Since you are arguing with orbitals that cannot be assigned energies, magnitudes would have no meaning. The anomeric effect is therefore a concept, that allows us to understand delocalisation in terms of a much simpler concept - it is not the true reason. – Martin - マーチン Sep 2 '14 at 4:01
• @Martin Thanks for the comment. Your point is well taken. But as for "... something that is not there. Since you are arguing with orbitals that cannot be assigned energies, magnitudes would have no meaning": Are you sure that there is not a little more reality in there than you give it credit. I mean, the procedure of looking at the interaction of the $n(\ce{e-})$ and the $\sigma^{*}(\ce{C-Cl})$ orbitals is basically the usage of perturbational MO-theory on the problem of what happens if I take a $\ce{{}^{+}CH2-Cl}$ fragment and let it interact with a $\ce{Cl-}$ fragment... – Philipp Sep 2 '14 at 19:21
• ...of course this treatment is far from being exact and additional symmetry consideration would have to be made but in principle I think it should be possible to assign energies to the fragment orbitals used here (although I would never go so far as to really do that with any accuracy as MO-theory usually serves only a qualitative purpose). – Philipp Sep 2 '14 at 19:25
• @Philipp There are many schemes that decompose energy based on orbital interactions, but they all work in the same way, i.e. using the original AO. I guess you can go ahead and do a full Valence Bond Theory approach, to find out which resonance structure contributes how much to the total structure and then you get an idea of how strong the $\pi$ and $\sigma$ interactions are - then this approach would be (at least semi) quantitative. The unfortunate thing is, that you can only observe the total bonding directly and not the singular contributions. – Martin - マーチン Sep 3 '14 at 2:02
• Thinking about it a little more, the EDA-NOCV may give you exactly what you are looking for. (I do not have access to this program, so I cannot help out here.) I would suspect your analysis would be correct. On the other hand, I have grown quite fond of the simplicity of Bent's rule, which explains the shortening of the bonds. So maybe there is no use overanalyzing it. As for many things, there is no definite truth when it comes to interpretation, we always put some of our own to any analysis. Your analysis is well founded and (from my point of view) a valid approach to this problem. – Martin - マーチン Sep 3 '14 at 2:11

The anomeric effect where σ(C−C) orbital overlaps with the σ∗(C−Cl) should be considerably less than the overlap of n(e−) and σ∗(C−Cl). Therefore by comparing the relative rates of substitution of DCM, and a primary alkyl chloride, whose alkyl group is similar in size as the chlorine atom, against MeCl should give us the answer. I would be thankful if someone could provide the data.

• Actually, I would say it is even justified to say that there can't be anything like the anomeric effect present in alkylchlorides. A $\sigma(\ce{C-C})$/$\sigma(\ce{C-H})$ orbital is much too low in energy compared to a $\sigma^{*}(\ce{C-Cl})$ orbital to show any appreciable interaction. As for the test you suggested: I think it will be hard to say whether a primary alkyl group/methyl group and a chlorine atoms exert a similar steric effect in the context of a nucleophilic substitution reaction. – Philipp Aug 31 '14 at 17:03
• Maybe one could go about it by employing the van der Waals radii of a chlorine atom and a methyl group. According to Wikipedia, the radius of a chlorine atom is 175 pm while the radius of a carbon atom (without any hydrogen atoms around) is 170 pm. If those lengths mirror the real spacial requirements of the groups in the context of a nucleophilic substitution reaction then a chlorine atom wouldn't exert a larger steric hinderance compared to a methyl group and the anomeric effect would clearly be the determining factor when explaining the unreactiveness of DCM. – Philipp Aug 31 '14 at 17:28
• The A-values for methyl and chloro are 1.7 and 0.4 respectively, suggesting that methyl is sterically larger than chloro. – jerepierre Aug 31 '14 at 18:56
• @jerepierre I am not sure if it is valid to apply the A-values to these acyclic molecules, as A-values are derived from cyclohexane conformations. – RBW Sep 1 '14 at 9:56