Why is cis-1,2-dichloroethene more stable than trans-1,2-dichloroethene?

Usually, trans compunds are more stable than cis ones, due to less strain and its non-polarity. But, in this case it's quite the opposite. I tried to search the reason behind it, and found that it's supposedly due to the cis effect?


Here's a pre-MOT rationalisation of the cis form being more stable than the trans form for 1,2-dihaloethylenes:

The lone pair of chlorine atoms is involved in resonance with the double bond, as it does so positive charge appears on one chlorine and negative on another. A cis geometry allows for a stabilizing interaction (attraction) between these positive and negative charges, which makes this resonance form a greater contributor to the resonance hybrid than it is for the trans isomer.

Resonance structures of 1,2-dichloroethene

MOT-based explanation:

Eyring's work suggests that there is maximum delocalisation possible in trans form.

Quoting, from the same paper:

In a recent paper Binghams proposes that the main effect responsible for the “cis effect” of ethylenes is an effect which tends to destabilize the trans form rather than provide extra stabilization for the cis form. In short, this is an unfavorable conjugation effect which stems from the fact that the conjugation in these cases is between two filled orbitals (halogen lone pair and $\pi$ bond). The interaction gives rise to two new orbitals, one bonding and one antibonding, the latter being more antibonding than the former is bonding. Consequently, the total interaction is energetically unfavorable, in contrast to interaction in hydrocarbon polyenes. This would be equally true for both the cis and trans forms but, following Eyring’s argument, the delocalization is greater in the trans form. Consequently this form will be the less stable one.

Reference: Structure of 1,2-difluorocyclopropane and the "cis effect"


In this answer, I will point out an inadequacy of the reasoning based on resonance structures and I will also provided another MO perspective, which I believe is more convincing, on the issue.

A flaw in the resonance argument

As presented by Abcd, the phenomenon can be explained without use of MO theory and purely based on resonance structures. Please refer to his answer for the argument. Based on this argument, we may actually expect the $\ce {C=C}$ double bond to become weakened. Since if the two zwitterionic resonance structures were to be significant, the central double bond should naturally have less double bond character and more single bond character. However, data presented by Skancke & Boggs (1979) show that the $\ce {C-C}$ bond length is in fact shorter in the cis isomer ($\pu{1.311 Å}$) than in the trans isomer ($\pu{1.320 Å}$)$^1$. This suggests that the resonance stabilisation may not be the cause of this phenomena.

Another MO perspective

Fleming (2009) provides an explanation which I feel is more convincing. Essentially, the idea is that extra stabilisation in the cis form is derived from the interaction between the antiperiplanar $\ce {C-H}$ $\sigma$ and the low-lying $\ce{C-Cl}$ $\sigma^*$ MOs. Jan provides more details on this here when he explains the cause of a similar effect responsible for the gauche conformation being favoured in 2-fluoroethanol. Another phenomenon that Fleming touches on which he uses the same explanation for is the preference for the gauche conformation in 1,2-difluoroethane. It is also explained in the same paragraph on p. 89.

The same explanation can be used for 1,2-dichloroethene. This is because chlorine is also a rather electronegative atom, although less so than fluorine. Thus, we can still expect the $\ce {C-Cl}$ $\sigma^*$ MO to be low enough in energy to interact with the the $\ce {C-H}$ $\sigma$ MO.


  1. Skancke, A.; Boggs, J. E. Structure of 1,2-Difluorocyclopropane and the "Cis Effect". J. Am. Chem. Soc. 1979, 101 (15), 4063–4067. doi:10.1021/ja00509a009
  2. Fleming, I. Molecular Orbitals and Organic Chemical Reactions. John Wiley & Sons, Ltd. United Kingdom, 2009.

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.