In my exam, I was asked why cyclopropane could decolourise bromine water (indicating that it reacted with the bromine).

All I could guess was that it is related to the high angle strain in cyclopropane, as the C–C–C bond angle is $60^\circ$ instead of the required $109.5^\circ$. No book I have read mentions this reaction. What is the product formed, and why does it occur?

  • $\begingroup$ If cyclopropane gives the bromine water test, does it also decolourise a cold solution of potassium permanganate? If yes, why? If no, why not? Also, (ncjindalps.com/pdf/CHEMISTRY/…) this book says it doesn't. $\endgroup$ Commented Aug 19, 2022 at 15:50

2 Answers 2


The following ring opening reaction will occour:

Reaction of cyclopropane with bromine to form 1,3-dibromopropane

You are quite right about the angle strain. Because orbital interactions are not optimal in this geometry. Consider p-orbitals, then a natural bond angle would be $\theta\in [90^\circ; 180^\circ]$. A mixing of s- and p-type orbitals allows a wide range of angles $\theta\in (90^\circ,\dots, 180^\circ)$.

In cyclopropane $\ce{C3H6}$ - which you can also describe as trimethylene $\ce{(CH2)3}$ - bonds have to be bent to overlap at all. A possible way of describing the bonding situation is regarding each $\ce{CH2}$ entity as $\mathrm{sp^2}$ hybridised. Two of these orbitals are used for $\ce{C-H}$ bonds (not shown) and one forms an inner two-electron-three-centre σ bond (left). This leaves p-orbitals to form some kind of degenerate π-like orbitals (middle, right).

orbital models

This very general approach can be derived from a Walsh diagram. Schwarz et.al. {@academia.edu} and Hoffmann {@roaldhoffmann.com} described bonding quite similar and it is in quite good agreement with a calculation (BP86/cc-PVTZ, $D_\mathrm{3h}$) I have done. From this I have prepared a chart of all occupied molecular orbitals formed from valence orbitals and the LUMO. Here is a preview. Each orbital is viewed from three different angles:

chart preview

Especially the symmetrical orbital 8 resembles very well the schematics. A quite rigorous approach for this theory can also be found here.

It is noteworthy - as mentioned by ron - that there is no notable increase in electron density in the centre of the ring. This may be due to the fact that there are much more orbitals having nodes in the centre than there are without.

total density

Now bromine is known to be easily polarised $\ce{{}^{\delta+}Br-Br^{\delta-}}$ and may intercept at any point of the ring causing a bond break and relaxation to a less strained structure. It will most likely attack at the the $\pi$ type orbitals since bromine is an electrophile. The mechanism is analogous to the addition of bromine to ethene, which is nicely described at chemguide.co.uk. The essential part is the attack of the bromine at the HOMO(s).

attack of bromine

The ring opening reaction can be reversed by adding sodium.

However, when there are bromine radicals present (UV light) then substitution will occur: \begin{aligned}\ce{ Br2 &->[\ce{h\nu}] 2Br.\\ &+(CH2)3 -> (CH2)2(CHBr) + HBr }\end{aligned}

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    $\begingroup$ Your proposed model for the hybridization of cyclopropane suggests significant electron density within the ring - any evidence for this? $\endgroup$
    – ron
    Commented May 12, 2014 at 15:10
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    $\begingroup$ @ron I read your answer and I cannot provide anything substantial at that point. I might come back to this. I would be very appreciative if you could provide an alternate bonding model that would avoid this kind of hybridisation. (I'd like to add, that I am generally no big fan of hybridisation in the first place, but I thought it would fit well here.) $\endgroup$ Commented May 12, 2014 at 15:24
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    $\begingroup$ In my answer I have a bonding model that seems consistent with the known data for cyclopropane. The key is that the C-C bonds are made of orbitals that are very high in p-character. They combine to form "bent" bonds lying off (outside) the C–C internuclear axis. The high p-character allows cyclopropane to behave analogously to an olefin. It also explains why the C–C bond in cyclopropane is shorter (1.51 Å) than C–C bonds in alkanes (1.54 Å). $\endgroup$
    – ron
    Commented May 12, 2014 at 16:08
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    $\begingroup$ @ron I cannot quite fit your explanation into the symmetry constraints imposed by $D_{3h}$. I have done some calculations and I agree that there is less electron density in the ring than my bonding picture would suggest. If you could elaborate a little more, that would be very helpful - I am rather puzzled right now. I might have to revisit the answer and consider deleting it. $\endgroup$ Commented May 12, 2014 at 17:05
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    $\begingroup$ Sure, what would you like me to elaborate? $\endgroup$
    – ron
    Commented May 12, 2014 at 17:19

The $\ce{H-C-H}$ angle in cyclopropane has been measured to be $114^\circ$. From this, and using Coulson's theorem

$$1 + \lambda^2 \cos(114^\circ) = 0$$

where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Now, using the equation

$$\frac{2}{1 + \lambda_{\ce{C-H}}^2} + \frac{2}{1 + \lambda_{\ce{C-C}}^2} = 1$$

(which says that summing the "$\mathrm{s}$" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the $\ce{C–C}$ bond is $\mathrm{sp^{3.74}}$ hybridized.

We see that the $\ce{C–C}$ bond in cyclopropane is very high in $\mathrm{p}$-character. It is this high $\mathrm{p}$-content that allows cyclopropane to behave in a similar fashion to an olefin in terms of stabilizing adjacent charge, absorbing bromine, etc. By the way, an x-ray study of a cyclopropane derivative1 shows significant electron density only exterior to the cyclopropane ring. There is no significant electron density interior to the ring, consistent with this analysis.

(1): Hartman, A.; Hirshfeld, F. L. Structure of cis-1,2,3-tricyanocyclopropane. Acta Crystallogr. 1966, 20, 80–82. DOI: 10.1107/S0365110X66000148.

  • $\begingroup$ According to your reference there is no significant increase in electron density in the ring - assuming they modelled their deformation density to a spherical atoms model. (I am not an expert on crystal structures.) $\endgroup$ Commented May 13, 2014 at 12:04
  • $\begingroup$ I just noticed, you are missing a square root in Coulson's theorem, i.e. $1+\sqrt{\lambda_i\lambda_j}\cos(\omega_{ij})=0$. Probably a typo as the numbers work out that you have included it. $\endgroup$ Commented Apr 17, 2017 at 1:44
  • $\begingroup$ @Martin Not a typo, just depends whether you define the hybridization index as $\lambda$ (Coulson) or $\lambda^2$ (me); I think the latter way made it a bit easier for students to follow. $\endgroup$
    – ron
    Commented Apr 17, 2017 at 2:21

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