I understand why the $\ce{CCC}$ bond angles in cyclobutane and cyclopropane are less than ideal $109.5^\circ$ due to the geometric restrictions they encounter in a cyclic structure. However, it is not clear to me how the $\ce{HCH}$ angles (which turn out to be greater than ideal) are affected by this constraint.

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    $\begingroup$ This earlier answer may be helpful. $\endgroup$
    – ron
    Oct 2, 2017 at 17:08

3 Answers 3


If one angle is smaller than the ideal, the others have more space to spread out. They won't spread out much, because the hydrogen atoms are small (not much repulsion) and rather happy at 109.5, which is also the inherent angle between bonds at an sp3 hybridised carbon atom.

The latter point changes a bit when the constrained bond angle is really far off, so the whole thing is no longer properly sp3 hybridised.


Computational investigations of geometry, aromaticity and bond energies at HF/6-31G** theory level [1] revealed the following deformed geometrical ($\alpha$) and interpath angles ($\beta$ -- the angle between two bond paths) for $\ce{H-C-H}$ bond, which are slightly deviating from the strain-free $109.5^\circ$ due to varying Baeyer strains:

\begin{array}{lr} \hline \text{molecule} & \alpha,~^\circ & \beta,~^\circ \\ \hline \text{cyclopropane} & 114.1 & 117.0 \\ \text{planar cyclobutane} & 107.9 & 109.7 \\ \text{puckered cyclobutane} & 108.3 & 110.0 \\ \hline \end{array}

enter image description here

P. S. Unfortunately, this data cannot be taken from single-crystal x-ray experiment as this method doesn't allow precise determination of H-positions, and I didn't find a neutron diffraction data on solid cycloalkanes mentioned.


  1. Cremer, D.; Gauss, J. J. Am. Chem. Soc. 1986, 108 (24), 7467–7477 DOI: 10.1021/ja00284a004.
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    $\begingroup$ Thanks for your help. I'm wondering if there was a way to explain this in an introductory organic chemistry course level $\endgroup$
    – Diracc
    Oct 2, 2017 at 15:07
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    $\begingroup$ @Diracc Well, for the introductory level the best would be to take off with Karl's answer; if you want to illustrate with the numbers, maybe you want to use $\beta$ values, as the paper claims they reflect the real geometry better. Also, maybe introduce Baeyer and Pitzer strains at some point. $\endgroup$
    – andselisk
    Oct 2, 2017 at 15:22
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    $\begingroup$ I believe NMR together with qc modelling should allow to determine the bond angles quite accurately. $\endgroup$
    – Karl
    Oct 2, 2017 at 15:25

One may approach it from a hybridisation standpoint. Take cyclopropane for example: the C–C–C bond angle is 60°, as is required by an equilateral triangle.

If the AOs at each carbon were simply four sp3 hybrids, the ideal bond angle between them would be 109.5° - which creates a great deal of angle strain. Consequently, there is some redistribution of s- and p-orbital character between the C–C and C–H bonds. Like hybridisation, this is not a physical process whereby the carbon atom "moves" p- and s-orbitals around. All it means is that the hybridisation deviates from the ideal 4 × sp3 hybrid case, where each sp3 hybrid has exactly 75% p-character.

When the s-character is decreased and the p-character increased, then the ideal bond angle will go down (pure p orbitals would have an ideal bond angle of 90°). So, it makes sense that there is more p-character in the C–C bonds. However, it's also a zero-sum game: since we only have three p orbitals to use, if we put more p-character into the C–C bonds, then the C–H bonds have to have less. And since there is less p-character, one can predict that the H–C–H bond angle will be increased.

According to ron's earlier answer, it turns out that the C–H bonds in cyclopropane can be described using a sp2.46 hybrid (i.e. 71% p-character), and the C–C bonds with a sp3.74 hybrid (i.e. 79% p-character).

Now, cyclobutane is exactly the same as above, except that the deviation from ideality is smaller.

As an addendum: the hybridisation of the carbon orbital used to form the C–H bond can be related to the one-bond C–H coupling constant in NMR. The larger the s-character, the larger the coupling constant; and Gunther's NMR Spectroscopy (3rd ed., p 424) states that for hydrocarbons an empirical correlation has been found:

$$^{1}\!J_\ce{C-H} \approx 500\cdot s$$

where $s$ is the fractional s-character. In methane, the coupling constant is 125 Hz, which corresponds to exactly 25% s-character, as expected for an ideal sp3 hybrid. On the other hand, in cyclopropane, the coupling constant is 161 Hz. The correlation with the s-character calculated above isn't perfect, but the fact that it is larger is already indicative of the increased s-character. For cyclobutane, it is 136 Hz. (NMR data taken from Hans Reich's tables.)


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