Is hybridization of the cyclopropyl anion sp5?

Question

I have read some answers about the $\mathrm{sp^5}$ hybridisation of carbon atoms in the $\ce{C-C}$ bond in cyclopropane as well as the $\sigma$-bond resonance in methylcyclopropyl cation.

Overall, the answers say that the $\ce{H-C}$ bond is due to overlap of $\mathrm{sp^2}$ orbital of carbon and $\mathrm{1s}$ orbital of hydrogen, while the bent bonds of the ring are formed by overlap of $\mathrm{sp^5}$ hybrid orbitals.

But, overall true separation of orbitals in such cases is not defined i.e. we cannot say that some orbital is completely $\mathrm{sp^2}$ hybridised.

Similarly, if the loss of a proton occurs from cyclopropane then the following:

In such an ion, will the hybridization change and $\mathrm{s}$ and $\mathrm{p}$ characters of the orbitals be redistributed?

My assumptions for this are:

• In the $\mathrm{sp^2}$ bond the lone pair due to loss of proton will pose increased steric hinderance as it occupies greater volume. This will cause increased repulsion and decreases stability. Moreover, shouldn't the $\mathrm{s}$-character increase to accommodate the increased negative charge?

• Accordingly, wouldn't the $\mathrm{sp^5}$ bonds will now have greater $\mathrm{p}$-character to ease even greater strain produced?

Then, what will be the change in the hybridisation in the overall molecule on formation of this ion?

References:

• Sp5 hybridization in cyclopropane?
• Are the triply-bonded carbons in pyridyne (and benzyne) sp-hybridised?
• Non-integer hybridization
• https://www.ch.imperial.ac.uk/rzepa/blog/?p=14548

Answer 1

Background

Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist. It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated. The key equation is $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$

where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

See these earlier answers for interesting examples where the theorem is applied:

• estimating the hybidization in cyclopropane
• angle of attack of a nucleophile at a carbonyl
• the lone pairs in water are non-equivalent - the resultant hybridization of water

The Question

The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°. 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. 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 "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 C–C bond is $\mathrm{sp^{3.74}}$ hybridized.

Now, if we remove a proton from cyclopropane and generate the cyclopropyl anion, we move from a situation where we had a pair of electrons shared between carbon and hydrogen in a $\ce{C-H}$ bond to a situation (the anion) where we have a pair of electrons residing entirely in an orbital on the carbon atom. In other words, we have increased the electron density in this carbon orbital. Bent's Rule tells us that the molecular geometry will change so as to lower the energy of this pair of electrons and that it will lower the energy of these electrons by increasing the s-character of the orbital they are in. In order to increase the s-character in this orbital, we will take some s-character away from the 2 $\ce{C-C}$ bonds and 1 remaining $\ce{C-H}$ bond. Hence, $$\lambda^2_{\ce{C-C}}>3.74$$ $$\lambda^2_{\ce{C-H}}>2.46$$ and $$\lambda^2_{\ce{C-electron pair}}<2.46$$ Said differently, the interorbital (not internuclear) $\ce{C-C-C}$ angle at the anionic carbon will decrease and the $\ce{C-C-H}$ angle will decrease making the substituents about the anionic carbon appear more puckered.