Skip to main content

Donald Clarke

Donald Clarke

Professor of Chemistry
NMR Spectroscopy, Biochemistry

Contact Information

Research Interests

Electronic Geometry, Molecular Shape, and Hybridization
The Valence Shell Electron Pair Repulsion Model (VSEPR Model)

The guiding principle: Bonded atoms and unshared pairs of electrons about a central atom are as far from one another as possible.

4 0 4 tetrahedral tetrahedral 109.5° sp3
Examples: CH4, CH3Cl, CH2Cl2, CHCl3, CCl4, NH4+

3 1 4 tetrahedral pyramidal 109.5° sp3
Examples: NH3, H3O+

2 2 4 tetrahedral bent 109.5° sp3
Examples: H2O, SCl2

Pauling introduced the idea of hybridization of C-H bonds in methane in 1930 to provide a quantum chemical explanation of the long known fact that its four C-H bonds are equivalent.

Henry Bent reviewed this topic in detail in Chemical Reviews in 1960 and wrote a parallel article in J Chemical Education advocating its teaching. From 1960 onward most Organic textbooks have adopted Bent’s idea. However, the discussion in chemistry courses is based on predicting C-H hybridization rather than measuring it experimentally.

Ethane, ethylene and acetylene are the common textbook examples of molecules with sp3, sp2 and sp hybridization respectively, but as gases, they are impractical for use in undergraduate lab experiments.

4-t-Butylphenylacetylene [C12H14], a readily available liquid with all three types of C-H hybridization, is suitable for experiments to show the relation of 1JCH coupling constants to C-H hybridization. Also it has quaternary C atoms, i.e. with no Hs attached.

The 9 H atoms of the t-butyl group are equivalent and have a chemical shift at 1.30 ppm. The 4 phenyl ring H atoms form two equivalent sets and show signals at 7.33 and 7.43 ppm while the single ethynyl H has a signal at 3.01 ppm. The solvent here is CDCl3.

The 3 equivalent C atoms of the t-butyl group are sp3 hybridized and appear at 31.4 pm. The four phenyl ring C atoms with attached Hs form two equivalent sets which are sp2 hybridized and have chemical shifts of 125.4 and 131.9 ppm while the acetylenic CH, which is sp hybridized, appears at 76.5 ppm. The remaining quaternary C atoms have shifts at 34.8, 83.8, 119.1 7 & 152.1 ppm. These chemical shifts are predicted reasonably well by ChemNMR, except for the ethynyl CH [76.5 vs 81.4 ppm; d = 4.9].

Comparison of Predicted versus Observed NMR Spectra

Comparison of Predicted VS Observed NMR Spectra

When the solvent is changed from CDCl3 to DMSOd6 the delta value for the ethynyl H increases from 3.01 to 4.10 ppm, indicating that it is H bonded to the solvent DMSO. In C6D6 this H appears at 2.76 ppm which is a closer to the value of 2.63 ppm, calculated using Gaussian, for the gas phase. Thus CDCl3 is not simply a spectator solvent.

In the 1H coupled 13C NMR spectrum of 4-t-butylphenylacetylene the 1JCH values are ~250 Hz for the ethynyl proton, ~167 for the protons on the phenyl ring and ~125 Hz for the methyl protons of the t-butyl group. An sp hybridized bond has 1/2 s character, an sp2 has 1/3 s character and sp3 has 1/4 s character. A plot of 1JCH vs the fraction of s character is linear and can be described by the equation 1JCH = 500 x (fraction of s character).

These 1JCH coupling constants are very little affected by changes in solvent. Accordingly, values calculated in the gas phase more closely agree with experiment than do the calculated chemical shifts using Gaussian. 2JCH and 3JCH are much smaller than 1JCH and generally 10 Hz or less. However, 2JCH for the ethynyl C is ~ 50 Hz.

The theoretical basis for this dependence of 1JCH on fractional s character is that an electron in an s type orbital has a finite chance of being located in the nucleus of an atom, while an electron in a p type orbital has a node at the nucleus and a zero probability of being located there. This is called the Fermi contact interaction.

Theory tells us 1JCH = 6.51 x [1JCD] This factor is the ratio of their gyromagnetic constants. It can be derived experimentally by running the 1H coupled 13C spectrum of C6H6 added to C6D6. Extensive tables of 1JCD of common NMR solvents are widely available. These can be used to show that electronegative elements cause fractional hybridization of C-H bonds.

If we consider cyclopropane we find contradictory statements regarding the hybridization of C in this molecule. This is due to the fact that C-H hybridization is not the same as that for C-C. What this tells us is that there is no such thing as the hybridization of an atom. It is the bonds between nuclei that are hybridized, not the atomic nuclei. Hence the appropriate questions are what is the C-H hybridization and what is the C-C hybridization. These are very different things.

Similarly, the statement that C in CHCl3 is sp3 hybridized because it is tetrahedral contradicts the chemical properties of this molecule. The proper question is what is the hybridization of the C-H bond in CHCl3.

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:

1JC−H≈500⋅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.)

Diatomic molecules, e.g. H2, O2, N2, HX, CO, have no bond angles. Triatomic molecules, e.g. H2O, CO2 have interbond angles. These ae generally referred to as bond angles, even though bonds have no angles. VSEPR theory predicts interbond angles to a reasonable first approximation. Thus H2O is predicted to be tetrahedral and CO2 linear. The measured bond angle in H2O is actually 104o rather than 109.5o, the angle for a regular tetrahedron. Thus while H2O is tetrahedral it is not a regular tetrahedron. The repulsion between H nuclei and lone pairs is different which causes deviation from a regular tetrahedral arrangement.