Intermediate Organic Chemistry

Consider carbons electronic configuration once again.

We have 4 unpaired electrons in this carbon atom, it is obvious then that the atom will give 4 bonds with other atoms. Consider CH₄ , C - H distance is 0.111 nanometres for all of the bonds, surely this cannot be the case as we have 2 different types of bond?! The p orbitals are weaker than s orbital overlap so the bond should be longer! What is happening?

This is a case of hybridisation this makes four equivalent bonds called sp_³ molecular orbitals Figure 8(a); this indicates that 1 s and 3 (2p) orbitals have mixed:

Hybridisation means mixing together of the orbitals to form equivalent bonds, the number of orbitals formed must be equal to the number mixing.

If we apply this to CH₄ methane we get the result shown in Figure 8(a). Notice the Hydrogen atoms have moved to the farthest point away from each other; this gives a bond angle of 109.50 {Figure 8b}. This is known as the tetrahedral angle.

This is not the only way that hybridisation can occur with the s and p orbitals, i.e. 1(s) and 2 x 2(p) orbitals can mix to form 3sp² molecular orbitals.

The sp² hybridised orbitals have a similar shape to the sp³ orbitals but we only have 3 orbitals to arrange as far away as possible from one another. The angle they choose is 120⁰ , what we get this time is a trigonal shape {figure 10};

This only leaves the 2p_z orbital to be arranged as follows:

The 2p z orbital is now orthogonal to the 3 sp 2 orbitals {figure11}. What we have now is the set-up for the formation of a double bond in an alkene. Two sp 2 orbitals and 2 p orbitals for the carbon atoms need to come together; the resultant is the formation of a s bond and a p bond (figure 12).

The resultant molecule which just requires the overlap of four hydrogen 1s orbitals with each of the remaining 4 sp² orbitals and we have the first of the alkenes – ethene:

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