Trigonal sp2 hybridization

We can now go on to apply the same ideas to some other simple molecules. In boron trifluoride, for example, we start with the boron atom, which has three outer-shell electrons in its normal or ground state, and three fluorine atoms, each with seven outer electrons. As shown in the upper diagram, one of the three boron electrons is unpaired in the ground state. In order to explain the trivalent bonding of boron, we postulate that the atomic s- and p orbitals in the outer shell of boron mix to form three equivalent hybrid orbitals. These particular orbitals are called sp2 hybrids, meaning that this set of orbitals is derived from one s-orbital and two p-orbitals of the free atom. 

Boron trifluoride has a plane trigonal shape a 2p orbital on each fluorine atom overlaps with a boron sp2 hybrid. In general, we can expect that all molecules in which a central atom uses three equivalent sp2 hybrid orbitals will exhibit plane trigonal geometry, since this represents the most symmetrical, and hence equivalent , arrangement of the three bonds. 

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Figure 3.17 Geometry of hybrid orbitals, (a) digonal sp hybrids oppositely directed along the same axis (b) trigonal sp2 hybrids pointing along three axes in a plane inclined at 120° (c) tetrahedral sp3 hybrids pointing towards the comers of a regular tetrahedron. (Reproduced with permission from R. McWeeny, Coulson s Valence, 1979, Oxford University Press, Oxford.)...

Changes in I strain bring about changes in the hybridization of the carbon atom during the reaction. Thus in the reduction of the ketones the trigonal sp2 hybridized carbon changes to tetrahedral carbon atom. 

The SHMO theory was originally developed to describe planar hydrocarbons with conjugated n bonds. Each center is sp2 hybridized and has one unhybridized p orbital perpendicular to the trigonal sp2 hybrid orbitals. The sp2 hybrid orbitals form a rigid unpolarizable framework of equal C—C bonds. Hydrogen atoms are part of the framework and are not counted. The Hiickel equations (3.3) described in the first part of Chapter 3 apply, namely,... 

The secondary electronic effects in the ester function are essentially similar to the anomeric effect discussed previously for the acetal function, involving an n-o interaction. The only difference is that the central carbon is trigonal (sp2 hybridized) in esters and tetrahedral (sp3 hybridized) in acetals. 

Hydrocarbon skeletons are built up from tetrahedral (sp3), trigonal planar (sp2),-or linear (sp) hybridized carbon atoms. It is not necessary for you to go through the hybridization process each time you want to work out the shape of a skeleton. In real life molecules are not made from their constituent atoms but from other molecules and it doesn t matter how complicated a molecule might be or where it comes from it will have an easily predictable shape. All you have to do is count up the single bonds at each carbon atom. If there are two, that carbon atom is linear (sp hybridized), if there are three, that carbon atom is trigonal (sp2 hybridized), and, if there are four, that carbon atom is tetrahedral (sp3 hybridized). 

If you had drawn the molecule more professionally as shown in the margin, you would have to check that you counted up to four bonds at each carbon. Of course, if you just look at the double and triple bonds, you will get the right answer without counting single bonds at all. Carbon atoms with no 7E bonds are tetrahedral (sp3 hybridized), those with one ft bond are trigonal (sp2 hybridized), and those with two Jt bonds are linear (sp hybridized). This is essentially the VSEPRT approach with a bit more logic behind it. 

Bromine is more electronegative than carbon and so the C-Br bond is polarized towards the bromine. If this bond were to break completely, the bromine would keep both electrons from the C-Br bond to become bromide ion, Br, leaving behind an organic cation. The end carbon would now only have three groups attached and so it becomes trigonal (sp2 hybridized). This leaves a vacant p orbital that we can combine with the n bond to give a new molecular orbital for the allyl system. 

Now the end carbon has a single unpaired electron. What do we do with it Before the bond broke, the end carbon was tetrahedral (sp3 hybridized). We might think that the single electron would still be in an sp3 orbital. However, since an sp3 orbital cannot overlap efficiently with a Jt bond, the single electron would then have to be localized on the end carbon atom. If the end carbon atom becomes trigonal (sp2 hybridized), the single electron could be in a p orbital and this could overlap and combine with the 7t bond. This would mean that the radical could be spread over the molecule in the same orbital that contained tire cation. 

A carbon with four bonded groups is tetrahedral, sp3-hybridized, and has 109.5° bond angles. A carbon with three is trigonal, sp2-hybridized, and has 120° bond angles. A carbon with two bonded groups is linear, sp-hybridized, and has 180° bond angles. 

A carbocation has three bonded groups and is trigonal, sp2 hybridized, and has 120 °bond angles. The empty orbital is the unhybridized p-orbital. 

The carbonyl C of RCOOH and RCOOR is trigonal sp2 -hybridized, but that of the intermediate is tetrahedral j/i3-hybridized. If R in R OH or R in RCOOH is extensively branched, formation of the unavoidably crowded transition state has to occur with greater difficulty and more slowly. 

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