Biirgi-Dunitz angle

Now that we have decided which are the important conformations, how do we know which gives the product We need to decide which is the most reactive. All we need to do is to remember that any nucleophile attacking the carbonyl group will do so from the Biirgi-Dunitz angle—about 107° from the C=0 bond. The attack can be from either side of C=0, and the following diagrams show the possible trajectories superimposed on the two conformations we have selected, which are in equilibrium with one another. 

It is a common misunderstanding to think that the Biirgi-Dunitz angle implies that the angles are acute. They can be sometimes, but they are not usually—the... 

Fig. 5.2 The Salem-Klopman equation applied to the Biirgi-Dunitz angle...

Allow the nucleophile to attack along the least hindered trajectory, taking into account the Biirgi-Dunitz angle. 

In a classic 1976 paper, Baldwin used the Biirgi-Dunitz trajectory to define the well-known rules for the design of cyclizations (the Baldwin rules). However, stereoelectronic factors for a bond formation to alkynes have been controversial. Originally, the rules for alkyne cyclizations were based on the assumption that nucleophiles add to alkynes along an acnte trajectory, instead of the obtuse Biirgi-Dunitz angle of attack. Subsequent experimental and computational analysis suggested that this trajectory is stereoelectronically... 

Attack by a nucleophile breaks the n-bond, and the electrons reside on the oxygen atom. This is energetically favourable, as a strong a-bond is formed at the expense of a weaker Jt-bond. Examination of crystal structures has shown that the nucleophile approaches the carbonyl carbon at an angle of around 107°. This is called the Biirgi-Dunitz angle (see Section 4.10). 

The same angle of attack is seen for addition reactions to an alkene. The reason for the Biirgi-Dunitz angle is that the adding nucleophile needs to attack the empty it orbital of the C=0 or C=C bond. Looking back at Sections 1.3.2 and 1.3.3, or in Appendix 3, we find that the character of this orbital on C points back and away from the bonding region. 

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