CHAPTER 3: Question 6

 

2D explanations

These compounds all contain a stereogenic axis which is indicated by the red line in the diagram shown below. It is rotation around this stereogenic axis which gives rise to the atropisomers and which causes compounds a-d to be chiral. Note that the other N-Ar bond (shown in blue) does not give rise to atropisomers since the two ortho-substituents on the aromatic ring are identical, as are the two meta-substituents.

Rotation around the stereogenic axis is restricted since the transition state for the rotation will have one of the two planar structures shown below. In these transition states, there is a steric repulsion between the R-group and one of the carbonyl oxygens. As the size of the R-group increases from ethyl (compound a) to isopropyl (compound b) to trifluoromethyl (compound c; NB fluorine atoms are significantly larger than hydrogen atoms) to tert-butyl (compound d), this steric repulsion will increase. Hence, the energy of the transition state will increase on going from a-d. The energy of the atropisomers themselves (i.e. of the non-planar form as drawn above) does not change significantly as the size of the R-group increases however, since there is no steric repulsion when the structure is non-planar. As a result, the activation energy for rotation will increase and the rate of rotation (i.e. the rate at which the atropisomers interconvert) will decrease as the size of the R-group increases.

No rate constant is quoted for R=H, since this compound is not chiral and so does not exist as a pair of atropisomers. The reason for the achiral nature of this compound is that the two ortho-substituents on the aromatic ring are now identical, as are the two meta-substituents.

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