Figure 1. Geometries of the stationary points in the activation of 1a to intermediate 1a’ by triethylamine.
The dihedral angle Cl(16)–C(12)–N(13)–O(14) in 1a is coplanar. N(13) has a strong induced electron-withdrawing effect owing to its lone pair electrons, making the H atom of the hydroxyl group connected to it acidic. Through activation by the triethylamine solvent, a weak organic base, the hydroxyl H atom leaves 1a . The triethylamine N atom provides a lone pair electron, while the H atom provides an empty orbital. The H atom migrates to triethylamine, resulting in an increase in the charge on the hydroxyl O atom. Because the C(12)–N(13) double bond is adjacent to the Cl atom, the C–Cl bond polarity is relatively high. The lone pair electrons on the N atom attack the C–N double bond, leading to sp hybridization and departure of the Cl atom carrying the negative charge. The C(12)–N(13) bond is gradually shortened from 1.283 to 1.161 Å and the N(13)–O(14) bond from 1.349 to 1.241 Å. The intermediate 1,3-dipole, 1a’ , which has a partial C–N double bond, is formed. The C(12)–N(13)–O(14) bond angle increases from 122.42° to 180.00°. The charge on the N atom decreases, while that on the O atom increases, and the molecule as a whole has no net charge.
Transition state TS1 was traced by IRC calculation, which confirmed that it is a first-order saddle point on the potential energy surface of the reaction. TS1 leads from 1a to 1a’ , which is the reaction intermediate. The ΔE a for this step is ΔE a1 = 22.1 kcal·mol-1. Using the energy of the product, [1a’ +Et3NHCl], as the reference, the energy level diagram of elementary reaction 1 was derived and is shown in Figure 2.