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.