Figure 10. Geometries of the stationary points in H+-assisted cyclization to generate product3a’ .
During the reaction, the H+ in solution migrates to O(29) to form intermediate med2, which was confirmed to be stable after geometry optimization. In theory, this process occurs spontaneously without a transition state. In med2, the N(13)–C(16)–N(28) bond angle is 124.88°, C(7) and O(29) are not bonded (bond distance of 2.940 Å), and the carbonyl C(7)–O(14) bond shows double bond character (bond length of 1.221 Å). As the reaction proceeds, C(7) and O(29) approach each other. The H atom bound to O(19) migrates to carbonyl O(14), and the C(7)–O(14) bond length increases to 1.316 Å. As the reaction crosses transition state TS3, the C(7)–O(14) bond length increases to 1.346 Å, and C(7) and O(29) forms a bond with a C(8)–C(7)–O(29) bond angle of 125.87°. As the molecule continues to rotate, the C(8)–C(7)–O(29) bond angle decreases to 113.11° and the configurational energy continues to decrease, resulting in the formation of 3a’ . In 3a’ , the N(13)–C(16)–N(28) bond angle is 113.29°, the C(7)–O(29) bond length is 1.431 Å, and the carbonyl C(7)–O(14) bond length is 1.378 Å.
Transition state TS3 was traced by IRC calculation, which proved that it is a first-order saddle point in the potential energy surface of the reaction. The ΔE a for this process is ΔE a3 = 26.92 kcal·mol-1. With product 3a as the reference, the energy level diagram of elementary reaction 3 was calculated and is shown in Figure 11.