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.