Question

The thiazole derivative A (structure shown in Figure 2) is acidic due to the resonance effects of the thiazole ring. Draw a good resonance structure or structures of the resulting carbanion B (structure shown in Figure 2), and briefly explain in words why your resonance structure supports the observation that A is acidic. (Recall, a strong acid has a stable conjugate base.) Briefly explain.

Answer 1

It should be noted that because of the thiazole ring's resonance properties, the thiazole derivative A (Figure 2) is acidic.

When there are numerous Lewis dot structures that can represent the same molecule, resonance develops. Each Lewis dot structure approximates the electronic structure of the molecule, and the real structure is a blend of all the resonance forms. This indicates that the carbanion B, the deprotonated form of A, is stabilized by resonance, making A more likely to lose its proton.

Because of the involvement of two rings' resonance, the H atom is very acidic. One is a thiazole ring resonance, and the other is a benzene ring resonance. As a result of this wide resonance, the carbanion formed following H+ removal is highly stable, and the H-atom is highly stable.