For the attached question, I thought the most acidic molecule would be 2 because Sulfur can stabilize the electrons better than oxygen, since it is larger. However, the correct answer had the first molecule as the most acidic. I was wondering how that was the case.
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Both compounds are deprotonated at the sp³-hybridized α-carbon, not at oxygen or sulfur. The resulting conjugate base is stabilized by resonance, with the negative charge delocalized between the α-carbon and the heteroatom.
In compound I, the negative charge is delocalized onto oxygen, whereas in compound II, it is delocalized onto sulfur. Because oxygen is more electronegative than sulfur, it stabilizes the negative charge more effectively. Therefore, the conjugate base of compound I is more stable, making compound I more acidic than compound II.
Although sulfur is larger and more polarizable, that factor is more important when comparing negative charges located directly on oxygen and sulfur, such as RO⁻ versus RS⁻. Here, the strength of the carbonyl-like resonance stabilization makes the oxygen-containing compound more acidic.
So, the correct conclusion is:
Compound I is more acidic than compound II because its conjugate base is better stabilized by resonance involving oxygen.