Acid/Base

[Temperature101]

Which of the following compounds would be most acidic?

A. H3CSCH3
B. H3COCH3
C. H3CNHCH3
D.H3CCH2CH3

I pick B as the answer, but the book says it's C. Can someone explain why?

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[milski]

Which of the following compounds would be most acidic?

A. H3CSCH3
B. H3COCH3
C. H3CNHCH3
D.H3CCH2CH3

I pick B as the answer, but the book says it's C. Can someone explain why?

pKa for the alkyls is 60 or so. The presence of O/S won't change that much.
The answer C compound has a N atom which can be de-protonated much easier (pKa about 35).

 

[SaintJude]

Is it also because strength of the bond? C-H bond which is stronger than the N-H bond. That's simpler for me to understand. I don't think I would remember pKa values.

Edit: Sorry, I don't why the heck I was talking about carbonyl bonds earlier...Thanks milski.

 

[milski]

Is it also because strength of the bond? C-O is a carbonyl bond (a double bond), which is stronger than the N-H bond. That's simpler for me to understand. I don't think I would remember pKa values

Sure, that's another way of saying the same thing. Btw, C-H bond, not carbonyl bonds. The C-O are not being broken.

 

[SaintJude]

Actually, looking at this problem again I think it's easiest to solve this problem just considering that nitrogen is more electronegative than carbon. And thus the proton will be much more likely to be pulled off from the nitrogen...I guess it's essentially the explanation of pKa values.
Is that correct rationale?

 

[milski]

Actually, looking at this problem again I think it's easiest to solve this problem just considering that nitrogen is more electronegative than carbon. And thus the proton will be much more likely to be pulled off from the nitrogen...I guess it's essentially the explanation of pKa values.
Is that correct rationale?

Still saying the same thing. N-H is weaker than C-H which makes it easier to pull the H atom which lowers the pKa.

I think they're trying to make you forget that there's an H attached to the N, unlike the O and S, and mislead you into thinking about inductive effects.