Negative Charge Stability: Oxygen vs. Carbon/Nitrogen

[Zayn]

Why is it that it is more stable for a negative charge to be spread over 2 oxygens than one oxygen and 3 carbon atoms? Following that logic, why is it more stable to spread a negative charge over 3 oxygen atoms than over 1 oxygen atom and one nitrogen atom? Is it because the difference in electronegativity is much greater between the oxygen and carbon than it is between the oxygen and nitrogen?

 

[Zayn]

Just to extend this idea, at what point does a hybrid carbon atom stabilize charge better than a more electronegative atom? For example, according to my textbook, an sp-hybrid C atom stabilizes negative charge better than a nitrogen atom. At what point does hybridization's effect overcome electronegativity?

 

[TeethWhitener]

Why is it that it is more stable for a negative charge to be spread over 2 oxygens than one oxygen and 3 carbon atoms?

It's clear that you have an example in mind, but for the life of me, I can't figure out what it is. I don't think you're asking what you think you're asking.

Here's what it appears you're asking: Why is the electron affinity for O₂ higher than for C₃O? and why is the electron affinity for O₃ higher than for NO?

Is this what you're asking, or do you want to clarify?

EDIT: the reason I ask is because these are very strange questions about some very strange species (C₃O in particular, which only really exists in outer space or maybe transiently in flames).

 

[Zayn]

It's clear that you have an example in mind, but for the life of me, I can't figure out what it is. I don't think you're asking what you think you're asking.

Here's what it appears you're asking: Why is the electron affinity for O₂ higher than for C₃O? and why is the electron affinity for O₃ higher than for NO?

Is this what you're asking, or do you want to clarify?

EDIT: the reason I ask is because these are very strange questions about some very strange species (C₃O in particular, which only really exists in outer space or maybe transiently in flames).

You're right, I'm asking about a specific example from my organic chemistry textbook, but I didn't want to say that as I'm not asking for homework help; I'm just wondering why this is the case. As I understand it, these questions are here as examples of exceptions to the ARIO (atom, resonance, induction, orbitals) rule learned in the textbook.

The question asks to determine which proton is the most acidic

Here is an imgur album that shows the questions with their answers

 

[TeethWhitener]

I’m still not getting how the problem you linked to in post 4 is related to the questions you asked in posts 1 and 2. If you want to know why a terminal alkyne is acidic, that’s fairly straightforward. It’s because the electrons in the sp hybridized orbital on the carbon are less shielded from the nucleus than in sp² or sp³ orbitals (because of the higher s character). So the acetylide conjugate base is more stable than an alkenylide or alkylide carbanion. In contrast, the NR₂^- anion is quite unstable.

Why the amide is so much more basic than the acetylide, well, unfortunately that’s just something you’ll probably have to memorize. I don’t think there’s any systematic reason why that’s t he case. If it helps you to remember, the acetylide salt is usually made by combining an alkyne and an amide (e.g., acetylene + sodium amide gives sodium acetylide + ammonia).

 

[mjc123]

In addition to the electronegativity difference, there is the fact that in the carboxylate and sulfonate anions there are 2 or 3 resonance forms that are identical (assuming free rotation about the ring-substituent bond), and therefore equal in energy and contributing equally to the resonance hybrid. By contrast, in the phenolate anion, the resonance forms with negative charge on carbon are of higher energy than the one with negative charge on oxygen, and therefore make less of a contribution to the overall resonance hybrid.

 

[Zayn]

I’m still not getting how the problem you linked to in post 4 is related to the questions you asked in posts 1 and 2. If you want to know why a terminal alkyne is acidic, that’s fairly straightforward. It’s because the electrons in the sp hybridized orbital on the carbon are less shielded from the nucleus than in sp² or sp³ orbitals (because of the higher s character). So the acetylide conjugate base is more stable than an alkenylide or alkylide carbanion. In contrast, the NR₂^- anion is quite unstable.

Why the amide is so much more basic than the acetylide, well, unfortunately that’s just something you’ll probably have to memorize. I don’t think there’s any systematic reason why that’s t he case. If it helps you to remember, the acetylide salt is usually made by combining an alkyne and an amide (e.g., acetylene + sodium amide gives sodium acetylide + ammonia).

I understand that, but my question was more along the lines of: Why is an sp2 hybrid C less able to stabilize the charge than nitrogen, but sp is better than nitrogen (I understand the principle that hybrid orbitals are closer to the nucleus and thus better able to stabilize -ve charge); at what point is the effect of C hybridization > electronegativity. I guess it's just been experimentally determined to be so, and, as you said, it's something that should just be memorized. And as for the other two examples, I was trying to understand how charge spread over 1 O + 1 C + 1 C + 1 C < 2 O, but from what mjc123 has said, it seems like it's not something that be understood as an equation.

 

[TeethWhitener]

It can be calculated using quantum chemistry programs, but unfortunately, there’s no simple heuristic way to calculate it. Slater’s rules aren’t really much help, for example, in determining how shielded the excess electron is.