t3rrr
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estimating acidity
I decided to improve my chemistry knowledge and joined courserra course on organic chemistry. Last week tasks include finding most acidic molecule
from given. My reasoning follows,
All charges in task are the same - null, as well as proton donating element - nitrogen, all donating orbitals are sp3. It leaves us with
1) Resonance
2) Dipole Induction
I can draw resonance forms for all elements, but more stable will be those which resonate with higher electronegativity element. And this is nitrogen,
and only in two bond distance, that is answer A and it's incorrect.
After failing that tast I used MarvinSketch for pKa calculation, and got wrong answer(D) second time.
I found a hint on forums that I should scope atom size(I guess, referring to E), and not just electronegativity. Could someone please describe me
process of comparing acidity in organic molecules, on this example?
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Justin Blaise
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For this question, you can compare the stability of the conjugate bases. For A, remove the hydrogen from the nitrogen and draw the resonance forms.
You will find that it has one resonance form with the negative charge on the nitrogen.
For B, once you remove the hydrogen, you don't have any good resonance forms.
For C, you can draw a resonance form where the oxygen gets the negative charge, and this is a good resonance structure due to the electronegativity of
oxygen.
D has the same sort of resonance as C, except you have 2 oxygens and therefore 2 additional resonance structures.
E is where it gets interesting. The resonance forms are the same as D, but the sulfur atoms are bigger and can disperse the negative charge better, so
these resonance forms are more stable than in D.
This is how I would work this problem. I hope it's of some use to you
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DraconicAcid
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Quote: Originally posted by Justin Blaise  | For this question, you can compare the stability of the conjugate bases. For A, remove the hydrogen from the nitrogen and draw the resonance forms.
You will find that it has one resonance form with the negative charge on the nitrogen.
For B, once you remove the hydrogen, you don't have any good resonance forms.
For C, you can draw a resonance form where the oxygen gets the negative charge, and this is a good resonance structure due to the electronegativity of
oxygen.
D has the same sort of resonance as C, except you have 2 oxygens and therefore 2 additional resonance structures.
E is where it gets interesting. The resonance forms are the same as D, but the sulfur atoms are bigger and can disperse the negative charge better, so
these resonance forms are more stable than in D.
This is how I would work this problem. I hope it's of some use to you |
Actually, in B you can disperse the negative charge around the aromatic ring. This would be more acidic than an ordinary amine (although that's not
saying much) in the same way that a phenol is more acidic than an ordinary alcohol. It would still not be as acidic as the amide, though (in the same
way that a phenol is not as acidic as a carboxylic acid).
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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t3rrr
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| Quote: Originally posted by Justin Blaise | For this question, you can compare the stability of the conjugate bases. For A, remove the hydrogen from the nitrogen and draw the resonance forms.
You will find that it has one resonance form with the negative charge on the nitrogen.
For B, once you remove the hydrogen, you don't have any good resonance forms.
For C, you can draw a resonance form where the oxygen gets the negative charge, and this is a good resonance structure due to the electronegativity of
oxygen.
D has the same sort of resonance as C, except you have 2 oxygens and therefore 2 additional resonance structures.
E is where it gets interesting. The resonance forms are the same as D, but the sulfur atoms are bigger and can disperse the negative charge better, so
these resonance forms are more stable than in D.
This is how I would work this problem. I hope it's of some use to you |
That was quite helpful. Thanks.
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Justin Blaise
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@DraconicAcid, you are correct. That was an oversight, but as you said, the imide is still more acidic.
@t3rrr, no problem
What is that functional group in E called? A thioimide?
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