Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Nozon on June 03, 2020, 08:55:46 AM
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Hello
So I am trying to understand how to determine relative pKa based upon the substituent bound to the carboxylic acid.
The 3 molecules is: Acetic Acid, (E)-3-(dimethylamino)acrylic acid and Triflouroacetic acid. (picture attached to show molecular structure)
My approach has been on eletronegative substituent and how relative close it is to the carboxic acid. Also if any had a benzene ring that would interfere and lower the substituents increasing acidity (which none had).
But I also know there's something about resonances of stability of the anion that would increase the acidity? (Which also is one of the reasons of the benzene's interference)
I believe that's the point in putting in (E)-3-(dimethylamino)acrylic acid, because acetic acid is clearly easy to determine low acidity compared to triflouroacetic acid.
So how should I analyze the amino-acrylic acid? Or is the polarization all cancelled out due to its two methyl groups?
My best regards
Jacob
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Besides inductive effects, what other effects are very common in organic chemistry?
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Besides inductive effects, what other effects are very common in organic chemistry?
Inductive effects also include sp^2 / π-bonds? if not - are you thinking about Zaitsev's rule?
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I don't see how Zaitsev's rule comes into play here. I was thinking about resonance.
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You're right. my mind went a bit out there (realized Zaitsev's is part of Elimination reactions while going through my notes).
Yes - That's what I spotted could be a possibility. I'm not really good at determine where the pi bond can move, besides maybe create a tetranitrogen cation which may create a very strong electron withdraw-substituent?
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Can you show us what you mean?
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I was thinking if it could formate into a zwitter ion?
(But have no idea if this is actually possible) - I have a real struggle with understanding laws of organic chemistry and I know I am only touching the surface of it in my class.
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You are moving in the right direction, but the drawing you showed is not correct. However, I will offer some ideas. First, I think it will be slightly easier to draw resonance forms for the carboxylic acid; therefore, you want to have a hydrogen present on one of the oxygens. Second, when one draws resonance forms, charge is conserved. Third, only sp3-hybridized atoms cannot participate in resonance. Fourth, remember that you cannot move or gain or lose hydrogens when you draw a set of resonance structures.
If you move the lone pair on nitrogen so that it forms a double bond to the adjacent carbon, where can you place the electrons that had been part of the C=C double bond.
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So I tried to follow your lead as far I can understand.
I know if a pi bond is created on a carbon the two others will be forced a bit closer. That's why I drew an extra version, but I am not sure if Oxygen is able to actually share hydrogen (never seen that in my textbooks).
I am not sure if I succeed in conserving charge, but I can't figure out another way to draw it. The netto charge will be conserved so to say.
I havn't lost any atoms - my alternative (second picture) kinda move a hydrogen.
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You are missing one resonance structure, one in which the lone pair is on the carbon atom adjacent to the carboxylic acid group. However, even with the resonance forms that you have, you can make a prediction about pKa. I would focus on the resonance form on the right: Do you think that it will be more favorable or less favorable for the proton to dissociate?
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I would think it be less favorable for the proton to dissociate, due to the already negatively charged oxygen right next to it. If the proton dissociate I believe it would just bind to the other oxygen, which is the same molecule in the end.
Therefore the pKa is even higher than Acetic Acid.
So pKa would be relative to: (high to low)
(E)-3-(dimethylamino)acrylic acid > Acetic Acid > Triflouroacetic acid
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That would be my prediction as well.
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Thanks a lot - it has been a huge help.
Have a great day.
My best regards.