Question about finding the most acidic proton from OCHEM Destroyer

[CherryBlossom1412]

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Hi! I have a question on #126 from DAT destroyer 2017 edition.
I'm having trouble identifying the most acidic proton here. I was trying to follow Chad's CARDIO mneumonic
(C - charged, A - atoms, R- resonance , D - dipole induction, and O - orbital)
so for this question, all protons attach to the same atom which is C which has no charge so i focused on resonance.

But then I got confused since A, D and E can form resonance. I was trying to use the logic that the most stable conjugate base is the strong acid but then I got stuck deciding which one is the best resonance.

the explanation at the end of the book is saying that vinylic and aldehylic protons aren't very acidic so D and E are eliminated, but I don't understand why. I got this question right (i believe it's by luck) because i remember other question from destroyer that carboxylic acid H is very acidic.

If anyone could explain how to approach this problem using CARDIO logic or any other logic, I would really appreciate it. Thank you!

 

[SuxDrugs&Rocuronium]

Protons D and E are not very acidic because the conjugate base will put a negative charge on carbon, which is not very stable and neither of the two have resonance. C would put a negative charge on Carbon as well, or with resonance it would put a positive charge on the oxygen which is not a good resonance structure because oxygen is electronegative. Between A and B the biggest factor would be that B is next to another CH2 group. Both A and B could resonate into the carbonyl group, but B has a CH2 group that is and electron donating group. The carbon for B would have a partial negative charge, and the adjacent CH2 group would add to that negative charge, which would make it a less stable conjugate base. Hope that explanation helps!

 

[SuxDrugs&Rocuronium]

Sorry i was mistaken. D would be able to resonate into the aldehyde carbon, but that would put two double bonds next to each other (C=C=C) which is not very stable. That's why it's not very acidic

 

[DAT Destroyer]

Hi! I have a question on #126 from DAT destroyer 2017 edition (picture is attached).
I'm having trouble identifying the most acidic proton here. I was trying to follow Chad's CARDIO mneumonic
(C - charged, A - atoms, R- resonance , D - dipole induction, and O - orbital)
so for this question, all protons attach to the same atom which is C which has no charge so i focused on resonance.

But then I got confused since A, D and E can form resonance. I was trying to use the logic that the most stable conjugate base is the strong acid but then I got stuck deciding which one is the best resonance.

the explanation at the end of the book is saying that vinylic and aldehylic protons aren't very acidic so D and E are eliminated, but I don't understand why. I got this question right (i believe it's by luck) because i remember other question from destroyer that carboxylic acid H is very acidic.

If anyone could explain how to approach this problem using CARDIO logic or any other logic, I would really appreciate it. Thank you!

First ....protons C, D, and E are NOT acidic. Not a chance. An alkene proton has a pKa in the 40s, and the aldehyde hydrogen is even higher. Only two protons are acidic and they are A and B. Recall that in carbonyl chemistry, removal of the alpha hydrogen yields the resonance stabilized enolate. OK,,,,,Both are acidic, so who wins ? It turns out that when two alpha protons exist, kinetics wins out most of the time. It is EASIER for the base to remove proton A than for B. Refer to any organic t textbook and look at the Haloform reaction. This very same situation arises. The first proton is removed from the outside preferentially. The base has a much clearer path. Now.....these protons would have pKa values around 17 and 18 respectively. You should know a few other pKa values, and forget CARDIO ...which usually ends up having you more confused and getting a problem wrong. I teach my students the pKa values for the main functional groups..... Carboxy acids are 5... a diketone around 9...phenols and thiols around 10. alcohols around 16.....aldehydes and ketones around 17 and 18.....alkynes 25.....alkenes 45...and alkanes 55.

Hope this helps.

Dr. Romano

 

[CherryBlossom1412]

First ....protons C, D, and E are NOT acidic. Not a chance. An alkene proton has a pKa in the 40s, and the aldehyde hydrogen is even higher. Only two protons are acidic and they are A and B. Recall that in carbonyl chemistry, removal of the alpha hydrogen yields the resonance stabilized enolate. OK,,,,,Both are acidic, so who wins ? It turns out that when two alpha protons exist, kinetics wins out most of the time. It is EASIER for the base to remove proton A than for B. Refer to any organic t textbook and look at the Haloform reaction. This very same situation arises. The first proton is removed from the outside preferentially. The base has a much clearer path. Now.....these protons would have pKa values around 17 and 18 respectively. You should know a few other pKa values, and forget CARDIO ...which usually ends up having you more confused and getting a problem wrong. I teach my students the pKa values for the main functional groups..... Carboxy acids are 5... a diketone around 9...phenols and thiols around 10. alcohols around 16.....aldehydes and ketones around 17 and 18.....alkynes 25.....alkenes 45...and alkanes 55.

Hope this helps.

Dr. Romano

Thank you so much!

 

[CherryBlossom1412]

Protons D and E are not very acidic because the conjugate base will put a negative charge on carbon, which is not very stable and neither of the two have resonance. C would put a negative charge on Carbon as well, or with resonance it would put a positive charge on the oxygen which is not a good resonance structure because oxygen is electronegative. Between A and B the biggest factor would be that B is next to another CH2 group. Both A and B could resonate into the carbonyl group, but B has a CH2 group that is and electron donating group. The carbon for B would have a partial negative charge, and the adjacent CH2 group would add to that negative charge, which would make it a less stable conjugate base. Hope that explanation helps!

thank you!