Which proton is most acidic?

[FROGGBUSTER]

Destroyer says that protons A & B are both more acidic than C, with the answer being A (I do get that A should be more acidic than B).

I chose C though because de-protonation of C would access the more conjugated system. I would think there is more electron delocalization if you de-protonate C compared to A or B.

Why are A & B more acidic than C?

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[fujindesu]

If you depronate proton C, the carbon would bear the negative charge, it has no where to delocalize since the carbons next to it are all sp3 hybridized.
Where as carbon A and B can delocalize the extra lone pair to form a C=C bond, and kick the pie electrons up to the oxygen.

Beat Cal

 

[AlbinoPolarBear]

Either way, there will be resonance.

In one case (C), the electrons would resonate back and forth between 2 carbons.

While, if you deprotonate a hydrogen alpha to a carbonyl, the electrons would resonate back and forth between a carbon and an oxygen. The oxygen is more electronegative and will be better at holding the negative charge associated with the extra electrons, compared to the carbon.

 

[mh0000]

Either way, there will be resonance.

In one case (C), the electrons would resonate back and forth between 2 carbons.

While, if you deprotonate a hydrogen alpha to a carbonyl, the electrons would resonate back and forth between a carbon and an oxygen. The oxygen is more electronegative and will be better at holding the negative charge associated with the extra electrons, compared to the carbon.

👍. It's not always about the number of resonance structures. Remember, each resonance structure is a "fraction" of the electronic distribution. With an enone, a smaller fraction of the electrons charge is distributed to the oxygen. But with the aldehyde (or was it a ketone), with only two resonance structures, half of it is distributed to the oxygen.

 

[FROGGBUSTER]

So I'm getting from all 3 of you is that the electrons would stay localized on the carbon, but when I push electrons I see otherwise. When I draw it out, I see 2 resonance structures; the major resonance form would be the one with the negative formal charge on oxygen, just like you would get if you de-protonate A or B.

My rationale was that if A, B, and C each have 2 resonance structures as well as "equal-quality" resonance structures (negative on most electronegative atom), we would go with the structure with most conjugation, which is C.

Did I push electrons incorrectly or something?

 

[FROGGBUSTER]

If you depronate proton C, the carbon would bear the negative charge, it has no where to delocalize since the carbons next to it are all sp3 hybridized.
Where as carbon A and B can delocalize the extra lone pair to form a C=C bond, and kick the pie electrons up to the oxygen.

Beat Cal

Actually I am pretty sure there would be re-hybridization at that carbon --> sp2, someone correct me if I'm wrong.

 

[AlbinoPolarBear]

So I'm getting from all 3 of you is that the electrons would stay localized on the carbon, but when I push electrons I see otherwise. When I draw it out, I see 2 resonance structures; the major resonance form would be the one with the negative formal charge on oxygen, just like you would get if you de-protonate A or B.

My rationale was that if A, B, and C each have 2 resonance structures as well as "equal-quality" resonance structures (negative on most electronegative atom), we would go with the structure with most conjugation, which is C.

Did I push electrons incorrectly or something?

I think you only read the first response.

Anyways, I missed the aldehyde carbonyl, but it really doesn't make too much difference.

The carbonyl carbon is a greater electron density withdrawing group as compared to the alkene.

The hydrogens alpha to the carbonyl will have less electron density, and therefore will be easier to remove.

The pKa of a hydrogen alpha to carbonyl group is around 20, while the pKa of a hydrogen alpha to an alkene group is closer to 40.

And also, the electron sink is more prominent the closer the electrons are to the electron withdrawing group.

 

[FROGGBUSTER]

I think you only read the first response.

Anyways, I missed the aldehyde carbonyl, but it really doesn't make too much difference.

The carbonyl carbon is a greater electron density withdrawing group as compared to the alkene.

The hydrogens alpha to the carbonyl will have less electron density, and therefore will be easier to remove.

The pKa of a hydrogen alpha to carbonyl group is around 20, while the pKa of a hydrogen alpha to an alkene group is closer to 40.

And also, the electron sink is more prominent the closer the electrons are to the electron withdrawing group.

Thanks APB.

 

[FROGGBUSTER]

This was bothering me, so I ended up emailing my orgo TA from last semester. The answer is C.

 

[AlbinoPolarBear]

This was bothering me, so I ended up emailing my orgo TA from last semester. The answer is C.

Wow really. Okay. His reasoning was the conjugation? Hm.. weird..

I'll try to ask some friends about this. I'm pretty confident it's not C.

 

[AllahIsGreater]

This was bothering me, so I ended up emailing my orgo TA from last semester. The answer is C.

lol Dr. Romano would run over you organic ta, no offense. He just recently submitted a proposed mechanism to harvard to get published. the man would steam roll any organic chemist in the world. TRUST ME. if you are really still confused, just email him, he will be more than happy to respond. the answer in the destroyer is correct.

 

[FROGGBUSTER]

lol Dr. Romano would run over you organic ta, no offense. He just recently submitted a proposed mechanism to harvard to get published. the man would steam roll any organic chemist in the world. TRUST ME. if you are really still confused, just email him, he will be more than happy to respond. the answer in the destroyer is correct.

Should I use the Contact Us form on his website? Or do you have his email? You seem like you know the man.

Lol btw, my TA is pretty good himself, I wouldn't bet against him.

 

[FROGGBUSTER]

Wow really. Okay. His reasoning was the conjugation? Hm.. weird..

I'll try to ask some friends about this. I'm pretty confident it's not C.

Yeah, the conjugation and he also wrote this:

The only thing I can think of is that A is less sterically hindered than C, so if your question was asking about which spot was going to form the kinetic enolate I'd say A. But the way it's phrased ("acidity" generally refers to thermodynamic acidity, or which spot will be most deprotonated at equilibrium), I think C should be the answer.

 

[AllahIsGreater]

Should I use the Contact Us form on his website? Or do you have his email? You seem like you know the man.

Lol btw, my TA is pretty good himself, I wouldn't bet against him.

[email protected]

and yes i know Dr. Romano, he is very nice and will not hesitate to help! Please let us know how it goes!

 

[FROGGBUSTER]

[email protected]

and yes i know Dr. Romano, he is very nice and will not hesitate to help! Please let us know how it goes!

Sent it out, hopefully he is not too busy and will respond promptly.

 

[AlbinoPolarBear]

Yeah, the conjugation and he also wrote this:

Alright, I asked some PhD organic chemist friends and they came to the same conclusion I did, with the same reasoning. Ask your TA what would be the pKas of each hydrogen, because I'm getting pka B~20, while pka C~40.

 

[FROGGBUSTER]

Alright, I asked some PhD organic chemist friends and they came to the same conclusion I did, with the same reasoning. Ask your TA what would be the pKas of each hydrogen, because I'm getting pka B~20, while pka C~40.

I don't want to bother my TA anymore.

But are you getting the pka of 40 from the fact that it's an allylic hydrogen?

It doesn't really make sense to me how you guys are getting such big differences in pka values for the 2 sites because de-protonation of C is clearly (at least to me) more thermodynamically stable. While the pKa of an allylic hydrogen is about 40, I would think it would be much much lower in this case due to the Beta-unsaturated aldehyde system it is a part of.

 

[needzmoar]

Alright, I asked some PhD organic chemist friends and they came to the same conclusion I did, with the same reasoning. Ask your TA what would be the pKas of each hydrogen, because I'm getting pka B~20, while pka C~40.

I actually stopped studying just to join this chem party.

APB: Are you sure C's pKa ~ 40 ? I would suspect that it's closer to about 17.

I actually don't understand why A would be more acidic than C

 

[AlbinoPolarBear]

I actually stopped studying just to join this chem party.

APB: Are you sure C's pKa ~ 40 ? I would suspect that it's closer to about 17.

I actually don't understand why A would be more acidic than C

What are you basing the 17 on?

I looked at numerous pka charts and couldn't find information on the pka of the C hydrogen.

The people from my lab said they wouldn't be able to measure the pka of this species, because after the A comes off (which it will come first because it's the most acidic), trying to get B and C off will be extremely difficult as you end up with dianion and trianion, which are extremely unstable.

 

[FROGGBUSTER]

What are you basing the 17 on?

I looked at numerous pka charts and couldn't find information on the pka of the C hydrogen.

The people from my lab said they wouldn't be able to measure the pka of this species, because after the A comes off (which it will come first because it's the most acidic), trying to get B and C off will be extremely difficult as you end up with dianion and trianion, which are extremely unstable.

APB can you please explain why you think the structures that result from de-protonation at C is thermodynamically less stable than the structures that result from A & B? Thermo stability is the definition of acidity after all.

 

[needzmoar]

What are you basing the 17 on?

I looked at numerous pka charts and couldn't find information on the pka of the C hydrogen.

The people from my lab said they wouldn't be able to measure the pka of this species, because after the A comes off (which it will come first because it's the most acidic), trying to get B and C off will be extremely difficult as you end up with dianion and trianion, which are extremely unstable.

Definitely true if C isn't the most acidic. I too checked several handbooks for the pKa of that particular function group and couldn't find anything. I figured it's closer to 17 than 40 because the pKa of an aldehyde is about 17. Is there a reason to believe that the electron pair from C will not delocalize into the aldehyde?

 

[needzmoar]

btw, thanks frogbuster for bringing this up.

 

[AlbinoPolarBear]

Definitely true if C isn't the most acidic. I too checked several handbooks for the pKa of that particular function group and couldn't find anything. I figured it's closer to 17 than 40 because the pKa of an aldehyde is about 17. Is there a reason to believe that the electron pair from C will not delocalize into the aldehyde?

Well, the electron pair will delocalize to the aldehydes if C gets deprotonated first. So this is trying to explain your reasoning after having your conclusion. I admit that I'm doing the same haha.

But the aldehyde is pretty far away, while the alkene is the closer group to C.
The 17 you got would be from the hydrogens alpha from an aldehyde group, not delta to it (which is in the case of C), so I don't think you can use that as a model compound.
How much pull does the aldehyde have as a group to pull electron density away from the C carbons, with an alkene in the middle? Certainly it cannot go from ~40 to under 20...

Frogger: I didn't like your TA's explanation at all, that's why I wanted to ask him those questions. Just to confirm if he knew what he was talking about. My argument is for the area for least electron density, as it would be easier for atoms to hold onto it's electrons. I'm not sure about the thermodynamic data, because electrons are not shared equality in any of these cases. The oxygen will act as an electron sink in either case. There's no resonance to speak of, and the product of the deprotonation will be unstable and will react right afterwards (aldol reaction).

 

[FROGGBUSTER]

Well, the electron pair will delocalize to the aldehydes if C gets deprotonated first. So this is trying to explain your reasoning after having your conclusion. I admit that I'm doing the same haha.

Acidity is a thermodynamic property though. When you speak of explaining the reasoning after having the conclusion, it makes it sounds like you're treating this as a matter of kinetics. Like my TA said, sterics could favor de-protonation of protons A and/or B at lower temps, but at higher temps, I am finding it very difficult to believe the extensive conjugation of the Beta-unsaturated aldehyde system doesn't make it more thermodynamically favorable.

I'm not sure about the thermodynamic data, because electrons are not shared equality in any of these cases.

What kind of thermodynamic data would you need? It's just (or should be, if I'm thinking about this correctly) a matter of de-protonation of proton C accessing the more conjugated and stable system vs. de-protonation of A & B not doing the same.

 

[AlbinoPolarBear]

Acidity is a thermodynamic property though. When you speak of explaining the reasoning after having the conclusion, it makes it sounds like you're treating this as a matter of kinetics. Like my TA said, sterics could favor de-protonation of protons A and/or B at lower temps, but at higher temps, I am finding it very difficult to believe the extensive conjugation of the Beta-unsaturated aldehyde system doesn't make it more thermodynamically favorable.



What kind of thermodynamic data would you need? It's just (or should be, if I'm thinking about this correctly) a matter of de-protonation of proton C accessing the more conjugated and stable system vs. de-protonation of A & B not doing the same.

Your TA is making it seem like these 2 deprotonated compounds are stable and are comparing which one is the most stable.

You should be aware that both of these compounds (after deprotonation) are extremely unstable and won't remain in that form for long. That's why I'm not even looking at it thermodynamically. Kinetically, the hydrogens alpha to carbonyl (with A being more acidic), will be deprotonated (reasons mentioned earlier), and will perform the aldol reaction even before the C gets a chance to be deprotonated.

Either case, I don't think anyone of us is going to be happy unless there are concrete pKa values to support our arguments. And thanks for bringing this up, this question really made me think today 😉

 

[LetsGo2DSchool]

I wouldn't trust TAs too much either. For example my Orgo lab was taught by a recent Ph.D. grad and he was a fool. I remember during the Thin Layer Chromatography lab he taught a bunch of things which were contradictory to what I just read in the textbook hours earlier. When I brought it up asking for clarification, even after reading a paragraph verbatim, his response was that the textbook must be wrong. Oooookay buddy... what a douche lol.

 

[FROGGBUSTER]

I wouldn't trust TAs too much either. For example my Orgo lab was taught by a recent Ph.D. grad and he was a fool. I remember during the Thin Layer Chromatography lab he taught a bunch of things which were contradictory to what I just read in the textbook hours earlier. When I brought it up asking for clarification, even after reading a paragraph verbatim, his response was that the textbook must be wrong. Oooookay buddy... what a douche lol.

Maybe, but this guy is pretty good I think. You have to be super to be accepted into Berkeley's graduate school for Chem. We'll see what Romano says.

 

[LetsGo2DSchool]

Maybe, but this guy is pretty good I think. You have to be super to be accepted into Berkeley's graduate school for Chem. We'll see what Romano says.

Actually, my OC lab teacher obtained his Ph.D from UPenn -- Ivy League. He must be one of those who fell through the cracks due to a misleading college GPA -- he attended a dinky college that I never heard of before. Go figure.

 

[FROGGBUSTER]

The answer is A. Dr. Romano sent me 2 emails.

Hi Tom. This is a great question. The answer is C. Your logic is correct. Removal of C would result in greater delocalization of the negative charge, hence greater anion stability. A would be next. Keep up the great work. Dr Romano NYU

This made me happy, but unfortunately there was a follow-up email...

Tom.....I have sent you 2 other emails. I went to the literature to double check this. I apologize for such a close call. C.....which I agreed with you on...being most acidic...has a pka of 20. It is a gamma proton. The proton in A.....has a pKa slightly less. It appears that speed......and kinetics have won the race !!!!! You wont see something this tough on the DAT. The DESTROYER answer stands correct.......but not a very good question. I apologize again.....for such a close call. !!!! Respectfully, Dr Romano.

I guess kinetics does factor in to acidity more than I thought it did.

 

[AlbinoPolarBear]

The answer is A. Dr. Romano sent me 2 emails.



This made me happy, but unfortunately there was a follow-up email...



I guess kinetics does factor in to acidity more than I thought it did.

Thanks for the follow-up Tom! It was a nice discussion haha.

 

[FROGGBUSTER]

Thanks for the follow-up Tom! It was a nice discussion haha.

No problem APB.

 

[needzmoar]

No problem APB.

Good stuff, man.