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Author Topic: COOH Very Acidic  (Read 6818 times)

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qw098

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COOH Very Acidic
« on: February 07, 2012, 07:47:22 AM »

Hi guys,

I am a third year student... and I got thinking...

What exactly makes the hydrogen on a COOH group so acidic? What is the underlying chemical reason?!

Thanks!
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Arkcon

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Re: COOH Very Acidic
« Reply #1 on: February 07, 2012, 08:08:03 AM »

Can you draw out the structure?  Does that give you any hints?  This is important, you can use those hints to predict other proton donors.  Is there another class of organic compounds that is a fairly strong acid?
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qw098

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Re: COOH Very Acidic
« Reply #2 on: February 07, 2012, 10:26:30 AM »

Ahhhh, is it because once the H is gone the corresponding COO- is now stabilized by resonance!! So losing the H is very favorable!
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Arkcon

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Re: COOH Very Acidic
« Reply #3 on: February 07, 2012, 04:34:25 PM »

Good work.  Now, go investigate the acidity of phenol.  That's also an important one.
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orgopete

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Re: COOH Very Acidic
« Reply #4 on: February 07, 2012, 07:07:40 PM »

Ahhhh, is it because once the H is gone the corresponding COO- is now stabilized by resonance!! So losing the H is very favorable!

The poster has stated exactly what I don't like about this explanation. Because the product is more stable, the proton is more acidic. Am I the only one to whom that this sounds wrong? I posted about a similar question with phenols and I argued it was an inductive effect and I think I picked up about five flags for it.

This is my thinking.
Compound (pKa), CH4 (50), CH3OH (16) - inductive.
CH4 (50), CH2=CH2 (44), HC≡CH (26), N≡CH (9) - inductive.
CH3CH2CH3 (50), CH2=CHCH3 (44), N≡CCH3 (26), O=CHCH3 (16) - inductive.
CH3COOH (4.75), FCH2COOH (), F2CHCOOH (), CF3COOH () - inductive.
O=C(CH3)CH3 (20), O=CHNH2 (15), O=CHOH (3.75) - really, we're going resonance here?

To me, resonance is the method by which neighboring properties are extended, not ones in which they are created.
RC(=O)CH=CHOH (8-10?), RC(=O)OH (4.7) - the first is a vinylogous acid. It is a better resonance structure, three contributors than the carboxylic acid. This is a resonance effect. The pKa is much lower than a ketone because the double bond extends the electron withdrawing properties of the carbonyl group to the enol OH. The pKa is much lower, but not as low as the carboxylic acid. Two oxygen atoms are much better electron withdrawing groups. If you took a methyl polyene, the extended conjugation and resonance potential would not overcome the low inductive effect of a CH3 attached to a C=C group. If you replace the carbon with a nitrogen or oxygen, it will greatly increase the acidity. I argue it does so because of induction or rather the electron withdrawing effect of more protons in the nucleus of the neighboring heteroatom.

Okay, I'll watch my flag count grow again, but I really have a hard time seeing how the proton knows what the anion is going to be like? Does it really think, "Oh, the carboxylate is resonance stabilized, I can leave now?"
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Proton

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Re: COOH Very Acidic
« Reply #5 on: February 08, 2012, 12:41:46 AM »

@organopete
Resonance must make some contribution to acidity. Not only does electron delocalization spread out the negative charge, it also imparts thermodynamic stability. So let's say two different molecules have a similar tendency to deprotonate resulting in two anions; one of which is resonance stabilized, the other is not. The activation energy of the reprotonation of the resonance stabilized anion will be higher than that of anion without electron delocalization, and it will therefore be less likely to reprotonate, driving the position of the equilibrium forward and leading to increased acidity.

Resonance stabilization is undoubtedly one of the contributing factors to acitidy but not the only one.
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fledarmus

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Re: COOH Very Acidic
« Reply #6 on: February 08, 2012, 03:02:20 AM »

Orgopete, I think the problem that you are having is seeing only the two different structures and not the continuum of possibilities in between. Rather like visualizing discrete electron orbitals versus electron clouds.

If you consider the reaction of a carboxylic acid group to from a carboxylate anion and a proton, in water so that both are hydrated, there will be an energy difference between the energy of the protonated carboxylic acid group and the energy of the deprotonated carboxylic acid group. The distribution of your total population of material between the protonated and deprotonated form is a statistical distribution based on the energy difference. If the deprotonated carboxylic acid group is stabilized, by resonance, induction, or any other means, more of the distribution will be in the deprotonated form and consequently there will be more protons - the compound is "more acidic". No, the starting material doesn't need to know that it's product will be more stable to decide whether or not to get rid of the proton, but the energy difference between the starting material and the product will determine how much of the material at equilibrium will be in each form. The additional stability of the carboxylate anion by resonance explains the experimentally perceived increase in the equilibrium amounts of deprotonated vs protonated carboxylate, which is described as increased acidity of carboxylic acids relative to other -OH groups.
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orgopete

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Re: COOH Very Acidic
« Reply #7 on: February 10, 2012, 08:24:17 PM »

Three flags!

Proton, I think we have different opinions as to what constitutes resonance and inductive. I think of resonance as what makes benzene more stable than predicted from three isolated double bonds. It is also what allows electrons to shift rather than remain stationary such as cyclohexatriene implies. I don't think of it as creating or stabilizing a charge. It allows a charge to be distributed to atoms than can stabilize a charge. For example, CH2=CH-CH2-, CH2=CH-O-, and O=CH-O- are all identical resonance structures, but not of equal stability. The difference is the heteroatoms that provide stability. If one were lower in energy than another, I would argue it is an inductive effect. I am not saying resonance is not involved. I am saying the stability you are seeing is due to the heteroatoms. Formic acid is a stronger acid than propene because of the oxygen atoms, or an inductive effect. If you had the anion of a pentadienyl anion or analogous longer chain, you could have a super resonance anion, but would it ever approach the acidity of acetaldehyde?

I thought I had included several examples of this in my original post. This included the vinylogous acid, which should give a greater resonance structure with the negative charge over three atoms, but less stable than a carboxylic acid can over two. While resonance is improved, inductive is still greater.

Okay, try this, acetaldehyde and its enol (1-hydroxyethylene). If resonance determines acidity (they have the same resonance structure), why is the aldehyde more stable?

For those of you who disagree with me, please re-read my post. Note how many instances of inductive effects I give. Then think about it. If you think I am wrong, then let me tell you acetaldehyde and its enol have different pKa values. The enol is much lower. If the acidity of a carboxylic acid is determined by the resonance stability of the anion, then acetaldehyde and its enol ether should have the same pKa as they have the same anion. 

I have written a blog about carboxylic acid acidity, inductive or resonance effect? (http://www.curvedarrowpress.com/wpblog/?p=74) It is much too long to post here.
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OrgoTutor

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Re: COOH Very Acidic
« Reply #8 on: February 17, 2012, 01:35:29 PM »

orgopete, I'm still not sure I completely understand your complaint--primarily because it seems you use different definitions of inductive effect and resonance than I'm used to. I take inductive effect to be the disturbance of electron density of an atom which is propagated through the sigma bond framework. And I take the resonance to be the disturbance or delocalization of electron density through a conjugated pi system (or adjacent p orbitals, more generally). Pi electron delocalization stabilizes anions because a dispersed negative charge is a lower energy state of affairs than a concentrated, localized negative charge. Anyway, I take you to be claiming that for any example of acidity where inductive and resonance explanations can be given, the inductive explanation is entirely sufficient, and so resonance is superfluous in explaining acidity. Otherwise, I thought you might be claiming that inductive effects are always stronger than resonance effects (assuming it affects acidity). Both are strong claims, and I'm not sure if either accurately characterizes your position, but assuming they do, I believe I have at least one exception to such claims.

Take the example of ascorbic acid. It has two hydroxyl groups on the ring. The interesting thing about the acidity of these two hydroxyl groups is that the hydroxyl group ("A"--the one attached to the carbon alpha to the carbonyl) which one would assume would have feel the greatest cumulative inductive effect is actually less acidic (pKa 11.57) than the other hydroxyl group ("B", pKa 4.17). This can be explained by the fact that resonance structures can be drawn for the conjugate base of hydroxyl "B" which delocalize the charge on oxygen, but not so for hydroxyl "A." To me this example demonstrates that resonance stabilization can influence acidity and that in at least some cases resonance effects are stronger than inductive effects. Why couldn't resonance be of equal or greater importance than inductive effect for the acidity of carboxylate anions, etc.?

I think we need a physical organic chemist to put some numbers to the inductive effect and resonance effect as applied to particular cases in order to sort all this out, rather than resorting to a qualitative discussion of acidity trends.

[reference for ascorbic acid pKas: http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/special2.htm]

Okay, try this, acetaldehyde and its enol (1-hydroxyethylene). If resonance determines acidity (they have the same resonance structure), why is the aldehyde more stable?

For those of you who disagree with me, please re-read my post. Note how many instances of inductive effects I give. Then think about it. If you think I am wrong, then let me tell you acetaldehyde and its enol have different pKa values. The enol is much lower. If the acidity of a carboxylic acid is determined by the resonance stability of the anion, then acetaldehyde and its enol ether should have the same pKa as they have the same anion.

One explanation for the difference in acidity could be that while the resonance stabilization is equal for both the inductive effect is greater for hydroxyethylene, so it is more acidic. This would show that resonance alone doesn't determine acidity--but I don't think anybody even claimed that. So the example seems to be beside the point. However, there is something wrong with the resonance effects explanation that people sometimes give. As my Solomons and Fryhle Ochem book tells me, resonance stabilization isn't determined solely by the conjugate base resonance hybrid, rather it is determined by the difference in energy between the resonance hybrid of the protonated acid form and the resonance hybrid of the deprotonated form. Thus, the difference in acidity of acetaldehyde and hydroxyethylene could also be explained by the likely fact that the conjugate bases of acetaldehyde and hydroxyethylene actually don't have the same resonance stabilization energy.

My Solomons book also tells me this article has more info on the inductive/resonance debate: Bordwell, F. G.; Satish, A. V. J. Am. Chem. Soc. 1994, 116, 8885-8889. I don't have access to it, so I can't relate its contents.
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orgopete

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Re: COOH Very Acidic
« Reply #9 on: February 18, 2012, 12:06:28 PM »

Let me go through this backwards.
Re: acetaldehyde v hydroxyethylene
Because they share the same resonance structure as the enolate, the acidity must be due to the energy levels of the aldehyde and enol. I argue an OH is more acidic than a CH is an inductive effect. This is true whether resonance is present or not.  

I use resonance to explain the energy difference where pi or non-bonded electrons can interact. You can see the effect in reactivity or energy measurements. Increasing the number of electrons I argue increases the resonance effect. You can see this spectroscopically with UV and IR.

Compounds can have resonance and inductive effects. If you were to ask, which had a greater resonance effect, an amide or a carboxylic acid, I would argue an amide. I would argue the decrease in carbonyl stretching frequency shows how the electrons of nitrogen are interacting with the carbonyl group. Resonance is consistent with an increase in electron density on oxygen.

Cyclohexenone forms an alpha anion due to greater acidity, if equilibrated forms the gamma-anion, resonance stabilized form. Conjugated enones are more stable than non-conjugated ones. I see these as a resonance effect, the manner in which pi-electrons interact with one another. Their interaction may stabilize an anion. Acidity is generally an inductive effect. A carboxylic acid is more acidic than an amide even if an amide were more resonance stabilized. A ketone uses what some have referred to as a non-resonance effect. 

Re: ascorbic acid
Ascorbic acid is a good problem to examine. I really like this example. It is data. I do not doubt the values. Because organic chemistry is really about electron movements, it is always intriguing to learn about how and what factors result in their movement and effects. Ammonia is an inductive electron withdrawing group to form an amide anion, but a resonance donating group in aniline.

What should the values for ascorbic acid be? Acetoacetate is about 11, acetylacetone is about 9, 3-oxobutanal is about 6, dimedone is about 5. I drew 3-oxobutanal as 4-hydroxybut-3-en-2-one as I believe that is its predominant form. This is the vinylogous acid I was originally noting. In that instance, I was arguing that although a greater number of resonance structures can be drawn for the anion, its pKa was less than a carboxylic acid. Even though resonance exists, separating the OH group from the more electron withdrawing C=O results in lower acidity. I think increasing conjugation will make an enol/aldehyde behave more like an isolated enol-aldehyde.

Why is ascorbic acid more acidic than dimedone or acetoacetate and which OH is more acidic? As pointed out, it could have been either one. "A" is closer to the electron withdrawing C=O group. "B" bears the resonance effect noted earlier. What if there wasn't a "B"-OH? Which carbon of an enone is more electron deficient? The beta carbon is. Is this resonance or inductive? It actually is both. The electrons can interact to deliver the pi-electrons toward the carbonyl group. Because they can interact does not lead me to argue the carbon should be electron deficient. The electron withdrawing property of the oxygen does. The actual acidity appears to be a combination of effects. The dimedone is lower than acetylacetone, so a ring probably increases the acidity. The OH at "A" may also increase the acidity slightly.

What is probably more interesting is the second pKa. Why should the pKa of "A" be so low? Presumably, all of ascorbic acid becomes an enolate as the pH increases. The enolate places a negative charge alpha to the OH as well. Presumably, our resonance isomers do not reveal the true electron withdrawing character of the vinylogous enolate to accept another negative charge, because 11 is surprisingly acidic for this dianion.

My "complaint" was a simple one. I do not agree that a proton is acidic because you can draw a resonance structure of its conjugate base.
« Last Edit: February 18, 2012, 12:49:48 PM by orgopete »
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Kran

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Re: COOH Very Acidic
« Reply #10 on: February 27, 2012, 12:23:05 PM »

Acidity is not a measure of how easy a proton can leave. It is about relactive stability of the protonated and desprotonated form. If something stabilizes the deprotonated form, it makes the protonated form more acidic.

Look at mechanism of the aromatic nucleophilic substitution. The intermediate arenium ion is very acidic, able to protonate even a clorine anion. But the peroxiacetic acid is not very acidic at all. I'm sure you can see why.
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SirRoderick

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Re: COOH Very Acidic
« Reply #11 on: February 27, 2012, 12:49:29 PM »

Acidity is not a measure of how easy a proton can leave. It is about relactive stability of the protonated and desprotonated form. If something stabilizes the deprotonated form, it makes the protonated form more acidic.

Look at mechanism of the aromatic nucleophilic substitution. The intermediate arenium ion is very acidic, able to protonate even a clorine anion. But the peroxiacetic acid is not very acidic at all. I'm sure you can see why.

Ehm...how easy it is for a proton to leave is exactly the same as the relative stability. They are basically the same thing, but different ways of reasoning about it.

The way I think about it, let's take a COOH group. In my mind the O atoms are drawing electrons away from the O-H bond, thereby weakening the protons attachment. It become easier for this proton to leave because this bond has become more unstable, less populated by electrons. That is a causal reason for the acidity. Now when this proton leaves, it turns out that the resulting situation is more stable and you can argue that the resonance is also a causal reason for the proton's ability to leave and therefore the acidity. (because that proton is less likely to attach back on)

But I always felt that using the argument that the product's stability promotes a certain reaction as a sort of retrograde reasoning. The real reason that that proton left is not because a hypothetical future state is stable, it has much more to do with the fact that the current state is unstable. And I believe that is roughly what orgopete is trying to say. That the reason that proton leaves is not the stabilisation of the product, but the destabilising effect of the induction coming from the O. That is the principal cause from a purely logical, sequential standpoint.
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fledarmus

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Re: COOH Very Acidic
« Reply #12 on: February 27, 2012, 03:25:09 PM »



But I always felt that using the argument that the product's stability promotes a certain reaction as a sort of retrograde reasoning. The real reason that that proton left is not because a hypothetical future state is stable, it has much more to do with the fact that the current state is unstable. And I believe that is roughly what orgopete is trying to say. That the reason that proton leaves is not the stabilisation of the product, but the destabilising effect of the induction coming from the O. That is the principal cause from a purely logical, sequential standpoint.

This isn't a "hypothetical future state" - this is a current state that exists in some equilibrium, with individual molecules rapidly cycling between states. Both the product stability and the starting material instability contribute to the energy difference between product and starting material, and it is that energy difference that determines the relative proportions of the two in an equilibrium. Logical, sequential standpoints really don't exist in a qaussian distribution world.
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Re: COOH Very Acidic
« Reply #13 on: February 27, 2012, 10:03:42 PM »

Because the product is more stable, the proton is more acidic. Am I the only one to whom that this sounds wrong?

But I always felt that using the argument that the product's stability promotes a certain reaction as a sort of retrograde reasoning. The real reason that that proton left is not because a hypothetical future state is stable, it has much more to do with the fact that the current state is unstable.

I am firmly in fledarmus's camp on this. The acidity of an acid is directly related to the basicity of it's conjugate base. If you can argue that the conjugate base is more stable, and thus less basic, it follows by definition that the acid is more acidic.

On the resonance/induction debate

Quote from: orgopete
My "complaint" was a simple one. I do not agree that a proton is acidic because you can draw a resonance structure of its conjugate base.

This is good point. I think the debate here is mainly down to the imprecise question. Acidity is a relative - you can't meaningfully explain why something is acidic without a frame of reference. We should ask why is a A more acidic than B, not why is A acidic.

That acetic acid is more acidic than acetamide is not a resonance effect (as orgopete has explained). But, that acetic acid is more acidic than ethanol is primarily a resonance effect.
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orgopete

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Re: COOH Very Acidic
« Reply #14 on: February 28, 2012, 09:40:27 PM »

I am firmly in fledarmus's camp on this. The acidity of an acid is directly related to the basicity of it's conjugate base. If you can argue that the conjugate base is more stable, and thus less basic, it follows by definition that the acid is more acidic.
I disagree with this. If this were true, then a ketone and its enol should be equally acidic if their acidity were determined by the stability of their identical conjugate bases. I argue the acidity is due to the proton-electron pair distances (as well as the nuclear proton field). I was arguing an enol is more acidic than a ketone because oxygen pulls its electrons further from the proton than a carbon of a ketone does. This is an inverse square argument embodied in Coulomb's law.

On the resonance/induction debate

Quote from: orgopete
My "complaint" was a simple one. I do not agree that a proton is acidic because you can draw a resonance structure of its conjugate base.

This is good point. I think the debate here is mainly down to the imprecise question. Acidity is a relative - you can't meaningfully explain why something is acidic without a frame of reference. We should ask why is a A more acidic than B, not why is A acidic.

That acetic acid is more acidic than acetamide is not a resonance effect (as orgopete has explained). But, that acetic acid is more acidic than ethanol is primarily a resonance effect.

This has to be my last post on this subject. It is clear than I have a different definition of resonance than others if acetic acid is more acidic than ethanol because of a resonance effect. I concede this is a difficult topic. I have not tried to enforce any conclusions about resonance. I find it sufficient that benzene is more stable than a cyclohexatriene (if the double bonds did not interact). This difference in energy is given the term resonance. Organic chemists use the interactions of pi electrons to draw structures reflecting these stabilizations as resonance structures. Resonance structures allow pi and non-bonded electrons to interact in ways we are familiar with drawing.

I've provided examples in which I argued the electrons are interacting in ways typical of resonance structures, but ones in which I have chosen the groups to either donate or withdraw electrons. I have argued it was the electron donating or withdrawing properties that altered the proton-electron pair distances and ultimately the acidity. I argued that a vinylogous acid (RC=OCH=CHCHOH) has a greater resonance contribution than a carboxylic acid yet is a weaker acid is due to a decrease in the inductive electron withdrawal. I argued that the kinetic enolate of cyclohexenone is an conjugate base of the stronger acid, while the conjugated enolate is the thermodynamically more stable enolate and likely due to resonance stabilization.

If you compare ethanol and acetic acid, the only consideration for acidity should be the proton-electron pair distance. They both contain an OH bond. Why are the electrons of acetic acid closer to the oxygen or further from the proton in acetic acid? Acetic acid is connected to a carbonyl group. It is electron withdrawing.

Let's go back to the ascorbic acid example. Note the pKa values. Let's do a titration. At pH 5.17, 90% will be the monoanion and at 6.17 99%. Draw the resonance structures. Predict the pKa of the OH next to the carbonyl group. Compare it to ethanol. Do the resonance structures enable you to predict an increase in acidity? Its acidity is very much like phenol, despite the fact it is an anion (I am assuming this data is correct).
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