[omegasynthesis999]

First off, is it possible to generate a Carboxylic Acid Anhydride using acid catalysis?

Second off, I was wondering why one oxygen (the double bonded oxygen of the carbonyl) acts as the base and gets protonated while the other one (the single bonded oxygen of the -OH portion) acts as a nucleophile and attacks the electrophilic carbon of the protonated carboxylic acid? Shouldn't one type of oxygen act as both the nucleophile and the base? Please see my attached document for the mechanism as I understand it.

Thanks very much.

[spirochete]

That reaction is correct in terms of electron book keeping, but in reality most carboxylic acids can't be converted directly to anhydrides like that.  The -COOH that adds is a much better leaving group than the hydroxide that gets booted out, so the reaction is very uphill in that respect.  One way to get around this is to convert the acid to an acyl chloride with SOCl2, and then add carboxylic acid.  The second step can be either acid or base catalyzed.

As to why the carbonyl carbon doesn't act as the nucleophile in the second step of anhydride formation, my best guess would by that the resulting product is less stable thermodynamically.  I tried drawing out a mechanism with CH3COCl plus acetic acid, and came up with a central carbon bonded to -OH, -COOCH3, -CH3 and -Cl.   

Another possibility is that under acidic conditions most or all of the carbonyl carbon is protonated, so it can't act as a nucleophile.  And under  basic conditions this is all a moot point because the deprotonated carboxylic acid has 2 equivalent nucleophilic oxygens.

As a side note, anhydrides can be made directly from DI-carboxylic acids in a cyclyzation reaction the involves a loss of water.  This requires a high temperature, because the reaction is favorable only entropically.

[omegasynthesis999]

Thanks for your insights.

But wrt to your answer to number 1, it is not the hydroxide that is being booted out. Between step 2 and 3, a molecule of water comes in and removes the H+ from the positively charge oxygen and then protonates one of the other -OH groups causing them to become OH2+ and therefore a good leaving group. Since the OH2 leaves, the electrons from the neighboring oxygen come in to form the double bond. But yes, the SOCl2 method would probably be a better pathway. I was just trying to test all combinations. 

Wrt to your answer to number 2, it makes sense. What I was trying to get at in my original question was why the -OH of acetic acid wouldn't get protonated. I think the carbonyl O might get protonated because it is more negatively charged and more basic than the other -OH oxygen. Furthermore, the resulting intermediate would have more resonance structures in addition to the fact that the partial positive charge would be more equally shared by the two electronegative oxygens, instead of a full on positive charge on one of the oxygens if it had two Hs attached.

This of course brings up an interesting observation according to my mechanism: in acetic acid, the carbonyl oxygen is more basic than the oxygen of OH while the latter is more nucleophilic than the carbonyl oxygen. Is this true?

[spirochete]

Thanks for your insights.

But wrt to your answer to number 1, it is not the hydroxide that is being booted out. Between step 2 and 3, a molecule of water comes in and removes the H+ from the positively charge oxygen and then protonates one of the other -OH groups causing them to become OH2+ and therefore a good leaving group. Since the OH2 leaves, the electrons from the neighboring oxygen come in to form the double bond. But yes, the SOCl2 method would probably be a better pathway. I was just trying to test all combinations. 

You're right I mispoke relative to the acidic mechanism you showed.  But it is still true that the direct conversion of acid-->anhydride is very difficult/impossible via undergrad organic organic chem, because of the relative reactivity of the two compounds.  There are two main reasons:  The anhydride has a better leaving group, making it more reactive for this reason.  And the acid benefits more from resonance stabilization:  The carbonyl carbon has less partial positive charge because of resonance donation from the other oxygen.  In the anhydride the two carbonyl carbons must share the resonance electron donation from a single central oxygen.  Thus the anhydride is more susceptable to nucleophilic attack.

Wrt to your answer to number 2, it makes sense. What I was trying to get at in my original question was why the -OH of acetic acid wouldn't get protonated. I think the carbonyl O might get protonated because it is more negatively charged and more basic than the other -OH oxygen. Furthermore, the resulting intermediate would have more resonance structures in addition to the fact that the partial positive charge would be more equally shared by the two electronegative oxygens, instead of a full on positive charge on one of the oxygens if it had two Hs attached.

This of course brings up an interesting observation according to my mechanism: in acetic acid, the carbonyl oxygen is more basic than the oxygen of OH while the latter is more nucleophilic than the carbonyl oxygen. Is this true?

That's a good observation about basicity of the carbonyl oxygen.  But I think if it's more basic then it's also more nucleophilic in this case.  My last post I was really trying to rationalize why the reaction goes as it does even though the carbonyl oxygen probably is more nucleophilic.  Don't forget that nucleophility is a kinetic property.  Even if the carbonyl oxygen is able to attack more easily, it leads to a less stable product than an attack by the
-OH. 

Try drawing a mechanism where the carbonyl oxygen attacks first.  I can't figure out a way to get from there to the product we would expect.  You get an ugly looking tetrahedral carbon bonded to two oxygens, like I said before.

Also remember that this discussion is only potentially relevant under acid conditions like I said before, and under acidic conditions the carbonyl oxygen is at least partially protonated.  And clearly a neutral -OH is more nucleophilic than a protonated carbonyl.   

[spirochete]

In case you were wondering I did just find a way to (sort of) convert two acids to an anhydride.  It involves heating one acid to 750 degree celsius with a catalyst so it dehydrates, forming a very high energy cumulated molecule called a ketene http://en.wikipedia.org/wiki/Ketene.  Ketene plus acid gives an anhydride.