[sammyjo06]

We are having a test soon on reactions with ketones and aldehydes.

I've been studying the mechanisms, and I think I know them pretty well, but I just realized recently that I don't know something that's probably pretty fundamental to understanding these reactions.

We learned about how ketones and aldehydes can be treated with acids or bases to form hemiacetals, and then with additional acid to form acetals.

We've also learned that ketones treated with acids can form enols.

So....Which is it? When you mix a ketone with acid, how do you make it do what you want it to and either form an acetal (and leave the alpha hydrogens alone) or to form the enol? Does it just have to do with having OH- and OR- groups in there when you want to form the acetal, or what?

Is it like this:

1) Acid, alcohol, and OR- group plus ketone leave the alpha hydrogens alone and attack the carbonyl carbon (by protonating the O first) to make hemiacetals, and then further react to make acetals
2) When you just have the acid but not alcohol or OR- groups, it still protonates the O, but then the deprotonated acid attacks the alpha hydrogens instead of the carbonyl carbon?
3) If there's a base in with a ketone, it'll attack the alpha hydrogens before the carbonyl carbon, unless there is an alcohol there?

I really don't get this, and would appreciate any help in explaining this. Thanks !

[azmanam]

You have it right - it all depends on the reaction conditions.  Your three scenarios are exactly right, let's just flesh them out a bit.

Before we begin, we need to start by recognizing that every step in the formation of an acetal under acidic conditions is reversible.  Carbonyl + 2 eq. alcohol + cat. acid --><-- acetal + water.  Every step in that mechanism is reversible.  The reaction can go in either direction with the same reaction conditions.  Second, every step in the formation of an enol under acidic conditions is also reversible.  Carbonyl + cat. acid --><-- enol.  Every step in that mechanism is reversible.  Sounds trivial, but it is not.

Ok, so.  how to predict what's going to happen.  Reactions only occur one way (in this chapter, at least)... nucleophiles attack electrophiles.  That's it.  So, lets examine the reactions: formation of an (hemi)acetal from a carbonyl, and formation of an enol from a carbonyl.

Case 1: Formation of a hemiacetal under basic conditions.

Let's first examine formation of a hemiacetal under basic conditions.  Carbonyl + sodium alkoxide.  For example, acetaldehyde + sodium methoxide.  We know the carbonyl carbon is a good electrophile.  We know the methoxide oxygen anion is a good nucleophile.  (the carbonyl oxygen is also a nucleophile, but a much weaker nucleophile than the  methoxide ion).  Nucleophiles attack electrophiles.  Based on the limited reaction conditions, the only plausible first step is nucleophilic attack of the alkoxide on the carbonyl carbon atom, and the pi electrons in the C=O double bond go up on to oxygen to form the anionic intermediate (see schemes for pictorial representations throughout).

From here, there are really only two options.  The electrons on the newly formed O^- atom can collapse back down to reform the carbonyl and kick off a good leaving group.  The only good leaving group is CH₃O^-, and that proves the first step in the mechanism is reversible.  The only other thing that can happen is a proton transfer.  The O^- atom is nucleophilic now and can act as a nucleophile.  But to do so, it would need an electrophile.  The only electrophile around is the acidic proton on solvent (methanol in this example).  The O^- atom can deprotonate a molecule of solvent to form the neutral hemiacetal.  The reasons for both of these steps take into consideration the possible nucleophiles and electrophiles in solution and proceeding accordingly.  This is what happens (and why) under basic conditions.

Case 2: Formation of an enol under acidic conditions.

To start, we should again examine the reaction conditions for possible nucleophiles and electrophiles.  Let's say the reaction conditions are acetaldehyde and H₃O⁺.  Under acidic conditions, the best electrophile by far is the acidic proton.  The carbonyl oxygen, as wehave mentioned, is a nucleophile (but a weak one).  The oxygen atom can pick up a proton from the acid (i.e. the carbonyl oxygen can deprotonate the acid), and the carbonyl becomes protonated.

Now, we have made the carbonyl a much better electrophile (we have activated the carbonyl).  The carbonyl is pretty unhappy with the positive charge on oxygen and would like to relieve itself of that charge.  The conjugate base of the acid (water in this case) can take a proton from the alpha carbon.  The electrons in the former C-H single bond can form the C=C double bond, and the electrons in the C=O pi bond can go up on to oxygen to relieve the positive charge.  This is how the enol is formed.  Again, it occurs because of the nature of the compounds... are they nucleophiles?  Are the electrophiles?  If they're nucleophiles, what electrophiles might they react with? 

The enol is itself a nucleophile.  It is nucleophilic at the alpha carbon atom (a resonance structure puts a full minus charge on the alpha carbon atom).  There aren't really any electrophiles to react with.  The only good electrophile is the acidic proton on the acid.  The enol can react with the acidic proton on the acid to regenerate the protonated carbonyl compound, proving that the enol-forming step is reversible.

Case 3: Formation of an acetal under acidic conditions.

The reaction conditions here look remarkably similar to the conditions for the formation of the enol.  the only difference is the presence of the alcohol.  Now, the alcohol is not a good enough nucleophile to attack the carbonyl straight away (it's nowhere near as good a nucleophile as the alkoxide from before).  So we need to make the carbonyl a better electrophile before the nucleophile can attack.  That's why we activate the carbonyl by protonating the carbonyl oxygen atom.  The carbonyl oxygen atom deprotonates the acid to form the protonated carbonyl compound. 

This is the same first step as the formation of the enol.  what gives?  Why couldn't it just form the enol under these conditions, too?  Here's the key: it does.  There's nothing stopping the formation of the enol.  It probably happens a lot under these conditions.  But why do we recognize these conditions as formation of an acetal... when the acetal mechanism doesn't involve an enol?  Here's where it's important to remember that all these steps are reversible.  The enol is a nucleophile.  The only electrophile it can react with is an acidic proton on the acid.  If (and when) the enol forms (and it probably does), the enol doesn't have anything useful to do except react with an acidic proton to reform the protonated carbonyl compound.  To the extent that it occurs, the enol formation leads to an unproductive intermediate that funnels back into the productive pathway.  In essence, we don't see the enol in the mechanistic pathway, so we don't concern ourselves with drawing it... but it does form.

Ok, so we know the enol does form under the conditions needed to make an acetal.  But we know the enol will decompose back into the protonated carbonyl.  Now, we also know the protonated carbonyl is activated and a much better electrophile than the neutral, unprotonated carbonyl.  Now that it's such a hot electrophile, the weakly nucleophilic alcohol molecule can attack the protonated carbonyl carbon to form the protonated hemiacetal.  From here, I'll assume you know the rest of the steps in the formation of the acetal under acidic conditions.


...


so, to answer your question.  how does the reaction know it's supposed to form an acetal instead of an enol?  it doesn't.  Reactions don't sit around thinking, "ok... There's acid nearby. but wait! I also sense an alcohol nearby.  I'd better not form an enol, because I'm supposed to form an acetal."  No.  The react to whatever reaction conditions are put in front of them.  In the acetal case, the reaction continues on the acetal pathway because the presence of the the alcohol molecule provides the reaction with a nucleophle to attack the activated carbonyl to begin our journey toward the acetal.

That was long winded and a bit rambling.  Did I help any or just make it more confusing?

[sammyjo06]

Wow! Thanks for the help, azmanam. I read it all and yes, it did help clarify that for me.

I assume the enolate also forms under basic conditions with an alcohol, but the net reaction is a hemiacetal since the enolate just goes back to the carbonyl.

I like how you worked out the mechanism rather than just reciting memorized steps, too (which is what I sometimes do  ).

I appreciate you taking the time to write all of that out for me; it definitely helped a lot !

[azmanam]

I assume the enolate also forms under basic conditions with an alcohol, but the net reaction is a hemiacetal since the enolate just goes back to the carbonyl.

Exactly right.  Good recognition.  The enolate is a nucleophile (and a better nuc than the corresponding enol).  Again, under these reaction conditions, there are no good electrophiles except for the acidic proton on methanol.  The enolate grabs a proton from methanol to reform the carbonyl and the alkoxide.

[orgopete]

I agree that azmanam has explained this well. It was also important that he started with Le Chatelier's principle, which generally controls these reversible reactions. That is, for ketone + alcohol  ketal + water, this is a reversible acid catalyzed reaction. If water is removed the reaction forms a ketal. If a ketal is reacted with water, the ketone forms.

I believe this is also important for enolization steps. If additions to carbonyl groups are unproductive, then the results of enolization may be found among the products. Herein is what I believe to be the case (I couldn't find kinetic results to verify this however). Additions to carbonyl groups (Azamanam Case 1 and 3) under acid or base catalysis are fast or faster than enolization steps. Because those additions are reversible, the addition may not be noted.

For example, in a haloform reaction, it is only at the last step that the trihalomethane is cleaved from the oxy-anion intermediate. Although it should form faster or to a greater concentration than in the unhalogenated precursor, hydroxide was adding to the carbonyl groups in all of the earlier steps. The isotopic exchange of oxygen is fast for ketones under acid or basic conditions. That is, we only note the addition to the carbonyl group because of the loss of a trihalomethane anion.

Under acid catalysis, I use a different analysis. A protonated ketone has two acid hydrogens than can be removed, an O-H and a C-H. An O-H acid should be a stronger acid, therefore enol formation should be relatively slow. An addition to the protonated carbonyl group can compete with the deprotonation steps. Because additions to carbonyl groups is relatively facile (formaldehyde is >95% as the hydrate), enol formation under acid conditions should be a slower process.

Accordingly, examples of self-condensation reactions of aldehydes and ketones under base catalysis are much easier to find. An acid catalyzed aldol condensation is further hampered by the low concentrations of enol AND protonated ketone or aldehyde.

I argue that the rate of additions are much faster than enolization. You might consider that enolization were much faster than addition. What or how would that affect reactions? For which case do the rates match the results? What are the rate limiting steps?

[sammyjo06]

^ I'm not quite sure I understand what you're asking, Pete.

I think I get what you said about enol formation being slower under acidic conditions, since the pka of the protonated carbonyl oxygen is -8, which is much more acidic than the pka of the alpha hydrogen, which is between 16 and 21 (although for this course, my professor wanted us to memorize 20 as that pka value for some reason). So water would be more likely to add to the carbonyl carbon attached to the protonated oxygen than it would be to deprotonate the alpha H.

So why do you want me to consider enolization being faster than addition if it's probably not? If I understood what you were asking, lol sorry if I got that wrong .

I don't really know about acid-catalyzed aldol condensation, since we just have to know the base-catalyzed mechanism. I could probably guess what it might be because of all the other acid-catalyzed reactions I have to know, though. We are covering very specific reactions and cutting out some of the topics that may have been in a 15 week course since I am taking this during a fast-paced 4 week course, so my prof's not making us memorize reactions that don't happen as often (I think that's her reasoning, at any rate).

When we were shown the haloform reaction in lecture, we had to show it going back and forth between adding the OH- and "kicking it out" by reforming the double bond on the oxygen until all of the alpha hydrogens were exchanged for Cl and then finally the OH- stayed and reforming the double bond kicked out the 3 chlorines attached to a carbon without a hydrogen instead, and then that anion took the hydrogen from the OH group.

But anyways, I think I get what you were saying about how when we're showing equilibrium reactions, we don't always show it going back and forth at each step, since there is really no net reaction there.

[orgopete]

I'm not quite sure I understand what you're asking, Pete.

Since I do not cite any data, I was stating an opinion that additions are faster than enolization reactions. The heart of your original question was what do carbonyl groups do, add or enolize? I was trying to indicate that they can do both, but not necessarily equally.

Since I was stating an opinion, I was asking a rhetorical question. How would the chemistry differ if enolization was faster than additions? Does it make sense from reactions that enolization is the slow step?

I don't really know about acid-catalyzed aldol condensation, since we just have to know the base-catalyzed mechanism.

Again, the original question was about acid catalyzed enolization. The acid catalyzed aldol is an enol reaction, but one in which enolization must be slow. I believe that is a reason that few acid catalyzed aldol examples are reported in the literature.

To clarify, the pKa of a protonated carbonyl oxygen may be -8, but the acidity of the alpha-hydrogens of a protonated carbonyl group will now be much lower than 16. The higher values are for deprotonation of a neutral carbonyl compound with a base.