On Acetals and Hemiacetals

by James

in Alcohols, Aldehydes, Ketones, Organic Chemistry 2, Organic Reactions

(Part VI on a series of posts on the reactions of neutral nucleophiles with carbonyl compounds)

When I started this series on reaction mechanisms of carbonyl compounds, it was in an effort to justify a little statement I made on the side of this summary sheet. I said that by understanding just 6 mechanistic steps in detail, you can understand greater than 95% of the 110 reactions on the sheet. Today, in the penultimate post in this series, I’m actually not going to introduce any new concepts, because all the reactions have been talked about before. The only thing that is changing here is the identity of the nucleophile and the electrophile.

First of all, let’s identify what I’m talking about here: I”m going to describe how hemiacetals and acetals are formed (and destroyed).

1-acetalsOn the left is what is probably the most well-known hemiacetal there is: glucose. Glucose can also be drawn as a straight-chain aldehyde, but the reaction between the C-5 hydroxyl and the C1-aldehyde to form the 6-membered ring is so favorable that there is only a miniscule (~0.003%) amount of the aldehyde is actually present in solution. Still, the aldehyde is present in a high enough concentration that treatment of glucose with a reducing agent like NaBH4 will eventually completely reduce it to the alcohol, hence its classification as a reducing sugar. Now, when you treat a hemiacetal with an alcohol in the presence of acid, the effect is to exchange OH for OR. The group on the right there is called an acetal (sometimes you’ll also see it called a ketal).

Let’s start with a discussion on how hemiacetals form.  It’s essentially just a [1,2]-addition of an alcohol to an aldehyde or ketone. The C=O π bond breaks, and we’re left with a tetrahedral compound. Like all [1,2]-additions to carbonyls with neutral nucleophiles, the reaction is faster in the presence of an acid, but it’s not an absolute requirement here: hemiacetals form quite nicely under neutral conditions. Thus, you can draw it via the neutral pathway or through the acid-catalyzed pathway:

2-acetals

However, in order for a hemiacetal to reach full acetal-hood, there is no getting around the requirement for a painful initiation ritual:  it must be subjected to treatment with acid to undergo further reaction. As we’ve seen before in imine formation and in the Fischer esterification, protonation of the hydroxyl group of the hemiacetal transforms a poor leaving group (hydroxyl – the conjugate base of water, pKa = 15.7) to an excellent leaving group (water – the conjugate base of hydronium ion, pKa = -1.7). This 17-orders of magnitude increase in the leaving group ability allows for the lone pair of oxygen to expel water from the carbonyl carbon in a 1,2-elimination reaction, leading to the formation of the highly reactive oxonium ion. [note again – positively charged species tend to end with -ium]. Just like with protonated carbonyls, the oxygen of the oxonium ion is positively charged, meaning that the C=O π bond is significantly weaker than in a neutral carbonyl. Thus, the highly electrophilic carbon is able to combine rapidly even with weak nucleophiles like alcohols, leading to a highly facile 1,2 addition.

3-acetals

Now since we’re interested in isolating the neutral product, not its protonated version, the final step is simply a deprotonation.

So to summarize, the 6 steps (5 if you make the hemiacetal through the neutral pathway) are: protonation, 1,2-addition, proton transfer, 1,2-elimination, 1,2-addition, and a final deprotonation. 6 steps – but if you’ve been following this series, it’s just the same old reactions, over and over. The lyrics change, there are subtle differences in tone, but it’s essentially the same song.

Again, this is an equilibrium reaction, so all the steps are reversible. To go from the acetal back to the aldehyde or ketone, one simply heats the acetal in aqueous acid. As with all equilibrium reactions, Le Chatelier is king: you get out of it what you (don’t) put into it.

In the next (and last) installment of this series, I’ll talk about the reaction that is probably the most involved you will ever be asked to draw: the acid-catalyzed aldol condensation. A 7-step beast that actually includes some (kind of) new reactions! And then I can put this series to rest.

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{ 7 comments… read them below or add one }

Jennifer

Wondering if you could post something explaining just how to look at one and tell if it is a hemiacetal or acetal. I understand that the hemi has an alcohol and an ether attached to the same carbon. But it would be nice to see each part of it labeled out and explained to us. Thanks this infor was amazing!

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James

I put a box around the relevant atoms that make up the acetal and hemiacetal functional groups. Does this help at all, or does it make it look worse?

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Adriel Williams

Quick question, in your structure of glucose, I’ve never seen the OH groups on Carbons 3 & 4 drawn that way. It actually looks like you’ve drawn the sugar Gulose there. Am I missing something? Cheers.

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james

Fixed. Thanks!

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Alfonso

Great explanation. The University lost a great professor

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Susan

Just wondering, does the pH of the solution affect the equilibrium between the hemiacetal and aldehyde form of glucose?

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James

Yes – it makes the OH into O- which is a better nucleophile (and worse leaving group) than OH, and it will increase the amount of ring-closed product.

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