(Part VII of VII on the reactions of neutral nucleophiles in carbonyl chemistry)

As I said at the beginning of this series, the thing with reactions of neutral nucleophiles is that their mechanisms tend to involve a lot of intermediate proton transfer steps, especially where acid catalysis is involved. Acid catalysis is extremely useful, but it also leads to a lot of arrow pushing.  Today’s reaction is a prime example of that.

Remember that we always look for three important roles of acids in carbonyl chemistry:

1) they activate carbonyls toward nucleophilic attack

2) they make OH, OR, NR2, etc. better leaving groups, through protonation

3) they increase the rate of interconversion of the keto and enol forms.

 Today we’re going to see all three of those effects in action.

The acid-catalyzed Aldol reaction is a [1,2]-addition reaction between an enol and an aldehyde or ketone. It can be used to join two identical aldehydes or ketones together (this is called dimerization), or to join two different aldehydes/ketones (this is called the cross-aldol). If the two different aldehydes/ketones aren’t chosen carefully, however, it’s very easy to obtain a BFM [note 1]: in other words, multiple products can form.

Decent yields of cross-Aldol products can be obtained if one uses an enolizable ketone (i.e. it has α protons) and a non-enolizable aldehyde. There are two reasons for this design. First of all, only the ketone can form an enol, which is the nucleophile in the Aldol reaction. Secondly, employing an aldehyde as the second component will help to minimize the self-Aldol of the ketone, since aldehydes are more reactive than ketones for steric reasons.


The reaction begins with two separate events: 1) tautomerization of the ketone to form the enol, and 2) protonation of the other carbonyl, which activates the carbon towards attack. The nucleophilic enol then adds to the carbonyl in a [1,2]-addition. Finally, we obtain the neutral Aldol product by deprotonation, which gives us the dimer as the end product.

 

So far so good. 4 steps, and nothing that hasn’t been covered before.

However, if the aldol product is heated in acid for long enough, we will start to see a new product growing in. This is the aldol dehydration product, which is formed by the loss of water from the aldol product. The mechanism of this process merits some discussion, so let’s look at in in further detail.

First of all, there is a proton transfer  to the hydroxyl group. This should look familiar:  protonation is like serving a deportation notice to the hydroxyl group, telling it that expulsion is imminent. However, the subsequent elimination step can cause confusion because depending on which source you refer to,  you may see two different pathways written.

Wikipedia, for instance, shows this elimination as proceeding through an E2 reaction, with water acting as a base. This has the advantage of simplicity, as the eliminated product is achieved in one step. While not impossible, it seems somewhat unlikely, saying as we are operating under acidic conditions and H2O isn’t exactly a spectacular base – after all, that’s also our leaving group here, right? Nevertheless, you might see the mechanism written this way in your course notes.

An alternative mechanism has the advantage of being more likely but is also considerably longer. In the first step, the ketone tautomerizes to the enol form. The rate of tautomerization is increased by the presence of acid in solution. This is followed by a 1,4-elimination to provide the protonated enone. This reaction is the exact reverse of the [1,4]-addition reaction (conjugate addition reaction) that we see in the Michael reaction. Finally, deprotonation provides the neutral product.

I happen to prefer this latter mechanism.   However,  if you find conflict on this point, I’d advise you ask your instructor for their opinion. Many textbooks show both pathways – Reusch, for instance, shows both. [Reusch’s online textbook is a beautiful labor of love, by the way – I highly recommend checking it out.]. I wish it were otherwise, but there will always be some inconsistencies in how organic chemistry is taught and how it actually works.

And with that… this series is finally complete. I will finally start writing about some Org 1 topics soon. Also, I will be preparing a summary sheet so that all the main reactions of these things are all on one page. This should be fun… I live for putting together summary sheets!

[Note 1]: (BFM = Big….Effing… Mess)

Related Posts:

{ 4 comments… read them below or add one }

Yuji Miyahara December 26, 2010 at 8:08 pm

The 1,4-elimination mechanism is new to me and appears to be a good idea. But , the usual 1,2-elimination mechanism is not so bad. Of course water as a base is not good if water is not used as the solvent [March (6th Ed page1349) says water acts as a base even in concentrated H2SO4]. The -OH2+ is a good leaving group and the H2O (or ROH when an alcohol is used) accepts the proton to form an oxonium ion. This may be viewed as an E1 elimination rather than the E2 elimination. Actually I find a discussion of the two mechanisms in Ingold (page1004-1005). Some people think that Cl- ion as the base instead of water to capture the proton to form HCl. But this is rather unlikely in hydroxylic solvents. I do not know which mechanism is the most probable one. Unfortunately, no chemists today will do experiments to elucidate the mechanism, because such project will not attract money.

Reply

James December 28, 2010 at 10:39 pm

If it was a straight E1 one might expect to see rearrangements, especially where a primary alcohol is concerned. Since the elimination of water from those substrates is facile, a mechanism where enolization occurs first followed by donation of a pair of electrons from the enol seems more reasonable to me rather than an E1 type pathway, but depending on the substitution pattern more than one mechanism might be operational. For a tertiary allylic alcohol, for example, this might be the dominant mechanism.
Thanks for taking the time to comment and also to add the annotations from March and Ingold.

Reply

uttam patil April 28, 2011 at 11:50 am

dear Sir,
i am research fellow and my research area is organic chemistry. the present site provides excellent information about organic reactions and mechanism. i am very much impressed and would like to mention lot of thanks for this presentation. by following this site, i am getting expected information and knowledge.
thanking you,
yours sincerely,
uttam patil

Reply

James April 29, 2011 at 10:02 am

thank you.

Reply

Leave a Comment