Sciencemadness Discussion Board

Using the Acetone enolate from Sodium Hydroxide

FireLion3 - 13-4-2014 at 15:46

The acetone enolate is useful for a variety of reactions, though when searching for instances of its usage, often encountered is people using Sodium of Lithium metal to form the enolate. Sodium Hydroxide seems to be avoided.

After looking into this, acetone is prone to self-aldol condensation in the presence of Sodium Hydroxide. This would obviously be a concern. Looking more deeply into the reaction it seems that this self-condensation is reversible, highly unfavored, and the equilibrium largely favors acetone. Even at elevated temperatures the condensation is hard to occur, and at very high temperatures, a variety of unwanted by products form. On the other hand, at cold temperatures, it appears the self-condensation of acetone hardly occurs, and would hardly be an issue. With this, could the Acetone Enolate from Sodium Hydroxide be used as long as the reaction is kept cold/not heated?

If this is the case then why do people go out of their way to handle dangerously reactive sodium or lithium metal? Is there simply no other option when the reaction using the enolate has to be heated?

ScienceHideout - 13-4-2014 at 16:05

I am interested in this very much as well. The enolate is actually sort of a keto-enol tautomer type-thing, isn't it?

CH3--C=O <-----> CH3--C--OH
| .................... |
CH3 ................ CH2


From what I understand, a carbonyl is MUCH more stable than any enolate, and the C=O bond is strong compared to the C-OH bond.

The only thing I can think of is to take advantage of LeChatelier's principle. What is the heat change of the forward reaction? The reverse reaction? Maybe the equilibrium favors the enol at a higher temperature... Or perhaps we could screw the equilibrium and just use a catalyst to go a route other than the tautomerism...

jwpa17 - 13-4-2014 at 16:27

Not sure if this is relevant to your discussion since the reaction doesn't use sodium hydroxide, but the self-condensation of acetone isn't apparently difficult:
http://www.orgsyn.org/demo.aspx?prep=cv1p0345
Sorry if this is irrelevant to your interests.

ScienceHideout - 13-4-2014 at 16:44

Thank you, jwpa17! Is that indeed the self-condensation of acetone, though? FireLion and I were specifically talking about forming a tautomer of acetone... the reaction mechanism certainly does show a double bond forming, but the carbonyl is unaffected because, the way I understand it, the OH group completely leaves the molecule, forming a carbanion, where the electrons move to form a double bond?

FireLion3 - 13-4-2014 at 16:44

Hmmm just read a few articles that say the condensation product is favored at lower temperatures, yet the reaction is exothermic, and a higher temperatures weird condensation products are formed... strange.

ScienceHideout - 13-4-2014 at 16:54

Well, that makes sense to me...

Ketone ----> Enolate + Heat

Therefore, according to LeChatelier, adding heat would shift toward ketone, therefore less heat would shift in favor of enolate.

AvBaeyer - 13-4-2014 at 17:51

You can use sodium hydroxide to form the enolate of acetone provided the reaction of the enolate with a condensation partner is faster than the self-condensation of acetone with its enolate. For example, the condensation of acetone with benzaldehyde using sodium hydroxide as the base to form benzalacetone is a well known reaction. (see, for example, Vogel). Even in this reaction, attention has to be paid to the exact conditions since acetone self-condensation products can easily predominate.

AvB

FireLion3 - 13-4-2014 at 17:53

Quote: Originally posted by AvBaeyer  
You can use sodium hydroxide to form the enolate of acetone provided the reaction of the enolate with a condensation partner is faster than the self-condensation of acetone with its enolate. For example, the condensation of acetone with benzaldehyde using sodium hydroxide as the base to form benzalacetone is a well known reaction. (see, for example, Vogel). Even in this reaction, attention has to be paid to the exact conditions since acetone self-condensation products can easily predominate.

AvB


Indeed. Though it doesn't help when I can't find the pKa values of the compounds in question.

leu - 13-4-2014 at 18:28

The pKa of acetone is about 19 :P Chemists have been calculating ionization constants for a very long time ;) Perhaps some reading is in order as opposed to posting queries that can be answered using Gooogle :cool:

Attachment: cr60287a002.pdf (599kB)
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FireLion3 - 13-4-2014 at 18:41

Quote: Originally posted by leu  
The pKa of acetone is about 19 :P Chemists have been calculating ionization constants for a very long time ;) Perhaps some reading is in order as opposed to posting queries that can be answered using Gooogle :cool:


I wish I were talking about acetone's pKa :p, that would make life easier. I am looking for the pKa values of a number of halide compounds, though I am unable to find them in any databases, wiki, etc.

[Edited on 14-4-2014 by FireLion3]

Etaoin Shrdlu - 13-4-2014 at 19:48

Quote: Originally posted by ScienceHideout  
Well, that makes sense to me...

Ketone ----> Enolate + Heat

Therefore, according to LeChatelier, adding heat would shift toward ketone, therefore less heat would shift in favor of enolate.

This is the reason that unlike most chemicals, a lot of aldehydes go bad more quickly when stored at low temperatures.

FireLion3 - 13-4-2014 at 22:38

Interestingly, I keep finding things like this:
Oraganic Name Reactions Reagents and Molecular Rearrangements - Aldol Condensation
Quote:

"The reaction between two ketones is not very successful. The equilibrium lies far to the left and is thus not favourable. The reason is that the carbonyl carbon of ketone is less positive because of the +I effect and more sterically hindered as compared to aldehydes. This reduces the nucleophilic attack on the carbon. However, it is possible to obtain diacetone alcohol in good yield by boiling acetone with solid Ba(OH)2 in a specially devised apparatus."


This is interesting. This makes it sound like the self-condensation of the acetone is difficult to occur. Thoughts?

Here's an additional "Link" saying the same thing.

And Another
Quote:

"Diacetone alcohol is thermodynamically less stable than acetone and hence the equilibrium for this reaction lies far to the side of the acetone. Another reason why diacetone alcohol formation is difficult is that it involves attack of a ketone upon a ketone. The steric hindrance from the alkyl groups raises the activation energy. Hence, the major product of diacetone alcohol in NaOH is acetone."



Many of these findings indicate to me that the acetone enolate of NaOH can effectively be used in various substitution reactions without the need for a "strong base" that causes complete deprotonation. While Diacetone Alcohol may form in very small quantities, it itself is more sterically hindered and less activated than acetone, so whatever small amount that is formed is unlikely to undergo any such desired substitution as the acetone enolate. An excess of acetone (being very cheap) can easily remedy such problems.

I can see why people would think otherwise, because at the surface judging based strictly off the brief literature portrayal of these reactions, side reactions may seem obvious, but taking a deeper look at the steric and equilibrium properties of said reactions, it stands out that these concerns are not as bad as would be initially assumed.


[Edited on 14-4-2014 by FireLion3]

adamsium - 14-4-2014 at 01:48

I actually did the aldol condensation of acetone and benzaldehyde to produce dibenzalacetone in my organic lab class recently.

I'd expect that the equilibrium isn't a huge issue here as the enolate reacts very quickly with the benzaldehyde (and, of course, more favourably than with the ketone), producing a quite stable product and removing the enolate from the equilibrium, driving the reaction forward. I suppose it would depend on the stability of the beta-hydroxy carbonyl product being produced and what its equilibrium is like. Of course, one adds the acetone last in order to minimise the acetone self-aldol.

The acetone/benzaldehyde aldol adduct is an example of one that will dehydrate spontaneously in basic conditions (via an E1cB mechanism), but most will not and require heating in acidic conditions (the aldol reaction is also acid-catalysed, whereby it proceeds via the enol tautomer rather than the enolate) to form the condensation product, proceeding via tautomerisation and a 1,4 elimination for overall loss of H3O+.

[Edited on 14-4-2014 by adamsium]

FireLion3 - 14-4-2014 at 02:44

Quote: Originally posted by adamsium  
I actually did the aldol condensation of acetone and benzaldehyde to produce dibenzalacetone in my organic lab class recently.

I'd expect that the equilibrium isn't a huge issue here as the enolate reacts very quickly with the benzaldehyde (and, of course, more favourably than with the ketone), producing a quite stable product and removing the enolate from the equilibrium, driving the reaction forward. I suppose it would depend on the stability of the beta-hydroxy carbonyl product being produced and what its equilibrium is like. Of course, one adds the acetone last in order to minimise the acetone self-aldol.

The acetone/benzaldehyde aldol adduct is an example of one that will dehydrate spontaneously in basic conditions (via an E1cB mechanism), but most will not and require heating in acidic conditions (the aldol reaction is also acid-catalysed, whereby it proceeds via the enol tautomer rather than the enolate) to form the condensation product, proceeding via tautomerisation and a 1,4 elimination for overall loss of H3O+.

[Edited on 14-4-2014 by adamsium]


That reaction is somewhat of an exception. Don't confuse Diacetone Alcohol with Dibenzylacetone ^.^. When I made this thread I had in mind primarily the reaction of acetone enolates with various halides. The reluctance some people had faced with using NaOH to deprotonate was because of the fear of aldol products, but it seems this is unwarranted in most cases.

The reaction you described happens very rapidly because benzaldehyde is a good electrophile and in any case the deprotonated acetone reacts much more rapidly with benzaldehyde than it can with its self. Additionally, this reaction happens at low temperatures spontaneously. The dehydration also spontaneously occurs because the electrons are delocalized due to conjugation between the aromatic ring and the carbonyl group. The formed double bond even further enhances the conjugated resonance of the molecule, facilitating a very easy deprotonation, and thus second reaction with the aldehyde.

This reaction is interesting because there are so many factors contributing to the conjugated resonance that enhance the reactivity.

With normal acetone, or even with normal alkyl ketones(lacking double bond on chain), there are not these factors to be concerned with.

leu - 14-4-2014 at 04:28

Chemists have been devising methods involving equations to predict dissociation constants for many decades ;) There's software that does that nowadays:

Abstract One of the most important physicochemical properties of small molecules and macromolecules are the dissociation constants for any weakly acidic or basic groups, generally expressed as the pK(a) of each group. This is a major factor in the pharmacokinetics of drugs and in the interactions of proteins with other molecules. For both the protein and small molecule cases, we survey the sources of experimental pK(a) values and then focus on current methods for predicting them. Of particular concern is an analysis of the scope, statistical validity, and predictive power of methods as well as their accuracy.

which are described in the attached article:

J Chem Inf Model. 2009 Sep;49(9):2013-33. doi: 10.1021/ci900209w.
Predicting pKa.
Lee AC, Crippen GM

:cool:

Attachment: jcim49_2013_09.pdf (253kB)
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adamsium - 14-4-2014 at 04:40

Yes, I know this is an exception - I think I was pretty clear about that and that was much of the point of my post.

I have no idea why you'd think I was confusing diacetone alcohol and dibenzalacetone (i.e. 1,5-diphenylpenta-1,4-dien-3-one, which is what I'm assuming you meant rather than dibenzylacetone, and would be something else entirely), either; I wasn't.

I also know why the acetone/benzaldehyde aldol reaction and subsequent dehydration occurs the way it does; I'm also not sure how that wasn't at least fairly apparent from my post.

Some people (including yourself) were using the term 'aldol condensation' when they probably meant 'aldol reaction'. The terms are not really interchangeable, although some people use them, loosely, as though they are. I also subtly hinted at this.

Nicodem - 14-4-2014 at 05:36

Quote: Originally posted by FireLion3  
The acetone enolate is useful for a variety of reactions, though when searching for instances of its usage, often encountered is people using Sodium of Lithium metal to form the enolate. Sodium Hydroxide seems to be avoided.

Please cite one example where sodium or lithium metal have been used to deprotonate acetone.
Sodium hydroxide is probably the most common base for deprotonating acetone. Even K2CO3 can be used.

Anyway, this thread is headed nowhere for the obvious reason: You ask a specific question about a specific case, but stubbornly refuse do give sufficient data.

PS: In future, please open beginner topics only in the Beginnings section. See the forum guidelines for more information.

Metacelsus - 14-4-2014 at 05:51

Is sodium hydroxide a strong enough base (conjugate acid pKa = 15.7)? The alpha proton's pKa is 19.3, so I don't think sodium hydroxide will be strong enough. Sodium hydride would work very well, and probably give quantitative yields.

https://en.wikipedia.org/wiki/Carbonyl_Alpha-Substitution_Re...

Sodium is a strange choice, as it is not truly a base. It might react with the enol tautomer to form the enolate and hydrogen gas. Alternatively (and more likely), it could do the pinacol coupling. There is a thread on alkali metals' reaction with acetone here: http://www.sciencemadness.org/talk/viewthread.php?tid=16568


[Edited on 14-4-2014 by Cheddite Cheese]

FireLion3 - 14-4-2014 at 09:02

Quote: Originally posted by Cheddite Cheese  
Is sodium hydroxide a strong enough base (conjugate acid pKa = 15.7)? The alpha proton's pKa is 19.3, so I don't think sodium hydroxide will be strong enough. Sodium hydride would work very well, and probably give quantitative yields.

https://en.wikipedia.org/wiki/Carbonyl_Alpha-Substitution_Re...

Sodium is a strange choice, as it is not truly a base. It might react with the enol tautomer to form the enolate and hydrogen gas. Alternatively (and more likely), it could do the pinacol coupling. There is a thread on alkali metals' reaction with acetone here: http://www.sciencemadness.org/talk/viewthread.php?tid=16568


[Edited on 14-4-2014 by Cheddite Cheese]

Sodium Hydroxide does deprotonate acetone but not completely. What I cannot seem to find is any source stating specifically that a complete enolate must be used. The entire point of making the enolate is so that the acetone can act as a nucleophile. The complete enolates will be stronger nucleophiles than the NaOH-Acetone enolate, but they are still both nucleophilic enolates. If there is a strong electrophile in the system, it is going to react with whatever nucleophile is present. The acetone-acetone condensation is reversible, but the homolytic fission of a halo-carbon bond which quickly undergoes nucleophilic substitution is not, because the halogen immediately forms the acid, which in the presence of NaOH would immediately neutralize.

Does anyone have a reason as to why a partial enolate cannot be used? I imagine it would leave a greater possibility for more side reactions, but with acetone, even that is minimal. If such a reaction were UV Light catalyzed, the highly activated halides would have little choice but to react with whatever carbon nucleophile that comes their way, which NaOH-Acetone enolate would be.

Edit:

This like provides insight. I guess only Aryl-Halides will work with the NaOH-Enolate

http://www.biologie.uni-hamburg.de/b-online/library/newton/C...

It discusses a similar reaction of how using hydroxide to deprotonate acetone to react with methyl iodide won't be viable because there is such a small quantity of formed enolate compared to acetone and hydroxide presence. Making it far more likely for an SN2 reaction to occur with Hydroxide (forming an alcohol and a salt I assume).

What this means is that the only halides that will work with an NaOH-acetone enolate are aryl halides... because aryl halides cannot undergo SN2 reactions, even with some of the strongest nucleophiles..... If an aryl halide were used the only possible reactions would be self-aldol of acetone (unfavored), and the substitution/alkylation of the acetone.... too bad I don't want to make that particular molecule. Meh. I guess I will look into those stronger bases for total deprotonation.


[Edited on 14-4-2014 by FireLion3]