Ochem Question

[StudentDoc1234]

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Hi everyone,

So in my notes from Chad's Ochem Videos I have reactions where a base reacting with cyclohexanone will extract an alpha hydrogen which makes it an enolate that eventually attacks a haloalkyl. But then in Bootcamp there was a question that had cyclohexanone reacting with CH2CH2Li, EtO. This reaction attacked the carbonyl of the ketone as opposed to extracting the alpha hydrogen.

I guess my question is how do I know when the alpha hydrogen is going to get extracted or if the carbonyl is going to get attacked?

 

[FeralisExtremum]

Depends on the reagent used. We generally use a (strong) base like LDA when we went to deprotonate the alpha H's rather than attack the carbonyl, but there are all kinds of reagents and situations to look for. Worst case scenario, you should be able to use the answer choices as a guide to tell if they're looking for an enolate creation or a standard organolithium/Grignard addition.

In general, organolithiums and Grignards are pretty powerful bases - you wouldn't want to use one on a carboxylic acid unless you intentionally wanted to create a carboxylate, for example - but they are also very strong nucleophiles and in a situation of a ketone/aldehyde those groups specifically would likely go for the nucleophilic addition on the carbonyl.

 

[StudentDoc1234]

Yah I guess I should have been able to identify that since the reagent is very similar to Grignard it'll react the same way.

Going through my notes I see the main reagents used to get an enolate ion are alkoxides, OH, LDA, and Gillmans.... do you by any chance know of other main ones that get the enolate product?

 

[FeralisExtremum]

I don't think we use Gillman's to create an enolate - we typically use Gillman's to add something in Michael's addition/1,4 addition/beta addition to carbonyls. It basically acts like a weaker nucleophile version of Grignards/organolithiums that preferentially attacks the beta carbon, assuming it's conjugated with the carbonyl.

NaOH is not that great for deprotonating alpha Hydrogens but we intentionally use this in things like "self" aldol condensations (aldols attacking the same aldol molecules) to deprotonate only a small portion of them and get them to react with the other aldols in solution. NaOH can also be used effectively to deprotonate molecules like malonic ester where the alpha Hydrogen is exceptionally acidic, e.g. here due to double resonance stabilization:

LDA is highly effective at deprotonating alpha-H's, so if you use it in a solution where you want aldol condensation among the same aldol molecule, LDA would end up making all of them enolates and it wouldn't work very well (there wouldn't be any non-enolate aldols left to act as electrophiles). But, we also use LDA effectively by separating steps: use LDA on the first reactant to create the enolates that you want, then in a second step add in the electrophile you want to get attacked. If you see LDA used you can probably count on an enolate being created.

Alkoxides (that match the leaving group) work well to deprotonate the alpha Hydrogens of esters because even if the alkoxide were to perform a nucleophilic acyl substitution, it would end up forming the original reactant anyway. You'll see this used in Claisen condensations.

Those are the only ones I can think of off the top of my head for deprotonating alpha H's. But I have to emphasize, a lot of figuring out what's going to happen will depend on the context of the reaction. The answer choices will (hopefully) make it clear that you have an alpha addition reaction vs a nucleophilic addition reaction. The DAT is multiple choice, so use that to your advantage - there have been plenty of times where I've instinctively looked for a substitution reaction only to see from the answers that it was an aldol condensation or something along those lines.

Also, some of these reactions can be acid catalyzed as well, albeit less effectively - so you may not even see base reagents being used.

 

[StudentDoc1234]

This is everything I was looking for all in one post... as always I appreciate the time you took to clear this up for me, thank you!