Carbonyl chemistry: Anionic versus Neutral Nucleophiles

by James

in Carboxylic acids, Drawing Reaction Mechanisms, Esters, Organic Chemistry 2, Organic Reactions, Understanding Electron Flow

Let’s talk about carbonyl chemistry today (aldehydes, ketones, carboxylic acids, and esters). Specifically, reaction mechanisms.

If you look at the reaction mechanisms in detail you will see that they break down into two broad categories. Those two categories are reactions with anionic (negative) nucleophiles and those with neutral nucleophiles.

I put this chart together for a student and thought it would make a good post:

1-neutralmechs

The anionic nucleophiles, on the whole, are much stronger nucleophiles than the neutral nucleophiles. Let’s point out 4 main differences:

1) Acid catalysis. We never add acid to reactions containing anionic nucleophiles, for a simple reason : it will protonate the nucleophile. Since the conjugate base is always a stronger nucleophile, this will essentially retard the rate of reaction (and in many cases, like with metal hydrides and Grignard reagents, stop it altogether). The only exception is when the anionic nucleophile is a weak base – cyanide, for instance.

With neutral nucleophiles, acid is a necessity to get the reaction to go to completion in the first place. Acid does two things: 1) it protonates carbonyls, speeding up 1,2-addition, and 2) it protonates groups like OH, OR, and NH2 to make OH2(+), HOR(+), and NH3(+) , which are better leaving groups by factors of billions.

2) Reversibility: With anionic nucleophiles, reactions are often irreversible, especially with strong nucleophiles that are poor leaving groups like R, H, and NH2 . If you’ve stuck with organic chemistry to make it this far, hopefully you’ve noticed that the weaker the base, the better the leaving group. Reactions with anionic nucleophiles are in equilibrium only if the nucleophile and leaving group are similar to each other in base strength.

With neutral nucleophiles essentially every step is reversible, and the entire mechanism represents an equilibrium. If that’s the case, how can we drive it to the direction we want? Well, in addition to using acid, we can do it by choosing the solvent. For instance, when forming an ester from a carboxylic acid and an alcohol, we can choose to use the alcohol as solvent, which will vastly outnumber our water molecules. Conversely, when going from an ester to a carboxylic acid, we use water as solvent.

3. Proton transfers: If you’ve tried to draw the mechanism of, say, Grignard reagents adding to esters and then compared it to adding water to esters (using acid), you should notice one big difference – all those pesky proton transfers in between each step. In the latter case, notice how proton transfers always precede 1,2-elimination.

4. The last step: If you start with a negative nucleophile, your product will have a negative charge on it. That’s why the last step is always addition of acid or water – to get the neutral product. If you use a neutral nucleophile and add an acid, your product will end up with a positive charge. So the last step in this case will be deprotonation.

Did I miss anything? Was this helpful? Leave a comment!

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{ 2 comments… read them below or add one }

Anonymous

Thanks! Short and concise :)

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Sandra Huang

So helpful!

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