Ace Your Next Organic Chemistry Exam.

With these Downloadable PDF Study Guides

Our Study Guides

Carboxylic Acid Derivatives

By James Ashenhurst

Let’s Talk About the [1,2] Elimination

Last updated: March 26th, 2019

(part IV of a series on the reaction mechanisms of neutral nucleophiles in carbonyl chemistry)

As I mentioned n the previous post in this series, if I had to name the  #1 most important reaction of carbonyl compounds, I’d say it’s the [1,2] addition. Think of the Grignard reaction: you add magnesium metal to an alkyl halide, stir for awhile (with the Grignard, sometimes a long while, unfortunately) and then add in your aldehyde or ketone. There’s a vigorous reaction while the Grignard adds to the carbonyl carbon, and at the end of it all you obtain a new alcohol.

A recurring theme in organic chemistry is that a lot of common reactions are reversible.  If the  [1,2] addition is the most important, then the #2 most important reaction has got to be the [1,2]-elimination. This reaction is just the [1,2]-addition in reverse, except you’re expelling a leaving group from a tetrahedral carbon to give you back a new π bond (i.e. a carbonyl).


What makes the [1,2] elimination interesting is that there are multiple groups that could be eliminated from the tetrahedral carbon, and it doesn’t have to be the same group that attacked the carbonyl in the first place. Thus, [1,2]-addition / [1,2]-elimination allows for an interchanging of functional groups attached to the carbonyl.

The key to understanding the [1,2]-elimination is knowing the properties of leaving groups. There are 3 types of groups you need to worry about: The groups that leave without effort, the groups that require coaxing, and the groups that never leave.

Let’s look at each of them.

Groups that leave without effort (Halides, AcO(-), and the conjugate base of just about anything else that has “acid” in its name):  The leaving groups in these cases are the conjugate bases of strong/medium strength acids.  These are good leaving groups and stable anions.  In the presence of any kind of half decent nucleophile (even water) they will be displaced from a carbonyl compound (This also includes phosphate, the conjugate base of phosphoric acid –  Nature’s way of making an alcohol into a good leaving group).  The stronger the acid, the weaker the conjugate base, and the better a leaving group it will be. As always, know your pKa’s – they are the key to understanding this. 

Groups that require coaxing (OR, OH, NH2, NHR, NR2):. A strongly basic nucleophile like HO(–) will perform [1,2]-addition/[1,2]-elimination on an ester to displace RO(–). Under neutral conditions, however, water is simply not a good enough nucleophile to get RO(–) to leave.  It’s helpful to consult our pKa table here for a second. I think I say this at least once a post. 


In order to make this reaction happen, we have to add acid. By adding acid, we go from having to displace a strongly basic leaving group to displacing a neutral leaving group. Now as I’ve talked about previously, acid catalysis has tremendous power. It achieves 2 things: 1) activate the carbonyl toward 1,2-addition, and secondly it protonates the leaving group. The difference in pKas between H2O and H3O(+) is about 17.  This means that protonation increases its leaving group ability by 16-17 orders of magnitude. This is a really big deal!. The same goes for NH3 and NH4(+): protonation increases the stability of the conjugate base by 29 (!!!!) orders of magnitude. That’s a tremendously large effect. It’s  hard to comprehend the magnitude of that effect number. A million times larger than Avogadro’s number.

Let’s talk about an example of this that demonstrates both principles of acid catalysis. The acid-catalyzed conversion of an ester to a carboxylic acid is a 5-step mechanism. Here’s how the reaction works. Note that it’s  important to use water as the solvent – the high concentration of water helps to drive the reaction to completion.


Since this is an equilibrium reaction, it’s driven to the right by the fact that there is a high concentration of water and a low concentration of methanol. [H2O] >> [MeOH]. The cool thing about this reaction is that all the steps are reversible. Interestingly, when you use methanol as solvent and therefore [MeOH] >> [H2O] you can make it run in reverse. It’s the exact same reaction. 

 What about hydrolysis of amides? Here’s the thing. It’s also the same reaction mechanism. In this case, you’re protonating the nitrogen so it leaves as the amine (the neutral amine). If you take the reaction as drawn and just replace the OR group with NH2 (or NHR, or NR2) – congratulations, you’ve just drawn the mechanism for amide hydrolysis. 

If you understand this 5-step mechanism, you understand ester formation, ester hydrolysis, and amide hydrolysis. In all cases you’re protonating the group you want to replace (OH, OR, NH2) until it falls off as its conjugate acid (H2O, HOR, NH3)

Nitrile hydrolysis is a bit more complicated [I’ll talk about it some other day] but it also goes along simliar lines: essentially, you keep on protonating the nitrogen until it eventually falls off as NH3.

Groups that never* leave: R and H.  

The pKa of H (~35) and of R groups (>45) are too high to allow for these groups to be displaced as anions, so these groups never leave in [1,2]-eliminations. Never is a strong word, however, so let’s say never*. Never* means they never leave except under very specific conditions.

Probably the most important exception is the retro-Aldol reaction:


So if the R group can be stabilized by resonance (with an adjacent carbonyl for example) that would allow it to be a stable anion by itself (remember the pKa of an enolate is about 20) then it can act as a leaving group. The enolate is pretty much the only example you will encounter in Org 2, but there are others. (Bonus points if you can name another one.)

The other class of exceptions that come to mind are the Canizarro,  the Meerwein-Pondorf-Verley oxidation, and the Pinacol rearrangement. All of these reactions are special cases in that the R or H group is not really leaving as a free ion, but it is migrating to a nearby reactive site in an intramolecular reaction. So they don’t really break the rule.

Up next: all about the mechanisms for the formation of imines and enamines. And I will probably tell you again to get out your pKa table (that’s a given).

Related Posts:


Comment section

5 thoughts on “Let’s Talk About the [1,2] Elimination

  1. Would the haloform reaction which involves nucleophilic attack by OH followed by displacement of a CBr3 group also count?

Leave a Reply to OBear

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.