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Carbonyl Chemistry: Learn Six Mechanisms For the Price Of One

So at some point during Org 2, you will probably be expected to learn a whole slew  of mechanisms.

Like these:

  • Conversion of carboxylic acids to esters (Fischer esterification)
  • Hydrolysis of esters to carboxylic acids
  • Hydrolysis of amides to carboxylic acids
  • Formation of imines from ketones/aldehydes and amines
  • Hydrolysis of imines to ketones/aldehydes and amines.
  • Formation of anhydrides from carboxylic acids

Six different mechanisms. That means writing out and learning a whole bunch of different steps for each of these reactions.

Sounds like a lot of work, doesn’t it?

What if you could cut down the amount of time you spent doing these mechanisms by a factor of six? That would be worth learning, right?

Well, you’re in luck. The six reactions listed might look different, but they actually have EXACTLY THE SAME MECHANISM. So if you’ve learned how one of them works – you’ve actually learned them all.  Let’s break it down using the first reaction (formation of esters from alcohols and carboxylic acids) as an example.


Step 1: Protonation


The reaction begins with protonation on the oxygen of the C=O (or C=N) bond. This means the oxygen (or nitrogen) has less electron density to share with the carbonyl carbon. Being more electron-poor, the carbonyl carbon has now become a better electrophile, which sets up…

1,2-addition (step 2)


Whereby the nucleophile adds to the carbonyl carbon, forming a new bond and breaking the [PI] bond beween the carbon and oxygen. Note that the neutral nucleophile is positively charged and the (former) carbonyl oxygen is neutral. An acid base reaction can now ensue, a fancy name for which is….

Proton transfer (step 3)

3-Proton transfer

where a proton is transferred from the atom which previously acted as a nucleophile to a different atom on the molecule (either O or N). Once the proton has been transferred,  the group becomes much less basic, which also means that it will be a much better leaving group. So one of the lone pairs from an adjacent atom forms a new Pi bond with the carbon, resulting in…

1,2-elimination (step 4),


ejecting a neutral molecule (such as H2O or NH3) as a leaving group. There’s just one more step to go. The atom participating in the new Pi bond has a positive charge, which means we have to remove it in order to get a neutral species – a step we call…

Deprotonation (step 5)


And that’s it. Note that the acid is not consumed – it’s a catalyst here, and can meet up with another molecule of our starting material back in Step 1.

Although the material is spread out over several chapters, the hidden pattern in their mechanism should come in handy when it comes to exam time.

Just remember this: Protonation / 1,2-Addition / Proton transfer / 1,2-Elimination / Deprotonation.


I’ve been trying to come up with a good mnemonic for this.  Piranhas Attack People Every Day? Here’s a random mnemonic generator.

PS – Here are the six mechanisms I’m talking about, broken down this way:




Chapter 06 Amines
Chapter 07 Aromaticity
Chapter 9 Carbohydrates
Chapter 11 Conformations and Cycloalkanes


Comment section

0 thoughts on “Carbonyl Chemistry: Learn Six Mechanisms For the Price Of One

  1. Hi James,

    In McMurry’s ‘Organic Chemistry’, they give the imine mechanism (and similar ones) like this:

    1. Nucleophilic attack on the carbonyl carbon
    2. Proton transfer from the nitrogen to the carbonyl oxygen
    3. Protonation of the hydroxyl to form a good leaving group
    4. Elimination of water
    5. Deprotonation


    I’ve read both ways: yours appears in ‘Instant Notes’ and elsewhere.

    So which is correct? Or does it not matter? I’m just wondering, in my exam, if I give one or the other, will it matter?


    1. Good question – it depends on pH. At neutral pH it operates similarly to what you mentioned. It’s also slower. As you go to lower pH, you’ll start to protonate the carbonyl oxygen (through the mechanism mentioned in this post) which results in a faster reaction.

  2. I always assumed that the making of anhydrides was a little more trickier than what has been presented here. I assumed that for an eg, acetyl Chlorides were required due to the chloride being a better leaving group. – but hey, Im an undergrad and here to learn. But still surprised to see that latter reaction.

    1. Well, there are two big ways. The first is the way you mentioned – adding carboxylates to acid halides, and this forms anhydrides. The second is by heating carboxylic acids with acid, which leads to loss of water (hence, “anhydrides”)

      1. I’d love to experiment with this. I dare say separating water from the formed anhydride from the mix would be necessary. The densities are pretty close, and under reflux concoctions DS would be out, so perhaps just a HCl catalyst, with acetic acid under reflux with a (DRY) CaCl2 drying tube fitted at the end. I wish my chem Professors would allow me to explore this, say with butanoic acid – as stinky as it is, but bypasses the issue of anhydrides and their legalities with respect to acetic and propanoic. Anyway, thankyou for your response.

        1. It’s not a very broadly useful method since it can only make symmetrical anhydrides, but this almost certainly how commercial acetic anhydride is made.
          Re: butanoic acid, why rule out Dean-Stark? Use benzene or toluene as co-solvent, the water will be driven off, allowing the reaction to go to completion. Although you will probably want to avoid social contact afterwards…..

  3. Thanks for the response James. So if i am getting this right, the co – solvent and toluene is somewhat bonding well, thus allowing the water to be separated via a DS trap. What I do not understand, and feel I need to, is how can carboxylates bond well with say methyl benzene. I could sort of understand this with butanoic acid,(4 carbons and their H’s, but for acetic acid, which has a methyl group and a carboxylic acid which would H bond with itself, and with water. I am confused. Obviously. Ill look up the use of co solvents(DS traps) with respect to driving of water, and see where I land. This is good, as I am learning via such a problem. etc. Sorry if this is annoying. !!

    1. Well, water and toluene (as well as water and benzene) form an azeotrope, and boil off together. The distillate ends up in the Dean Stark trap, and the water, being more dense than toluene, settles at the bottom and doesn’t come out. Excess toluene fills the trap, and trickles back down into the reaction flask.

  4. Yes, for water to of formed, the mechanisms must of worked. The DS reasonably quickly started to fill with two layers. So I’d take it the anhydride would be within the mix of the toluene, carboxylic acid used, and as stated the water is now removed. So the next step is using a good long column, (vigreux – maybe) to separate. Am I making sense, or got things backwards.

  5. what I meant was that water remained in the DS trap, and the mix started to drip back into the pot, like you said. Water forming, to me at least proves the mech, so I think this is what has happened.

  6. I have one question with this reaction though James. It seems imperative that only one of the carbonyl groups is protonated. I understand, I think, that the carbonyl is more basic compared to the OH group on the carboxylic acid, but it seems to me, having both molecules(carbonyl functional groups) protonated, would be a problem. So Ive tried to figure out how a chemist would optimize only one of the acids being protonated, and can only come up with a speedy addition of the acid catalyst, to an already under mild reflux conditions. It just seems to make sense, the quicker this reaction can happen, the better. The longer the acid catalyst remains with the molecule in question, the longer it has to protonate both of the carbonyls. Also the acid catalyst and the amount used must play a part in this. Theoretically, we only want half the molar amount of the carboxylic used, but considering how it is being regenerated, perhaps even a much smaller amount would be better, would reduce the amount of carbonyls being protonated, hence that is probably how a chemist would go about this. USE the minimal amount of catalyst. I also thought of what catalyst would be best used. I thought of H2SO4, but need to look at the pKa of this diprotic acid, as the goal is to remove the H2O formed, and not turn it back into an acid, that will play a further role. Anyway, just thinking this though. Thankyou. I have my butanoic to play with, and pretty sure I can work on this.

  7. James,

    The fischer esterification mechanism (acetic acid with ethanol, sulfuric acid, and heat) in my text book is described as the following:

    1. Protonation of carbonyl oxygen from ethyloxonium ion
    2. nuclephilic addition of ethanol
    3. deprotonation of oxonium ion to form neutral tetrahedral intermediate
    4. protonation of hydroxyl oxygen
    5. 1,2 elimination
    6. deprotonation.

    Why may step three and four be as shown above rather than your described mechanism with an intramolecular proton transfer?

    Thanks Josh

  8. On the proton transfer (step 3) why shouldnt it say electrophilic?

    “where a proton is transferred from the *nucleophilic* atom to a different atom (either O or N)”

    1. If you read it again it says “which previously acted as a nucleophile”.
      Electrophilic substances are either positively charged or have vacant orbitals which attract electrons(or they are attracted to electrons?).In this case proton transfer from oxygen doesn’t make it an electrophile.

  9. I’d go a bit perverted if it were one of my organic chemisty students, but I like the idea of an acronym. I try to use words that sound like what’s actually happening. So, for example, instead of PAPED I would use PATED (or RAPED) and make it the Penis (Proton) Added himself, Transferred himself, got shit on (Eliminated), and De-Penified (got cut off). So P for Penis (and Proton), Added, Transferred, got shit on (Eliminated), and cut off (Bobbit style). Would likely work well for guys but equally good for girls, so I have found.

  10. In the conversion of carboxylic acid to anhydride mechanism, the second structure isn’t right, the proton is on the wrong oxygen.

  11. Hi,
    I have a huge orgo 2 exam in two weeks. I’m a little confused about the mechanisms. I’m always confused on what will be protonated/ deprotonated in the different mechanisms. I’m still waiting for that lightbulb to go off haha. My professor wants us to memorize the mechanisms, but I’d like to know exactly why all the steps in each mechanism happen the way they do. I really want to understand this better. Help =(

  12. Hi James,

    I love this! Isn’t the anagram a bit misleading for the formation of imines? I get that you need to protonate the O at some point but surely as a 1st step it would inhibit the approach of the RNH2 (which would also presumably protonate before C=O)

    Thanks, Tom

  13. If it helps anybody, the mnemonic PAPER works for this mechanism; it’s not super specific, but IMO mnemonics are meant to help rejog your memory:

    Protonate Carbonyl
    Addition of Nucleophile
    Proton Transfer
    Eliminate H2O
    Restore Carbonyl via Deprotonation

  14. Great post as always!

    In Klein’s text, he explains that protonation of the carbonyl O is not likely in the production of imines or enamines. So that threw me a bit off.

    Secondly, in your mechanism for anhydride 1,2-addition, the arrow doesn’t make sense. (At least from my perspective)


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