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This is immediately following ron's answer from Why is a ketone more nucleophilic than an ester?

One of the most simplest questions you can ask, how can you rationalise the order of reactivity towards nucleophiles, which is given as
order of carbonyl compound versus nucleophiles
acyl halide > acid anhydride > aldehyde > ketone > ester ~ carboxylic acid > amide > carboxylate ion

There might be many concepts out there, that explain this order, the above linked question could only halfway satisfactory answer it in the kind of generality I am looking for.

I am looking here for any model that can be used to explain this reactivity.


I will also later on post an answer to this question using quantum chemical calculations, trying to generalise from example. The key points that will be included are stated in ron's answer linked above.

I like Martin's posts, they make me think - and that is always a good thing. Here is where my thinking has taken me:

  • I believe Martin's approach (basically frontier MO) is fundamentally correct. I would be more convinced that his actual numbers are meaningful if he were to calculate 1) the carbonyl carbon LUMO coefficient and 2) the HOMO-LUMO separation for the following series of compounds in order to see if his calculated values correlated with this order of reactivity of carbonyl compounds towards nucleophilic attack (we could use water as the nucleophile) - but Martin, I think such calculations might be a poor use of your time and resources.

While I agree with most of this, I do strongly disagree with the last statement. Which I will hopefully be able to prove soon.

In the meantime I welcome everyone to share their thoughts on the matter.

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  • $\begingroup$ Do you mean "nucleophilicity" or "electrophilicity"? I associate the power of a compound to act as a nucleophile with the term nucleophilicity whereas you included the "order of reactivity towards nucleophiles" in your question which is a measure of electrophilicity. Or did I misunderstand something? $\endgroup$ – Philipp Jul 25 '14 at 12:49
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Here is my "old school" explanation.

Below is a drawing of the reaction coordinate for nucleophilic attack at a carbonyl carbon. The energy well for the starting carbonyl compound is shown on the left. As the positively-polarized carbonyl reacts with (forms a bond with) the nucleophile we pass over a transition state and fall into a second potential well representing the tetrahedral intermediate. $\ce{E_{act}}$ is the activation energy required to pass over the transition state. To whatever extent a substituent might stabilize the starting carbonyl compound, the energy of the carbonyl compound will be lowered, and consequently $\ce{E_{act}}$ will increase. The central carbon (it was the carbonyl carbon) in the tetrahedral intermediate is fully saturated (just $\ce{sp^3}$ bonds) and without charge, so the substituent X cannot interact through resonance to stabilize the tetrahedral intermediate. In other words, while substituent X can stabilize or destabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.

enter image description here

Below are the resonance structures we can draw for our carbonyl compound.

enter image description here

If structure III contributes significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, lower in energy, than a carbonyl compound where structure III contributes little.

Let's start by comparing the ester to the amide. In both cases structure III involves a 2p-2p pi bond, but the resultant positive charge is placed on oxygen when we start with the ester and on nitrogen when we start with the amide. Clearly, for electronegativity reasons, the amide case is better with the positive charge on the less electronegative amide nitrogen. Our prediction: the amide is more stabilized than the ester and therefor should be the slower reacting of the two.

$$\ce{ester > amide}$$

In the case of the carboxylate anion, resonance structure III is equivalent to resonance structure I (note: there is no positive charge in structure III for the carboxylate case). Since they are equivalent, they carry high weight stabilizing and in describing the carboxylate anion. The carboxylate anion is so stabilized by resonance structures I and III that it is basically unreactive to nucleophiles

$$\ce{ester > amide >> carboxylate}$$

There's really no difference between the ester and carboxylic acid in terms of resonance structure III contribution so we would expect them to have similar reactivity towards a nucleophile.

$$\ce{ester \sim acid > amide >> carboxylate}$$

Let's compare the ester to the anhydride. Resonance structure III is equally effective for both compounds, BUT, there are two carbonyls in the anhydride for the oxygen to interact with. Therefor, the oxygen will only be half as effective at stabilizing any one anhydride carbonyl as it was for the ester carbonyl.

$$\ce{anhydride > ester \sim acid > amide >> carboxylate}$$

In the case of the acid chloride, resonance structure III involves a 2p-3p pi bond, not very effective overlap due to the size difference (also chlorine is electronegative so placing a positive charge on it is not desirable, but the poor pi overlap is what limits the contribution from structure III in the acid chloride case).

$$\ce{acid~ chloride > anhydride > ester \sim acid > amide >> carboxylate}$$

For the ketone, our resonance structure III has a slightly different look, it involves hyperconjugation as shown in the following figure.

enter image description here

Since there are 3 hydrogens, there are three such hyperconjugated structures, They are not super significant resonance structures, but there are 3 of them, they do place the positive charge on a proton, so they do count for some stabilization. To be honest, I would have guessed that it creates more stabilization than resonance structure III in the acid chloride case, but I wouldn't have known where it fits between the acid chloride and the ester. Wherever I put it, I'd put the aldehyde to its left. There is no resonance structure III for the aldehyde.

I wind up with $$\ce{acid~ chloride > anhydride > ester \sim acid > amide >> carboxylate}$$ and $\ce{aldehyde > ketone}$ fitting in somewhere between the acid chloride and the ester

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  • $\begingroup$ +1, a very good analysis, I think. For what it's worth, Fleming's book Molecular Orbitals and Organic Chemical Reactions basically agrees with your relative ordering. W/r/t anhydrides and aldehydes, Fleming says "[...] there is a more delicate balance with acid anhydrides, which are similar in electrophilicty to aldehydes," so perhaps anhydride ~ aldehyde > ketone is reasonable. $\endgroup$ – Greg E. Jul 26 '14 at 15:03
  • $\begingroup$ Thanks Greg, good info about aldehydes in your comment. It'll be interesting to see how Martin's calculations handle it. $\endgroup$ – ron Jul 26 '14 at 18:05
  • $\begingroup$ I finally got around to write up my answer, I would appreciate any comment. And while you are at it you could suggest a solvent which could be used for calculations. (I was thinking of DEE or THF maybe.) $\endgroup$ – Martin - マーチン Aug 18 '14 at 11:09
  • $\begingroup$ Why is acid above amide? $\endgroup$ – Abcd Jul 12 '18 at 17:10
  • $\begingroup$ @Abcd Because resonance structure III (as shown at the top of my answer) contributes much more in the amide case, than in the acid chloride case. This makes the amide carbonyl more stable \ less reactive as compared to the acid chloride carbonyl. $\endgroup$ – ron Oct 15 '18 at 19:18
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For this approach I am basically employing Frontier Molecular Orbital Theory (FMO) to predict the reactivity of carbonyl compounds towards nucleophiles.

For the purpose of this explanation I have chosen water as nucleophile. In principle we are looking at the addition of an electron rich particle to an electron poor system. In this case, water will attack the positively polarised carbonyl carbon to form the tetrahedral reactive intermediate. The activation barrier of following step, i.e. dissociation of the leaving group (or proton for aldehydes and ketones), must be lower in energy, because otherwise the reaction will not proceed. It is therefore sufficient to look at only the initial step of the reaction, and more rigorous we can even only look at the ground state properties of the reactants.

In FMO theory there are basically three main observations:

  1. Occupied molecular orbitals of the reactants repel each other
  2. Opposite (partial) charges in the reactants attract each other
  3. Occupied orbitals of one reactant interact with unoccupied orbitals of the other reactant (and vice versa) causing attractive interaction.

The first point is fairly obvious and will not cause much change in the interaction of the two molecules.
The second point is also straight forward. In all carbonyl compounds the carbon will be carrying a (partial) positive charge, while the nucleophile will be having a (partial) negative charge. The third point is not so obvious. It can be expected, the LUMO of the carbonyl compound (electron poor carbon) and the HOMO of the nucleophile (electron rich) will interact to form the new bond. This will also be the strongest interaction. The opposite combination is also thinkable, but the energy difference between these orbitals will be much larger, as we will see. In other words, the closer the energy difference between these orbitals will be, the more attractive the interaction will be. Since we keep the nucleophile constant, we can predict a order of reactivity towards that one specific nucleophile based on the energy of the LUMO of the carbonyl compound. (This comparison will hold, even if we change the nucleophile.)

The following table assembles the energies of the frontier orbitals in $\mathrm{eV}$ as well as the most important partial natural atomic charges. As a model system, the ethyl substituted compounds $\ce{Et-CO-R}$ have been chosen. The calculations were performed with Gaussian09 Rev. D1 employing Becke's exchange functional in conjunction with Perdew's correlation functional. The density fitted def2-SVP basis set was used. For charges the Natural Bond Orbital (NBO) analysis was used.
For convenience I have not included any solvent, so these calculations assume gas phase, vacuum and absolute zero.

\begin{array}{lrrrr} \ce{R =} &\ce{HOMO}&\ce{LUMO}&\Delta E &\mathbf{q}(\ce{C})&\mathbf{q}(\ce{O})\\\hline \ce{-Cl} & -7.00 & -2.11 & 4.89 & 0.55 & -0.46 \\\hline \ce{-O-OC-Et} & -6.35 & -1.85 & 4.50 & 0.82 & -0.52 \\ & & \text{(L+1)} -0.85 & & & \\\hline \ce{-H} & -5.79 & -1.79 & 4.00 & 0.41 & -0.51 \\\hline \ce{-Et} & -5.55 & -1.48 & 4.07 & 0.57 & -0.54 \\\hline \ce{-OH} & -6.33 & -0.88 & 5.45 & 0.81 & -0.58 \\ & & & & &(\ce{OH}) -0.68 \\\hline \ce{-OEt} & -6.15 & -0.74 & 5.40 & 0.81 & -0.58 \\ & & & & &(\ce{OEt}) -0.56 \\\hline \ce{-NH2} & -5.53 & -0.24 & 5.29 & 0.67 & -0.60 \\ & & & & &(\ce{N}) -0.83 \\\hline \end{array}

These kind of calculations cannot handle anions very well, so I have also omitted the carboxylate ion. For obvious reasons, I would say its tendency to react as an electrophile is negligible.

The HOMO of water is one of the lone pairs at oxygen, at $-6.35~\mathrm{eV}$. Correlating the LUMO energies from the table, the expected order of reactivity is $$\ce{acid~chloride > anhydride\approx aldehyde > ketone > acid\approx ester > amide}$$


With NBO we can look at localised orbitals, which all look very similar and also give very similar numerical results. The $\pi^*$ antibonding LUMO is composed of $67\pm2\%$ carbon $\ce{p}$ AO and $33\pm2\%$ oxygen $\ce{p}$ AO for all compounds. Here are the FMO shown for the anhydride. nbo pi orbital
nbo pi star orbital


Unfortunately I was currently not able to find any reaction barriers for this approach. My educated guess is, that the chosen nucleophile is too weak to coordinate to this compound in the gas phase. I believe I will have to include a solvent for transition state stabilisation, but I am not sure if I can afford the calculations at the moment.

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  • $\begingroup$ This is very nice work, very informative and it demonstrates how powerful your approach is (wish I could do it on my home computer - maybe someday). The one suggestion I have is that it would be nice to have a second table where you show ΔE for the water-carbonyl HOMO-LUMO and LUMO-HOMO interactions. $\endgroup$ – ron Aug 21 '14 at 17:41
  • $\begingroup$ Am I correct that the HOMO and LUMO numbers in your table reflect the carbonyl HOMO-LUMO? What was the HOMO and LUMO for the nucleophile? As you noted, "The opposite combination is also thinkable, but the energy difference between these orbitals will be much larger, as we will see." Where is the delta E for the nucleophile(LUMO)-carbonyl(HOMO) interaction? $\endgroup$ – ron Aug 22 '14 at 20:20
  • $\begingroup$ Oh, thanks for the comment, I forgot about that while writing. I will try to add that when I get back to the office, perhaps on monday. $\endgroup$ – Martin - マーチン Aug 23 '14 at 14:09
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There are several different things going on here in this rather long series. I will go through them in turn.

Acid chlorides and acid anhydrides

These are going to be reactive because they have a good leaving group. Chloride ion is so stable that it may break away from the molecule and give $\ce{RCO+}$ and $\ce{Cl-}$ spontaneously, not to a great extent, but sufficient for this to have no limiting effect on the reaction rate.

Similarly an acid anhydride may break down to $\ce{RCO+}$ and $\ce{RCOO-}$. We know that the anion is again quite stable, but not as stable as $\ce{Cl-}$. This is apparent from the fact that $\ce{RCOOH}$ is an acid, but not as strong an acid as $\ce{HCl}$.

$\ce{H}$ and $\ce{R}$ are not stable anions, so aldehydes and ketones are much less reactive than acid anhydrides and chlorides. Of course, the $\ce{H}$ and $\ce{R}$ groups remain attached to the molecule, which is not the case with the chlorides and anhydrides. Being an addition reaction instead of a substitution reaction, there is a decrease in entropy.

An interesting case here is the iodoform reaction where $\ce{R-CO-CH3}$ reacts with iodine to form $\ce{R-CO-CI3}$. $\ce{CH3-}$ is a poor leaving group, but the $\ce{CI3-}$ ion is a good leaving group and breaks away from the molecule, leaving $\ce{RCO+}$, which then reacts with water to form $\ce{RCOOH}$, while the $\ce{CI3-}$ reacts with water to form $\ce{HCI3}$.

The thing to note here is that there is significant thermodynamic drive here for the good leaving group to be substituted by a poorer one.

Aldehydes and ketones

An important factor that makes aldehydes more reactive than ketones is that there is less bulk blocking the approach of a nucleophile. There is less steric hindrance.

Esters and carboxylic acids

We are now back to substitution reactions. But $\ce{RO-}$ or $\ce{OH-}$ are much poorer leaving groups than the $\ce{Cl-}$ or $\ce{RCOO-}$ that we saw with acid chlorides and anhydrides.

There is another factor at play here. The $\ce{C=O}$ double bond is conjugated with the $\pi$ electrons on the other oxygen, which significantly stabilises the molecule. This causes a resistance to both $\ce{S_{N}1}$ reactions (where the $\ce{RO-}$ or $\ce{ROH}$ spontaneously detaches from the molecule leaving $\ce{RCO+}$) and $\ce{S_{N}2}$ reactions (where the nucleophile attacks the molecule, giving a tetrahedral intermediate before the $\ce{ROH}$ or $\ce{H2O}$ is kicked out.) Either way that $\pi$ system has to be broken down, giving a kinetic barrier to reaction (high activation energy.)

Amides and carboxylate ions

In certain reaction mechanisms OH- may spontaneously leave a carboxylic acid. $\ce{NH2-}$ is more basic and so is very unlikely to leave of its own accord. $\ce{O^{2-}}$ is even more basic (unheard of in solution.)

Other reaction mechanisms require that the molecule is protonated first. The next step may be nucleophilic attack ($\ce{S_{N}2}$ type mechanism) or spontaneous cleavage of the molecule into $\ce{RCO+}$ and a leaving group ($\ce{S_{N}2}$ type mechanism).

If $\ce{COOH}$ is protonated, the leaving group will be $\ce{H2O}$, a neutral compound.

If $\ce{CONH2}$ is protonated, the leaving group will be $\ce{NH3}$, a basic compound.

If $\ce{COO-}$ is protonated (to an acid) the leaving group will be $\ce{{}^{-}OH}$.

Generally the more acidic/less basic a group is, the more stable it will be when separated from the $\ce{RCO+}$ carbocation. Here we have a neutral molecule water (a good leaving group), a weak base ammonia (poor leaving group) and a strong base (very poor leaving group) so the reactivity order is as would be expected.

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