I have been told that electron donation decreases carbonyl bond strength and thus decreases the bond stretch frequency, but electron withdrawal increases it.

This does not seem to make any sense as I presume that increasing electron density between two atoms should increase the attraction of their nuclei to those electrons and thus to each other. Is this reasoning flawed? or does this just apply to carbonyls?

  • $\begingroup$ Decreasing bond strength decreases the stretching frequency, first of all. It is not clear to which bonds your sentences apply. $\endgroup$
    – Alchimista
    Mar 26 '19 at 11:50
  • $\begingroup$ I took the liberty to edit the question (in accordance with the suggestions of another reviewer). Please check! $\endgroup$
    – Buck Thorn
    Mar 26 '19 at 12:10
  • $\begingroup$ @Alchimista i'm referring to the c=o bond in the carbonyl. Now that you mention it i'm not sure if the strengthening of the bond is from the sigma or the pi bond of the cabonyl. $\endgroup$ Mar 27 '19 at 15:44
  • $\begingroup$ @NightWriter thanks for the edit $\endgroup$ Mar 27 '19 at 15:44
  • $\begingroup$ Your thinking isn't flawed. What is flawed is where/what are the effect you mention. The point is that electron withdrawing exerted by a group attached to the carbonyl reflects to the other side, but not subtracting electron density, opposite decreasing the weight of -O---C+. Bond order increases and thus frequency increases. I will have a calm look at the answer and if necessary comment or answer back. $\endgroup$
    – Alchimista
    Mar 28 '19 at 8:47

acyl chloride and ethyl acetate

Let's take the above molecules, acetyl chloride and ethyl acetate, as examples of acyl compounds with electron-withdrawing and -donating groups, respectively.

We know that Cl has a higher electronegativity than O as a halogen, so the electron density from the carbonyl carbon should be pulled toward the chlorine in acetyl chloride and pushed toward the carbonyl oxygen in the case of the ester. This might not be immediately obvious, as the two oxygens should have the same electronegativity, so I like to relate this latter point to electric force laws: $$ F=\frac{kq^2}{r^2} $$ Since C=O has a shorter bond distance than C-O, the C=O will exert a greater force on surrounding electrons than the C-O (this assumes electrons to be point charges, so it is only useful as a trend).

Therefore, we draw the following picture:

acyl chloride and ethyl acetate with dipole arrows

Meaning that there is more electron density on the carbonyl carbon for ethyl acetate than for acetyl chloride.

Now, that's somewhat of a problem for the carbonyl oxygen of ethyl acetate! It already has a lot of negative charge as an oxygen, so it will experience repulsive forces due to the increased electron density on its adjacent carbon. Remember that oxygen does not like having more than a charge of -2 formal charge, which is already satisfied by the double bond it has to the carbonyl carbon; a similar relation can be made for actual charge density.

These repulsive forces, well, repel the oxygen away from the carbon a bit, increasing the C=O bond length. And we also know from physics that increasing the bond length decreases the frequency of vibration. This should make intuitive sense: if the bond distance increases, then the atoms have more distance to travel to complete one cycle of vibration, thus decreasing the number of cycles that can be completed per unit time.

Now onto our example electron-withdrawing group:

Acyl chlorides pull electron density away from the carbonyl carbon, as mentioned earlier. Following the same logic, we can conclude that the C=O bond distance decreases(... to a point*). It follows that the C=O vibrational frequency should increase.

Well, that was a lot of words. But let's approach this from another, much simpler, way. In an acyl chloride, the chlorine acts as an electron-withdrawing group, destabilizing the already electrophilic carbonyl carbon. While this decreases the stability of the molecule, that does not mean that all bond strengths are diminished. In acyl chlorides, the C-Cl bond is the weak, labile bond, not the C=O bond. In fact, the C=O bond strength increases for the reasons outlined above. In our example ethyl acetate, the C-O donates electron density to the electrophilic carbon, thus stabilizing the compound. This does not mean that all bonds are strengthened. If anything, it might indicate a greater C-O bond strength than would otherwise be expected. The C=O bond however, is a different bond that is being affected inductively, so it must be considered differently.

Personally, I am a fan of the first explanation, as that provides an actual understanding of what's going on, whereas the second explanation leads to a more memorization-focused approach to chemistry. I hope this helps!

*Extra information--Let's consult the following graph (ignore the specific numbers on it):

energy vs radial distance

What this shows is the distance between two atoms and the potential energy of their "bond" (this is one of those areas where it should be noted that bonds are really just constructs to help us make sense of molecules... but this is not important for now). To the right of the minimum, attractive forces (nucleus --> electrons) dominate. To the left, repulsive forces (nucleus <--> nucleus) dominate. As two positive charges get further away from each other, their energy goes to zero (this can similarly be rationalized by simple physics II-type equations). As their distance approaches 0 (where a distance of 0 indicates superposition, or direct overlap), the energy between the two atoms goes to infinity. This is all just to say that there's a limit on how close the atoms can get to each other. You can see that the slope to the left of the minimum is much steeper and increases at a much faster rate than the slope the the right. It follows that there should be less difference in the C=O bond distance as compared with some fair reference molecule such as acetaldehyde than between the ester and that reference. Part of why I'm bothering with this at all is because of how useful this graph is in just about everything I have studied in some shape or form, so it is very useful to familiarize yourself with it.

  • $\begingroup$ I don't think that your analysis involving the forces of electrostatic interactions is sound. Could you cite a reference for the theory you have presented? $\endgroup$ Mar 30 '19 at 1:28
  • $\begingroup$ @TanYongBoon I cannot, that is just a nice way to rationalize the electron-donating effect of the ester's C-O bond that I came up with in undergrad. It is, however qualitatively accurate, which is all that is needed to answer this question. What exactly do you think is unsound about it? $\endgroup$ Mar 31 '19 at 17:20
  • $\begingroup$ After stating Coulomb's Law, you write: "Since C=O has a shorter bond distance than C-O, the C=O will exert a greater force on surrounding electrons than the C-O". This is not physically incorrect. The electrostatic force would be experienced by the oxygen and carbon nuclei. So yes, it is certainly correct to say that the force experienced by each nuclei is greater. But you seem to be trying to say that the force experienced by surrounding electrons is also greater? $\endgroup$ Apr 1 '19 at 1:14
  • $\begingroup$ I personally do not think it is wise to consider these electrostatic interactions using such physical laws in chemistry. I agree that it definitely can be done. In fact, high-level computational programmes do these sort of analysis, considering all the different electrostatic and quantum chemical interaction that occur. But for the student, this is beyond our abilities. And it can also be dangerous to do so as it is likely that you probably neglected some other important consideration since so many interactions are at play. $\endgroup$ Apr 1 '19 at 1:18
  • $\begingroup$ @TanYongBoon What I am saying with that statement is that the O on the C=O pulls electrons off the carbon more effectively than the O on the C-O, and the closeness of the C=O bond helps the oxygen suck in electrons around it; precise reasons are unnecessary. There are other effects to consider, but this is not the place to discuss that - it's just a simple way to qualitatively rationalize an effect. Physics is inseparable from chemistry in both concepts and in history, and everything presented in my answer is literally all stuff learned early on in an undergraduate chemistry degree. $\endgroup$ Apr 1 '19 at 4:55

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