So far in discussing resonance forms I’ve mentioned two important principles:
1. Minimize charges: the resonance form with the fewest charges will be the most stable.
2. How to break π bonds. If you absolutely must break a π bond to make a resonance form, do it in a way so that the pair of electrons end up on the more electronegative atom of the π bond.
That second rule is a handy one to live by, but it’s just a representation of a more general principle when it comes to resonance (and organic chemistry in general).
If your molecule must bear a negative charge, it’s most stable on the least basic atom.
You could argue that this definition is circular, because basicity is essentially a measure of the stability of an anion. OK, fine. The more stable an anion is, the less basic it is. So how do we go about finding out information on basicity?My suggestion is to visit your best friend, the pKa table. Or you could familiarize yourself with the key trends that affect acidity, and apply these concepts to resonance forms.
Let’s go these trends them one by one.
Factor 1: Electronegativity
Recall that basicity decreases as the electronegativity of an atom increases.
That means that in comparing two resonance forms of equal charge, we can determine which is more stable by examining the electronegativity. In the examples above, we can tell that the resonance forms on the left are more stable than the resonance forms on the right, since the electronegativity of O > N and also O > C. There are a lot of chemical structures that have resonance forms with alternating charges on C and N (or O), so this is a particularly useful rule.
Factor 2: Polarizability. The basicity of an atom decreases as the polarizability increases. This trend especially applies for atoms in the same column of the periodic table, such as O and S.
Factor 3: Inductive effects. Basicity decreases as the negative charge is stabilized by adjacent electron withdrawing groups. So in the cases below, the more stable resonance form is the one where the negative charge is on the atom with the extra electron withdrawing groups. [Note 1]
But it ain’t always straightforward. Organic chemistry being what it is, it’s common to have situations where we have to weigh the effects of multiple variables. Like in these examples.
What do we do here? We rely on experimental results. One body of work in this area is the pKa table, an experimentally derived guide to the multi-variable dependent phenomenon of acidity. Another is to compute the relative energies computationally. Or, a third is simply to apply the 4 principles we’ve discussed earlier to the situation at hand and see which act in the same direction and which oppose. At least 3 of the examples here should be fairly straightforward to figure out .
So to make a long story short, as I’ve said many times before, the pKa table is your friend. For three good reasons.
1. It tells you about the relative strengths of acids and bases (itself very important)
2. It gives you an idea of which atoms will bear negative charge the best (useful for determining the stability of resonance strutures)
3. When you learn about substitution and elimination reactions, it also is the Rosetta Stone for determining leaving group ability.
[Note 1] . it should also be pointed out that theres going to be some repulsion in charge between the oxygen & negative charge or the fluorines & negative charge. Furthermore once the electron withdrawing group can participate in resonance we have to think about π donation and π accepting.
Note 2: What about hybridization? Doesn’t that affect stability of negative charges too? After all, the reason why alkynes are more acidic than alkenes which are more acidic than alkanes is because the ability of orbitals to stabilize negative charge increases as increase the amount of s character [sp3 to sp2 to sp ]. So shouldn’t that apply, for example, in the instance below?
From examining these two resonance forms we might be tempted to believe that the nitrogen on the left (the one bearing a hydrogen) is sp2 in one version and sp3 in another. If you think about what resonance forms represent, however – different ways of depicting electron density – you will realize that this can not actually the case since that would involve moving atoms around [re-hybridizing from sp2 (trigonal planar) to sp3 (tetrahedral). ]. Both nitrogen atoms are, in fact, sp2 hybridized, and their stabilities cannot be distinguished based on this characteristic. [Thank you to reader Brian E. for correcting an earlier version of this post. In fact, Brian informs me that calculations indicate the right-hand resonance form makes a significantly higher contribution to the resonance hybrid than does the left-hand resonance form].