Lecture Outline Resonance

Name ______________________________________ Text section 3.6 introduces the concept of conjugation. 1 Some of the molecules below have lone pairs that are conjugated with π-bonds (just in case of a font substitution issue, that's "pi-bond" — it left my computer as a Greek "p"). Circle the conjugated part(s) of each molecule. (Not the whole thing, just the conjugated part.) N H O N O N O 2 These compounds have no lone pairs, but some have conjugated series of π-bonds. Circle the conjugated π-systems. Lecture outline Resonance — – Sometimes a single Lewis structure is not sufficient to represent a molecule's structure and e distribution, so we describe it as an average or superposition of the 2 or more best Lewis structures, and we call these resonance structures. "Resonance" is an unfortunate term for this because it conveys the impression that there is some sort of rapid switching among all the Lewis structures — this is not the case. The molecule always has one structure. It's just that we can't draw it easily. We can, however, draw a bunch of simple Lewis structures — each of which, by itself, is an inadequate representation of the molecule — and mix them together to get a pretty good description of what the molecule really looks like. A classic example is: HCO2 – (being closely related to "HCO2H", it's understood that both Os are bonded to the C) Experimentally, we know that each O has –1/2 charge and that the CO bonds are equivalent and have a length between C–O and C=O. But there's no way to draw a simple Lewis structure that's even close to being consistent with this. The best we can do is the following: And the actual structure looks like this: Note the use of a special type of arrow to indicate resonance — this should be used exclusively to indicate resonance structures, never to indicate any kind of rxn. Implies resonance — use only for resonance Implies a reversible reaction Implies an irreversible reaction And since we're on the subject of arrows we may as well learn to use curved arrows to show shifts of e–-pairs in reactions (a practice that we like to think has some connection to reality) and to convert one resonance structure to another (a practice that has absolutely nothing to do with reality — since individual resonance structures aren't meant to represent real things!) Fill 'em in... CH3 O O O C C C O CH3 H O CH3 +H O nothing "happens" (this is just us shuffling the dots and lines in our structural representations) something happens (this is a reaction) Of course, here's the actual reaction we're trying to show: O O CH3 C O H CH3 –1/2 C +H O –1/2 Can you illustrate that with curved arrows? No, you can't. Don't even try. You might injure yourself. Here's a critter called an "enolate ion" — (note that this has an embedded carbonyl, though it may be less obvious than the previous neutral examples) CH2COCH3 – Now those two structures are not equivalent. Which one is more important and why? Now, considering that the actual structure is more like the more important ("better") resonance structure, draw the actual structure of the anion below. When we don't know exactly how much charge is on an atom, it's customary to show + – partial charges as "δ " or "δ ". Too bad there's not a similar notation for "partial lone pairs" — we generally don't even try to show these. Resonance implies delocalization (spreading out) of e–s (and therefore charge) — delocalization is always energetically favorable. It's better for e–s to be spread around rather than localized in one place, so... Molecules are always more stable (i.e., lower in energy) if e–s and charge can be delocalized. In other words, what we call resonance, is really a type of delocalization that nature always takes advantage of. One more example — here's a delocalized cation. Draw the two resonance structures and the "actual" delocalized structure. CH2CHCH2 + Now let's return for a moment to geometries, hybridization, and bonding.... Let's start with the allyl cation (directly above). What are the atom geometries? N What's the hybridization of each C, and how do the orbitals overlap to make bonds? Here we'll want to forget about the uninteresting σ-bond (sigma-bond) framework and focus on how the p-orbitals overlap. Is the orbital picture consistent with the resonance picture? (Hint: it had better be (always) or something is very very wrong.) Now let's revisit that enolate anion from the previous page. Same questions... geometry, hybridiz, orbitals... this one isn't quite as straightforward, though... Now for a brief digression that's necessary to illustrate a very important concept... Ammonia, NH3, inverts very easily. The activation energy (the E-barrier) for the process is 9 about 5 kcal/mol. This is so low that an ammonia molecule inverts about 2 x 10 times per –9 second on average, or once every 0.5 nanoseconds (1 ns = 10 sec). That's fast. Sketch the transition state for this inversion. What's the hybridization of the N? Sketch the lone pair orbital. H H H N H H N H So the bottom line is (1a) it takes a 2nd-row atom of this type (3 σ-bonds + one lone pair) very little energy to change it's geometry from pyramidal (tetrahedral) to trigonal planar, and thus go 3 2 from sp to sp . This puts the lone pair in a p-orbital. (In other words, the VSEPR prediction that pyramidal is the lowest energy geometry is correct, but trig planar isn't much worse.) (1b) When flattening out puts the lone pair into conjugation with one or more π-bonds, the resulting resonance stabilization (E decrease — good) is usually enough to offset the small cost of the geometry change (E increase — bad). Thus, a neutral N or negative C of this type (3 σ2 bonds + 1 lone pair) in conjugation will almost always be planar and sp -hybridized! (2) When we're dealing with an O instead of an N, (now we're talkin' 2 σ-bonds + 2 lone pairs) the cost of this hybridization change is zero (there's no geom change). That is, it costs an oxygen 3 2 no energy to go from sp to sp ! It's just a matter of which orbitals we choose for the lone pairs! Remember that this is true even for the oxygen in water — a molecular orbital calculation gives 2 us one in-plane hybrid lone pair and one lone pair in a p-orbital (i.e., a nominally sp oxygen). If 3 we like, we can add and subtract these orbitals to get two sp -ish hybrid lone pairs ("rabbit ears"). So you can choose whichever one of the following statements you like for this type of oxygen: (i) It can always place a lone pair in a p-orbital, at no energetic cost, if doing so creates a conjugated system. (ii) One lone pair is already in a p-orbital. (3) We should probably also mention that going from bent to linear is a much steeper mountain to climb. So this sort of change is just too energetically costly, even in cases where the lone pair can be delocalized by resonance. CH2 sp2 N hard CH2 CH3 N CH3 sp 2 Same deal with oxygen. The O in methoxyacetylene, HCCOCH3 is bent (sp ), not linear (sp). Sure, a linear, sp-hybridized O could have both lone pairs in p-orbitals and conjugated with the π-bonds, but straightening out is just too hard for that O. So the O settles for having one lone pair in a p-orbital and conjugated with one of the π-bonds. (Remember going from the VSEPR3 2 tetrahedral-sp to sp for this O is a moot point, just like it is for the O in water, but to go from 2 sp to sp would require a significant and costly geometry change.) ***** Note — this is the end of the digression. What was the key concept that we need to take away from that? If an atom can put a lone pair into conjugation simply by flattening out, i.e. going 3 2 from -sp to sp , it will usually do that. Now back to resonance...