For each pair, choose the molecule that you expect to have the...

Question

For each pair, choose the molecule that you expect to have the highest wavenumber for its C=O stretch.

(d) Two molecular structures compared: a six-membered ring with C=O vs. a six-membered ring with C=O and NH.

Accepted Answer

All right. Hi, everyone. So this question is asking us which is expected to have the lower wave number for its carbon stretch X or Y compound X is four metal tetra hydro two HPN two on and compound Y is four methyl purine 20. Now recall that when it comes to Ir spectroscopy, right, the wave number is the exact location at which a particular bond appears in the IR spectrum. Now, the wave number is directly proportional to the frequency at which that particular bond is stretching at. So when it comes to understanding what is going to affect the frequency and therefore the wave number of the car bal stretch specifically, right, we have to ask ourselves what factors may contribute to its stabilization or its destabilization. Because when you consider factors that may destabilize the car bal overall, that is going to make the car bal more resistant to stretching, which means that it's going to do so at a higher frequency because more energy is required, right, more ir interaction is required. So if we're understanding what's going to stabilize or destabilize a car bal, right, we have to consider what factors are going to stabilize or destabilize the resonance form of the car bal in which our carbon is positive and our oxygen is negative, right. This is our Twitter ionic resonance form. So there are a few factors that we have to keep in consideration. The first of which is the resonance stabilizing effect. And the second of which is the inductive effect. Now, both factors, one and two are going to contribute again to the stability of the zitter ionic resonance form. And they are heavily dependent on what happens to be right next to the car bal because compound X is an ester. So we have an oxygen next to the carbonel, whereas compound Y is an amide, right? So we have a nitrogen here instead next to the carbonel. So when it comes to the oxygen of our ester, right, recall that oxygen is going to be more electron than our nitrogen. And that of course, is going to impact how our oxygen stabilizes or destabilizes the zitter ionic form of our car bal because oxygen being more electron oxygen is therefore going to be less willing to donate its non bonding electrons to the car bono itself. Now, it's not to say that resonance doesn't take place because it does. But the idea here is that oxygen being more electron is less willing to participate in resonance than nitrogen would be. So if we want to consider the first factor, which is the resonant stabilizing effect nitrogen in this case is going to win, so to speak, it's going to have more of a stabilizing effect because nitrogen being less electron is more willing to donate its non bonding electrons towards the carbon carbon, which will therefore stabilize the positive charge on that carbon in the zitter ionic resonance form. However, right, it's at this point that we have to take into consideration the inductive effect of both our oxygen and our nitrogen. Because the inductive effect refers to how electron was drawing the oxygen and the nitrogen happen to be. Right. So because the oxygen here of our ester is more electron than the nitrogen of our amide are oxygen is going to have more of a withdrawing effect, right? Because by its nature, it's going to direct electron density towards itself. And what that does is it's actually going to destabilize the carbo cion because you're removing electron density away from the positive charge on the car bono carbon in the residents for. So, if we're discussing the inductive effect, right, oxygen is going to dominate because oxygen is going to have more of a withdrawing effect as opposed to nitrogen due to its electron negativity. Now, if we were to take these factors into consideration by themselves, the answer that we would, that we would be drawn to is the compound wide, would have the lower wave number because the resonant stabilizing effect is going to dominate over its inductive effect. But there's one more thing that we haven't quite yet considered and its actually going to be the most important one to help us understand the answer to this question. And that is going to be ring string because ring strain is going to have the effect of shortening the car Bonal stretch. And as soon as you physically shorten the car Bonal stretch, right, you're actually going to make that bond a lot stronger because it's closer together, right? It's shorter. And that overall is going to make it more difficult to stretch, which requires a higher frequency of infrared radiation to actually stretch it. So our Esther here, our Esther is a six member ring or is in a six member ring, whereas our amide is in a five member ring. Now recall that the five member ring is going to in fact experience a ring strain because it's the six member ring, that's the most ideal, so to speak and actually has no strain whatsoever. So because of this factor alone, right, even though nitrogen is going to stabilize the zoo ionic resonance form, the compound overall experiences a ring strain due to the fact that it has a five member ring. So that is the key here, right? Because by introducing a ring strain, regardless of the first two factors that we discuss a higher frequency of IR is going to be required to counteract the effects of the ring strain itself. So our answer actually is going to be that compound X is expected to have a lower wave number for its carbon stretch and there you have it. So if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

For each pair, choose the molecule that you expect to have the highest wavenumber for its C=O stretch.
(d)

Key Concepts

Infrared Spectroscopy

Wavenumber and Bond Strength

Resonance and Inductive Effects

Watch next