The spectrum of trans-hex-2-enoic acid follows.

Question

1H NMR spectrum of trans-hex-2-enoic acid showing peaks and complex splitting around 7 ppm.

(a) Assign peaks to show which protons give rise to which peaks in the spectrum. (b) Draw a tree to show the complex splitting of the vinyl proton centered around 7 ppm. Estimate the values of the coupling constants.

Accepted Answer

Okay. Hey everyone, let's do this problem. And says the proton NMR spectrum of trans spent two in OIC acid is shown below. One assign all signals in the NMR spectrum to write a splitting tree for the proton with chemical shift about seven PPm determine the values of each coupling constants. So first let's draw our structure so that we can assign all the signals in the proton NMR. Right? We need to know what protons we have. So trans pent, two ene OIC acid. While pent is going to be five carbons And two in OIC acid means we're going to have a car box silic acid that's always on the end of our compound. Right? So that's carbon one. Carbon 2-3 will have the double bond. And it says our compound is trans. So we'll have these two large groups trans to each other. So let me redraw that in a trans configuration, highlighting our double bond. So there's our carb oxalic acid group and we have the rest of our carbon chain. Okay, so how many proton signals do we have? Our spectrum Shows 1, 2, 3, 4, 5 signals. So we're going to have a proton A on our car back silic acid, be on the same carbon as the carb oxalic acid group. And then on the other carbon C and D. N. E. In our carbon chain. So let's list out our protons. So we can assign these peaks. Alright, so the first one that's easiest to start with is our car box silic acid proton because that's very unique in its chemical shift. It's going to be the one all the way down here around 11 p. p. m. Alright. And proton B is B and C. Are going to be a little bit more downfield. So B and C are going to be these two signals here because they are closest to the carb oxalic acid group. And in this alki in here. So let's look at the number of neighboring protons because I see signal at six PPm is a double it. So that means it should only have one neighboring proton And the signal at around seven PPM has a lot of splitting going on. So that means it has several neighboring protons. Well, if I look at proton B, there's only one neighboring proton on the next carbon carbon three. And that's proton C. Right? If we go the other direction, carbon number one has the car back silic acid group. So there's no immediate proton. So proton B is going to be the one at six PPm. One neighboring proton plus one will make it a doublet. And we see that proton C. Has CH two on the neighboring carbon. And that proton B is also neighboring it. So that will be the weird splitting one at 7. PPm. And for part two, we'll have to make the splitting tree diagram for that one. So that leaves DND over here on this carbon and D. Is going to be more downfield because it's closer to this alkaline. So, that'll be the signal at two PPm. And that leaves E. To be the signal at one PPm Awesome. So that is it for part one assigning all the protons. So part two is a little more tricky right? The splitting tree for the proton with chemical shift at about seven PPm. And determine the value of each coupling constant. So we determined that the proton at seven PPm is proton C. So what protons are causing? This splitting it is being split by proton B. Right? That's one of the neighboring neighboring carbons and that one is trans that proton is trans two proton C. And it's also neighbored by proton D. Which is gem. Inal to proton C. So which one do you think is going to have a higher G value? Well, knowing the relationships and their effect on the splitting signal and the coupling constants. Rather, we know trans protons have a higher J. Value. And genital protons have a lower J. Value. And this is helpful to know because when we're making our splitting tree, we know that we split from highest to lowest hurts values. Okay, so we're going to do the splitting by HB first. So I actually added this handy dandy little splitting tree diagram and in our spectrum let me go back up, we are given this legend here of how many hurts our spanning our signal. So let me transpose that to our splitting tree. So it's ending at around 50 Hz. Mhm. And starting at about zero Hz. So we'll draw increments of 10, something like that. Alright, so the values of coupling constants, J. B C. And J C. D. Can be determined by finding the distance in hurts between the two corresponding signals or peaks in the NMR spectrum below. So, this is a zoomed in version of our proton NMR spectrum. So we see that our first split for J. B. C. Our trans proton is going to be over 10. It'll be about 15 hertz. Let me do. So that's for these two larger signals. And then when we split those signals, we have this overlapping signal in the center and then these two outer signals. So measuring the distance between each one, we get a little under 10 hertz. So it's about eight hertz. So the coupling constant for b. c. is 15 hertz. Which is what we know about trans protons. Right? And for germinal protons or J. C. D. It is It hurts. All right. So, that makes be the correct answer for this problem. That's it for this one. I hope this video was helpful. And thanks for watching

The spectrum of trans-hex-2-enoic acid follows.
(a) Assign peaks to show which protons give rise to which peaks in the spectrum. (b) Draw a tree to show the complex splitting of the vinyl proton centered around 7 ppm. Estimate the values of the coupling constants.

Key Concepts

NMR Spectroscopy

Chemical Shifts

Spin-Spin Coupling

Watch next