Draw the NMR spectrum expected from ethanol that has been shaken ...

Question

Draw the NMR spectrum expected from ethanol that has been shaken with a drop of D2O.

Accepted Answer

Everyone and welcome back to another video. In today's lesson, we are going to write the NMR spectrum expected from a sample which contains a mixture of methanol and a drop of D. Iterated water that have been mixed thoroughly. So what we're going to do, we're going to think about this proton exchange. That's what we are observing here. Methanol essentially corresponds to structure C O C 30 H. And basically we have a reaction between methanol and reiterated water, which essentially is D. O. D. Or D. T. O. Right, so you can essentially show that that's D. O. D. We are expanding these oxygen, hydrogen and oxygen deuterium bonds. Now this reaction is apparently at equilibrium, right? Because this is a rapid exchange. So we can essentially add equilibrium arrows and we can clearly see that you can think of this as a double displacement reaction. Right? Where we have that proton exchange, This proton can be easily replaced with the deuterium. So if you exchange them, you will get to new products. One of them will have C. H. 30. D. Right? So that will be our first product and the next product will be D. O. H. Or H O. D. Okay, so now notice that the products essentially allow us to understand how the spectrum would look like. We would see t observations after the exchange, the elation methanol. So we can say the ohh group in math. No, is now replaced with OD. Right? So we were essentially observing O. D. And essentially what that means is that the signal for the. Oh age group will disappear. Right? So we will not see the oh H. Peak. So this peak will disappear from our H. And M. R. And now the second observation is that we are producing H. O. D. H. O. D. Is produced which gives us an extra signal or basically a new signal. Now we were having this information, we can just sketch a possible H. And M. R. Spectrum for the scenario. So let's see that first of all, let's add our axes. Okay, so now we will add our PPm valleys. Let's make it started zero over here. Okay, so that be our zero. Let's suppose that here we have t That would be one then three four, five sex 78, nine And 10. So let's suppose it looks like this. Now those are our PPM values. Parts per million. Right? And we essentially want to add our peaks. First of all thinking about methanol, we won't see the O. H. Peak, but we will see that carbon hydrogen peak, right? Because we have a methyl group. So there are three hydrogen is, carbon is bonded to oxygen, and oxygen is bonded to the deuterium. So we simply understand that there will be this three hydrogen peak, right? Because hydrants are bonded to carbon, it will be single at right? Because the adjacent atom is not even carbon if we have a carbon and there are some neighbors, we essentially split it into something else. But in this case because there are no adjacent neighbors on that atom. Right? It's not carbon. We can essentially say that this would be this would correspond to a single let. And we understand that generally whenever you have an alcohol group bonded to an oxygen atom directly, you expect to see such a peak at around 3. PPm value. So that will be our first peek. Now, if we look at our spectrum, we essentially want to make sure that we add this signal later. But now let's think about H. O. D. That is produced first. So there are several things to keep in mind. First of all, we have to make sure that the peak is broad and the reason why the peak is brought is because of radiation damping effects. So that peak would essentially be broad, but it would correspond to a single because once again we're observing hydrogen bonded to oxygen. Right? So if if a single hydrogen is bonded to oxygen, we don't see any splitting patterns. So that be single. It, Right. We call that whenever you have two carbon atoms. Apparently the splitting pattern depends on the hydrogen atoms bonded to the neighboring carbon, but in this case of hydrogen is bonded to oxygen directly. That's a single it as well. So we would have a broad single it. And experimentally it has been determined that the peak occurs around 5.25 Parts per million. We also have to keep in mind because this is one hydrogen and these are three hydrogen. This peak. This peak will correspond to a higher integration value. So it will be essentially higher and this peak will be slightly smaller. So keeping all of this in mind, we can essentially sketch our spectrum. Let's start with 3.4 ppm. So what we're going to do for this one we want to add and 53 and essentially somewhere here, we have 3.5. So let's move on a bit to the right. Our peak would start somewhere here and we want to make sure that we are adding a 3.4 PPm single act. Right? So because it's a single, that we can essentially draw a straight vertical line because it's not split into more peaks. And then we understand that there will be a broad peak, write a broad peak again, single at at 5.25 PPm. So that's our 545 and about a quarter quarter of that would correspond to 5.25. So somewhere here, it'll be centered at 5.25. So let's draw a broad peak centered at 5.25 we can essentially enjoy it as follows. Let's actually edit this a bit. So that would be let's start here. Okay, And if it's centered at 5.25, let's make sure it's over here and then we're going to make it broad, right? So we're going to and A peak of that kind. So the center is at around 5.25. So this would be our two peaks. This one corresponds to age OD. And in particular we're looking at this hygiene. So it's a broad single at 5. PPm. For this hydrogen once again it's broad because of radiation damping effect. And for the other peak we have, I were carbon hydrogen. Right? We can say that we H two C. H. O. D. And we are in particular looking at any of these three hydrogen. That's that would be three H three hydrogen atoms right? Single at again because we don't have any carbon with hydrogen atoms here at 3.4 PPm. Our standard peak is higher because it has a higher integration. Three hydrogen versus one writes. We apparently must have a larger area for the Speak and that would be pretty much it. Right. So once again, let's recall what we did because of the rapid protein the exchange. Which we illustrated at the very beginning, we noticed that these the hydrogen atom from methanol can be replaced with deuterium and we produce two products after the proton exchange. We essentially understand that there are two observations, the O. H. Group and methanol is replaced with O. D. We won't see the O. H peak anymore. So we didn't even worry about it. And because H O. D. Is produced there will be a new signal which is broad single. Now, understanding these, we have plotted our H NMR spectrum and we can essentially build the spectrum as our final answer that would correspond to our final hmm, our graph now the correct answer choice. The small pool choice question is a however, if you have any questions, don't hesitate to leave your comment down below in the comments section and thank you for watching.

Draw the NMR spectrum expected from ethanol that has been shaken with a drop of D2O.

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Deuterium Exchange

Chemical Shift