All right, guys. Let's start off with the order of the protons. What we would see is that the most deshielded should be Hc because Hc is the closest to my electronegative atom, and the most shielded should be my order should have been here, Hb here and Hc here because I'm assuming that it is going to be the one closest to 0, and Hc is going to be the one furthest downfield because it's the most deshielded. That takes care of chemical shifts. Now, in terms of splitting, we have to use n+1. For Hc, I would say that n is equal to 2. So it's got 2 here. So that means that it should be a triplet. So I should draw a triplet for like that.
Now for Hb, what I see is a little bit more complicated. Hb is getting split by 3 on one side and by 1 on the other. Since we're assuming n+1, we can just add that altogether, and that's going to give it a second. That's going to be n equal to 4, which means that we would get a quintet. Remember that a quintet would be, I believe, the 1-4-6-4-1 pattern as predicted by Pascal's triangle. So then we would try to draw that as best we can. So 1, 4, 6, 4, 1. If you're not perfect, it's not a big deal. But we'll try to do it as best we can.
Then finally, Hc. So Hc is being split by how many protons? Well, 2 on this side and 2 on this side. So once again, for Hc, n is equal to 4. So we're going to get another quintet. So let's do the same exact thing. This one's not as nice. There we go. We have our triplet, quintet, and quintet, and now all we need is integrations. Okay. So Ha has how many hydrogens that are of that type? 6. This should be an integration of 6 H. Why? Because I've got these 3 here, but I've also got these 3 here. Remember there was symmetry? Otherwise, they would have had their own peak, but there's symmetry so they're both the same.
So that should be 6 hydrogens are of type 6 A. 6b, how many types of that are there? Well, that's going to be 4 H because I have 2 and 2. And then finally Hc, how many are there? Just one H. That is our filled-out problem.
Now notice that by making these boxes and putting it kind of in order, it made it easier to fill out. But this is all the basics of drawing your own NMR spectra. Really from here, now after you've done this practice problem, I should be able to give you a blank NMR a marked spectra, and you should be able to draw pretty much every single peak, every single signal, all these different splits. Maybe even take a stab at the integration just from all this information.
Now by the way, if you were to represent this as a ratio, the ratio would have been 1 to 4:6. So it's that easy. Now if you do get a ratio that has multiples of numbers, let's say that I'm just going to give a crazy example. Let's say this molecule had been twice as big. So it was actually 2 H, it was actually 8 H, and it was actually 12 H. Let's just say this is just theoretical. If that was the ratio, then it could still be expressed like this because you could simplify all the numbers. You could divide them all by 2, and then you'd say, well, there are more hydrogens but they're still in a one 4 6 ratio. So just trying to show you that your ratios don't always add up to the amount of protons you have. What they do tell you is the relative amounts of each type of proton that you have. Hope that made sense. Let's go ahead and move on to the next topic.