There's one more type of Proton relationship that we should be aware of, and that's called E and Z Dia stereo ice aneurysm. Big word. So how does that work? Well, this exists when you're Q test is used on a terminal double bond, and it yields what we call a new tribunal center. Now, this is a term we haven't been over in a long time. But if you guys recall, a tribunal center is basically like a CIS and trans Isom er okay, so a tribunal center would be like the relationship between assist double bond versus a trans double bond. That's what we call a tribunal center. So notice if I used the Q test on a terminal double bond, let's say a scratch out this age. Replace it with a Q notice that what I just did by putting the queue at the bottom would be I just drew the CIS version, right, sis? Or if you're using E Z notation, that would be easy, sis slash z right, But who says I have to put it at the bottom? What if I had put the queue at the top? Well, if I had put the queue at the top, then that would be Trans right? So that means that the fact that I could make two different I summers depending on where I put the Q, means that this easy relationship is possible. Okay, so these protons are always gonna be dia stereotype. OPIC. Okay, we don't have to look for prior Cairo centers anything like that. Anytime you could make an ear Z or assistant trans relationship with Q. With the Q test always die hysteria topic. Now, based on what you learned about dia stereotyping before, do you think that thes hydrogen will have share a peak share signal or do you think they'll get different signals? It says it right there. They're gonna be non equivalent, and they're going to get different signals. Okay, now I know that I've gone a lot through the rules of how to figure this out. But I just want to explain this really quickly intuitively, because you might be struggling to think, Well, why would they be non equivalent? We'll think about it like this. Let's say you've got this double bond right? And we draw one of my age is there? Can we draw one of my age is there? Okay, so it's basically the same thing. Same example that we have on the top. I just told you that because there's an easy relationship between these ages, they get their own peak. But how does that make sense in terms of shielding, Remember, in terms of putting on the wool coat, how does that make sense? Well, notice that this H is always going to be closer to that Ethel group because it sis to it. This age is always gonna be further from the Ethel Group because it's trans to it. So that means that the red age and the green age will be shielded slightly differently. One of them is gonna be a little bit more bundled up, and one of them is gonna be a little less bundled up because of how far they are away or how close they are to this group over here. So that means that they're gonna have to each get their own peak. They're gonna they're each gonna have to get their own signal. Okay, so in this example specifically, this would be a check A and this would be HB. They would each get their own signal because of their unique position on the molecule. Okay, so that said, we have a few more practice problems. Go ahead and try to figure out for a If, you know, you think that these two hydrogen on the end that double bond deserve toe have the same signal or different, and then tell me the total number of signals you would get. All right, so check it out.
2
example
Determining Diastereoisomerism
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So let's figure this one out. When you used the Q test on these h is what was your conclusion? Guys, they're actually exactly the same, and they're going to share a signal. The reason is because, let's say a replace that h replace it with a. Q. Do we get an easy relationship or assist trans relationship? No, we don't, because notice that if I make the queue on the right side, then it's gonna be next to unethical group. However, if I put the cute on the left side, it's also next. When Ethel Group. Since this double bond is symmetrical, since it's symmetrical, there is no CIS and trans possible. Okay, okay, Since there's no system trans possible, that means that they're going to share peak. Okay, so these air actually, Homo topic okay, Their home topic. So let's go ahead and then just count out the peaks like normal. It would be that thes are a um we've got this carbon. Well, I mean, we already counted that carbon. Those those ages. We've got this carbon, which you're going to scratch out because it doesn't have any h is. We've got this plane of symmetry, so That means that this is B and that means that this is C and that's it. We've got three different ones. The other side's air also B and C. So this would get three signals. All right, try that with the next one and let me know what you get.
3
example
Determining Diastereoisomerism
2m
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So were the hydrogen. Is on this one Homo topic or die a stereotype. OPIC. And after using the Q test, I've got h. I've got h. I use Q over here. And what I noticed after using the Q test is that actually does make a difference. Worked with the Q. If I put the queue at the top, I get trans. If I put the queue at the bottom, I get cysts. So actually, these hydrogen, our diet stereotype, OPIC. So I have a dia stereotypic relationship. And since they're dying stereotype OPIC, that means I would I would give each of them their own signal so I'd have h A and H B. Now I just have to count up the rest. This would be C. This would definitely be D. Now, what do you guys think about the last two? Did you guys give them the same signal? Good. Okay, I'm glad. I could tell a lot of you guys gave these the same signal because there's a plane of symmetry there. Okay, so both, um, would have the same. Now there is that one detractor out there. Okay? And it's okay. I'm glad you're thinking so hard. But I know you're out there saying, Johnny, but isn't this e? Isn't this one closer to the double bond? And isn't this one further away? Right? So shouldn't they actually different signals, right? Shouldn't they be different? And the answer is actually no guys, because the Onley Way that you can have dia stereotype pick in the sense is on a double bond because there's no free rotation. But notice that since thes two methyl groups are on a single bond, this thing can rotate as much as it wants. Okay, so that means that the e let's say this is E one. Let's say this is e to right. E one right now happens to be closest to the double bond. But after it rotates than e two could be Justus close. So actually there is no difference between e one and e to they can rotate. The only difference happens with H A and H B. Because they can't rotate. HB is always locked in the downward position. H A is always locked in the outwards position. So that's why they become dia stereotypic in that sense. Okay, I know the rest of you guys are like Johnny, you're over complicating it. But just letting you know you have to think about just the rules that I told you. Don't overcomplicate it. And don't try to think too hard about these because I've given you a pretty good set of rules already. Okay, so let's move on to the next topic.