Draw an isomer of dichlorocyclopropane that gives an ¹

Question

Draw an isomer of dichlorocyclopropane that gives an ¹H NMR spectrum

c. with three signals

Accepted Answer

All right. Hi, everyone. So this question is asking us what is the isomer of tetra bromo cyclo pentane that will give four signals in the proton N M R spectrum. Option A says 1234 tetra bromo cyclo pentane. Option B says 1123 tetra bromo cyclo pentane. Option C says 1122 touch A bromo cyclo Pentin and option D says 1223 touch a bromo cyclo pen. So on the bottom here, I've gone ahead and actually drawn all four isomers corresponding to options A through D when you go from left to right. Right. So the reason I did this was to understand our structures and therefore understand how many signals we would expect to see in a proton and M R spectrum. Right. So recall that when we're talking about proton N M R, the number of signals that you see it is going to correspond to the number of protons in your molecule, right, that are as I like to call it unique, right. So when I say that a proton is unique, what I mean by that is that it is not chemically equivalent to other protons in the same spectrum or other peaks, I should say. And the reason why it's important to bring up whether or not a proton is chemically equivalent to others is because recall that when you have a molecule that has an element of symmetry to it, that actually reduces the amount of signals that you expect to see in the spectrum itself. Right? Because what that means is that even though they may be on different carbons, certain protons may actually be chemically identical, right? So they would in fact correspond to the same signal in a proton N M R spectrum. And here it's important that I bring up symmetry, right? Because if, when we go ahead and we start off with our 1234 tetro bromo cyclopse, we do in fact see an element of symmetry because if I go ahead and I draw a dash line going through carbon five, I'll notice that I am seeing the same thing on either side of my split, right. So this indicates an element of symmetry because I'm seeing a mirror image on both sides. Yeah. So because I'm seeing a mirror image, I know that I'm going to have multiple protons that are identical to each other, right? For example, the protons that are on carbons two and three are going to be chemically identical. So I'm going to label them both as H A similarly, right, the protons on carbons one in four are also identical to each other but not to other protons within the same spectrum. And the protons on carbon five are going to correspond to their own signal because there are no others similar to it. Right. So they would correspond to their own signal. Now, 1123 tetra bromo cyclo pentane has no such symmetry to speak of. Mhm. But before we get into that, I just wanted to clarify that our 1234 touch of bromo cyclo pen is going to have three signals in the proton N M R spectrum. But because 1123 does not have symmetry, we should expect to see more signals in our proton and in more spectrum specifically, right, carbon one has no protons because it has two bromo substituent. So we won't expect to see any signals there. But carbons 234 and five each have protons which are chemically unique, right? Because because there is no symmetry in this molecule, right? Each of our protons here are going to be chemically distinct from each other. So I'm going to label the protons on carbon two here H A on carbon three H B on carbon 4h C and on carbon five H D, right. So because carbon one has no protons, I expect to see four signals from my remaining protons in my N M R spectrum. Now 1122 toucher bromo cyclo pentane once again has an element of symmetry to it, right? Because I can go ahead and draw a dashed line going through carbon four. And once again, I see mirror images on both sides of my split. So here carbons one and two have no protons each because they each have two bromo substituent, right? But the same cannot be said for carbons 23, sorry carbons 34 and five. And specifically because of our symmetry, because of our mirror image carbons three and five are chemically identical. So I'm going to label them both as H A and the protons on carbon four. I'm going to label as H B because they are not equivalent to H A. So 1122 te bromo cyclo should have two signals in our proton and M R spectrum. And lastly, right, 112, sorry, 1223 tetra bromo cyclo pentane does in fact have an element of symmetry when I go ahead and I draw a dashed line going through carbon two. Now, carbon two itself does not have any protons because it has two bromo substituent, but the protons on carbon one in three are chemically identical to each other because of our symmetry, right? Because of our mirror image effect. So I'm labeling them both as H A OK. And the protons on carbons four and five are once again identical to each other. So I'm labeling them both as H B and this tells me that 1223 te amo cyclo pentane has two signals in the proton and M R spectrum. So when we keep all this in mind, right, only one isomer has four signals in the proton and M R spectrum. And that is 1123 tetro bromo cyclo pentane. So our answer here is going to be option B in the multiple choice and that is going to be your final answer. So if you stuck around to the end, thank you very much for watching and I hope you found this helpful.

Draw an isomer of dichlorocyclopropane that gives an ¹H NMR spectrum
c. with three signals

Key Concepts

Isomerism

Nuclear Magnetic Resonance (NMR) Spectroscopy

Chemical Environment

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