Draw the ¹H NMR spectrum you would expect to see for each of t...

Question

Draw the ¹H NMR spectrum you would expect to see for each of the molecules in Assessment 15.63.

(d) Structural formula of 2-chloropentane, showing a five-carbon chain with chlorine attached to the second carbon.

Accepted Answer

All right. Hi everyone. So for this question, lets go ahead and draw the expected proton NMR spectrum for the molecule shown below. And here we have two meth oxy propane. Now recall that the number of signals in a proton and Mr spectrum corresponds to the number of sets of equivalent protons present in your molecule. So if we consider the structure here of two meth oxy propane, there should be three sets of equivalent protons, one from the methyl carbon in our meth boxy group and the other two from our propal chain, right from our three carbon chain because one of them is secondary attaching directly to the oxygen, whereas the other two are primary. So because we have three sets of equivalent protons, we should expect to see three signals in our proton and in our spectrum. But now we have to consider a few things, we know how many signals there are. So for each of them, we have to consider their multiplicity, their integration and their chemical shift. Now recall that the multiplicity has to do with how many protons are a maximum of three sigma bonds away from any given set of equivalents because those can couple together and therefore influence the splitting pattern of your actual signal. And recall that the multiplicity of a given signal follows the N plus one rule. So for example, right, if we consider the meth oxy carbon here that I've circled in green, because it's connected directly to the oxygen, it doesn't actually have any neighboring protons. So when I consider the N plus one rule, it has zero neighbors. So when I add 0 to 1, I get a total of one, which means that that signal should produce a singlet in my spectrum. No, when I consider my ch proton here on the other side of my oxygen, my C proton is connected to two metal groups. So because I have two metal groups next door, so to speak, my ch here is going to have six neighbors, which when I add one to, that means that I'm going to have a septet in my proton NMR spectrum. And lastly, right, my two methyl groups that are equivalent are each connected to the same secondary carbon, the same CH proton. So because of this, they correspond to the same signal and they're going to produce a doublet according to the end plus one rule. So that's multiplicity, right? And now when it comes to the integration, recall that the integration tells you how many protons correspond to the same signal. So for the methyl group highlighted or circled in green here, there's only one methyl group, which means that the integration should be three protons and my one ch proton should give me an integration of one. Whereas the two metal groups that make up my third signal should give me a total integration of six because all methyl protons here are equivalent to each other, right? So all six of them contribute to the same signal. So now let's consider the chemical shift because recall that the chemical shift deals with the location of each signal on the spectrum itself expressed in parts per million. So when it comes to each type of proton, that we can see in a proton, more spectrum, there are certain ranges that we should expect to see those signals in. And the reason for this has to do with how shielded or D shielded that particular proton happens to be because for example, right, if I consider for a moment, my meth oxy protons here, these meth oxy protons are considered to be relatively d shielded due to its proximity to our oxygen. Here. Now, electrons are the ones surrounding the protons inside of a nucleus, therefore shielding it, so to speak from the NMR effects. So by having an electron atom like oxygen in close proximity, oxygen removes electron density away from the meth oxy proton. Therefore exposing them more to NMR effects, meaning that they're going to appear more downfield or more to the left side of the spectrum. Now, meth oxy protons are still ocular protons. However, due to their proximity to our oxygen, right being oop protons in close proximity to an electron atom, they're going to appear between approximately 3 to 4 parts per million as an estimate. So in this case, right, that signal which indicate integrates to three and will appear as a singlet should appear at about three oops 3.3 parts for million. However, here we're dealing with an ether, right, which means that there is a second carbon in close proximity to the oxygen, which will make it appear de shielded for similar reasons. So my ch proton here, because it is also next to my oxygen is going to appear relatively de shielded as well, which means that its signal is going to appear close to that of the meth oxy proton or protons. So that would place my subset here at approximately 3.2 parts per million, very, very close to our singlet that's d shielded for a similar reason. And lastly, right, let's go ahead and talk about our doublet involving the methyl protons farthest away from the oxygen of two meth oxy propane. No, because they are relatively far away from our oxygen, they're going to be shielded in a way that's expected for oop protons just like it, right. So generally speaking, methyl protons are some of the most shielded signals that we can find in a proton in a more spectrum, meaning that they appear the most up field or the farthest to the right on the spectrum itself. So that's going to place these methyl protons here add or between 1 to 1.2 parts per million. So we can place our doublet here in the middle of that range at approximately 1.1 parts FMI. And with all this in mind, we now have the information necessary to actually go ahead and draw our spectrum. Now, generally speaking, a proton in a more spectrum tends to range between zero to about 13 parts per million. But in this case, because her highest value for the chemical shift is about 3.3 I'm only going to show the spectrum in between zero to about four parts per million. So here I can separate this like so and always remember to put your unit on the X axis here. So that's why I'm writing part 2 million on the bottom and I'm putting smaller tick marks in between to show the halfway point. So first, I'm going to start with my signals that should appear farther to the left in the spectrum. So that leaves me with my first singlet eight at 3.3 parts per million, meaning just above the three and just before the tick mark indicating that 3.5. And so that places that at around here and then adjacent to that just before the three parts per million mark is my Septet. So that gives me like so and now I can go ahead and add my final peak at 1.1 parts per million for the methyl protons that were farthest away from the oxygen. And because that gave me a doublet, I will be drawing two smaller split peaks corresponding to the same signal. Here's one split peek. And here is the uh so here is the approximation of the proton and a more spectrum of two meth oxy propane, we have one doublet at 1.1 parts per million with an integration of six, a singlet at 3.3 parts per million with an integration of three and a subset at 3.2 parts per million with an integration of one. And so with that being said, thank you so very much for watching if you stuck around to the end. And I hope you found this helpful.

Draw the ¹H NMR spectrum you would expect to see for each of the molecules in Assessment 15.63.
(d)

Key Concepts

¹H NMR Spectroscopy

Chemical Shift

Spin-Spin Coupling

Draw the ¹H NMR spectrum you would expect to see for each of the molecules in Assessment 15.63.(d)

Key Concepts

¹H NMR Spectroscopy

Chemical Shift

Spin-Spin Coupling

Draw the ¹H NMR spectrum you would expect to see for each of the molecules in Assessment 15.63.
(d)