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Lecture Comments (47)

1 answer

Last reply by: Professor Starkey
Sun Feb 12, 2017 10:08 PM

Post by Ay Ayy on February 7, 2017

I have this question
1. (a) Determine the digital resolution of a 12 ppm proton spectrum collected on a 300 MHz NMR instrument.  Number of data points collected for this spectrum was 1K (1024 bytes). [Hint: On a 300MHz instrument, 1PPM=300Hz]              
(b). What would be the resolution of the above spectrum if 2000 data points were collected?            
C. If we increase the spectral width from 12 to 20 ppm and keep all other acquisition parameters same as in problem 1(a) and collect a proton spectrum, what would be the acquisition time and resolution of the newly acquired proton spectrum?

1 answer

Last reply by: Professor Starkey
Fri Feb 3, 2017 9:52 PM

Post by Kaye Lim on November 30, 2016

I have a question regarding how NMR instrument works. This is what I thought, please check if it is correct.

So there is only 1 MHz value of radio wave applied on the sample (like 42.6 MHz for 1 Tesla magnetic field). That powerful pulse of radiowave excites all nuclei including nuclei of F,N,C as well. Then the radio receiver would tune into a correct MHz value to read the released energy from the nuclei (example, 42.6MHz to read proton nuclei, and other corresponding MHz to read nuclei of other atom type). Is everything above correct?

Thank you!

1 answer

Last reply by: Professor Starkey
Fri May 6, 2016 12:50 AM

Post by Tram T on May 4, 2016

For protons on Carbon table at 37:04, Why proton of Methyl is more upfield (more shielded)than methylene and methine proton?

I thought that since alkyl R is EDG, the more alkyl R group like in the case of methine proton, the more electron rich the area thus methine proton would give the most upfield signal instead of proton on methyl.

Please explain! Thank you! Great lecture!

1 answer

Last reply by: Professor Starkey
Wed Nov 11, 2015 9:10 PM

Post by Jeremy Cohen on November 11, 2015

Dr. Starkey, I didn't know where to put this but I just wanted to say thank you for all of your help this semester.  Your lectures have been incredibly helpful in getting me through orgo 1.

1 answer

Last reply by: Professor Starkey
Fri Jul 17, 2015 1:27 PM

Post by Akilah Futch on July 16, 2015

what if you are not given the formula of the structure and all you have is the H nmr.

3 answers

Last reply by: Professor Starkey
Mon Jul 7, 2014 12:05 AM

Post by Anhtuan Tran on July 1, 2014

Hi Dr. Starkey,
When it comes to calculate the chemical shifts for CH2 group, we use the formula: 1.2 + ΔR1 + ΔR2 and we look up the table for the values of Δ. My question is where those values are coming from and how did they calculate those values and what is the difference between the Δ values and the regular values that we use for H that has only one neighbor.
Thank you.

1 answer

Last reply by: Professor Starkey
Mon Feb 3, 2014 12:04 AM

Post by Andrea Cola on January 31, 2014

How many 1H NMR signals would 1,3,5-trimethylbenzene give?

5 answers

Last reply by: Professor Starkey
Tue Jul 8, 2014 12:03 PM

Post by brian loui on April 2, 2013

on example 2, (the one w/ the carbonyl) aren't the "e" methyls diastereotopic and therefore not equivalent? i made models... and they're not superimposable and aren't enantiomers.

1 answer

Last reply by: Professor Starkey
Sun Feb 17, 2013 5:29 PM

Post by Betty Vowles on February 17, 2013

Like Marina, I too am having difficulties with the last portion of the video. Have the technical difficulties been resolved?

1 answer

Last reply by: Professor Starkey
Thu Feb 7, 2013 10:58 AM

Post by Synthia Gratia on February 6, 2013

On the last example on example 5, when figuring out the number of signals in an NMR, I'm a little confused on how you designated the different protons. when you did the stereochemistry for the H and t-butyl group that's not a real stereocenter right? I mean that C has a t-butyl group a H and when you try to figure out the other 2 groups it is the same because the molecule is symmetrical. So how did you apply stereochemistry there? Or was that to explain the different H's?

1 answer

Last reply by: Professor Starkey
Fri Dec 14, 2012 11:21 AM

Post by Natalie Bossi on December 13, 2012

How can I move on ahead of what the lecturer is talking about?? It appears that I am stuck with wherever she is talking about, no matter what I click on in the contents. This is wasting a huge amount of time.
Please help.

2 answers

Last reply by: Amirali Aghili
Sat Apr 6, 2013 4:38 PM

Post by Marina Bossi on November 22, 2012

In addition to this, if the video reaches a certain point where the data hasn't been loaded yet, it goes back to the very beginning again!

2 answers

Last reply by: Marina Bossi
Tue Nov 27, 2012 6:50 AM

Post by Marina Bossi on November 22, 2012


The lectures are very helpful but why can't wait click on the exact position we wish to see? It is quite frustrating because I have to watch the whole lecture before I get to the bit I was up to. Thanks

1 answer

Last reply by: Professor Starkey
Fri Sep 21, 2012 12:11 PM

Post by fiorella alzamora on September 19, 2012

Why is Toluene 7ppm? y wouldnt it be 2.3 ? Thanks

2 answers

Last reply by: Gabriella Kaminer-Levin
Tue Jul 3, 2012 4:58 PM

Post by Gabriella Kaminer-Levin on June 29, 2012

Dear Dr. Starkey:

How come hydrogens bonded to an oxygen (say in an alcohol group) don't show up on an NMR (or do they)? At around 45 minutes in this video you are describing the approximate positions of hydrogens in an ester/ alcohol and you do not include the hydrogen bonded to an oxygen in an alcohol group in your analysis.

1 answer

Last reply by: Professor Starkey
Fri Feb 17, 2012 8:33 PM

Post by janine jones on February 15, 2012

trying to work a problem that I am stuck on about signals is there any way I can upload an image to you>

1 answer

Last reply by: Professor Starkey
Sun Feb 5, 2012 10:02 PM

Post by Kimberly McDevitt on February 5, 2012

Can you please inform me how to fast forward the lectures or to select the section that I previously left off on without having to watch the entire lecture over again?

2 answers

Last reply by: Sitora Muhamedova
Wed Jun 19, 2013 4:19 PM

Post by Jason Jarduck on October 17, 2011

Excellent lecture very detailed explanation.

Thank You

Jason Jarduck

Nuclear Magnetic Resonance (NMR) Spectroscopy, Part I

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  1. Intro
    • Purpose of NMR
    • How NMR Works
    • Information Obtained From a ¹H NMR Spectrum
    • Number of Signals in NMR (Chemical Equivalence)
    • Size of Signals in NMR (Peak Area or Integration)
    • Using Integral Trails
    • Location of NMR Signal (Chemical Shift)
    • ¹H NMR Chemical Shifts
    • ¹H NMR Chemical Shifts (Protons on Carbon)
    • Chemical Shifts of H's on N or O
    • Estimating Chemical Shifts
    • Calculating Chemical Shifts
    • Effects of Resonance on Chemical Shifts
    • Shape of NMR Signal (Splitting Patterns)
    • Understanding Splitting Patterns: The 'n+1 Rule'
    • Explanation of n+1 Rule
    • Summary of Splitting Patterns
    • Predicting ¹H NMR Spectra
    • Intro 0:00
    • Purpose of NMR 0:14
      • Purpose of NMR
    • How NMR Works 2:17
      • How NMR Works
    • Information Obtained From a ¹H NMR Spectrum 5:51
      • No. of Signals, Integration, Chemical Shifts, and Splitting Patterns
    • Number of Signals in NMR (Chemical Equivalence) 7:52
      • Example 1: How Many Signals in ¹H NMR?
      • Example 2: How Many Signals in ¹H NMR?
      • Example 3: How Many Signals in ¹H NMR?
      • Example 4: How Many Signals in ¹H NMR?
      • Example 5: How Many Signals in ¹H NMR?
    • Size of Signals in NMR (Peak Area or Integration) 21:23
      • Size of Signals in NMR (Peak Area or Integration)
    • Using Integral Trails 25:15
      • Example 1: C₈H₁₈O
      • Example 2: C₃H₈O
      • Example 3: C₇H₈
    • Location of NMR Signal (Chemical Shift) 29:05
      • Location of NMR Signal (Chemical Shift)
    • ¹H NMR Chemical Shifts 33:20
      • ¹H NMR Chemical Shifts
    • ¹H NMR Chemical Shifts (Protons on Carbon) 37:03
      • ¹H NMR Chemical Shifts (Protons on Carbon)
    • Chemical Shifts of H's on N or O 39:01
      • Chemical Shifts of H's on N or O
    • Estimating Chemical Shifts 41:13
      • Example 1: Estimating Chemical Shifts
      • Example 2: Estimating Chemical Shifts
      • Functional Group Effects are Additive
    • Calculating Chemical Shifts 47:38
      • Methylene Calculation
      • Methine Calculation
      • Protons on sp³ Carbons: Chemical Shift Calculation Table
      • Example: Estimate the Chemical Shift of the Selected H
    • Effects of Resonance on Chemical Shifts 53:11
      • Example 1: Effects of Resonance on Chemical Shifts
      • Example 2: Effects of Resonance on Chemical Shifts
      • Example 3: Effects of Resonance on Chemical Shifts
    • Shape of NMR Signal (Splitting Patterns) 59:17
      • Shape of NMR Signal (Splitting Patterns)
    • Understanding Splitting Patterns: The 'n+1 Rule' 1:01:24
      • Understanding Splitting Patterns: The 'n+1 Rule'
    • Explanation of n+1 Rule 1:02:42
      • Explanation of n+1 Rule: One Neighbor
      • Explanation of n+1 Rule: Two Neighbors
    • Summary of Splitting Patterns 1:06:24
      • Summary of Splitting Patterns
    • Predicting ¹H NMR Spectra 1:10:46
      • Example 1: Predicting ¹H NMR Spectra
      • Example 2: Predicting ¹H NMR Spectra
      • Example 3: Predicting ¹H NMR Spectra
      • Example 4: Predicting ¹H NMR Spectra

    Transcription: Nuclear Magnetic Resonance (NMR) Spectroscopy, Part I

    Hi; welcome back to Educator.com.0000

    Today, we are going to talk about NMR spectroscopy: NMR stands for Nuclear Magnetic Resonance, and it is an extremely important tool that chemists use to determine chemical structure.0002

    Now, the purpose of an NMR is: we take an unknown sample; we place it in an NMR tube, and it is placed inside of a superconducting magnet--a very, very strong, powerful magnet that needs to be cooled to very, very low temperatures in order to have its properties.0015

    We are going to apply some energy: so like all spectroscopic methods, we are going to analyze our sample by irradiating it with some kind of energy and then observing how it interacts with that energy.0031

    In this case, we are using radio frequency waves; so these are very, very low-energy waves, something on the order of 60 megahertz, all the way up to 900 megahertz.0043

    These instruments just keep getting more and more powerful.0056

    The larger the instrument, the more sensitive it is, the more detail you can get; but very often, a 60-megahertz NMR would be all you need to get a lot of very interesting spectroscopic data and structural data.0061

    We are going to use radio frequency waves to produce an NMR spectrum; and so, this is an example of what a spectrum looks like; so we are going to be learning today what all the details of a spectrum are and what you can learn about the structure by looking at a spectrum.0078

    And then, we are going to learn how to interpret that to come up with an actual molecular structure.0093

    NMR is an incredibly powerful tool for use in the laboratory; it is also used in the medical field--MRIs--that stands for Magnetic Resonance Imaging--it uses the same technology.0099

    It is not called NMR in the medical field, because the use of the word "nuclear" can be a little scary for patients to think about it; but if you have ever had an MRI, you get inserted into a tube and hear lots of clinking and noises going around as you are exposed to the radio waves.0112

    And then, the results are analyzed.0131

    We are going to see how we can use these spectra in the laboratory.0134

    Let's talk a little bit about the theory of NMR and what is going on: how the spectrum is generated.0139

    OK, and if you think of a nucleus of an atom, certain nuclei have a magnetic moment.0145

    That means they behave like little, tiny magnets; so they have a spin: the spin can be either +1/2 or -1/2; that is the way we can think of it.0156

    So, if you imagine just a collection of atoms...or you can have these nuclei have their spins in all random distributions--OK, all of these are equal in energy.0163

    Now, if you take that sample, though, and you apply an external magnetic field (remember, that is the first step in our NMR--to place the sample inside of a very powerful magnet), what is going to happen is: all of the nuclei with magnetic moments are going to align themselves in the same direction as that applied magnetic field.0176

    And then, there are only two possible orientations they can have: they can either be aligned with the field, pointing in the same direction, or aligned against the field, pointing in the opposite direction.0196

    These spin states, now, are only two possibilities.0206

    These two states have different energies; the ones in which they are aligned with the field--we describe that as the α spin state, and those are going to be lower in energy.0212

    Those that are aligned against the field are described as β, and those are higher in energy.0224

    Now, the populations of nuclei in these two states are very similar, but there is a slight excess of the nuclei with the α spin state, where they are aligned with a magnetic field.0231

    What we are going to do after placing our sample in the magnet is: we are going to pulse it with some energy--that radio frequency energy--and some of that energy is going to be absorbed in certain situations.0242

    What is going to happen is: we are going to induce a spin flip, in which a nucleus that is aligned with the field is going to accept that energy and absorb that energy (we call that resonating), and it is going to now have the β spin state.0256

    It is going to be higher in energy; we are going to pulse it with some energy, and then we are going to let that come back down to its original state.0276

    And when that energy is released, we call this process relaxation; after letting the sample relax, we observe which frequencies have been absorbed, and we can generate a spectrum.0283

    Now, in order for this process to work, you have to have a nucleus with an odd spin number; you have to have...like a 1/2 is the case that we have in these nuclei.0297

    And so, it turns out that protons--ordinary hydrogens--the normal isotope of hydrogen--has a spin state of 1/2, as does the isotope of carbon known as C-13, also certain isotopes of nitrogen and fluorine.0310

    So, this is very convenient, because if we are studying organic structures, we have in all of our molecules...we have carbon atoms; we have hydrogen atoms.0328

    And so, by investigating the hydrogen atoms and the types of carbons that we have in our structure, we can figure a lot out about the actual structure and how these atoms are connected to one another.0339

    So, when we take a look at a spectrum (here is an example of a spectrum), there is a lot of information that is buried in here; and one by one, we are going to break these down and learn more about them.0353

    OK, one of the things we are going to see is the number of signals--how many signals are there in the spectrum?0363

    Right here, here are two signals; here is another one; here is another one; here is another one; so the number of signals that we have is going to tell us something about the different types of hydrogens in our structure.0369

    Now, we are going to be starting with proton NMR, and just looking at the normal isotope of hydrogen; we describe that as proton NMR.0379

    And this is an example of what a proton NMR looks like.0395

    The first thing is the number of signals: that tells us how many unique types of hydrogens there are in the structure.0398

    The next thing we are going to look at is the integration: we are going to look at the size of these peaks--how big are they?--that is going to give us some indication of how many hydrogens there are giving rise to that signal or that resonance.0404

    The next thing that we will look at is the chemical shift: and that is asking the question, "Where on this range is the peak occurring?"0417

    OK, so you see our numbers here are kind of 0 to 10 or so, and the location on this spectrum tells us a lot about the structure, as well.0425

    These are called δ values; they are given in parts per million, and the location tells us something about the electronic environment of it (is it a very electron-rich environment?--a very electron-poor environment?--and so on).0433

    OK, and finally, we are going to look very closely at the peaks and the shapes of those peaks to identify what is known as a splitting pattern.0446

    Is it a single peak?--is it split up into 2 or 3 peaks?--that is going to tell us something about the neighboring hydrogens on the structure.0454

    So you will see, we are going to be able to put our pieces together, depending on these splitting patterns.0462

    OK, and we will explore each of these, one at a time.0468

    Let's start with the number of signals in an NMR: this is taking a look at something known as chemical equivalence, and that is where we decide which hydrogens are similar to other hydrogens.0473

    OK, so for example, if we were to ask how many signals you would expect in the proton NMR--we take a look at propane, and propane has 8 hydrogens.0484

    OK, but there are not going to be 8 signals in the NMR, because not all of these hydrogens are unique.0493

    For example, all three hydrogens on this methyl group (this CH3) are chemically equivalent, because you could simply rotate that tetrahedral carbon, and each hydrogen could take the place of the other; there is no way to distinguish between any of those three.0499

    We describe them as chemically equivalent, and all of those would give just a single peak in the NMR.0514

    OK, this CH2, though--these two hydrogens--the two hydrogens are identical to one another, but these are in a unique chemical environment from the CH3 group; so this would give a different signal.0522

    And so, we can label these as type b protons.0536

    OK, and how about this last CH3?--would this be another unique signal?0540

    Well, one thing we can do in looking for chemical equivalents is: we can look for symmetry in a molecule; and you can see that, in propane, this methyl and this methyl are chemically equivalent, because there is no way to distinguish between the two, right?0545

    If we just flipped the molecule over, you wouldn't know which methyl group was which.0556

    So, in fact, these are the same types of protons as in the first methyl; and so, we would label those as type a, as well.0559

    So, if we were to take the NMR of propane, we would expect to find two signals, or two resonances, in the NMR spectrum.0567

    OK, let's look at a more complex molecule.0577

    OK, on this first carbon, do we have any hydrogens?--remember, this is a proton NMR, so we are looking for the hydrogens in this structure (in this case, we need to figure out what our line drawing means, right?).0579

    Well, this line ending in space is a CH3, so all three of those hydrogens would give one signal.0592

    How about on this carbonyl--are there any hydrogens here?0600

    There are no hydrogens; all four bonds are something other than hydrogen here; OK, this carbon--does that have any hydrogens?0603

    It does: there is a hydrogen right here; we can draw that in there, so we can see it.0611

    We can call that type b, and this methyl...now, is this methyl, this CH3, identical to this CH3, or do you think it would be unique?0615

    Now, one test you can have for this is: you can imagine adding an atom, replacing the hydrogen here and replacing the hydrogen here, and looking at those two resulting structures.0630

    If those two resulting structures are identical, then that means that those two protons are in fact chemically equivalent.0641

    But no, this CH3 is attached to a carbonyl; this one is attached to a CH3; they would give unique structures; and so, this is a different kind.0648

    How many other unique protons do we have here?--does this carbon have any?--this does; right here, there is a hydrogen.0657

    So, that would be a unique one; this hydrogen is attached to two methyls; this hydrogen has just one methyl attached; so these are not chemically equivalent.0665

    OK, this last CH3 is going to be unlike any others that we have seen there, so that is e; and then, how about this CH3 up here--do you think that is f, or is that equivalent to something we have already drawn?0674

    Imagine these two methyl groups, if we were to rotate around this carbon-carbon bond; we would be able to swap those two methyls; and so, really, there is no way to distinguish between them.0691

    If I replace this one or replace this one, I would end up with almost the same compounds; we will talk about that slight difference in just a minute.0702

    And so...I'm sorry, excuse me: they are chemically equivalent, so this would also be type e.0714

    All 6 of those protons would give rise to a single signal; so we have 1, 2, 3, 4, 5; this would have 5 signals in the NMR; this is called chemical equivalence.0722

    Let's look at a few more.0734

    OK, how would we treat this benzene ring?--well, let's start over here.0738

    This CH3 is one kind; this CH2 is unique; and the next CH2 would not be the same as the first CH2, because this is not directly attached to an oxygen.0743

    OK, so those are three unique signals.0756

    This carbon has no hydrogens on there; but here is a hydrogen.0761

    In fact, on this aromatic ring--on this benzene ring--we have these five hydrogens; OK.0766

    So, this one we could call type d, because it is clearly unique from the other three; but how about this next one--is that type d--would this be equivalent to the first one?0773

    No, because this one is ortho--is right next door to the substituent, where this one is meta, or 1,3 to the substituent; so this is unique kind e, and in fact, this last one is also unique, because it is para, or 1,4 (across from the substituent).0785

    OK, but how about these last two aromatic protons--do we have some chemical equivalents here?0801

    Well, sure, because we have some molecular symmetry again; so this ortho proton is the same as this one, so this is also d, and this meta one is also e.0805

    So again, looking for molecular symmetry is going to be very important here.0817

    This has 1, 2, 3, 4, 5, 6 signals that we expect in the NMR.0821

    OK, how about our next one?--now, this one is a pretty symmetrical molecule; we have symmetry right here, and we have symmetry right here; and in fact, we also have the plane of the molecule as offering a plane of symmetry.0828

    So, if we imagine this CH2 one up and one down, these two would be identical to one another, because they are symmetrical.0845

    And so, these would be, let's say, type a; but then, those would also be type a over here; and because of this symmetry, these are also going to be type a, and these last two are going to be type a.0855

    So, this whole molecule is going to have just one signal in the NMR, because it is such a symmetrical molecule.0873

    Now, if you think very closely about replacing each of these hydrogens, one at a time, you would find that the two structures that you get--let's say we put a chlorine here and a chlorine here (we can even do that, so that you are not trying to imagine this, because this is an important topic), let's put a chlorine here, and let's put a chlorine here; OK, are those identical compounds?0883

    Those are not identical compounds.0914

    What is the relationship between the two of them?--they do have a relationship.0917

    If you...it looks like I have a chiral center here; I have inverted the stereochemistry of that chiral center; and if you flipped this over, maybe you could see the mirror image relationship between the two.0922

    These are enantiomers.0933

    It is not entirely correct to say that these two hydrogens are exactly equivalent; we describe them as being enantiotopic.0938

    They are enantiotopic because replacing one or the other leads to the production of a pair of enantiomers; but it turns out, in the NMR, that we cannot distinguish between enantiotopic protons.0949

    So, even in this case, they are, in fact, chemically equivalent, so they would all still give rise to one signal.0962

    OK, but this, now--when we come to this last structure, we see a case where it's a little more complicated; and thinking about that stereochemistry, let's first start with this carbon.0971

    This carbon--any hydrogens there?--there are no hydrogens here, but this is a CH3 and a CH3 and a CH3; and what do you think about those three CH3s--would they be equivalent?0985

    Sure, they are a t-butyl group; we could rotate around all three of those, and so these are all going to be one signal; we will call that type a.1000

    OK, when we come here, is there a proton here?--yes, it's a tetrahedral carbon; there is a hydrogen here; and so let's actually just look at one--let's show the stereochemistry, here, for this being a tetrahedral carbon.1012

    Let's think about that stereochemistry, because that is going to be relevant, in this case, when we are trying to determine equivalence.1028

    OK, this is unique; we will call that proton b; OK, and what is important now is: when we look over here, and we consider these two hydrogens, even though it's a CH2, just like when it was a CH2 up here, we said both of those hydrogens count as a single signal--will give rise to the same signal--because they are chemically equivalent.1034

    But that is not true for all CH2s; that is not true for all hydrogens that are on the same carbon; and this is a perfect example of that.1053

    If we were to place this wedged hydrogen with, let's say, a chlorine, and we showed this as a wedge; and we replaced the other hydrogen with a chlorine...1061

    I'm sorry; this is still a wedge--excuse me: if we just replaced the dashed one or the wedged one, we would get these two structures; those two structures are not identical; those two structures are not enantiomers; what is the relationship between those two structures?1080

    If you keep one chiral center the same, and you invert the other chiral center, we describe these as being diastereomers.1094

    So, the hydrogens on this CH2 are described as being diastereotopic.1104

    And diastereotopic protons do lead to different resonances in the NMR: these would give different signals.1114

    OK, and you might be able to see it, even without thinking of and learning about that terminology: you might be able to see how these are unique, because this hydrogen is cis to the t-butyl group; this hydrogen is trans to the t-butyl group.1124

    So, they do not have identical chemical environments.1135

    Let's call this proton c, and let's call this proton d; they are going to be unique.1140

    There is no plane of symmetry in this case; there is no plane of symmetry that reflects one molecule, one hydrogen, onto the other; and that is why they are not chemically equivalent; they are unique.1144

    OK, so we can think about that on our next CH2--we would still make that distinction.1157

    One hydrogen is cis to the t-butyl; one is trans to the t-butyl; so those are going to be unique, e and f.1163

    And how about on this last CH2?--let's think about the dash and wedge and keep them separate.1171

    The difference here is that we are dealing with something that has a chiral or prochiral carbon--something that has something to distinguish...takes away a plane of symmetry--takes away some of the symmetry from the molecule.1178

    That is what makes this molecule unique from the ones we saw before.1192

    OK, well, these two hydrogens are still unique, so this is g and h; but now we are coming across something--the rest of the molecule does have some symmetry, because this molecule has a plane of symmetry going right here.1197

    And so, when we look at this CH2 and compare it to this CH2, what do you think we will find?1215

    When I compare this...the wedged hydrogen here--have we seen any of these before?--well, yes, that one would be just like this one.1221

    In this case, these are enantiotopic; replacement would lead to different enantiomers (you can try that to confirm that).1231

    But, for NMR, that means they are equivalent; so this is type c, and this is type d.1238

    And then, we look over here, and we will find that we come back to another pair of e and f.1245

    What do we get?--1, 2, 3, 4, 5, 6, 7, 8 signals for this one.1252

    You can see: sometimes symmetry...symmetry will always simplify a spectrum and make it have a lot fewer signals, because we have this chemical equivalent.1260

    That is the first thing we are looking for, and we want to analyze and understand about the NMR: that each unique signal is telling us about a unique type of proton that is in the molecule.1272

    The next thing that we will be concerned with is the size of the signal: because each signal might have more than one hydrogen contributing to it, it turns out that that size is going to be corresponding to the number of hydrogens in the signal.1284

    What we are going to be measuring is the area under the peak; luckily, the computer is doing that for us.1301

    It will determine what the area is under the peak (how big the peak is, in other words); it turns out that that is proportional to the number of hydrogens.1307

    It is going to be given as a ratio; so all that the computer knows is that this peak is twice as big as this peak, for example.1315

    It doesn't know the exact number of hydrogens that are giving rise to it.1321

    And there are two ways that we are going to be given this integration data.1326

    One way--the simplest way--is: you will simply be told that this signal has one hydrogen contributing to it; this has one hydrogen; this has two hydrogens; three hydrogens; three hydrogens.1331

    So a lot of times, with NMR, you are just going to be given those numbers, and you can run with it.1342

    OK, and what this is saying: the reason you need to be given this data is: this is not something we can analyze visually, because all of the various peaks might have different shapes to them.1347

    So, this one--both of these are three hydrogens, but because this is split into two peaks and this is split into three peaks, this looks a little taller.1357

    OK, so what we are looking for is the mathematical computation of the area under the peaks.1365

    OK, sometimes we are given the numbers, like this; sometimes we are just given these--these are known as integral trails.1370

    This is another way that the computer can just tell us this data; and what it is going to do is: it is going to give a trace that corresponds to the area under the peak.1382

    All we are looking at is: we are looking at the height of where the pen starts--what level the trail starts at at the start of the signal--and where it ends up at the end of the signal.1394

    We are simply looking at the height of the integral trail, and how far it went up as a result of that signal.1408

    OK, and what you can see is: that distance matches this distance.1416

    You will notice that this is the same spectrum, just shown two different ways; and that is because these are both one hydrogen.1423

    OK, when we come over here, sometimes you will see that one integral trail kind of connects to another, and that is simply because the pen didn't have time to reset back to the baseline.1433

    That is OK; we just simply, again, always look at where the trail starts and where it stops.1442

    Where does it start and where does it stop?--simply the height is all we care about--where does it start for this signal?--it starts here and ends here.1449

    OK, and what we would need to do is: we would need to evaluate this and say, "Well, if this height is one inch"...I'm sorry, "If this height corresponds to one hydrogen, then this height must correspond to two hydrogens, and this height must correspond to three hydrogens."1458

    OK, now you can actually get out a ruler and measure it and kind of do the calculations, but it is really the easiest thing--what I do is: I just grab a piece of scrap paper and a pen, and I line it up next to the peak, and I just make that height.1481

    I say, "OK, if this is one hydrogen, then this is two hydrogens; this is three hydrogens"; I make myself a little scale, and then I could use that to very easily line up to the other peaks and determine how high they are.1494

    OK, but sometimes you will be given these integral trails, and you will need to know how to convert that into a ratio.1505

    Let's get a little practice with that.1514

    OK, if you are given integral trails, what you need to do is just look at the height of each peak; so again, in this case, for this peak, our height starts here and ends here; so that is what I care about.1517

    Here it starts here and ends here, and here it starts here and ends here--OK, something like that.1529

    One way to get started is to assume that the smallest peak is just one hydrogen; that is not always the case, but if we did that, this is our smallest peak, and so let's assume that that is one hydrogen.1537

    And then, we could use that as our measure to see how many hydrogens are in each signal.1549

    So, here we can look, and we can see that this one is about twice as big, and then this one--if we break it up and break it up, we kind of get...let's see, this is twice as big and twice as big; it's hard to do without my piece of paper here, but that might be 2, and 4, and 6.1555

    So, this works out nicely to be 6 hydrogens; OK, so we can figure that out ourselves.1580

    But in the end, we need to make sure that we adjust this as needed; and of course, we can only have whole numbers; there is no such thing as a half of a hydrogen, so we will see, sometimes, that we need to adjust that.1585

    And you also want to account for all the hydrogens in the formula; in this case, the formula for the molecule is C8H18O.1597

    And how many hydrogens have we shown here?--6, 7, 8, 9.1606

    So, what that means--how would we have this represent 18 hydrogens?--it means that we need to double every one of these numbers.1610

    The ratio stays the same, but this, in fact, is not just one hydrogen--it's two; this is four hydrogens; and this is twelve hydrogens.1618

    Now, we know how many hydrogens there are giving rise to each signal, and now we can begin to come up with the structure.1628

    OK, let's look at a few more examples.1638

    C3H8O; OK, so again, we start at the bottom (look for each signal); we start at the bottom and go to the top.1640

    That is for that signal, bottom to top; bottom to top; so those are our three distances we need to evaluate.1648

    And, if we assume that this is one hydrogen, when we go to line this up, it kind of comes out here...we would not get to 2; this would kind of be one and a half hydrogen.1657

    So, assuming that your smallest peak is one hydrogen is good to start, but that doesn't always work.1667

    Maybe this isn't one hydrogen; maybe this is actually two hydrogens.1672

    Now, if I look it that way and cut this in half and say, "OK, if this is 2..." now it works out quite nicely that this is 3, and this is also 3.1677

    We add up 2, 3, 6, 7, 8; we add up to the number that is required.1688

    OK, so sometimes the smallest peak might be two; it might be 3; it might be 4; it depends, but at least start with one and make adjustments from there.1693

    OK, this one is kind of tricky--how about C7H8?1702

    We have 8 hydrogens that we need to distribute over these two peaks: here is one big peak and one smaller peak.1705

    Now, that really can't be one hydrogen or even two hydrogens, but they are not equal, so we can't have four and four; so here, working it out and maybe envisioning it, we can say, "Well, this must be three, and this must be five," and that would give us a reasonable way to distribute the eight hydrogens over those two signals.1714

    OK, so get some practice with this; try it with some scrap paper; you shouldn't have to use a ruler and calculators--that is just a huge waste of time.1734

    You want to be able to do this pretty quickly.1741

    OK, the next thing we will take a look at is the chemical shift: the chemical shift is where on the spectrum our peak appears.1747

    OK, these are described as δ values (we use the Greek letter δ); it is given in parts per million (ppm), and so what that means is, regardless of which type of instrument you use (whether it's a 60 megahertz or a 500 megahertz), it is still going to show up in the spectrum at the exact same location, because we are describing this in terms of parts per million.1757

    This means our spectra will be consistent, no matter what kind of frequency we use.1779

    We'll talk about our reference: our reference at 0--we use a molecule called TMS to define our 0 point, so everything is with TMS as a reference.1785

    I'll show you that structure in just a moment.1795

    OK, and when we have a low number, like something that is closer to the right here--close to 0 ppm--we describe that as being "upfield," and when we have a high number, we describe that as being "downfield."1797

    Now, this always seemed counterintuitive to me, and it is, because our higher number means it's lower field.1810

    And so, what always helped me as a student is: I would imagine this little ramp underneath the NMR.1819

    This really helped me just wrap my head around this terminology; so now, when you see...as you are moving from left to right, you are moving uphill; you are moving upfield.1825

    And, as you are moving down to the left, you are going downfield.1836

    This terminology isn't always used, but a lot of times you will see it introduced in terms of things shifting upfield or downfield, and you want to be able to fairly quickly understand the meaning of that.1842

    Now, we will find that the protons that come further to the right (further upfield), we describe as being shielded.1854

    And the shielding is occurring by electrons; so things that are in electron rich environments are going to be occurring far to the right, with very low ppm's, further upfield.1865

    OK, the things that are further downfield, we describe as being deshielded; and that means there are not many electrons around; these are going to be regions that are electron poor.1880

    OK, now let's take a look at the structure of tetramethylsilane.1895

    This is a silicon compound; we don't see or work too many of those; it has four methyl groups on there.1898

    Now, silicon is right below carbon on the periodic table; so, just like carbon, it likes to have four bonds--this is a nice, happy, neutral silicon.1903

    OK, and if you think about these protons and whether or not they would be electron rich or electron poor, what we want to remember is the trends in electronegativity.1912

    Remember that fluorine is the most electronegative atom; so as you move up a row, you increase in electronegativity--like moving closer toward fluorine.1924

    So, it turns out that carbon is actually more electronegative than silicon; and so, each of these bonds--each of those electrons shared in that covalent bond--are being drawn over to the carbon, away from the silicon.1934

    And so, what does that do to these protons in the methyl groups?--it makes them very electron rich.1951

    We describe them as being shielded by those electrons; those are very, very shielded, and very likely, the most shielded electrons we will ever come across in an organic molecule.1957

    And so, they are so far to the right that we set that as 0; and that was very convenient, because then almost every other structure we are looking at ends up being to the left of that.1969

    OK, it doesn't mean you can't have negative numbers; it is possible to have a molecule with protons that are even more shielded, more electron rich, than TMS; it is just very rare.1979

    Most of our numbers that we're going to find are going to come in this range, somewhere between 0 and 10; so they are really easy numbers to work with.1991

    OK, so let's take a look at some of those ranges.2002

    How can we decide where on the spectrum our peaks are going to show up?2005

    This kind of gives you an overview of hydrogens that are attached to carbons--what will we find?2011

    Well, alkanes--so using the letter R (that just refers to some kind of a carbon chain)--an RH would just be a plain old alkane.2017

    An alkane is a fairly electron rich proton; it is pretty well shielded; these come pretty far upfield, pretty far to the right--somewhere around 1 ppm; so those are just where we would find our normal, ordinary alkanes.2026

    Let me skip over these just for a minute.2040

    If we were to attach something like an oxygen--an electronegative atom--to the carbon, we know that that oxygen would pull electron density toward itself; so what would that do to these hydrogens?2044

    It would make them less electron rich; it would have a deshielding effect, and therefore it shifts it downfield; we get a higher number somewhere closer to 4.2057

    OK, so attaching any electronegative atom would have that effect; this is going to be, now, deshielded.2068

    Now, if you have π electrons in your molecule, like you would for an alkene or a benzene ring--an aromatic ring--or maybe a carbonyl, the action of those π electrons when they are placed in a magnetic field is going to have an effect on the electronic environment.2077

    OK, so for example, if you take a look at an aromatic ring, you have these π electrons; and what happens when you place these in a magnetic field (let's say our magnetic field is here, pointing in the up direction)--these are going to start to circulate and create their own field.2092

    Their goal is to create a field that counteracts the external magnetic field.2115

    And it turns out that it just kind of circles around and makes this look here; and so, the hydrogens that are in this position end up being deshielded by this effect.2125

    This effect is called anisotropy, and this anisotropy causes the hydrogens that are on a benzene ring or an alkene to shift to the left--to shift downfield to higher numbers.2137

    So, aromatic protons--OK, the word "aromatic" means that the proton is on a compound like benzene; benzene is an example of an aromatic compound.2154

    It is not the only aromatic compound, but it is the one we are going to see most often.2165

    Aromatic protons show up very far downfield (somewhere around 7); alkene-type protons show up in the range of 5, 6, something like that.2169

    And because of this anisotropy effect, protons that are near π bonds (not directly attached to it, but next door--so if you are benzylic, meaning next to a benzene ring, or allylic, or α to a carbonyl)--these are all going to end up having a deshielding effect because of those π electrons.2179

    Those all, rather than showing up around 1 like a normal alkane--like a plain alkane--putting it next to those groups shifts it down closer to 2.2204

    OK, now the good news is: you are not going to have to memorize all of these numbers.2214

    As you work with them, you will become more and more familiar with them; but you will be given tables, and you will be able to work with those tables as you are solving NMR problems.2216

    Let's take a look at a sample: OK, so these are various protons on carbon, and they are typical chemical shift; they are typical δ value in parts per million.2224

    And so, again, what did we just see?--well, plain old alkyl protons are somewhere around 1; as it turns out, there is a little difference if it's just a methyl group of CH3; that is the most electron rich--that comes the furthest upfield.2234

    But, compared to a methylene, a CH2, or a methyne, a CH...those shift a little further downfield.2247

    Being allylic or α to a carbonyl or benzylic (meaning next to a phenyl group)--those all shift because of anisotropic effects; those all shift down closer to 2, 2.2, somewhere around there.2255

    Alkynal protons come around 2.5; again, the triple bond has an anisotropic effect as well; that looks a little different.2271

    But that also explains this deshielding effect.2281

    If we attach any kind of electronegative atom (like an oxygen or a halogen or a nitrogen), we see that that shifts it downfield to numbers 2 to 4--somewhere around there, depending on exactly which heteroatom we are dealing with.2287

    OK, being vinyl means that it is one a carbon-carbon double bond; that brings it somewhere around 5; being aryl, meaning on a benzene ring or any other aromatic rings, brings this around 7.2303

    And then, an aldehyde is a very special proton; that not only has anisotropic effects, but it also has the electron withdrawing effects of the carbonyl; so this is very deshielded, very electron poor.2314

    And so, this comes all the way up near 10; so it's a very, very far-to-the-left, very downfield proton.2325

    OK, so these are the kinds of ranges--the kinds of functional groups--that are going to affect the chemical shift of a hydrogen.2332

    Now, if we have a hydrogen on something other than carbon (we can have NH groups; we can have OH groups, for example), these, we are going to find, are quite variable.2342

    In other words, there is a wide range of chemical shifts that can be expected.2351

    Now, why is that?--the range is going to depend on things like the solvent we use to run the NMR, what the concentration is, what the temperature is...all of those things are going to have an effect on exactly where it's going to show up on a given day.2356

    OK, so these numbers are not set in stone; and that is because these can all hydrogen bond, and therefore they are exchangeable protons.2370

    They can leave one oxygen and move to another oxygen; that reactivity--that hydrogen bonding--is going to have an effect when we observe that proton on an NMR.2378

    OK, one thing it does--not only does it vary the range in which it's going to show up, but it also ends up being kind of a broad peak.2388

    So, instead of having something that is a nice, sharp peak or something, we will see that it is split into several peaks.2395

    You will usually get kind of a broad peak--maybe even a little lump, depending on just what the specific concentrations are.2402

    But these look pretty unique in the NMR, and we can usually pick them out that way.2409

    So, the NH ranges...again, having an amine is the most shielded; having an aniline derivative is going to be a little more deshielded, because of those anisotropic effects; having an amide is going to be, again, a big range, but pretty high numbers--pretty far downfield.2415

    Alcohols are somewhere 1 to 5; phenols are somewhere closer to 5 to 7; and look at this carboxylic acid, way down here over 10.2433

    So, most things show up between about 0 and 10; but carboxylic acids are unique--those can come at 10 or 11 or 13 or really high numbers.2442

    Again, you will be able to find all of this data in some kind of table that shows the difference between an alcohol and a phenol--whether the OH is on a plain old tetrahedral carbon or a benzene ring.2451

    Carboxylic acids are our highest number.2465

    And then amines versus anilines and amides and where they are going to show up...2467

    We can use some tables like that to estimate our chemical shift; we can look at a structure and kind of figure out what functional groups we have around.2475

    OK, we could start at this end: this is a plain old alkyl proton, and just-plain-old-alkyl-protons show up at about 1 ppm--somewhere around there--a little higher or a little lower...2487

    This next one--the same thing: about 1 ppm; maybe slightly downfield, because it's a CH2 compared to a CH3.2503

    This one is about 1 ppm; OK, all of those are plain old alkyl groups; so you see, if all you have are alkyl chains, your NMR becomes pretty messy, because all of these peaks are very similar in chemical shift, and so a lot of times, we get overlapping peaks.2510

    Sometimes, they are not nicely resolved, where you can see one peak before another peak starts up.2525

    So, a lot of times, our alkyl groups--if we have large alkyl chains, you will just get a giant mess in the area of 1 to 2 ppm, and you won't be able to get much data from that.2530

    OK, but this one we would describe as being benzylic: this CH2 is benzylic, meaning it's next to this benzene ring; and that is one of those groups we see shifting downfield to about 2.2.2542

    Now, these aromatic protons--we saw how, when we were predicting how many peaks to expect in an NMR, we said, "Well, actually, these ortho protons should be different from the meta and the para."2556

    We would expect three different peaks here; but because they are all so similar in the chemical environment, a lot of times their chemical shift is going to be coincidentally the same.2569

    And so, rather than seeing three unique peaks, a lot of times they all show up as one peak; but we are going to find them around 7 ppm.2581

    We will see some cases where the aromatic protons are distinguishable, and do come at significantly different peaks, but sometimes we will just see them all as--again, kind of a big lump around 7--at or near 7.2588

    OK, how about this next one?--we have this proton here, just attached to another plain old carbon group; so this is at about 1.2603

    OK, in reality, because the next atom over is an oxygen, and we know that oxygen pulls electron density away, this probably will be a little downfield--a little higher number than 1; but we expect to find it somewhere in that region.2614

    OK, right now, we are just making very, very general estimates.2630

    OK, this carbon is going to be significantly shifted downfield--significantly deshielded--because it's directly attached to the oxygen.2633

    So, when we look that up on our table, we see that, when it's directly attached to an oxygen, that brings us somewhere around 3.8--a really big shift downfield.2642

    CH's directly attached to oxygens are very easy to pick out on the NMR, because they are so unique.2652

    This carbonyl doesn't show up on the proton, because we are still looking at proton NMR; but we will see evidence of it, because this CH2 is α to that carbonyl; we call these α protons when they are α to a carbonyl.2659

    And again, that shifts it kind of about the same as a benzene ring does; these benzylic protons show up around 2.2, and those α-to-a-carbonyls show up around 2.2, as well.2679

    OK, and what do you think for this one--what does it have attached?--just a plain old carbon on this side, but it has an oxygen here.2691

    Any CH attached to an oxygen is going to bringing it way up to 3.8 again--somewhere around 3.8--OK.2699

    So, we can use that table...if all we have for each carbon is a single functional group, then we can use a table like that to estimate approximately where on the NMR spectrum our peak is going to show up.2707

    OK, but what if we have more than one functional group (which is the case in many organic molecules)--how would we begin to predict those chemical shifts?2720

    Well, it turns out that those effects are additive; OK, so if we look at an alkane, and we say it's somewhere around 1 (in fact, methane comes below 1--very far upfield)--OK, if this has about 1, and we add in a chlorine (that chlorine is electronegative, so it withdraws electron density; it has a deshielding effect), we see that it shifts somewhere to around 3.2729

    OK, we see that this adds about 2 ppm by having a chlorine on here.2751

    So, guess what happens when we have two chlorines: both of those chlorines pull electron density away from these hydrogens, and so this peak is going to come even further downfield; and we see that it increases about another ppm.2756

    So, each chlorine is causing that proton to have about a 2 ppm shift down.2771

    If you had 3 (like we do here in chloroform), 3 chlorines pulling electron density...we add another 2 ppm, and we bring it all the way into the aromatic region.2778

    Normally, when you see a peak at around 7, you think, "Hey, that might be a proton on a benzene ring," but in fact, if you have the right combination of other functional groups, you can bring those protons all the way down into that range, as well.2790

    Another example: here we have an ordinary alkane, somewhere around 2; this is a 2-hydrogen signal.2804

    That is because it is a CH2 here (right?--this is a CH2), so we'll have a 2-hydrogen signal at about 1 ppm (1.2 in this case).2811

    When we add a carbonyl, that brings it to 2.4; and when we add two carbonyls, it brings us to about 3.6.2820

    So again, we can see a shift: this added about 1.2 ppm; this added about 1.2 ppm.2829

    Each carbonyl is worth that much for a shift.2837

    It turns out that these chemical effects (the shielding or deshielding effects of various functional groups) can be quantified and can be estimated, and we can even calculate where we expect a proton to be, using some more sophisticated tables that have that information on it.2840

    Let's see an example of that.2857

    If we have a CH2, that means there are 2 groups attached: if there is just one functional group there, then we could use a table to estimate it, but if we have two functional groups, we can use this formula to determine about where that signal is going to be.2860

    We start at 1.2 ppm, because that is where a normal CH2 would show up (a plain old alkyl CH2), and then we add two numbers to it: one for group 1, and another one for group 2.2880

    We are going to get these δ values; we are going to get those adjustment numbers from a table.2894

    OK, if we have a CH, a methyne, we now have three groups attached that are affecting the chemical shift; here we had just two.2900

    And so, when we have three groups attached, we need to add three numbers to adjust: we need to adjust it with all three of those, and we are also going to start at a different place, because methynes, remember, usually come a little further downfield than methylenes.2909

    So here, we will start at 1.5, and we will add three numbers to it; here we start at 1.2; we add two numbers to it.2923

    Now, where do we get the δ values?--again, these come from experimentation to develop these tables.2931

    If we have just a plain old alkyl group attached, that has no effect on the chemical shift; it is neither shielding nor deshielding, and so there is no effect.2940

    But if it's allylic, next to a double bond, or propargylic, or α to a cyano (propargylic means it's next to a triple bond)--you can see those all add about one ppm shifted down.2948

    Benzylic, α to a carbonyl, or α to an ester carbonyl--so all of those kind of shift it about 1 ppm.2960

    Depending on the type of carbonyl (whether it's an ordinary ketone or aldehyde, or maybe an aromatic ketone or aldehyde), we have slightly different things; so you can get pretty nuanced, pretty specific, in the effects on the chemical shift.2969

    The different halogens have different effects, of course, depending on electronegativity: fluorine is the most electronegative, so it is the most deshielding; it causes it to move furthest downfield to the highest number.2985

    Being attached to oxygen always has a big effect: look at these numbers we are adding--2 or 3 to our original number--so a huge effect.2996

    But the type of oxygen makes a different effect; so if it's an alcohol versus an ether versus an aromatic ether, those have slightly different numbers compared to an ester or an aromatic ester.3004

    OK, and the nitrogens have effects--amines, nitros, thiols, and so on.3018

    OK, this is just a sampling, but you can get these really detailed tables from NMR books or some textbooks.3023

    Let's try an example: I have reprinted a portion of that table so we can use that in this case.3031

    Let's say we wanted to estimate the chemical shift of this proton here--the one that is in bold.3037

    OK, now we see that it has three groups attached to the carbon: we have this one; we can call that R1--that is our first group; we have this one--we'll call that R2; and we have this one--we'll call that R3.3042

    The formula we are going to use is that our chemical shift in parts per million starts at about 1.5 and then shifts according to the values for R1 and R2 and R3.3059

    It is 1.5, plus...now, how would you describe R1?--it's a benzene ring, and so we are going to call that an aromatic ring, like a phenyl.3070

    So, we will call this Ar for aromatic; or we could put this down here.3083

    And that number for a benzene ring is 1.4--just so you can kind of use the table to see where these numbers are coming from; it's an Ar group that is just meant to be a bond.3091

    An Ar group--an aromatic group, like benzene--is attached to that position; OK.3101

    How about R2--how would you describe that?3107

    It's just a methyl group; it's just a plain old alkyl group; so where does that show up on our table?--right here; just a plain old tetrahedral carbon; that is alkyl.3111

    And so, that has no effect on our chemical shift.3124

    And then, finally, R3 is an oxygen of some kind; which oxygen is it, though?--that is the detail we want to get to.3128

    It has an oxygen and then a carbonyl; so it's an ester, and it just has a plain old methyl group here; so it's just an R group attached to the ester, not an aromatic.3139

    It would be this one...O, C, O, R...oxygen of an ester, which adds 2.8.3151

    So, when we do the math, 1.5 plus 1.4 plus 2.8, we end up with 5.7.3161

    We can estimate that the peak should show up somewhere around 5.7, and these calculations can be, really, quite reliable.3168

    But again, they are estimates; so it is OK if it's off a little bit; but it means we are not going to expect it at 4; we are not going to expect this at 7; if it's that far off, we have done something with our structure, and we have an incorrect structure.3175

    OK, one more thing to talk about for chemical shift is how resonance might affect that chemical shift.3194

    If we take a look at this proton, it's a vinyl proton, so we expect it to show up somewhere around 5; OK, and just like we did the tables for an alkyl proton that is on a CH2 or a CH, you can look up those same tables for protons that are on double bonds, or protons that are on benzene rings.3201

    OK, and what you will find is: the groups attached to the double bond are going to have very dramatic effects--can have very dramatic effects--on this proton, either in a shielding fashion or a deshielding fashion.3221

    OK, and so, for example, in this case, it's attached to a double bond, so it's a 5.4.3235

    This is still attached to a double bond, but all of a sudden it shot downfield to 6.7.3240

    That is a really significant change; even though there is nothing attached to this carbon that is different, all the way over here we have a new group attached.3245

    OK, but the addition of this carbonyl is significant, because with that addition comes some resonance delocalization of those π electrons; this is an electron withdrawing group, and as an electron withdrawing group, it pulls electrons toward itself.3254

    And we now have a resonance form for this molecule.3269

    Now, we know, with resonance, that the actual structure is a blend of all of these different resonance forms; so when you think about that blend, that actual, true hybrid, what do you know about this carbon?3279

    It has some partial positive character; and so this is extremely electron deficient.3290

    This proton is in a very electron deficient environment, meaning it is deshielded--it doesn't have electrons around to shield it.3296

    What does that do to its chemical shift?--it shifts it downfield.3304

    We expect a higher number in this case.3307

    OK, here is an example where these two hydrogens on the same molecule--this hydrogen comes at 4.6 ppm; this hydrogen comes at 6.3 ppm.3311

    Interesting, because we expect alkene protons, vinyl protons, to come somewhere around 5; this one is shielded a little upfield; this one is deshielded a little downfield; what is going on here?3322

    Well, this one shifted downfield because of inductive effects, being next to that oxygen; that is going to shift it to be deshielded toward this one.3332

    OK, but the oxygen on a double bond is an electron donating group, and it also has resonance; that resonance looks like this--any time we have a lone pair next to a double bond, you can have resonance.3347

    That brings the lone pair in--that allylic lone pair--and you can draw an additional resonance form.3363

    So, what does that look like?--this now has 1, 2, 3, 4, 5; oxygen wants 6; this is an O+.3372

    This carbon has 1, 2, 3, 4, 5; carbon wants 4, so this is a C-.3388

    This resonance form is contributing to the structure; so what can you tell me about the environment of this hydrogen?3394

    It is on a carbon that has some partial negative character; so this is very electron rich, and therefore, it is very shielded, and we expect it to occur higher field, in a lower number, and it does get shifted to 4.6.3400

    Resonance is something that can really explain why a proton might be in a more electron rich or electron deficient environment.3420

    OK, another example are these two aromatic compounds.3430

    This is toluene--with toluene, we find that all these protons all show up--all 5 hydrogens show up--around 7 ppm.3433

    There is nothing interesting to distinguish one from the other, even though this one is ortho to a methyl, and meta, and para.3441

    The methyl isn't doing anything dramatic with the ring, and therefore they all have essentially the same electronic environment, and they all kind of show up at the exact same spot.3450

    OK, however, if you were to put an electron donating group (kind of like we had with an oxygen here)--if you put a nitrogen on that ring, then this has resonance, just like we saw up above, which puts a negative charge here.3460

    Any other resonance forms we can have for this compound?--let's look for those allylic lone pairs.3478

    Here is a lone pair that is allylic; so this can come in, and we can have a third resonance form here; and any more?--yes, we are still allylic; and we can have, in fact, another resonance form.3484

    What does this tell us?--it tells us that the environment of these three protons (here and here and here)--those three protons are quite electron rich; what does that do to the parts per million?3505

    If we are expecting it to start somewhere around 7, what do you do if you have extra electron density and you are more shielded?3519

    That is going to push it to the right; it's going to push it upfield to a lower number, and I think I have the data here...yes.3527

    These are at 7.2 ppm, and these are at about 6.6 ppm.3535

    It is, in fact, shielded; it is going to be shifted a little closer to 0--a little further to the right.3549

    OK, the last thing to consider is the shape of the NMR signal.3559

    So far, we have talked about "How many signals are there?", "In chemical equivalents, how many unique hydrogens are there?"; we talked about how big the peaks are, meaning how many hydrogens are in each signal.3564

    We just finished talking about where on the spectrum they are going to reside; now, we are going to look at the signal itself and look at its shape.3576

    We are going to find some different shapes, like we see in this NMR.3582

    Sometimes, this has 1, 2, 3, 4, 5, 6, 7, 8, 9 signals; so that means this molecule has 9 unique types of protons in it.3587

    And some of the signals--this signal is just a single peak: it comes up; it comes down--nothing special about it; we describe that as being a singlet.3600

    A singlet means that it is a single peak.3612

    OK, but sometimes, the peaks are split into different shapes; in this case here, we have two peaks at the top.3615

    You can see, it goes up and down; and so this is described as a doublet, when we have two peaks.3624

    We have a few cases here where they are split into three peaks; when they are split into three peaks, we call that a triplet.3633

    Oh, here we had another doublet.3643

    When they are split into four peaks (like over here--1, 2, 3, 4), we call that a quartet; and so on.3649

    OK, we also have something here--1, 2, 3, 4, 5 peaks--that is called a quintet.3657

    And this one, even--you can barely see these outside peaks, but if you look very closely, it has 1, 2, 3, 4, 5, 6, 7; this is called a septet.3664

    And it can go on and on; OK, but we are going to see these different patterns, and we are going to have to understand what that tells us about the molecular structure.3673

    OK, and it turns out that the amount it is split (or not split) depends on how many neighboring hydrogens that proton has.3685

    OK, and we describe the relationship between the number of hydrogens and the number of peaks as the n+1 rule.3697

    If you have one peak (meaning it's a singlet--so the peaks we saw on the last slide--the one that we described as a singlet, having just one peak), that means that that proton has 0 neighboring hydrogens.3703

    If you look to the next carbons over--the next atoms over--they have no hydrogens attached to them.3717

    Anything that is split into two peaks (it's a doublet) has one neighboring hydrogen; a triplet has two; a quartet has three; and a quintet has four; etc.3722

    OK, so what we are seeing is: for n neighbors, you get n+1 peaks.3735

    That is known as the n+1 rule.3746

    This is going to tell us: when we look at this shape, it is going to tell us how many neighboring protons we have; and that is going to allow us to decide which pieces of the puzzle are attached to which other pieces of the puzzle.3748

    Now, what is going on here--why does this splitting occur?3763

    Let's take a look at this molecule: this has two different types of protons--these are type a, and this is type b--we have two hydrogens over here of type a and one of type b.3767

    OK, so let's take a look at proton a, at signal a, and see what it's going to look like.3780

    First of all, let's ask: how many neighbors does HA have?3788

    What it means to be a neighbor is: we look at the neighboring carbon, or the neighboring atoms, and we ask how many hydrogens are attached to it.3791

    We are usually looking at carbon, because those are the only ones that will have hydrogens that are going to be doing the splitting.3798

    So here, we have only one neighbor; HB is our neighbor.3804

    What is the splitting pattern for HA?--if you have one neighbor, you are going to have n+1 peaks; so there will be two peaks.3810

    We call two peaks a doublet, so the signal for HA is going to look like a doublet.3821

    Why does it become a doublet--what makes it a doublet?3831

    OK, well, remember that each of the protons in our molecule behaves like a tiny magnet.3833

    The nucleus inside has a +1/2 or -1/2 spin, and that little magnet is going to align with the field or against the field (the plus or the minus one-half); it can have...with the field is our α spin; against the field is our β spin; and so, those are the two possible configurations that this neighbor can take.3840

    If I am HA, there are two possible configurations of my neighbor; if HB is aligned with the field, then that adds to the field that is being experienced by HA; this is a deshielding effect.3867

    In order to resonate--in order for the spin to flip for HA--it is going to require a smaller external field than it would if HB didn't exist, because if HB didn't exist, it would need a certain energy to flip; when you put HB in and HB is aligned with the magnetic field, that makes it an even stronger magnetic field; so it's going to be easier to cause the flip--to do the resonance.3882

    OK, but in the case where HB is against the magnetic field, and it has the β spin state, then this is now going to be slightly shielded, because this is going to counteract that magnetic field.3913

    And so, I'm going to need even extra energy to be put in, in order to cause the flip.3923

    So, this is a shielding effect.3927

    In some molecules, you are going to have a neighbor with a slightly shielding effect; in other molecules, you are going to have a neighbor with a slightly deshielding effect.3930

    So, because HA is actually exposed to two magnetic fields, then it is going to be split into two signals.3937

    We take HA (which normally would just be a single signal), and by giving it a neighbor, we split that into two.3949

    The result, then, is what we call a doublet--we are going to get our peak split into two, and each of those is going to have an equal size, because it's just as likely that you are going to have a neighbor...that our one neighbor, HB, is going to have an α state or a β state.3960

    That is equally likely; so you get two peaks of equal size.3979

    Now, let's take a look at an example where we have two neighbors.3984

    OK, let's look at HB now and ask the same question: how many neighbors does HB have?3988

    Well, it has two neighbors, right?--it has two neighbors; each of them is called HA.3995

    What is the splitting pattern we expect?--when you have two neighbors, you are going to get n+1, or 2+1, so we are going to have three peaks.4002

    We are going to call that a triplet; so we expect HB to show up as a triplet.4015

    Now, how do we get these three signals--where do these three signals come from?4021

    The same exact principle is going on: you have: each HA nucleus is acting like a tiny magnet.4023

    But now, we have multiple neighbors--we have two neighbors; and so, what are the different combinations we could have of those spins?4032

    Well, they could both be aligned with the field (right?), or we could have the first one aligned with the field and the second one against the field (so these are HA's that we are looking at)--both HA's could be with the field, or one can be up and down.4039

    Or, we could have the first one down and the second one up; or they could be both down.4056

    Those are the only four possible combinations we can have for those neighbors.4061

    What effect does that have with respect to the overall magnetic field?4069

    Well, you see, in this case, those two are going to be adding to the external magnetic field; and therefore, we have this deshielding effect.4073

    It is going to be a little easier for this HB to resonate.4086

    If they are up and down, then that is going to have no effect on the external field.4093

    And whether the first one is up or the second one is up--it doesn't matter; those would have the same effect on the external field.4100

    These are going to be equal, and they have no effect; it is just going to be like we have the plain old B0, the plain old external magnetic field.4108

    And then, we have a third possibility, where they are both aligned against the field; and that is going to be our shielding effect.4118

    When we ask ourselves, in this case, for HB, "How many different magnetic fields is B experiencing?"--the answer would be 3 different fields: 1, 2, 3.4128

    And, because it is exposed to three magnetic fields, it is going to give three signals; we are going to see a triplet.4140

    OK, now we can do that same trick here for HB; we can imagine it just being a plain old signal; then we can say, "Well, it has this one neighbor, HA, and HA causes it to split."4149

    But then, it has a second neighbor, HA, and that HA causes the peaks to split again; but the amount of this first splitting is exactly the same as the amount of the second splitting.4166

    So, we end up having these two occur at the exact same position.4180

    And so, we are going to get three signals: we are going to get a triplet with equal spacing between the three peaks; but this middle peak is going to be twice the size of those outer peaks, because it is twice as likely--it has twice the probability that we are going to have a neighboring arrangement where we have a plus and a minus, an α and a β; it is going be twice as likely as having two αs or two βs.4185

    And so, our triplet is going to have--not only three peaks, but they are going to be a specific ratio; it is going to be 1:2:1; this is what a triplet looks like.4219

    OK, and we can go on and on of having four neighbors and three neighbors and five neighbors, and the same things happening every time.4229

    Every additional neighbor that you add gives us the two possibilities (the α and the β) and causes a splitting of all the peaks that used to be there.4236

    OK, what we end up with is very predictable splitting patterns that also have predictable ratios of the various peaks.4245

    So, if we have just one peak, we call it a singlet; sometimes, you just use the letter s to describe a singlet--s for singlet, d for doublet, t for triplet, q for quartet.4255

    So, a lot of times, those are not spelled out; you should know what those abbreviations mean.4265

    OK, so if we have a singlet, our peak is just whatever height our peak is--it just depends on how many protons we have, right?4270

    If you have a doublet, though, you are now going to have two peaks, but they are going to be equal in height.4277

    Sometimes they are a little staggered, a little skewered; they are a little skewed, not exactly 1:1; but they essentially have a 1:1 ratio.4281

    OK, when you have a triplet, what did we see?--that was three peaks, and they had a ratio of 1:2:1.4293

    We can predict...these ratios are simply a matter of statistics, so we can predict for a quartet--if you have four peaks, what is the ratio going to be?--it is going to be 1:3:3:1.4301

    And what is happening here is, we are seeing a pattern; this pattern is called Pascal's Triangle.4315

    What we are doing to get each number is: we are adding the two numbers above it to be the number here; so this...it always starts with a 1, and then this is a 3, and then this is a 3, just like adding these two together came up with 2.4322

    What would you expect for a quintet, if we had 5 signals with 4 neighbors?4335

    We would expect a ratio of 1:4:6:4:1; so, what is going to happen is: that middle peak, or those middle peaks, are always going to be equal in height.4341

    So, for a sextet, if we had 6 signals, we would end up with 1:5:10:10:5:1; so we are going to end up with these symmetrical patterns, where the top peaks are significantly higher than the outer peaks.4357

    OK, and we can continue on; but notice that our outer peaks always have their ratio of 1, and these keep growing exponentially on the inside.4375

    What happens, sometimes, is: if we have many, many peaks--a lot of splitting--those outside peaks sometimes disappear; sometimes, we can't even resolve; they aren't resolved--it depends on how sensitive our instrument is.4384

    So, sometimes, even if you don't see those outermost peaks, by looking at the ratios of the innermost peaks, you can decide what kind of a pattern you have.4398

    We will maybe see an example of something like that.4405

    OK, so let's come down to predicting some NMR; that is our first step; now that we have all of the basics of the components of an NMR, let's see if we can look at a structure and predict what its NMR is going to be.4411

    We'll start by labeling each unique hydrogen; so first, we are going to determine how many peaks we expect, and label each one as a, b, c, etc.4427

    OK; then we are going to predict, "Well, what is the splitting pattern for that peak--is it going to be a singlet or a doublet or a triplet (and so on)? And about what chemical shift are we expecting?"4435

    OK, so the first thing: in this molecule, how many different signals are we going to expect?4445

    Well, we will start with just one end here, this CH3; all three of those protons are going to give rise to one signal, because they are all chemically equivalent.4452

    Let's call that type a; are there any other type a protons?--this is maybe a good approach: look for symmetry, and look for any other type a.4460

    And sure enough, we see that these two methyls are attached to the same carbon; and so, this isopropyl group can rotate; we can interchange each of those CH3s, so those would be chemically equivalent; so this is also type a.4469

    OK, our next proton is here; this CH is unique, so this is b; and our last type of unique proton are the three hydrogens on this methyl.4484

    So, we expect to have three signals: a and b and c.4495

    What do you expect of the splitting pattern?--let's look at proton a now.4502

    OK, we can put a little note here that how many hydrogens...how big is that peak going to appear--how many hydrogens are going to be resonating wherever we determine signal a to be?4506

    We expect 6, because there are three from each methyl; so it's going to be a 6-hydrogen signal, and how do we decide the splitting pattern?4517

    Is it 6+1?--no; when we look at n+1 (remember, the splitting pattern is about the n+1 rule), the n represents the number of neighboring hydrogens.4527

    We look to the next carbon over: how many neighbors does it have?--just 1, so this signal a is going to be split into a doublet (or we could just put a "d," for short--we could abbreviate with "d" to be a doublet).4538

    And about what chemical shift do you expect?4557

    Now, we take a look at: is it an electron rich environment?--is it an electron deficient environment?4560

    Now, we look at what kind of functional groups are attached to these methyls; and here, we see, we just have a plain old carbon.4565

    OK, now there is an oxygen somewhere down the line; and in fact, when we do those calculations, if you want to have a very precise calculation, you not only look at the functional groups that are directly attached to that carbon; you look at the ones that are β and so on--that are further down the line, to get really precise.4572

    So yes, this is going to have an effect; but for our very rough calculations, we would just say, "You know what, this is a plain old alkyl group it's attached to; so where does that show up?--somewhere around 1 part per million."4590

    Just somewhere around 1--it's a plain old alkyl, nothing too special.4601

    OK, how about proton b--how many protons are going to be in that signal?4607

    It is going to be just a 1-hydrogen signal, so it is going to be a very small signal, and what is its splitting pattern going to be?4611

    Now, we look to the atoms attached: we have an oxygen, carbon, carbon; we ask how many hydrogens are attached to those, and we see that there are 3 here, 3 here; we have 6 hydrogens total, so this is going to be split into a (6+1, remember, so this is going to be a) septet.4618

    We are going to have seven peaks here--a lot of splitting there.4639

    And where would you expect that to be?--well, now we look at the carbon, and we ask, "What is that carbon attached to?"--nothing special; nothing special; here is an oxygen.4645

    Oxygen is one of the most ubiquitous elements we are going to see; it has a very dramatic effect on the chemical shift--brings it all the way down to about 3.8.4653

    So, let's put that as a note...maybe 3.5; it depends on the molecule as a whole--but somewhere around there.4664

    And then, finally, how about this methyl group--this type c proton?4674

    How many protons are going to be in that signal?--it is going to be a 3-hydrogen signal, so remember, we are either going to be given 6 and 1 and 3 attached to the peaks, or they are going to show integral trails with that ratio of 1:3:6.4678

    But that is how big that peak will be; and what will its splitting pattern be?4694

    We look at the next atom over; we ask how many hydrogens are attached; and we see there are none.4699

    So, if you have 0 hydrogens, what is your splitting pattern going to be?--n+1 means you have just one signal; it's going to be a singlet (or s for short, for singlet).4705

    So, it's going to be a 3-hydrogen singlet, and where is it going to be on your spectrum?4719

    It is attached to an oxygen; and so, we are going to expect that somewhere around 3.8 as well; so this is called a methoxy group.4724

    A methoxy group is going to be 3 hydrogens, so it's going to have the same general look, right?--it's going to be a 3-hydrogen singlet at around 3.8 ppm.4731

    This is our first step: being able to predict what a spectrum is going to look like.4740

    Let's see another example.4745

    OK, so here we can, again, decide how many unique protons we have; this would be type a; this would be type b; this would be type c; there are only three different protons here: a, and b, and c.4750

    The ratio of those peaks: a will be a 3-hydrogen signal, b a 2-hydrogen signal; c will be a 1-hydrogen signal.4774

    And how about our splitting patterns--any neighbors for our CH3?--we are next to a carbonyl; you have no neighbors here, so this is going to be a singlet.4786

    0 neighbors--remember, this is our n+1, where n is the number of neighboring hydrogens.4800

    b--how many neighbors do we have here?--again, on either side, no neighbors; this is another singlet.4806

    And, when you have an OH, it is always a singlet; remember, we get those broad signals, and because of its exchangeable nature, we do not observe splitting.4814

    Even if there were neighbors on this carbon, it typically does not split with that; you almost never see splitting.4824

    This is going to be a singlet--no splitting for NH or OH, typically; on rare occasions, you might see it, but almost never; you expect it to just be a singlet.4832

    So, this wouldn't split a neighbor, and the neighbor wouldn't split it.4848

    OK, so they are all singlets; and what about their parts per million?--our first type a proton here is α to a carbonyl, so that brings it somewhere around 2.2.4853

    We can take a look at a table to look that up--somewhere around 2.2, when you are α to a carbonyl.4864

    This one--we are α to two carbonyls, so this--we would want to calculate it; if we have at our disposal those tables, where we can calculate it, then that is great.4871

    If you are not given access to those, you need to be able to estimate them based on just a simple table; you will almost always be given a table to work with.4883

    OK, so what do you think we might have for it?4895

    If one carbonyl gave us 2.2, then a second carbonyl--what do you think that might bring it to?4899

    Remember, we started out at about 1, being an ordinary alkane; so this brought it up a little over 1; so this maybe brings it up another 1 or 2.4909

    So, maybe 3.4--bring it up another 1.2...something like that; OK, we can calculate it; I'll see if I can look up those numbers--I don't have them in front of me.4918

    OK, but somewhere around 3, let's say...and then, how about this OH?--remember, OH's are very, very wide ranges, so we are not going to know precisely where it is; but what kind of an OH is it?4928

    Is it an alcohol or a phenol or a carboxylic acid?4942

    Being next to this carbonyl makes it a carboxylic acid; carboxylic acid is what we are going to look up on our table; and those are the ones that are the furthest downfield, somewhere like 10 to 13--really, really far off to the left.4946

    Sometimes, they don't even show them--they are so far offscreen; they might just report that there is an additional peak there.4961

    Let me see if I can calculate this: to calculate it, we have a CH2; so we said that is going to start at 1.2; and then, what do we have on one side?--we have just a carbonyl with an R group (so a plain old alkyl group).4968

    And then, on the other side, we have a carbonyl, but it's a carboxylic acid.4985

    Those are actually going to have different values associated with them; so we have 1.2, plus an ordinary ketone is another 1.2 or so, so kind of like we said here, it would bring you to about 2.2; maybe 2.4 is a little closer.4993

    And then, next to the carboxylic acid is another 1.1--so not a really big difference; and what did we get?--1, 2, 3.5.5010

    So, with our calculations and our tables, it was about where we guessed in the first place.5018

    This was one that we could make an educated guess and come out pretty close to the actual number.5023

    OK, let's try this one.5032

    Now again, we have these 5 hydrogens; although they are technically not chemically equivalent, a lot of times they all show up at about the same; because this group can have no resonance effects on the benzene ring, it is not going to be something that really makes a big difference.5036

    And they are all going to be essentially the same; so we can call all of these type a.5056

    A lot of times, what we get is: we get a 5-hydrogen signal for an alkyl benzene, let's say--a benzene with just an alkyl group attached.5063

    We get a 5-hydrogen signal, and what we could describe this as is an apparent singlet.5070

    It is not really a singlet, because each of these has neighbors--it would have splitting; but because they are all coinciding in the same parts per million, we are not seeing that chemical shift.5080

    We are not seeing that splitting; and so, a lot of times, in our aromatic, you might just get one giant lump of 5 ppms.5090

    Sometimes you might get a multiplet; a multiplet means it is not clearly a doublet or a singlet, or you have complex splitting; and so, you might see that; but most often, you are just going to see one big peak there.5099

    OK, and where do you expect it to show up--what is our chemical shift for an aromatic proton?--somewhere around 7.5116

    OK, how about this CH2?--that is a unique kind of proton; that is type b.5126

    That would be a 2-hydrogen signal.5130

    And what splitting pattern do you expect here?--well, we have an oxygen on one side, so there are no hydrogens; we have a carbon with no hydrogens on the other side; so when you have no hydrogens, you expect a singlet.5134

    You just put an s here, for singlet.5147

    And what kind of chemical shift would we expect?--well, here is one where we certainly would want to calculate, if we can, because we have both an oxygen and an aromatic ring; and so, those are both going to have a shift on here.5150

    So, let's see if we can estimate it: we would start at 1.2 for our methylene; and then we have an aromatic ring attached on one side, and what do we have on the other side?5168

    We have an oxygen; oxygen, and then a carbonyl, and then an R group, right?--we have an ester attached--the oxygen of an ester.5183

    So, this comes out to 1.2, plus our aromatic ring, is another 1.4; being benzylic brings it down to about 2.6; and then, attached to the oxygen of an ester is 2.8; so we have 2.6+2.8--what is that, 5.4?5195

    1, 2, 3, 4; 1, 2, 3, 4, 5.4; yes, so about 5.4 ppm; so again, 5 is where we might expect a vinyl proton, a proton that is on a double bond; but if you have an oxygen pulling it downfield and a benzylic pulling it downfield, we can add up to those higher numbers.5217

    OK, what other protons?--we have this; it's type c; this is a 2-hydrogen signal, and what splitting pattern do we expect here?5239

    A 2-hydrogen signal: how many neighbors does it have?--we have 3 neighbors, so 3+1 is 4, so we get a quartet here (q stands for quartet).5250

    s stands for singlet.5264

    And where do we expect that peak c to be?--because it is next to a carbonyl, that is very similar to being just an ordinary benzylic--somewhere around 2.2.5269

    And then, this last methyl is also unique--type d; that will be a 3-hydrogen signal, and how many neighbors does it have?--it has 2 neighbors, so that means it will be a triplet (t for triplet).5283

    And where do we expect it to be?--well, it's just next to an ordinary CH2, so there is no significant deshielding effect here, and so we expect it to come...this would be the peak that is furthest upfield, closest to 0, somewhere around 1.5304

    OK, and notice: we have this little...this is what we would find for an ethyl group, right?--a CH2, CH3 is going to have this pattern.5319

    So sometimes we can pick this out; we have a 2-hydrogen signal that is a quartet (because it's next to a CH3), and then we must have this 3-hydrogen signal that is a triplet because it's next to the CH2.5327

    So, when it comes to coupling, what we are going to find...for splitting, we describe it as being coupled when one neighbor splits another neighbor, and so if neighbor a sees neighbor b, we are going to see that splitting by a, and we must--we must--see the same magnitude splitting of the other one, going back.5340

    OK, so sometimes we will see groups like this, that we describe as an ethyl pattern in the NMR.5361

    OK, last but not least, let's try this one: how many different signals do we expect?5368

    This is a highly symmetrical molecule; we have symmetry here; we saw a similar molecule earlier; we see symmetry here, so in fact, every single proton in this molecule is chemically equivalent (either identical--homotopic--or enantiotopic).5374

    And so, we expect just one signal (proton a), and it doesn't matter how big it is, because there is only one signal, so there is no way of quantifying how many protons are there, because it only gives a ratio of the size of one peak relative to another peak.5396

    So, there is just one peak: but here is the next question--well, we can estimate its ppm; let's do that first.5414

    What do you expect for the chemical shift?--well, it is attached to an oxygen; remember, oxygen has a huge deshielding effect, and so that would bring it pretty far downfield; it brings it to about 3.8.5421

    And what do you expect for its splitting?--well, you could say that it has two neighbors; OK, but because these neighbors are also type a, and they occur exactly where the first ones are, we are going to observe no splitting whatsoever.5435

    So, here is our little extra note that we need to make sure we know about the n+1: this represents the number of non-equivalent neighbors.5452

    OK, and that is usually not an issue; usually, your neighbors are always different from yourself; but if you have a case like this, where the immediately adjacent carbon is equivalent to the first one, you are going to observe no splitting.5474

    So, this is going to appear just as a singlet.5488

    So, we will find sometimes these splitting patterns; in fact, in the next lesson, we are going to get into some more advanced splitting patterns where you have several neighbors, but they don't all have the same magnitude splitting.5492

    They are not all equivalent; and so, how do we handle those?5505

    Also, in the next lesson, we are going to be getting into the more challenging goal of interpreting an NMR spectrum.5508

    When you take a look at a spectrum, how can you go about to come up with a structure?5516

    We have done, so far, the easier part of going from a structure to predicting a spectrum; and in the next lesson, we are going to learn some techniques to solve NMR problems.5520

    Thanks very much for coming to Educator.com; see you soon.5532