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Lecture Comments (55)

1 answer

Last reply by: Professor Starkey
Fri Mar 31, 2017 10:06 PM

Post by Ay Ayy on March 28, 2017

Hello Dr.
I have question  why for 13C there are not integration and splitting?

thank you

1 answer

Last reply by: Professor Starkey
Wed Mar 22, 2017 11:55 AM

Post by Ay Ayy on March 21, 2017

Hi Dr.
Thank you so much,
I have question, in the last practice, in the ring,why carbon double bond attached with ch2 and there are not close together in chemical shift?  

1 answer

Last reply by: Professor Starkey
Thu Mar 31, 2016 2:11 PM

Post by Adiam Ghebre on March 28, 2016

Good lecture! Is it possible for you to do an example with alkenes and the double bond is between two carbons?

1 answer

Last reply by: Professor Starkey
Mon Oct 12, 2015 1:14 AM

Post by sania sarwar on October 10, 2015

Hi Dr Starkey,
thanks for the lectures, they are really helpful.A question that I wasn't sure of is that in example 2 the benzene ring is a singlet so if CH2 is attached to it, wouldn't that make it a triplet?

1 answer

Last reply by: Professor Starkey
Sat Jul 19, 2014 10:28 PM

Post by John Subaitani on July 17, 2014

So are we saying that the H-NMR for C10 H14 is para or can it be para, meta. or ortho?

2 answers

Last reply by: Professor Starkey
Sat Jul 12, 2014 10:32 AM

Post by Francesco Frigo on July 10, 2014

Hello dr. Starkey, at minute 93:41 you define the signal given by proton A as a "clear triplet". But wouldn't proton A couple with Hb with some J value and with Hc with some different J value (since Hb and Hc are chemically different)? So in the end wouldn't this signal look like a doublet of triplets ? Maybe it wouldn't actually be easy to notice since the J values are so similiar but at least we could see some jagged peaks, right?

1 answer

Last reply by: Professor Starkey
Sat Mar 22, 2014 12:08 AM

Post by in gi seo on March 20, 2014

Are we not going to cover DEPT NMR????

1 answer

Last reply by: Professor Starkey
Fri Mar 14, 2014 11:59 AM

Post by saima khwaja on March 13, 2014

Hello Dr. Starkey,

If it wasn't for your lectures there is no way I would understand this part of organic chemistry.  My professor goes really fast and these lectures help me to clarify things in my head.  Thank You!

2 answers

Last reply by: Calin Cochran
Sun Mar 9, 2014 8:40 PM

Post by Calin Cochran on March 4, 2014

Hi Professor Starkey!

I know I've told you before, but your lecturing abilities are phenomenal! I can't thank you enough for all the help you have given me!

I do have a quick question. My professor for Organic 2 has clumped the second half of NMR, Ketones, Aldehydes, Carbohydrates, and Carboxylic Acids together for our upcoming exam. Your lectures have helped tremendously with all, but I'm a little lacking on the carbohydrates. I wasn't sure if I may have missed it somewhere on this website or we just don't cover it on here.

If you have any guidance, I would appreciate it tremendously! Thanks so much!

Calin Cochran

1 answer

Last reply by: Professor Starkey
Sun Feb 23, 2014 5:27 PM

Post by Jia Cheong on February 22, 2014

You are the best!!!!! Be my lecturer! :)

3 answers

Last reply by: Professor Starkey
Mon Feb 17, 2014 11:37 PM

Post by Udoka Ofoedu on February 17, 2014

hey dr. starkey ,
Why did u choose hb for only a germinal j value. it has a vicinal j value too ? Please why did u not split that ? Thanks

1 answer

Last reply by: Professor Starkey
Sun Jan 12, 2014 12:23 AM

Post by Mike Anderson on January 11, 2014

Is there a way to select a lecture and be able to listen to just a part of it?  For example it seems if I want to go back the next day and listen to the second part of a lecture, I have to listen to the whole first half of it first.


Mike Anderson

1 answer

Last reply by: Professor Starkey
Tue Oct 29, 2013 10:30 PM

Post by Joel Barrett on October 28, 2013

Professor Starkey, you are wonderful. I saw your YouTube videos as well. Your hard work is appreciated. I love o-chem just a little bit more because of your videos ;)

1 answer

Last reply by: Professor Starkey
Sun Jul 21, 2013 10:52 PM

Post by Amy Lin on July 21, 2013

Hi Dr Starkey,  I don't have a question. I finished all the lectures and I can't express how much of a help it has been. Your lectures have been really concise and you break it down in a way I can finally understand and work through. I have always had a horrible time with Chemistry and this is the first time I actually feel like I can do this.  (And that is a lot given how many times I had to retake Chem...) I am done my subscription and I just wanted to tell you what a big help you have been.  Thank you so much.  !!!

1 answer

Last reply by: Professor Starkey
Wed Jun 5, 2013 10:44 PM

Post by Heidi Schmeck on June 5, 2013

Dr. Starkey:

Just a quick note to thank you for your informative and engaging lectures. I used and your lab tutorials (Cal Poly Pomona) as supplemental sources to reinforce my understanding of my Organic Chemistry II coursework. Your detailed and clear explanations of complex concepts helped me earn an "A" in both lecture and lab. Thank you! :)

1 answer

Last reply by: Professor Starkey
Mon Feb 25, 2013 10:24 PM

Post by Ryan Rod on February 25, 2013

Hi, did you also cover(including IR and NMR of them), Ethers , Epoxides, and Sulfides? how about Aromatic compounds?
Sorry I have an exam and panicking!

1 answer

Last reply by: Professor Starkey
Wed Feb 20, 2013 9:42 PM

Post by José Menéndez on February 19, 2013

Hello Dr. Starkey, I wanted to know if you have a mass spectrometry lecture? Thanks.

1 answer

Last reply by: Professor Starkey
Wed Feb 20, 2013 9:44 PM

Post by Ryan Rod on February 18, 2013

What about Carbon NMR?? did I miss it, or have you not covered it?


Your are a AMAZING!!

1 answer

Last reply by: Professor Starkey
Sun Feb 17, 2013 5:35 PM

Post by Amirnikan Eghbali on February 17, 2013

Thanks and a suggestion, it's better if you label doublet triplet etc with small letters (s,d,t) and the other A, B, C, D... that you use for identification with capital letters to avoid confusion.

1 answer

Last reply by: Professor Starkey
Fri Dec 14, 2012 11:23 AM

Post by Marina Bossi on December 12, 2012

Hi Professor Starkey,
I am confused about labelling some functional groups in certain areas of the spectrum where multiple groups can be found. For instance, methyl groups are found at 10-30 and methylene groups at 15-55. How would I know the difference?


2 answers

Last reply by: alister guerrero
Wed Nov 7, 2012 4:02 PM

Post by alister guerrero on November 4, 2012

Hi, I just got my account. i was wondering if there is a way to download all the slides together? Thank You

1 answer

Last reply by: Professor Starkey
Fri Jul 13, 2012 1:33 PM

Post by Gabriella Kaminer-Levin on July 5, 2012

Dear Dr. Starkey:

How can one distinguish between two closely spaced singlets, and a doublet with a large coupling constant (J value)? At 73:30 you examine a doublet with a large coupling constant, but how can one be certain that it is a doublet with a large coupling constant rather than two closely spaced singlets (since in this case examining the ratios does not help)?
Also, do you have any lectures on Mass Spectrometry? I wasn't sure whether they are included in the course and I just wasn't able to locate it in the Table of Contents.
Thank you again for your clear presentation of the material!


1 answer

Last reply by: Professor Starkey
Mon Apr 9, 2012 11:34 PM

Post by Ghazal Fata on April 7, 2012

Dear Professor,
First of all, I wanted to thank you for the great lectures you provide.
Second, I was wondering if the 13C-NMR is going to be thought in a more complex way in Educator. The organic chem I am enrolled in emphasizes a lot more on how to read 13C-NMR without the help of IR or 1H-NMR and it's really complicated and I'd appreciate it if you provide more videos for us, or just refer me to a place which I can find helpful information.
My course actually emphasizes on how different fragments of molecules appear on NMR and which bonds are cleaved and molecular ions are created.
Thank you so much,
Ghazal F

1 answer

Last reply by: Professor Starkey
Tue Mar 27, 2012 11:23 PM

Post by Robert Shaw on March 23, 2012

Dr Starkey, My NMR table has aldehydes listed as 9.5 to 9.9 yet yours shows it at just over 8. Which is correct?

1 answer

Last reply by: Professor Starkey
Sat Jan 21, 2012 1:06 PM

Post by Jason Jarduck on January 20, 2012

Hi DR starkey,

I have a question about an NMR.

I only have hydrogen NMR with a d20 shake.
The question is stating that 3200 - 3500 which specifies an alcohol group.
NO C NMR spcetrum is given
Find compound A, only H NMR spectrum
FIND COMPUND B, only H NMR spectrum
This is an assignment question!!!

I also have the multiplicities for the hydrogens for compound A and B. How many carbons?


What would be the best strategy to solve this problem!!! Also where can I find a chart with all energy levels in Joules.

I enjoy your lecture alot and I'm in a hurry because I have a lab to do for Wensday and I must have this question done for Monday.

Thank you

Jason Jarduck

Nuclear Magnetic Resonance (NMR) Spectroscopy, Part II

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.

  • Intro 0:00
  • ¹H NMR Problem-Solving Strategies 0:18
    • Step 1: Analyze IR Spectrum (If Provided)
    • Step 2: Analyze Molecular Formula (If Provided)
    • Step 3: Draw Pieces of Molecule
    • Step 4: Confirm Pieces
    • Step 5: Put the Pieces Together!
    • Step 6: Check Your Answer!
  • Examples 9:17
    • Example 1: Determine the Structure of a C₉H₁₀O₂ Compound with the Following ¹H NMR Data
    • Example 2: Determine the Structure of a C₉H₁₀O₂ Compound with the Following ¹H NMR Data
  • ¹H NMR Practice 20:57
    • ¹H NMR Practice 1: C₁₀H₁₄
    • ¹H NMR Practice 2: C₄H₈O₂
    • ¹H NMR Practice 3: C₆H₁₂O₃
    • ¹H NMR Practice 4: C₈H₁₈
  • More About Coupling Constants (J Values) 57:11
    • Vicinal (3-bond) and Geminal (2-bond)
    • Cyclohexane (ax-ax) and Cyclohexane (ax-eq) or (eq-eq)
    • Geminal (Alkene), Cis (Alkene), and Trans (Alkene)
    • Allylic (4-bond) and W-coupling (4-bond) (Rigid Structures Only)
  • ¹H NMR Advanced Splitting Patterns 1:05:39
    • Example 1: ¹H NMR Advanced Splitting Patterns
    • Example 2: ¹H NMR Advanced Splitting Patterns
    • Example 3: ¹H NMR Advanced Splitting Patterns
  • ¹H NMR Practice 1:22:53
    • ¹H NMR Practice 5: C₁₁H₁₇N
    • ¹H NMR Practice 6: C₉H₁₀O
  • ¹³C NMR Spectroscopy 1:44:49
    • ¹³C NMR Spectroscopy
  • ¹³C NMR Chemical Shifts 1:47:24
    • ¹³C NMR Chemical Shifts Part 1
    • ¹³C NMR Chemical Shifts Part 2
  • ¹³C NMR Practice 1:50:16
    • ¹³C NMR Practice 1
    • ¹³C NMR Practice 2

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

Hello; welcome back to Educator.0000

Next, we are going to continue with our NMR lesson to learn about some more advanced NMR facets, to learn how to interpret an NMR spectrum, and also to get into a little of carbon-13 NMR spectroscopy.0001

Let's talk about some problem-solving strategies we can use when given a proton NMR--how can you go about determining a structure from it?0020

Three basic things we need to accomplish: our first--determine what pieces are present; OK, what we are going to be doing is very much like building a puzzle.0030

And so, the first thing we have to do is figure out what puzzle pieces there are; then we have to figure out how to put those pieces together; and when we're done and we have a structure that we would like to propose as an answer, what we are going to do is: we are going to confirm that the structure does, in fact, match the spectrum.0039

We are going to use those skills that we did in being able to predict an NMR spectrum, given a structure, and confirm it.0055

What is great about NMR is: when you have the right answer, you know you have the right answer, and you can breathe a sigh of relief and move on.0061

OK, so here are the steps we are going to take: the very first thing we need to do is: if we are lucky enough to have an IR spectrum provided to us for our molecule, then we are going to analyze that spectrum for any functional groups that might be present.0071

OK, a lot of times, spectroscopy problems are given in combination, so you will get maybe an IR spectrum and a proton and a carbon NMR.0083

And so, the more data you are given, great--work with it.0092

Let's take a look at the IR and see if we have things know, a carbonyl is really obvious in the IR (somewhere around 1,700 we find that very strong signal); OH groups are very obvious in the IR (those broad peaks around 3,300).0095

All right, we can look for the presence of double bonds, hydrogens on those triple bonds, and so on.0112

OK, so those are going to be pieces for our puzzle; those have to be incorporated to our final answer, and to the structure that we have.0119

OK, the next thing we are going to do is: we are going to look at our molecular formula.0126

If it's provided, if we are lucky enough to have our molecular formula, we are going to analyze that and determine how many degrees of unsaturation we have.0130

Now, what we are going to do is: we are going to look for the formula's the formula we expect if we have a saturated molecule.0137

If we have any nitrogens present, we need to account for those by...for every one nitrogen, we need one extra hydrogen in our structure; so we can modify our formula a little bit to accommodate those.0148

I'm sorry, 2n+2...I see that I missed my n there--so twice the number of hydrogens, plus 2.0161

OK, and every two missing hydrogens gives us a clue to our structure, because it tells us we have a degree of unsaturation; that means we must have a double bond, or we must have a ring, to account for those missing two hydrogens.0168

OK, if we happen to have four or more degrees of unsaturation, then it might be possible to structure as a benzene ring.0183

If you take a look at a benzene ring, we have a 6-membered ring with alternating π bonds; OK, so that means we have three π bonds, plus one ring; that equals 4 degrees of unsaturation.0190

That means, if we have 8 missing hydrogens, then it is possible that we have a benzene ring in our structure; that is a great way to account for a lot of degrees of unsaturation very quickly.0205

If we do not have four degrees of unsaturation--if we have fewer than that--it is impossible to have a benzene ring in our structure.0218

So really, the molecular formula is telling us something very significant about our structure.0222

OK, the next thing we are going to do is: we are going to draw out the various pieces to our molecule.0230

We are going to be using the peak integration, the size of each peak, to tell us how many hydrogens there are in each.0235

OK, so for example, if you see a signal that has three hydrogens in it, let's assume that is a CH3, because it is possible to have three hydrogens coinciding some other way (other than a methyl group), but if off in the distance, you hear something galloping--you hear hooves beating against the ground--you should probably think in terms of horses before you imagine that they're zebras.0241

It is possible that they are zebras, but let's go with the more obvious choice, the more likely choice, OK?0267

A 3-hydrogen signal means we probably have a CH3; a 2-hydrogen signal means we have a CH2; a 1-hydrogen signal means we have a CH, or maybe an OH or an NH, if we have those elements in our formula.0272

OK, a 4-hydrogen signal is a little trickier--not as common--but obviously, there is not a CH4 puzzle piece, so maybe we have 2 CH2s that are chemically equivalent, therefore giving rise to one signal.0286

Or maybe a CH3 and a CH...that can't be chemically equivalent, but they might just be overlapping because they have the same chemical shift.0300

So, there may be some more complex things that we'll come across; but we will deal with those when we come across them.0308

If we have peaks around 7 parts per million, we are going to assume there is an aromatic ring, a benzene ring.0315

OK, it might just be a single peak--very common; if they are very similar in their chemical environment, then they are going to have the same chemical shift.0321

Or, you might have several signals in the region around 7; OK, we are going to look at all the signals that are around 7, and we are going to see how many total hydrogens there are.0333

If we have 5 hydrogens around 7, that means we have a monosubstituted benzene ring; so, in other words, the piece to our puzzle is a benzene ring with something attached.0342

OK, that means, if you attach one if you have a monosubstituted benzene ring, that means you have 1, 2, 3, 4, 5 hydrogens left, and those are going to show up in your NMR.0354

OK, if you have a total of 4 hydrogens around 7, that means you have a disubstituted benzene ring; so this is monosubstituted; disubstituted means that you have a benzene ring with two things attached.0363

Where are those two things attached?--they could be ortho or meta or para, but you have only four aromatic protons, meaning two of those positions are occupied by something other than hydrogen.0374

OK, so we are going to do this to come up with all of the pieces in our puzzle; and then, the next thing we are going to do is: we are going to confirm our pieces.0387

And make sure--before you start building your molecule and putting your pieces together, you had better make sure you have come up with all of the correct pieces.0394

What we are going to do is: we are going to add them up, and we are going to make sure that they match the molecular formula.0402

OK, the way that you could do that is to make sure: have you accounted for all the degrees of unsaturation?0409

If you said there were two degrees of unsaturation, do you have two double bonds in your pieces?0414

If not, know that you still have to incorporate a ring when you put those pieces together to get that second DU.0420

Have you accounted for the functional groups in the IR?--if the IR told you you have an OH, then one of the pieces in your puzzle needs to be an OH.0429

OK, so we are going to confirm that we have all of the pieces present; now we are ready to start putting them together.0436

OK, how do we decide which two pieces are connected?--well, we are going to consider their chemical shift (is it electron rich? electron deficient?); we are going to consider the splitting pattern (does it have any neighbors?).0444

OK, and we need to consider both of those.0456

The easiest thing is to start with an end piece (like a methyl group) and start to see what things we can attach to it.0457

OK, for example, if it is next to an oxygen...I could attach it to an oxygen if its chemical shift is around 3.5.0465

Maybe it's next to a carbonyl or a benzene ring, if its chemical shift is around 2.2.0474

We'll consider where it is on the spectrum to decide who it's next to; and we are also going to consider the splitting patterns.0479

If it's a triplet, that means it must have two neighbors, so I'm going to find a CH2 piece, and I'm going to attach it to that.0487

OK, so bit by bit, we are going to take all of these different pieces and start hooking them together until we come up with a complete structure.0494

Before we are done and we pat ourselves on the back, we need to go back and check our structure: this is the last step in our methodical process.0503

Does it match the molecular formula?--sometimes, when we start working, and we spend a lot of time on a spectrum, it's easy for a piece to fall off and never get added back on; and then you think you have the solution, and you realize you are missing some carbons or oxygens or nitrogens.0513

So, does it match the molecular formula?--does it match the functional groups in the IR that you predicted?--does it match the NMR?0527

Remember, we worked on how to predict an NMR spectrum: how many signals are you expecting?0534

Is there the symmetry in the molecule that is required to get the right number of signals?0539

What would the integration be for each peak?--what would the splitting pattern be?--what would the chemical shift be?0544

OK, so it's our very last step: to confirm that we have, in fact, determined the right structure.0549

Let's try some problems.0556

OK, here we have a structure with the formula C9H10O2, and here is our NMR data.0558

So, instead of giving you a spectrum, it's presented to you just in numerical form here; OK.0564

And the problem with NMR that I see again and again, working on NMR problems, is: there is so much information, you could just get blown away by that information; and it's very easy to spend 30 minutes working on an NMR problem and go nowhere, because you just are grabbing at bits and pieces, OK?0572

So, we are going to develop a systematic approach; we are going to use that same approach every time; and we are going to hopefully find success in that approach.0588

OK, so what was our first step?--our first step was to look at the IR spectrum and see what we can figure out from that; but we are not going to be looking at any IR's today.0595

A lot of the problems you are going to see in your courses or in the books will have IR combined with NMR, so you are lucky in those cases--you get that as an additional clue of what functional groups you have.0604

But in all of these cases, we are going to be skipping the IR.0615

OK, do we have a molecular formula?--well, you won't always be given a molecular formula, but if you are lucky enough to get it, let's use it.0618

We do have a molecular formula: it's C9H10O2.0625

What can that tell us about the structure?0629

What we need to do is: we need to calculate the DU.0631

OK, the way we calculate the DU is: we ask ourselves, "If it was saturated, what would the formula be?"0635

We have nine carbons: it would be C9H...remember, it was CnH2n+2, so its 9x2, is 18, plus 2 is 12 if it was saturated, it would be C9H20.0642

What do we have?--we have C9H10; OK, so what do we have?--we have 10 missing hydrogens, and every 2 hydrogens is a degree of unsaturation, or a site of unsaturation; so we have 5 DU.0658

So, our formula, when we are done, has to account for all 5 of those.0677

OK, now what do we do?--we figure out what pieces we have to our puzzle.0683

We use the integration for that; so we have 5 hydrogens here, somewhere around 7; what piece does that tell us?0687

What things come at around 7?--that would be aromatic protons.0696

We have 5 of them, so that tells us that we have a monosubstituted benzene ring; we have a benzene ring with one...something...hanging off of it, attached to it.0700

OK, we have a 2-hydrogen signal: what does a 2-hydrogen signal mean?0713

It means we have a CH2; and when I draw this piece, look what I'm going to do: I'm going to draw two arms on it.0717

And that is because we know every carbon is going to have four bonds in our structures when we are done, because we are only looking at stable structures.0723

A CH2 must be attached to two other things to get those four bonds.0733

And what does a 3-hydrogen signal tell us?--that means we have a CH3; a CH3 has just one arm, because that would be the fourth bond.0737

These are our pieces; are they all of our pieces?0748

Well, what do we have so far?--we have C...6, 7, 8--we have C8H...5, 6, 7, 8, 9, 10.0752

OK, so we have accounted for all of our hydrogens, of course, because they show up in the NMR; it's hard to miss those.0764

But we only have 8 carbons shown; so we still have a carbon; we still have 2 oxygens that we need to account for (oxygen has 2 arms, right?--oxygen likes to have 2 bonds); and what else are we missing?0771

Let's check our formula: our formula says we need 5 degrees of unsaturation--how many have we shown so far in these pieces?0785

We have only shown a benzene ring, which has 1, 2, 3, 4 degrees of unsaturation; so we still have 1 degree of unsaturation that needs to be in our structure when we're done.0793

Now, this looks like a lot of random pieces, but take a look at this: there is actually a very nice functional group we can imagine, that takes care of all three of those.0805

What is a way to have a carbon and an oxygen with a degree of unsaturation?0813

How about a carbonyl?--a carbonyl would very nicely account for all three of those pieces in one nice functional group that we see all the time--carbonyls are very, very common in organic molecules.0819

And of course, a carbonyl has 2 arms.0834

OK, so these, now, are our pieces; so we have taken care of this, so we have a phenyl ring, CH2, CH3, a carbonyl, and an oxygen.0836

Those are our pieces; now we are ready to put them together.0845

We start with an end piece, and we decide what it's attached to.0848

So, let's start with our CH3: what are our choices?0853

We could either attach it...could we attach it to the benzene ring?--we can't attach it to a benzene ring, because that is an end piece; if we attach the methyl and a benzene, we would now have toluene.0859

Our structure would be done; we couldn't fit in any of the other groups.0870

OK, but I could attach it to a CH2 or an oxygen or a carbonyl; those are the other possible pieces.0874

How do I decide which one it belongs to?0882

Well, now I take a look...where is my CH3?...I take a look; it must be a singlet, so what does that rule out, if it's a singlet?0885

"Singlet" means that I have zero neighbors, right?0893

I have 0 protons on my neighboring carbons; so I can't attach it to the CH2; that would split it.0900

But it could be the oxygen or the carbonyl; how do I decide which one it is?--well, I take a look at the chemical shift.0907

Is that the chemical shift I would expect when I'm next to an oxygen, or when I'm α to a carbonyl?0913

That means I am α to a carbonyl; if I put it next to an oxygen, that brings me too far downfield; it brings me closer to 4.0919

So, I must have this attached to the carbonyl; OK, that took care of this piece; that took care of this piece; what comes next?0926

What are my options?0934

I can either have an oxygen, or I can have a CH2.0937

OK, now, it turns out, in this case...let's take a look at that: let's put the oxygen first, and then the CH2, and then the last piece I would have is the benzene.0942

Or, the other possibility: we are down to just two possible structures--we could have the CH2 and then the oxygen.0957

OK, and we can use...would that explain the chemical shift and the splitting pattern for the CH2?0967

It would, actually, because this is a singlet, which is true in both cases; and it comes around 5.0973

How does it get all the way to 5?--well, it is both next to an oxygen, and it is next to something else (in this case, a benzene ring--it's benzylic--and in this case, it's next to a carbonyl).0979

So, actually, both of these answers are reasonable; the way I would distinguish between the two of them (if I had some tables, maybe I could calculate it a little more precisely to see which one best matches 5.1, but)--the other thing that I would recognize (it's a little more sophisticated) is: I know that having an oxygen on a benzene ring is going to have some resonance, which is going to make some of these protons much different--give them a much different chemical, electronic environment--than the others.0990

So, because it is a 5-hydrogen singlet, I am expecting all of these hydrogens to be very similar, and therefore just attached to a regular CH2; and so, this is the best match.1023

This is the best match; occasionally we might have a case where there is more than one possible answer, but this is a pretty rare situation.1034

OK, let's try a different one.1046

It turns out we have the same formula, and we have the same pieces; so just like we did our process before, we would find that we have 5 degrees of unsaturation.1049

So, we have a 5-hydrogen aromatic signal; we have a CH3; we have a CH2; we have a carbonyl; and then, we have an extra oxygen--same 5 pieces as we did in the previous problem.1061

How are we going to put them together?1082

Well, again, let's start with our methyl group; let's start with this end group.1083

Tell me about this methyl group; it's a singlet again, so it has the same splitting pattern as before; but now it's a 3.6 ppm; so which piece would make sense to attach to the methyl, that would bring it to 3.6?1089

This would bring it to 3.6 if we put an oxygen next to it; OK, so that takes care of our oxygen, and it takes care of our methyl.1105

What do we have next--or what other pieces can we put together?1114

Well, let's think about our CH2--what is our CH2 going to be attached to?1119

Our CH2 is at 3.5; OK, so if we put (let's think about our possibilities) that here--if we put our CH2 next--we would expect that to be a 3.5, just like the first one.1124

But then, when we attached it to another piece, the carbonyl or the benzene ring, what would happen to that 3.5?1143

That would deshield it even more, and we wouldn't expect it to be there any longer.1151

We can't put the CH2 here; instead, it's the carbonyl and then the CH2.1157

Don't be afraid to look at your options; rather than having it all swirling around in your head, and trying to imagine all the possibilities, just write out the two possibilities that you are considering, and then see which one fits the data best.1168

OK, right?--so what I'm saying here is that if this were the carbonyl (if we looked at this option), then this would be much more downfield--this would be much higher than 3.5.1180

OK, but now, if we were to predict this, where do we expect this to be?1194

Well, it's going to shift a little downfield because of the carbonyl, and it's going to shift a little downfield because of the benzene ring; that would put it at 3.5.1199

3.5 ppm's is very reasonable for this one.1207

We would expect it to be a 2-hydrogen peak; is this a singlet?--it is a singlet, just like we predicted.1211

I went through this a little bit on the previous one, but we want to make sure that we double check our work: this CH3...what do you predict for that?1220

It splitting pattern would also be a singlet; and where do you expect it to occur--what would its chemical shift be?1228

It's attached to an oxygen, so again, 3.8...we would expect that here; it's 3.6; that makes perfect sense.1235

These hydrogens...all our 5 hydrogens are around 7 (in this case, 7.4).1243

OK, so we can propose a structure and then double-check it, and we know we have the right answer.1249

OK, let's try a different molecule, a little we have some spectral data here.1258

Because we have no IR information, let's go straight to the chemical formula, C10H14.1266

What do we know about C10H14?--let's check whether there are any degrees of unsaturation.1275

Here is the question we asked: rather than doing some kind of formula to randomly come up with an answer--magically come up with an answer for degrees of unsaturation, it is much more reliable to kind of think about it in terms of, "Well, if it was saturated, what would the formula be?"1282

If it was saturated with 10 carbons, how many hydrogens can possibly fit on 10 carbons?--2n+2, 20+2; so it would be C10H22 if it was saturated.1298

So, when we compare the actual formula to the saturated formula, we have 6, 7, 8; we have 8 hydrogens missing.1312

And how does that translate to degrees of unsaturation?1321

Every two missing hydrogens tell us we have a double bond or we have a ring; we have a DU; so there are 4 DU in our structure.1325

OK, and that is a very important clue; it is going to help us determine what our pieces are.1333

OK, so then, the next thing we do is: we look at our peaks, and we decide what our pieces are, based on the integration.1338

OK, and here, rather than integral trails, they are just giving you the number of protons; so we can start anywhere.1349

6 hydrogens--how would we get 6 hydrogens?1357

It can't be a CH6; what do we have--what would be the most likely thing we could have here?--probably a CH3 and another CH3 that are equivalent, and therefore giving rise to a single signal.1360

OK, so 2 CH3s: a 3-hydrogen signal here means that we have a CH3; another CH3; a 1-hydrogen signal means we have a CH.1375

Now, a CH has 3 arms, right?--to come up with 4 total bonds.1388

How do I know it's not an NH or an OH?--because I only have carbons and hydrogens; I just have an alkane, a hydrocarbon, with my formula; so it must be a CH.1393

And then, what do we have over here?--we have 4 hydrogens, and they are around 7, so what does that tell you?1404

We probably have a benzene ring; how do we make a benzene ring with only four hydrogens on it?1414

That means that we have 2...there are six hydrogens to start with on benzene, so we must have two of those hydrogens replaced by something else; so this is another piece to our puzzle--having two groups.1423

Now, do I know that they are para to each other?--no; they could be para; they could be meta; they could be ortho; but we'll just pick one pattern to start, and then when we're done, we'll confirm to see if that seems like a reasonable substitution pattern.1434

OK, so those are our pieces; let's double-check before we start putting things together--make sure we have accounted for everything.1447

Do we have C10H14?--we know we have H14, because we did the 6 and 3 and 1 and 4; we have all the right number of hydrogens, and we have 6, 7, 8, 9, 10 hydrogens.1455

OK, so we have confirmed with our formula that we have all of the right pieces; now we can start to put them together.1467

OK, it is best if you start with an end piece; OK, and we have several: we have these CH3s; we have this CH3; and we can decide how to attach them.1473

Let's take a look at this signal: it has 6 hydrogens total, and how would you describe that signal--what is the splitting pattern?1486

It has two peaks; if you look all the way up at the top, sometimes it's a little easier to see what kind of shape it is, in fact.1495

And this is a doublet; and so, what does it tell you about how many neighbors we have?1503

It means there is one neighbor.1510

So, which piece can I attach it to that would give rise to that splitting?1513

Well, I could attach it to this CH: if I attached this CH to 2 CH3s, and if I put both CH3s on that, would that account for the signal I see?1519

It does, because now these would be equivalent, because I could rotate around and those would be chemically equivalent.1530

And that would be a 6-hydrogen signal; that is a doublet, so what I have done is: I have taken these and these little pieces and connected them to be a bigger piece--a larger piece.1535

OK, what pieces can I connect next?--how many oxygens do I have?1547

This is an end piece; this is an end piece; they each have just one arm, and then I have this benzene ring with two arms.1554

So, what can I do?--I disconnect them all; I put the methyl on one side, and I put the isopropyl on the other side; this is a CH with a CH3 and a CH3.1560

And now, I have put all of my pieces together.1578

OK, let's see if it makes sense--let's predict if that makes sense.1582

What do you expect this peak to look like?1586

It should be a 3-hydrogen signal; what should the chemical shift and the splitting pattern be?1590

How many neighbors does it have?--it has no neighbors, so this should be a singlet.1595

And what chemical shift do we expect for being next to a benzene ring?--about 2.2, and where is it?--there it is: about 2.4, maybe--somewhere around there.1600

That is good; if you want, you could label your peaks; this is a and this is a; that would be a very great way to check your work at the end, and demonstrate that you really know for sure that the structure matches the spectrum.1613

OK, we said these methyl groups were together; we already analyzed that to make sure that makes sense; that would be signal b, let's say.1628

We said it's a 6-hydrogen signal that would be a doublet, because it has one neighbor.1638

How about this one?--let's call this type c, and what do you expect for that?1642

For that signal, it should be one hydrogen; and how many neighbors does it have?--it has 6 neighbors (3+3), so we expect that to be a septet.1648

And it's kind of hard to see that, so we have blown it up here; this is kind of common, to see smaller peaks blown up and expanded.1658

And what do we have?--1, 2, 3, 4, 5, 6, 7 peaks; there it is--it's a septet.1665

So, we must n+1; we must have 6 neighbors.1673

So, that makes sense to be peak c's chemical shift; where do you expect it?--well, it's benzylic, so that's 2.2; it's also a CH--that brings it a little further downfield, as well; so 2.8 is not unreasonable for that, where it was--that makes sense.1678

And then, finally, these 4 hydrogens--it looks like...what do we have here?1699

It is kind of tough to see, but it looks like we have a doublet right next to another doublet; and because they are coupling with each other, because they are splitting each other, a lot of times you will see the doublet kind of leaning, skewing toward this other signal that it is splitting with.1704

And so, that is why we have these really high peaks on the outside and these short peaks on the inside; they are just slightly different chemical shifts that are being resolved into two different signals.1723

But, a lot of times, if this were a less sensitive instrument, this would just show up as our typical singlet in this region.1735

But should we expect two signals?--well, yes: this hydrogen is different from this hydrogen, and we are at c; so we could call one of these d, and one of these e.1742

This is e, and this is d; so we have 2 hydrogen-type d's (these are doublets--I don't want to confuse my d for doublet with hydrogen d).1755

So, this is hydrogen d and e, and we have two doublets here.1768

So, if we wanted to narrow it down a little, this symmetry would explain that splitting pattern and the characteristics of that shape; so that does look fine.1774

OK, let's try another one.1787

OK, here we have, again...let's start with our formula, C4H8O2; when we are doing our degrees of unsaturation, we ignore the oxygen.1792

So, we ask: if it was saturated, our formula would be C4H....2n is 8, plus 2 is 10; our actual formula is C4H8 (we ignore the oxygens, because they have no impact on our degrees of unsaturation).1803

And so, we find that we have just 2 hydrogens missing; and that means that we have 1 degree of unsaturation, so we need to account for that in our puzzle pieces.1819

OK, so let's figure out what pieces we have: we can start anywhere; we have a 1-hydrogen signal here, so we have a CH with three other bonds.1834

We have a 2-hydrogen signal, which is a CH2.1848

We have another 2-hydrogen signal that is blown up here to see a little more detail, but we have another 2-hydrogen signal; that is another CH2.1855

And then, we have a 3-hydrogen signal; that is a CH3.1863

OK, so those are some of our pieces; are those all of our pieces?1868

Well, we haven't accounted for our degree of unsaturation yet; we also have 2 oxygens.1872

OK, so is there anything we can do to consolidate some of these pieces into something recognizable?1882

Well, this hydrogen is kind of unique; having a hydrogen, a proton, resonating all the way out at 8 is a very, very far downfield signal, and we have only seen a few of those.1888

A carboxylic would come this far, an OH, or an aldehyde, CH.1900

A carboxylic acid comes even further (usually higher than 10); about 8 is where we expect our aldehydes; so guess what?--if I take this CH and attach them to the carbon as a carbonyl, then that would be a very reasonable puzzle piece that is consistent with all our data.1908

This would be a singlet this case, 8; it could be even 9 or 10 ppm.1929

And the carbonyl gives us one of our oxygens and accounts for our degrees of unsaturation.1941

OK, so I think this is a reasonable puzzle piece to have: we still have a second oxygen, and then we have the CH2, CH2, and CH3.1947

So, we have 1, 2, 3, 4 carbons; we have 3, 4, 5, 6, 7, 8 hydrogens and 2 oxygens; OK, you absolutely have to double-check before you start putting things together and jumping all over the place.1954

If you don't have the right puzzle pieces, you cannot get the right answer.1969

OK, so this is a really important step we don't want to skip.1973

OK, now we are ready to put them together.1976

Where is a good place to start?1979

Really, the only place we can start is with this methyl group, because that is an end piece, and we have to figure out who it is attached to.1981

So, tell me about that methyl: it is up here, away at 1--very close to 1; and it is a triplet--it has 1, 2, 3 peaks (see this 1:2:1 shape?--that is very typical that we see for a triplet).1989

So, what does that tell you it is attached to--what are your choices?2003

It could be attached to an oxygen or a CH2; those are the only choices--I can't attach it to the carbonyl, because that is taking my two end pieces and connecting them, and I leave out everything else.2005

So, it must be attached to one of the CH2s.2016

Why not an oxygen--what would that do?--well, two problems with that: it would make it a singlet, and it would bring it all the way down to 4 or so.2020

So, this is now consistent.2029

OK, so I used a methyl group; I used my other CH2.2031

OK, now I have a CH2, and we can see where that goes, or we can kind of look at another point of view.2036

We have these two CH2s; I don't know yet which is which; I don't know if it's this one or this one; so let's take a look at that.2046

It has how many neighbors?--it has these three neighbors, so it is going to be a quartet.2057

This one is only a triplet; so this can't be this CH2, because it doesn't have a big enough splitting pattern.2064

So, let's just label this: if this is a, and then this CH2 is b, it must be this, because it has at least three neighbors.2072

What gives it this much splitting?--let's take a look.2081

It has 1, 2, 3, 4, 5, 6 peaks--6 peaks means it has 5 neighbors, so in addition to these three neighbors, it also has another two neighbors; we must have the other CH2 on the other side.2083

OK, even though these three hydrogens are not exactly equivalent to these two hydrogens, their spatial relationship to protons b here are similar enough that they have the same splitting constant, the same coupling constant.2099

And so, we end up with this very nice signal, like it has 5 neighbors all of one type.2117

OK, so that is...this must be proton c here; that is the other CH2 signal; now tell me about this CH2 signal.2125

It's a triplet, and it's up here at 4.2; what does that tell you that it's attached to?2135

If it's all the way down at 4, that means it must be attached to the oxygen.2142

So now, I attach my oxygen; and then, I only have one piece left--my carbonyl.2146

So, bit by bit, I attach it; sometimes we can keep working linearly; sometimes we group things.2155

You could take a look at this CH2; you know it's a CH2; and you could say, "You know what, it's all the way at 4, so I know it's attached to an oxygen."2161

So, you can start to group your pieces that way.2169

Or, you can group them in a different fashion.2172

So, when it comes to putting the pieces together, that one is a little more individualistic on how you are going to solve it, versus someone else.2177

OK, but the key is: start at an end point (like with a methyl group), and see who it's attached to.2184

Slowly start to bring your scattered pieces into larger pieces, and then see how those larger pieces can finally come together into a complete molecule.2191

There will not be 1,000 choices; it might feel like the possibilities are endless, but there is really a finite number of choices.2200

And so, sometimes, you come down to just a couple of choices; draw them out and see which one works, OK?2208

Try and be as systematic...and keep moving forward as best you can.2214

OK, does all the data match, if we were to look at this structure and confirm that it matches the NMR?2218

Do we have an aldehyde proton in our NMR?--we do, up here at 8, so let's call that proton type d.2226

Aldehydes are always singlets, because they are always attached to a carbonyl; so we expect that somewhere really far downfield.2237

What do you expect for this carbon?--now, cover up the NMR: don't look at the NMR at all; look at your structure and predict it.2247

OK, if you are doing it while you are looking at the NMR, you can miss things; you can confirm things; you can lose sight of some evidence.2256

So, let's predict this.2262

What do you expect this signal to look like--what is its splitting pattern?--how many neighbors does it have?2263

It has two neighbors right here, so n+1--we expect this to be a triplet.2269

Where do you expect this to show up?--it's next to an oxygen, so that's 3.8; in fact, in this case, it's 4.2--that is certainly consistent, certainly reasonable; that makes sense.2274

OK, what do you expect for b--where do you expect it to show up...or what is the splitting pattern, first?2290

We could do that: it has 3, 4, 5 neighbors, so we expect this to be a sextet, and it is, in fact: 1, 2, 3, 4, 5, 6.2297

Where do you expect to have it?--nowhere special; it just has alkyl groups on either side, so somewhere around 1 is what we expect.2307

Why is it up here, all the way closer to 2?--well, because not too far away, it has an oxygen.2314

So, that does have some effect; but as usual, these inductive effects decrease with distance.2319

It is not as much as if it was directly attached to an oxygen, but the presence of that oxygen brings it down a little bit; so, yes, 1.8 is fine for that.2324

And this last CH3...what do you expect its splitting pattern to be?--it's a triplet, because it has two neighbors.2333

And where do you expect it to be?--now it's really far away from that oxygen, so we expect it to be somewhere around 1; in fact, it's even a little below 1, in this case.2342

One by one, we confirm that all of the data is there; we are guaranteed to have the right answer; well done.2352

OK, let's do one more.2360

C6H12O3: so what do we do first?2362

Since we don't have an IR to analyze, we will go straight to the formula, and we will ask ourselves, if it was saturated...?2367

The formula would be C6H...2n+2...that is 12+2; it would be C6H14.2374

The actual formula is C6H12, so we have 2 hydrogens missing.2384

That tells us we have one degree of unsaturation.2395

So, what does that mean?--our structure has to have a ring, or our structure has to have a double--it must have one of those two in order to come up with the right number of hydrogens.2399

There is our formula; now we go to find our pieces.2409

We have a 3-hydrogen and another 3-hydrogen signal; that means I have two methyl groups.2414

OK, these are very close to each other, but (can you see?) one is a triplet (1, 2, 3), and then this one is a doublet.2422

We can point those out; they are not quite overlapping; they are very close to one another.2431

What other pieces do we have?--we have a CH3 (you can write them down here, or you can write them up here--wherever you are most comfortable--eventually you have to get them kind of close to each other, so you can work with them).2435

And then, what do we have here?--we have a CH.2450

Remember, a CH has three arms to make four bonds.2460

And then--two-hydrogen signal--we have a CH2.2466

It looks like this is pretty far to the left--pretty far downfield--but notice, I'm just showing 0 to 5 here; so this is not as far down as it might appear.2470

We have a CH2--good; any other pieces?2480

We have 1, 2, 3, 4, 5 carbons, but we need 6, so there is an extra carbon; we need 3 oxygens; and what else do we need?2485

We need a degree of unsaturation; so I think I have an idea of how to consolidate some of these little pieces.2499

What can you do with an extra carbon, and an oxygen, and a degree of unsaturation?--I think we can turn it into a carbonyl with two arms.2508

All right, a carbonyl itself doesn't show up in the proton NMR; all we see are the protons; so we have to work with the formula to deduce the presence of a carbonyl, in this case.2521

But we still have 2 oxygens (remember, oxygens each have 2 arms); OK, where do you think those oxygens are attached?2532

What pieces do you think have oxygens on them?2539

Well, we can look down here, and we see that an oxygen would cause it to be pretty far downfield, right?--so, I'm thinking that this methoxy group probably is a methoxy group; it's a singlet, and it's all the way down past 3, so this looks like an OCH3.2543

And these two look like they are attached to an oxygen; we only have 2 oxygens to work with, so we can't put one on every single piece.2566

OK, but I know those are attached to oxygens, as well; so that is something that we can keep in mind.2576

OK, but to really put things together: I know I have an OCH3, so we can cross those out and make that its own piece.2582

We have an OCH3 group, for sure, as one piece.2591

And let's take a look at one of these other methyl groups and start building from there.2596

So, what do we have for this?--this one CH3 is a triplet--what does a triplet tell you?2602

What piece?--here are our pieces; it could be attached to another...well, it couldn't be attached to another methyl, because that is bringing two end pieces together, but we could attach it to the carbonyl, the CH, or the CH2.2609

What is it attached to?2620

It must be attached to the CH2; that is what makes it a triplet.2622

OK, so now, we have taken away that piece; we have taken away another methyl.2627

And then, let's jump to this second methyl group.2631

This other methyl group is a doublet; what makes it a doublet?2636

This doublet must be next to the CH (just one neighbor), so it has two other arms.2642

OK, so now, we have come up with 1, 2, 3, 4, 5 pieces; how can we put these together to have them make sense?2652

Remember, we said that this CH2 needs to be next to an oxygen, and this CH needs to be next to an oxygen.2662

So, there are a few ways we could put this together.2673

One of them is: we can have this oxygen, then attach to our CH; and we said this has a CH3; so that way, this oxygen is shared both with the CH2 and the CH.2677

All right, so that would take care of all those, and then we have an OCH3 and a carbonyl--so the carbonyl must come next.2701

We could have that, and there is actually one other possibility that would also work in this case, because you have this ethoxy group and this methoxy group.2715

They can be kind of interchangeable; so we could have a CH3 over here, and then a carbonyl, and then the OCH2CH3.2730

This would not really have much effect on the NMR, because we would still have the same functional groups present, and still have the same splitting patterns.2743

OK; now, if we had our tables, we would be able to calculate them more precisely; and it turns out that our correct answer is the one with this methoxy group here, so it's actually the ethyl ester with a methoxy in this position.2760

And you would be able to calculate for this CH2 (for example): is this CH2 next to the oxygen of an ester, or is this CH2 next to an oxygen just of a plain ether?2776

So, if you were given NMR tables, where you can calculate those, you would be able to more precisely approximate that this is coming at 4.2; this is going to be more consistent with a 4.2.2789

OK, but let's confirm the rest of these, OK?--let's label these a, b, c, d, e (put d and e up here), and let's see if we can label them one at a time and confirm that we have the right answer.2804

a...we have a CH2...we can start with our structure; sorry, I'm jumping around a little bit.2823

We could start with our structure, so here we have a CH3; what would we predict for this spectrum?2828

A 3-hydrogen signal--it is a splitting's a singlet; where do we expect it to come--what chemical shift?--it's next to an oxygen, so that brings it around 3.8.2834

And do we have that?--that is right here; it's a little closer to 3.3.2849

So, we can modify that; and these are protons c.2854

How about this CH3--what do you expect for that signal?2861

It has just one neighbor, so we expect that to be a doublet; it's attached to an ordinary carbon; even though there are some groups attached to that, we would expect it to shift a little bit, so somewhere around 1.2865

It turns out that it's a little closer to 1.4.2877

That makes sense: this is a doublet, and so this is peak d.2883

This proton up here has how many neighbors?--it has 0 neighbors, and...I'm sorry, it doesn't have 0 neighbors; you look at the carbon (that's why it wasn't making sense!).2892

The carbon--what carbons are attached to that?--there are 0 neighbors over here, and 0 neighbors over here, but this carbon has 3, so that is why we expect that to be a quartet.2913

Luckily those neighbors showed up, because I am looking over here, and our 1-hydrogen signal is a quartet, so it must have 3 neighbors.2924

Of course, that means it is attached to a methyl; so this is peak b.2929

Does a chemical shift make sense?--yes, because that carbon is attached to an oxygen, and it is attached to a carbonyl; so we expect that to be pretty far downfield.2934

This CH2 is a 2-hydrogen signal; what is its splitting pattern?2948

How many neighbors does it have?--it has 3 neighbors, so this we expect to have a quartet.2954

And what chemical shift do we have attached to that oxygen?--we expect it to be about 3.8; that oxygen and a methyl ester brings it even further down, so that is evidence of this structure being the better one.2961

This is going to be a, peak a.2972

And finally, this last methyl group has 2 neighbors, so we expect it to be a triplet; and it's just attached to an ordinary carbon, although there is an oxygen after that; so, rather than at 1, we are seeing it at somewhere...where is our triplet?--it is this first peak, so 1.3 or so.2977

And that is peak e.2997

OK, so one by one, we can confirm it; in certain cases, if we are not given NMR tables to do precise calculations, we might find more than one possible answer, like in this case.3000

But if we are given those more precise tables, we could really narrow it down to which is the best match.3011

OK, and let's try this one: C8H18 is our formula.3020

If it was saturated with 8 carbons, what would we expect?--we would expect this is asking, if saturated...?--it would be 16+2; it would be C8H18.3030

And guess what?--our formula is C8H18.3041

So, there are no degrees of unsaturation; that means this molecule has no double bonds, and it has no rings; so it's just a plain old hydrocarbon--carbons and hydrogens.3045

So, what pieces do we have?3057

We have a 1-hydrogen signal; that is a CH.3060

It looks like a multiplet here...well, not a multiplet; it looks like there are many, many signals; this is blown up--so a lot of splitting here--a lot of neighbors; but we will worry about that later.3067

A 2-hydrogen signal means I have a CH2; a 9-hydrogen do you think we can come up with a 9-hydrogen signal?3074

It must be 3 methyl groups that are all equivalent, if they are all coming out to be one nice signal.3084

And how about 6 hydrogens?--6 hydrogens could be a CH2 and a CH2 and a CH2--that would require a lot of molecular symmetry--but a more likely option, and one that is going to work in this case, is 2 methyls, 2 CH3s.3092

OK, so that is all of our hydrogens; that is 10; that is 18 hydrogens; how about our carbons?3110

We have 1, 2, 3, 4, 5, 6, 7 carbons, but actually, we need 8.3114

Another piece to our puzzle--it's very important that we check that, because in addition, if we just tried to put those pieces together, it is never going to work, because we are missing a carbon.3120

OK, so now, let's try and put those together.3129

We can start anywhere: our 2 CH3s here are a doublet (we have two peaks there, so that means they must be attached to this CH; that makes them a doublet).3133

That takes care of this piece and those pieces.3157

This CH2 is also a doublet; so what does that mean it's attached to?3163

That must be attached to this CH, to have just one neighbor.3170

And then, these three methyls--how do we get three equivalent methyls that end up being a singlet, to have no neighbors?3177

Well, we'll put them all on this same carbon (let me draw them separately so we still have our pieces together).3183

OK, and this can't be a hydrogen, because this carbon had no hydrogens on it; it was just a carbon attached to all carbons.3200

And so, how do we finish off this molecule?--well, we simply attach the two pieces, because we have already used all of our pieces; so it's this half connected to this half, and then we're done.3206

I can draw it as a line drawing.3219

We have a CH3CH3CH, and then a CH2, and then a carbon with no hydrogens on it, but three methyl groups.3226

So, let's make sure this makes sense: here we can label these a, b, c, d, as our four different types of protons in this molecule.3234

Ha is a 1-hydrogen...I'm sorry, we'll go backwards here, just so we can label them.3249

Let's look at this one first: these two methyl groups we expect to be equivalent, so we're going to get a 6-hydrogen signal.3256

What splitting pattern do we expect?--it has just one neighbor, so we expect this to be a doublet; and notice that all of these chemical shifts...this is 1 (I'm sorry that's so small), 2...all of these chemical shifts are very similar, coming around 1, because there are no π bonds, there are no heteroatoms, there is nothing to cause any deshielding.3264

So, they all have very similar...the chemical shift isn't going to help us, in this case, on deciding who gets connected to whom; it's simply based on the splitting patterns.3288

We expect this to be a doublet, and it is; so where is our 6-hydrogen doublet?--right here, peak d.3296

And what do we expect for this hydrogen?--it's going to be a 1-hydrogen signal, and how many neighbors does it have?--a lot of neighbors!3306

It has 3, 6, 7, 8; it has 8 neighbors, so it's going to be 9 peaks.3317

It's going to have 9 peaks; now, take a look at this spectrum, and if you look very, very carefully, you can see, actually, there is a little bubble down here: 1, 2, 3, 4, 5, 6, 7, 8, 9; so you can maybe, maybe see those.3329

In all likelihood, you are not going to be able to see those; but you can see how tiny, how shallow, that peak is, because there is only one hydrogen accounting for it, and the area is spread out over those 9 peaks, so they are all going to be extremely short and small.3342

And those satellite peaks, those end peaks, are probably going to be indistinguishable.3356

But you just would have to trust that there are many, many peaks there, even if you can't number them all--even if you can't pick them all out.3361

OK, so that is good; so that is peak a--that is signal a.3371

How about this CH2?--it would be a 2-hydrogen signal, and how many neighbors does it have?--no hydrogens on this carbon, and one hydrogen on this carbon, so we expect it to be a doublet.3376

Sure enough, we have a 2-hydrogen doublet right here, signal b.3388

And then, finally, we have all three of these methyls; they're equivalent, so we have a 9-hydrogen signal here; and then, that is going to be...we take a look at how many neighbors, and here is the carbon they are attached to; this has 0 hydrogens on it, so that is going to be just a singlet--no splitting at all for those 9.3394

And where do we have a 9-hydrogen singlet?--right here, peak c.3417

OK, so bit by bit, we are going to determine our pieces; we are going to start to put them together; eventually, we are going to come up with a complete structure--a complete molecule.3423

Now, let's talk a little bit more about some advanced patterns; we are going to see advanced splitting patterns, and in order to do that, we need to have a better understanding of these coupling constants (or they are called J-values).3433

The magnitude by which one proton is split by its neighbor has to do with the spatial arrangement of those two neighbors: it's called their dihedral angle.3446

So, when you are looking at a typical neighboring situation here, of just vicinal hydrogens (hydrogens that are on neighboring carbons), that is going through three bonds; that is a typical kind of splitting.3457

That is the only kind of splitting we have seen so far, right?--looking at neighboring hydrogens.3473

That coupling constant is on the order of about 7 hertz.3477

Now, what we have here is: we have rotation around this carbon-carbon bond; so these two hydrogens have various relationships between the two of them.3481

And so, what you see is just an average effect.3490

And we get a splitting pattern of about 7 hertz; so that is just kind of a medium number--all of the peaks we have seen so far, the splits we have seen so far--those are typical splitting patterns.3493

OK, but you can have some splitting that is higher or lower; when you have two non-equivalent protons on the same carbon...3506

Now, most CH2s are going to be identical, and so therefore, the two hydrogens would not split each other, because they are chemically equivalent.3515

But remember, we saw some cases where this hydrogen and this hydrogen are going to be diastereotopic, where replacement of this one would give a diastereomer compared to when you replace this one.3523

So, we saw some examples where it was cis and trans to another group.3534

Anyway, it turns out that if these are non-equivalent, then that means they can split each other; and because they are so close in space, and because of their relationship to each other, what we get is very large splitting, 10 to 15 hertz.3539

So, instead of two peaks very close to each other, you see two peaks that are spaced further apart.3553

OK, so this distance between the peaks in a doublet or a triplet or a quartet--between the peaks in a signal that is split--this is the J-value.3559

This is the distance between the peaks.3578

Distance between peaks is what I'm referring to when I say "the magnitude of the splitting."3580

OK, if we take a look at a cyclohexane molecule that is kind of locked in a given if I put a t-butyl group on the ring, that means it can't undergo ring flips anymore, because that t-butyl group locks it with the t-butyl in the equatorial position.3591

OK, if it was flipping, again, we would just kind of get an averaging out of all the different coupling constants; but when it's fixed, what we now have is a fixed relationship between these various hydrogens, OK?3608

And just looking at a and b and c, realizing that there may or may not be additional protons--but just looking at that splitting, having one axial and another axial (that is a dihedral angle of 180 when you are looking down that bond--one is straight up and one is straight down; that is being anti to one another, 180 degrees), that is a very large (10 to 13) coupling constant.3621

There is large splitting there.3645

OK, but looking at b to c (I'm sorry, I have a typo here: rather than a to c, if we are looking at b to c), so one that is axial and one that is equatorial (or if you have two equatorial hydrogens near each other), those dihedral angles are smaller or closer to 90.3650

And when they are 90, you don't have any splitting at all; and so, we get very, very small splitting when you have either two protons that are both equatorial or one that is axial and one that is equatorial.3677

OK, but this trans splitting, we see, can be very large--this trans-diaxial.3690

OK, so when you take a look at Ha, if you were to ask, "What is the shape of Ha?", if you just looked at its relationship with b and c and discounted the fact that you might have additional splitting by other neighbors...remember, we said that geminal (meaning on the same carbon)--this is also a very large splitting; these would not be equivalent protons.3695

One is cis to the t-butyl group; one is trans to the t-butyl group; so this is a case where we would experience splitting between these two hydrogens on the same carbon.3720

And this axial is also a large splitting; so what we might get is an apparent triplet, because it's like we have two neighbors--one that is on the same carbon, and one that is on the neighboring carbon.3729

We have two neighbors; both have the splitting about the same magnitude; so it ends up looking as if we had two neighbors that are both the same, and therefore have the same J-values.3745

OK, so let's see some other examples.3758

What if we have hydrogens on a carbon-carbon double bond--what kind of splitting do we have for those neighbors?3761

Well, again, if these are non-equivalent (a and b), they will split each other; but in this case, when it's an sp2 hybridized carbon, this is very, very small splitting--1 to 2 hertz.3768

1 to 2 hertz is so small that, depending on your instrument...if you have a very high-field NMR, you can see that splitting; and a lot of times, it just kind of looks like a little jag.3778

Each peak has a little, tiny splitting--a little jagged edge--that is one to two hertz; it's very, very small.3792

But, if you have a lower-field instrument, you might not even see that splitting at all; it might just look like a single peak.3800

OK, so this is very, very small; it was big when it was sp3 hybridized, when it was a tetrahedral carbon, but it's very, very small when it's sp2, when it's on an alkene--small or even nonexistent, depending on your instrument.3806

OK, the cis relationship is kind of a medium splitting, 9 to 12 hertz.3820

And the trans is going to be a very large splitting--14 to 16--very, very large.3828

OK, and again, it has to do with the dihedral angle; this is kind of like the diaxial; we are seeing they are 180 degrees from each other, and that causes a very large splitting.3835

Some other splitting we can have occurs with remote protons; so they don't have to be on exactly neighboring protons.3847

If we have an allylic relationship--so this hydrogen can couple with this allylic proton--again, very, very small for this long-range coupling, but this is going through four bonds.3855

These are pretty far away; this is not neighboring--it's on the next carbon over--but it is still possible to have some splitting.3870

So again, you just sometimes might see that little doubling, and that is not something to be shocked about; that is maybe some allylic coupling.3876

OK, by the same token, you can have splitting between two protons that are meta to each other, that are not on adjacent carbons, but separate.3884

This is for rigid structures; we describe it as W-coupling, because you kind of have this shape, this W shape.3895

You can have it for some other rigid structures; here is our W; so when it has that relationship, that spatial arrangement, it's something that can cause an interaction, and therefore a little splitting.3901

And again, in a cyclohexane situation, where you have a locked conformation, these two equatorial protons--even though they are not neighbors for each other, we can see a little of this long-range coupling, because we have that W shape.3913

OK, the same thing as a and b: a prime and b prime can do some splitting.3927

OK, so we can have some long-range splitting; we'll see one example of that, coming up.3933

OK, so knowing that these J-values, the coupling constants, are not always the same, how does that manifest itself in looking at an NMR?3940

What do we see?3951

OK, this n+1 rule that we have seen is only applicable if all of the neighbors that you are looking at all have the same relationship with the signal in question.3953

OK, if they all have the same coupling constant--let's say they are all 7--then, for every neighbor you have, you have n+1 peaks.3965

So, that is OK; so for example, if we take a look at this hydrogen, and we see how many neighbors it has...well, it has one neighbor over here; it has one neighbor over here; and in this case, they are equivalent.3976

But even if this molecule were a little different, if these were close enough, you are still going to observe a triplet if they both have the same coupling constant.3990

So, what is going to happen is: your H is going to be split; it is going to be split by one neighbor, by 7 hertz; and then, it is going to be split by that second neighbor, again by 7 hertz.3998

And that is how we end up with this very nice triplet peak, because those middle peaks line up exactly, because the second splitting is exactly the same as the first splitting.4016

OK, so this happens, in this case, because we have two neighbors that are exactly the same; even if they were slightly different, we would describe this as an apparent triplet, because the coupling constants are so close.4026

So, where is this proton, this...we can label that...Ha?4039

It is going to be a triplet; here we have our triplet, coming at about 1.6, because it has some oxygen somewhere nearby, but not directly attached to its carbon.4046

OK, while we are here, let's look at the rest of this and see if we can assign these peaks, just for a little more practice.4057

OK, it's a symmetrical molecule, so we are only seeing half the number of peaks as the number of protons here.4063

So, that is actually...there are two hydrogens here; so this would be a 2-hydrogen signal; where is this proton, Hb?4073

What is Hb going to look like?4083

That carbon is directly attached to an oxygen; how many neighbors does it have?4086

It has 3 neighbors; we typically do not couple with OH's or NH's, so it just has these three neighbors, so we expect that to be a quartet at...I don't know...3.8 or something.4094

Do we have that?--we have...oh, I'm sorry, it doesn't have just these neighbors (thank you): I'm looking at this, saying, "That doesn't look like a quartet!"4105

It has three neighbors over here, plus two neighbors over here; so this has five neighbors, so this should be a sextet.4115

It has 1, 2, 3, 4, 5 neighbors; and sure enough, 1, 2, 3, 4, 5, 6; OK, that makes sense; so that is signal b.4124

And what do we expect for this methyl group?4133

This isn't showing any integration, so we don't worry about how many signals are in each, but what do we expect for its chemical shift and its splitting pattern?4137

It has just one neighbor, so we expect it to be a doublet and somewhere around 1 (in this case, it's 1.2); and there is c.4145

OK, so the rest of this makes sense; but just kind of looking at...I'm sorry, what is this mystery peak here at 3.6?--see this kind of broad peak here?--that is exactly what our OH looks like.4155

It is typically not going to be split by anything, and very often broad; so it will be a singlet, and sometimes it's wide like that.4170

OK, so if your neighbors all have the same...4178

The same thing here--this is a sextet: even though these three neighbors are not the same as these two neighbors, they have essentially the same spatial relationship, and so we count them all together, and we apply the n+1 rule, and we end up with six signals.4181

OK, let's see an example where that is not the case--where the J-values are not the same.4198

OK, again, let's just look at this signal, at this proton resonance, for comparison.4203

OK, this still has two neighbors: it has one neighbor to which it is trans on a double bond--remember, that is where we found it was a very large coupling constant, and the coupling constants are given here (this has a J of 15.1).4212

And it has a second neighbor over here, but this is now just kind of a more typical relationship of two protons, so this has a splitting pattern of 6.2.4230

So, let's take a look at what happens in this case.4245

We have our signal; let's call this proton a--we have our signal for a; now, that signal is going to be split by a large coupling constant of 15 to make it a doublet.4247

And then, it's going to be split a second time, but with a smaller coupling constant of only 6.4267

So, what does the final pattern look like?--it is a doublet and another doublet: we describe this as a doublet of doublets, or a dd for short.4277

We have a large split and then a smaller split; where is this on our spectrum?--it's right here--I blew it up, and you can see it.4294

It's two peaks here and two peaks here; this distance is the 15, and this distance in here--the smaller distance--is the 6.4303

So, we call it a doublet of doublets.4324

Now, when you look at that peak--when you look at that signal--why is that not a quartet?--why would you not look at that peak and call it a quartet?4325

It has 4 peaks; what makes it not a quartet?4334

Well, there are two things: the spacing is not even between all of the peaks--sometimes it might end up a little closer, so that isn't always true, but tell me about the ratio of the peaks.4340

In a quartet, do you expect to have four equal-sized peaks, or do you expect to have a 1:3:3:1 ratio?4352

That is why it is not a quartet: when we look at that signal, we must recognize it's a doublet of doublets.4361

It has one neighbor of a large coupling constant, and a second neighbor of a smaller coupling constant.4367

OK, so we can have these really advanced patterns.4374

Let's take a look at some other protons here.4378

If Ha is splitting this proton by 15 (let's call it b), then Hb must be split by Ha by the exact same amount; that is why it is called a coupling constant.4383

It is a coupling constant between two protons; so they split each other by that same magnitude.4395

So somewhere, we must have a doublet with a large splitting pattern--with a large splitting--and that is right here; this is Hb.4400

And it is a nice, simple doublet; and this spacing is as wide as this spacing.4411

OK, that is a trans coupling--that large coupling that we can see: very nice.4420

OK, let's take a look at another interesting example of this: this is a spectrum that I acquired myself, so we have a scan of the actual spectrum.4426

You can see that you can get the peaks to be picked, and then you can do some math, find the difference in this, and multiply it by the strength of your magnetic field--what instrument you are looking at.4437

And then, you can actually calculate those J-values; you can calculate them.4450

OK, but I see here...I see a doublet here and a doublet here; let me ask one question, right away.4455

Do you think this proton is a neighbor of this proton?4462

In other words, they each have one neighbor, and they are splitting each other: is that something that makes sense?4466

Do you see that this is a shorter distance--this is a smaller splitting than this?4473

Because that distance is not equal, they cannot be splitting each other.4479

These two protons have other neighbors to which they are splitting.4484

OK, so let's take a look at these vinyl protons: we have these protons.4490

Remember that this geminal splitting is very, very small; in fact, it is not observed at all in this spectrum, because the NMR was not powerful enough.4498

OK, so what do we expect for each of these peaks?4507

Let's call this a and b and c; what do we predict for the NMR--what do you expect for this one?4510

How many neighbors does it have?--well, it has two neighbors, but it has one that is cis and one that is trans; so we expect, not a simple triplet, but a doublet of doublets.4521

It is going to be split into a doublet by one proton and split into a doublet by the other proton, and we are going to expect a large splitting--a large J and a medium J--the large J for the trans and the medium J for the cis.4536

OK, how about Hb?--Hb also has two neighbors, but remember, this geminal splitting is...let's just make a little note that this geminal...J equals 0 in this case, essentially, because we are not able to resolve that.4559

So, how many neighbors does it have?--it just has one neighbor, Ha, so this is going to be a doublet.4575

And is it going to be a large J-value or a smaller J-value?--it's going to be a medium one, because this is cis--a medium J-value.4582

And how about Hc--this is going to have how many neighbors?--we ignore this neighbor; it is not going to be doing the significant splitting; but here, we have this trans neighbor.4592

So, this is going to be a doublet, and it is going to have a large J, because it's trans.4604

Can we find these protons?4614

It is going to be these three protons down here; you can see the integral trails.4618

OK, do you see that this here is one height?--here is one height; here is one height; here is one height; and here is another.4624

This is actually one hydrogen, one hydrogen--these are all equal, and these peaks down here are two hydrogens each.4638

You can kind of split it in half; you see this is about twice the height; two hydrogens, two hydrogens.4653

OK, so these are our vinylic protons here, in the range of 5.4 to 6.8; which is which?4661

Well, we have two that are just simply doublets, and that is going to be protons b and c; the one with the smaller coupling is Hb, the one that is cis to Ha; the one that is trans to Ha is going to have the larger coupling, so this is Hc, a doublet with the larger coupling.4669

And then, here, this is a doublet of doublets; again, we have four signals, all the same height, so this is a doublet of doublets; so that must be Ha.4693

And so, what has happened is: we started with Ha; we split it by Hc (it's a little easier to do the larger coupling first--this is Jac, which is the larger one).4705

And then, we split it by Hb, which isn't that much smaller, so they probably overlapped a little bit, like this, to get something like this.4721

See how these middle two peaks are kind of closer to each other?4732

We get that pattern; so this second splitting is Jab, which is our medium splitting.4737

OK, so I know this is a lot to go through all at once; I just want to make sure you have some exposure to these more complex splitting patterns, so that you can tackle them when you come across them.4747

Before I leave this spectrum, let's take a look...I didn't show the rest of the spectrum for this carboxylic acid proton; that was missing.4757

This extra peak is just my solvent; I used deuterochloroform here, so this is a little left-over chloroform; that is not uncommon to find; that comes around 7.26.4766

In fact, sometimes you can even use that as a reference, if you are expecting that, depending on your solvent; so that is why I am ignoring that peak.4778

But let's take a look at these aromatic signals: we have four hydrogens here, and this molecule has some symmetry; so we expect...there is some symmetry here, so we expect these two to be equivalent.4785

Let's see, let's call this proton d and proton e.4802

But take a look: there is a huge difference in their chemical shift.4815

One is at 7.5, and one is 8.1--a huge difference in chemical shift.4819

Do you think you can figure out which is which?--how would you begin to assign these?4826

Now, actually, if you have some access to some spectroscopy books, you can come across some tables that will help you calculate these, just like we did for methane or methylene peaks, alkyl peaks...4831

You can look up tables for variously-substituted alkenes like this, and you can look up protons that are on benzene rings.4845

If it's ortho to this kind of functional group or meta to this kind of functional group, para to this kind of functional can add up all those various effects and estimate this.4854

OK, but let's assume we don't have access to this: how could we begin to describe the difference in these electronic environments?4862

Well, we have a carboxylic acid here; we have a very strong...this is going to be explained by resonance, and so we see that we have a very strong electron withdrawing group attached to the benzene ring.4870

So, how is this carbonyl interacting with the benzene ring?4882

It is withdrawing electron density: this has resonance.4885

We talked about the effects that resonance can have on chemical shifts; and so, let me just draw this resonance form and see if we can explain the difference here.4892

OK, that resonance form does what to proton d?--does that make it an electron rich environment or an electron deficient environment?4905

It makes it very electron poor; and what does that do to your chemical shift--is that shielding? deshielding?4914

If you are electron poor, that means you are deshielded; and what does that do to your chemical shift?4922

That moves it downfield.4928

So, Hd is going to be the one that is all the way here at 8.1, and He does not have that effect with the electron withdrawing group; that is not impacted by it, and so that has the more typical closer-to-7 for its chemical shift.4934

Splitting pattern for those--we many neighbors does Hd have?4953

It has just one neighbor, so we expect it to be a doublet; He, same thing.4958

Notice that the coupling constants--the spacing of the doublet for Hd must match the spacing for the doublet of He.4963

OK, let's wrap up with a few more predictions and see if we can come up with a structure by deciphering an NMR with some more complex splitting, in this case.4976

OK, we are going to go all the way back to our beginning steps, though: the first thing we are going to do is look at our IR data, if we are lucky enough to have that, and look at our chemical formula.4990

Here we have C11H17N; so we are going to ask ourselves, "If it was saturated, what would the formula be?"4998

C11H...2n+2, but remember, if you have nitrogens, it's CnH2n+2+the number of nitrogens; we need to adjust that count for every nitrogen.5006

So, 2n+2 would bring us to 22, 23, 24; and then, plus your nitrogen: it would be 25.5023

2n is 22; so this is 22+2+1; that is how we came up with 25.5033

So, if it were saturated, it would have that formula; our actual formula is C11H17, so we are missing 8 hydrogens.5044

8 missing hydrogens means how many degrees of unsaturation?--every two missing hydrogens is one DU, so we have 4 DU.5060

OK, we will keep that in mind when we do our pieces.5070

And what I see right away: we have these scattered four peaks all around 7; what happens when we have peaks around 7?--what do we think it might be?5074

It might be a benzene ring; if we have 4 degrees of unsaturation, then that would allow for a benzene ring; if we have four aromatic hydrogens, what does that tell us about our benzene ring--what does it tell us about our puzzle piece?5085

We must have two groups attached.5098

Now again, we could just attach those two groups anywhere to start; we will look more closely at these peaks--these have been expanded here--we will look more closely at those to decide what the actual substitution pattern is, here.5101

OK, but that is one piece to our puzzle.5114

Four hydrogens--how do you have four hydrogens?5117

It looks like a pretty clean peak, doesn't it?--it doesn't look like overlapping peaks; it looks like a very nice quartet, so that probably means I have a CH2 and another CH2 that is equivalent to it.5120

Three hydrogens means I have a CH3; 6 hydrogens means I have a CH3 and another CH3 that is equivalent to it.5134

OK, any other pieces?--we have 6, 7, 8, 9, 10, 11 carbons; we have taken care of that; we have our 17 hydrogens shown; and we need a nitrogen.5147

There is also a nitrogen--how many arms does nitrogen have?--it forms three bonds; so that is our other piece.5159

OK, so these are all of our pieces; we have accounted for our DU; we have matched our formula; now we are ready to start putting them together.5165

OK, of all these pieces, who do you think is attached to the nitrogen?5173

Well, that is going to be the one that is furthest downfield; so these CH2s are probably attached to the nitrogen--that would make sense.5180

That takes care of these pieces.5191

And then, what is the CH2 attached to?--we can continue the piece that way, or we could start with the methyls and work backwards.5195

This CH2 is a quartet (1, 2, 3, 4); so it must be attached to a CH3; and in fact, those are the only pieces left (CH3s), so this is one where we really don't have many options.5201

Once you get all of your pieces, there really are not that many options on how to put them together, because there are so many end pieces.5214

So, in fact, we have these two ethyl groups; and now we have two end pieces, and we have a benzene ring needing two pieces; so we attach them all together.5221

OK, it turns out--let me just show you the actual pattern; they are actually meta to one another, like this (sorry...pen mistake).5232

They are actually meta to one another, and we will see that when we examine these aromatic peaks.5249

OK, but let's just double check to make sure everything makes sense.5256

What do we expect--what do we predict?--remember, this is our last step before we are done; what do we predict this methyl group should look like in the NMR?5261

Should it be a 3-hydrogen signal?--what would its splitting pattern be?5268

It's attached to just a benzene ring, so there are no hydrogens there; it's a singlet.5274

And what chemical shift do we expect for a benzylic hydrogen?--we expect it to be about 2.2 (where is it?--it's 2.4: perfect).5279

OK, so that is good; we can label these, if you want: a, b, c, d, e, f, g; let's label those, so we can identify them all.5291

So that is signal f.5305

And how about this CH2--what do you expect for that CH2?--how many protons should be in this signal?5312

Well, there are two equivalent ethyl groups here, so we expect a 4-hydrogen signal.5318

What should its splitting pattern be?--it has 3 neighbors; it is next to this CH3 group; so we expect it to be a quartet.5326

And its chemical shift--well, it is attached to a nitrogen, so we could look that up: not quite as deshielding effect as the oxygen, because it is not quite as electronegative, but still, we see it coming somewhere around 3.5; so that looks like signal e.5335

That makes sense.5354

And how about this methyl group?--this, now, would be how many protons?--this would be 6 hydrogens.5355

How many neighbors does it have?--it has two neighbors, so it would be a triplet.5362

And where do you expect it to be?--nowhere special, just kind of like an alkane--somewhere around 1.5367

And, in fact, that is right where it is: 1.2 ppm.5372

So, this looks like it is proton g, very nicely.5375

OK, so these make sense: let's see if we can figure out--let's see if we can assign these other hydrogens (these aromatic hydrogens), based on their splitting pattern.5380

Now again, their chemical shift--we could do that if we had those tables; we can see what effect this nitrogen has, and this methyl group, and we could also estimate that.5389

But, just based on the splitting patterns, let's see if we can predict it.5399

OK, what splitting pattern do you expect for this hydrogen?5403

How many neighbors does it have?--it has none here and none here, so we expect that to be a singlet.5407

Do we have any singlets?--now, you will notice: see, this is a little messy; this has some little extra jags, but this one is the closest to a singlet; this is essentially a singlet.5417

This is going to be hydrogen c.5431

How about this hydrogen--how many neighbors?5437

It has just one neighbor, so we expect that to be a doublet.5441

This hydrogen has two neighbors; they are not identical (chemically equivalent), but they are very, very similar; so we expect this to look like a triplet.5447

And this last one looks like a doublet.5459

OK, so where do we have a triplet?--this one is a triplet; doublet; singlet; doublet; so the triplet we can pick out as being Ha, because that is a triplet, very nicely.5463

These last two doublets, though--proton b and proton d--it is going to be hard to distinguish those.5476

But if we think about resonance effects, resonance effects are going to take this nitrogen as an electron donating group.5482

It is going to add electron density; so what does that say about this proton here?5489

Is that going to shift it closer to 0 or further down?--this makes it electron rich (right?--because it has a partial minus there): what does that do?5494

That shields it; that pushes it higher up; so this is going to be d, proton d, and this one is proton b.5505

I'm sorry, I'm mixing my d's...this is our doublet, and this is our doublet; so this is proton b, and this is proton d.5518

OK, so that one is a little more subtle.5528

But the other thing I wanted to point out, that is interesting about this, is not just the splitting patterns.5530

So this explains the substitution; but then also, if you take a look at...where does this extra splitting come from?--why do we have some extra splitting here?5535

Why is this one...this was proton b; why is b not just a plain old doublet?5545

OK, that is because b also has this meta relationship with two other protons; and remember, we can have some of that W-coupling with the meta--a very, very small coupling, if you have a sensitive enough instrument.5552

So, because it has two neighbors with this very, very small J-value, that is why we get this extra splitting.5566

Do you see that those kind of look like triplets?--you can maybe see it down here, even, before it is expanded.5573

OK, so what we have here is a doublet of triplets, with a larger J-value and a very, very small J-value.5579

OK, and we have the same thing over here, because proton d...proton d...same thing: it has those same two neighbors, so we get that same splitting.5589

This small splitting is the same as this small splitting; and why is this singlet--this was proton c; why does proton c have a little extra splitting?--well, because that has two meta ones, as well.5599

That has an extra little splitting; so all three of these meta ones with meta protons (meaning 1,3) have that long-range W-coupling that you can see.5612

Why doesn't Ha have that?--Ha is this nice, beautiful, clean triplet; why is it?5621

It only sees these two neighbors to be a triplet; it has no additional splitting; that is because in this position there is no H; in this position there is no H; it's a nice, clean triplet, so we don't see any of that long-range, very, very small couplet.5626

This is a really interesting NMR.5640

OK, and let's try one more and see if we can predict this spectrum.5645

C9...I've blown up a little bit of it--this portion has been enlarged; let's increase our ppm's (I'm sorry these numbers are so small) so you can get a feel for our chemical shifts.5652

Let's start by looking at our formula: what can you tell me about C9H10?5666

Well, if it was saturated, C9 (9 carbons) would have how many hydrogens?--2n+2, 18+2, is 20.5671

In fact, we have C9H10; so there are 10 missing H's.5682

This formula tells me that I have 5 degrees of unsaturation; that is a lot of unsaturation here.5689

OK, what pieces do we have?--well, we have around 7...between 7 and 7 1/2; we have a 3-hydrogen signal; we have a 2-hydrogen signal.5697

And what do you think all of that adds up to?--I think I'm going to combine those 5 into an aromatic ring, a benzene ring with just one group attached.5710

This is an aromatic; it's a phenyl group with just one group attached.5724

I haven't blown that up; these are actually resolved--they are not overlapping, but we don't need to worry too much about that; we just looked at some nice splitting patterns for an aromatic ring.5729

OK, so that is one piece to our puzzle; what do we have here?--we have a CH; another 1-hydrogen signal is another CH; 2-hydrogen signal is a CH2; 2-hydrogen signal--CH.5740

OK, let's count up: we have 6, 7, 8, 9, 10 carbons; there is a problem.5763

We are only allowed 9 carbons; so where could I have made a mistake?5770

Well, I think this one hydrogen, instead of being a CH--do we have any other options?5776

We have an oxygen, so it must be an OH instead; and looking at it more carefully, I see, "Well, actually, that does look like an OH peak, because it has that broad shape to it--a broad singlet; so that is pretty reasonable."5782

Of course, it could be anywhere in this range--1 to 5--but the chemical shift is reasonable; the shape is reasonable; so it's an OH.5795

Of course, if I were lucky enough to have an IR, I would have seen that right from the beginning; but if you are working without it, you need to be able to come up with that piece on your own.5803

OK, so now, let's double-check the formula: we have 6, 7, 8, 9 carbons; we have our oxygen; we have 5 DU's, and we have only 4 shown, so we also need to have a degree of unsaturation.5812

Now, what is that degree of unsaturation?--it could be a double bond; it could be a ring; but when I take a look over here, and I take at the chemical shift of these two hydrogens, I see they are all the way down between 6 and 7.5834

That is really far downfield, and we only have one oxygen to attach--we only have an OH that can attach somewhere.5847

In fact, where do you think that OH is attached--what other piece do you think has the OH attached to it?5854

It must be this CH2, because it is above 4; so we can bring those together to account for this chemical shift.5861

So, what is bringing these carbons so far downfield--what else could there be?5877

Well, we have a degree of unsaturation; those hydrogens must be on a carbon-carbon double bond.5882

We must have a CH and a CH.5892

Now, let's take a look at our...we could either have a trans or a cis; let's just go ahead...rather than think about it too much in our minds, let's just draw out our two options.5894

These are the two puzzle pieces we can have: we could either have two H's that are cis to each other, or that are trans to each other.5909

And then, what are the two groups that are attached here?--well, we only have two other pieces to put together: we have a phenyl group, and we have a CH2OH.5917

So really, the only choices we have are whether it's cis or trans; otherwise, there is only one to put these different pieces together (again, assuming we have made this connection knowing that the CH2 has an OH on it).5927

OK, so what do you think--how could we decide between the two?5942

Well, we could take a look--what is the difference?--we expect this...when the two hydrogens are close to each other, we expect to have kind of a smaller coupling constant.5947

The cis is kind of a medium coupling constant/a small coupling constant, where, if they are trans, we expect them to have a large coupling constant.5957

I just noticed we have an extra line here; sorry for that.5970

We have a large coupling constant there; so if we take a look at the hydrogens, how would we explain these two signals--what is the splitting pattern here?5974

This one--we have a doublet; and how would you explain this one (it's blown up a little)?--do you see that we have 1, 2, 3 peaks, and another 1, 2, 3 peaks?5984

It looks like we have a doublet of triplets: we have a doublet that has been, now, split into triplets.5995

So, I guess we need to decide whether this looks like we have a large splitting here or a smaller splitting here.6002

Now, we don't have the numbers provided to calculate it; but let's take a look at our other splitting patterns.6012

See, this is kind of a normal splitting; that is on the order of 7 hertz or so--we would expect most coupling constants to be about 7.6018

And then, this one looks like it's larger than that, doesn't it?--this looks like it's greater than 7 hertz.6030

And so, what is that consistent with?--that is consistent with having a trans carbon-carbon double bond.6036

Let's put those pieces together, then.6044

Let's put these hydrogens trans to each other; and then we can put a benzene ring in one position, and then our CH2OH in the other position.6050

OK, so there is our molecule: let's see if we can assign our peaks.6062

We will just call all of this a; we are not going to get into distinguishing all of those--the aromatics; so we could just look at all of these and say, "Well, we have some multiplets; 5 protons there..."6067

We could split it out into the three, the ortho and meta and para; you really can see those details; but we'll skip that for now.6079

OK, let's call this b and c and d and e.6087

OK, who is who?--b and c are the two protons on the double bond; what do you predict for the splitting pattern and the chemical shift?6091

Now, we haven't gotten into calculating chemical shifts for protons on alkenes, but you can do that; and so, we would be able to distinguish which one would be at 6.3 and which would be at 6.8.6108

But we could just make the assignment based on the splitting pattern; what do you expect the splitting pattern to be for this proton?6122

How many neighbors does it have?--there are none over here; there is one over here; so we expect this to be a doublet.6129

And because this is trans, we expect this to have a very large J.6136

And so, b or c; what do you think that matches?--I think that matches b; let's circle these, so we can distinguish between our d's and our doublets.6142

OK, so this is proton a; the doublet with the large coupling constant is b; and it's vinyl, so that is why it is with this high ppm.6154

How about this proton--what do we predict?--how many neighbors does it have?6167

Well, it has one neighbor over here; that makes it a doublet with a large J; so for the same reason that this had a large splitting, the other proton would have to see that same large splitting.6172

But what other neighbors does it have?--it also has these two hydrogens, and two hydrogens makes it a triplet with a medium J--with a more typical J-value.6183

We are expecting a doublet of triplets, and here it is: here is our large J-value; if we kind of look at the top of this peak, the middle peak to the middle peak, we can see that large split, which matches this split.6194

And then, in addition to that, we have our triplet, which is smaller, that will match the other splitting we see.6207

OK, so this is a doublet of triplets; so that corresponds to proton c and the splitting pattern we see there; excellent.6215

CH2--what do we expect for this CH2?--it's a 2-hydrogen signal; this was just a 1-hydrogen signal; this was a 1-hydrogen signal; this was a 5-hydrogen signal.6226

This is a 2-hydrogen signal; how many neighbors does it have?--well, it does have this OH, but typically we don't see splitting through oxygen, so we discount that neighbor; it has just one neighbor, so we expect that to be a doublet.6240

And sure enough, it's a doublet with just normal splitting; that is proton d, and our OH is proton e--a singlet (a broad singlet, in this case) at about 3.7.6255

This is a pretty complicated molecule, but simple pieces; but sometimes it's a little tricky to put it together when you know you have a degree of unsaturation.6268

How do I bring these CH's together?--again, the best idea is to just jump in, try some possible options, and then sort through to see which ones best match the spectrum.6277

OK, finally, let's...this whole time, we have been talking about proton NMR, looking at the hydrogens in our molecules; but that is not the only nucleus that can be observed by NMR.6290

Another very commonly observed nucleus is the carbon nucleus; now, the only isotope of carbon that is visible in the NMR, that can do that flip of the nuclear spin, is C-13--that isotope.6301

So, it reads 13C NMR, but we read it as C-13 NMR; that is how we say it.6318

OK, here is a typical C-13 NMR, or one we might see.6325

As we saw with proton, what does it have in common with proton?--well, we still get one signal for each unique carbon.6331

So again, if there is symmetry, that makes one carbon chemically equivalent to another; they will coincide on the spectrum.6337

OK, our chemical shifts are a little different numbers; the protons were kind of 0 to 10; here we have about 0 to 200, or maybe a little higher--so just a much wider range of chemical shifts.6345

OK, the signals in most C-13 NMR you will see are typically all singlets; that is described as a proton-decoupled spectrum--that is the most common experiment that we have.6359

So, we just see a line representing each unique type of carbon.6370

OK, but it is possible to determine how many hydrogens are on each carbon by doing a different experiment called a DEPT experiment.6376

And using a DEPT analysis, you can learn for each signal--is that a CH3, CH2, CH, or just a carbon with no hydrogens on it?6386

So, it is possible to get that information; and very often, that information is provided to you, either by just telling you straight out, or by providing the DEPT spectra and having you interpret the DEPT spectra and just knowing the rules for that.6396

But I'm not going to get into that here.6410

OK, one last thing to note, though, is: since all we can see is the C-13 isotope, that only accounts for about 1% of the carbons in any structure; and so, you either have to have more sample if you want to acquire a C-13 have to have a larger-sized sample, or maybe just spend a longer time acquiring more scans on the spectrum.6412

So a lot of times, it takes longer in order to bring down the noise of the signal and get a good C-13 NMR.6436

Let's take a look at what affects the chemical shift of C-13.6445

It has the same concept as proton NMR, in that, if you are electron rich and shielded, those are things that come upfield and closer to 0; or down here, we have electron deficient carbons that are deshielded.6451

OK, so that concept still holds.6470

What we find on the far right, then, are our plain old alkane carbons; and if we attach a halogen or a nitrogen or an oxygen or something electronegative, we would expect that to cause a downfield shift.6473

Remember our little triangle for the upfield versus downfield; so going to the left would be a downfield shift.6485

Triple bond comes more around 80 or so, in that range; carbon-carbon double bonds--either for alkenes or for aromatic...for benzene rings--those kind of come in this range (100 to 150--something like that).6492

And then, at the very far left, what we have on the most deshielded are the carbonyls; so carbonyls are very easy to pick out on a C-13 NMR.6508

They are somewhere around 200; carboxylic acid derivatives are a little more shielded, because these groups all add electron density by resonance.6518

So, these are more electron rich carbonyls; that brings them a little to the right, a little upfield, where aldehydes and ketones are going to be a little higher than 200.6528

OK, so to look at some numbers: again, you would typically be provided with tables for these.6539

We see a slight difference, depending on whether it's a primary, secondary, tertiary, or quaternary alkyl group.6547

So, the primary alkyl groups are the most shielded, just like the methyl is the most shielded for a proton NMR; the same is true for C-13.6553

So, those are the ones that are furthest to the right, furthest upfield; they move down a little bit as we add carbon groups on it.6561

Attached to an iodine, bromine, nitrogen, chlorine, oxygen--depending on what heteroatom we have--we have slightly different ranges.6568

OK, but as you might imagine, those shift them all downfield a bit.6574

OK, then we have alkynyls a little further downfield still; alkene and benzene--not so easy to distinguish here, because they all kind of come at a very similar range (100, 150, somewhere around there).6579

And then, like I said, the carbonyls--depending on whether you are a carboxylic acid derivative, with a group on here that will add electron density, this is more electron rich, and that shields it and brings it to a lower number.6591

And aldehydes and ketones bring it to just usually a little above 200--at or above 200.6607

OK, and we will just work through a couple of problems; typically, C-13s are provided in conjunction with a proton NMR, and we saw how many proton NMRs we solved without even needing a C-13.6617

So, a C-13 just kind of adds some additional information; but it really should just confirm what you have already found, using the IR and the proton NMR data.6632

But it is possible, using the same strategies we had before, to solve a C-13 NMR; but there is a lot less information in here, so it's a little harder to put the molecules together if you are given only a C-13.6642

But let's try it in a couple of cases.6655

This first one--we'll start as usual: we will work with the formula, since it is given to us: C6H14.6657

If this was saturated and we had 6 carbons, how many hydrogens could we expect?--2n+2 is 12+2; it would be C6H14, and we have C6H14.6666

So, that tells us it's saturated; that means there are no π bonds; there are no rings (right?--that is what that lack of degrees of unsaturation tells us).6678

OK, and what we see here is: I have drawn, on top of each peak, what many hydrogens are attached to each; so this has already taken that DEPT information and incorporated it into the C-13 spectrum.6693

And so, this kind of tells us our pieces: we have a CH; we have 2 CH2s.6707

OK, notice, I am always drawing them with the appropriate arms, so I know what they could be attached to.6714

We have 3 CH2s; we have a CH3; we have 2 oxygens; OK, and then we add up.6720

Do we have our pieces?--we have 1, 2, 3, 4, 5 carbons; we have C5; we have 3, 4, 5, 6, 7, 8, 9, 10...H10O2.6731

So, actually, these pieces have not accounted for all of our formula: we are missing a carbon, and we are missing four hydrogens.6746

So, how could we make this work?6759

Well, one thing we can do is: we can have another CH3.6764

So, you see how this peak is a little bigger: sometimes that can give us a clue--sometimes the size of the peak is proportional to how many hydrogens are on there.6773

You see a little difference there; a quaternary carbon is usually a pretty small peak.6781

But, in these generated spectra, this is a bigger peak because there is more than one methyl group that is contributing to it.6786

So, just like the magnitude of the area under the peak is proportional in the proton NMR, the same is true for the C-13 NMR; but we are not given integrations in this case.6795

So, you wouldn't know for sure, other than maybe seeing a difference in the size, exactly how many methyl groups there are.6806

What we could do is: we could say we have two CH3s, and then we only have one extra hydrogen that we need to account for.6815

We have one extra hydrogen: now we have C5H10O2 adding up.6825

Now, where can we attach that hydrogen?--we can't put it on any of these carbons, because we have already indicated how many hydrogens are on each; so what we could do is: we can combine this into an OH group.6831

That hydrogen must be on an oxygen, and that is why it wasn't showing up in our C-13 with our DEPT analysis.6844

OK, so let's start putting them together: we have some oxygens to play around with--who do you think is attached to an oxygen?6853

Which of these pieces does it look like are attached to an oxygen?6860

It looks like we have a CH and a CH2 and a CH2 that are all pretty far downfield.6863

OK, so I think these all have oxygens attached.6871

And we have these CH3s; we have two CH3s that are equivalent; how could we have two CH3s that are equivalent to each other?6879

Well, we can put them on the same carbon; and which piece can they attach to?--they can't attach to a CH2, because those would be putting two end pieces onto a CH2.6888

So really, the only way we can attach these two CH3s is by using the CH that is given.6903

This CH...we put two methyl groups on it, and then that accounts for the fact that both of these are chemically equivalent, and they show up as just one signal.6913

OK, now what did we say about that CH--what do we think that CH is attached to?6924

I think it's attached to one of the oxygens, so that will take care of one of our oxygens; I can't put an OH here, because again, that stops the molecule.6928

I have to put on a piece that has more arms.6936

So, what can come next here--how many choices do I have?6940

There is only one choice: I can only put a CH2, and then the other CH2, and then the other CH2, and then, finally, the OH; there is only one way to put these pieces together.6945

So, the trickiest part here know, that also accounts for the chemical shift.6957

We can't use coupling; we can't use those splitting patterns anymore to decide who is attached to whom; we can only use chemical shifts when we have a significant functional group that alters that.6963

So really, a C-13 is really difficult to put a molecule together, except in certain circumstances like this (we can do a pretty good job there).6974

So, before we leave, though, let's just double-check: if you had to predict what the C-13 NMR would look like for this, what would we have?6982

Well, we would have...we can go ahead and label these: a, b, c, d, e; and make sure we can identify them and bring them together.6992

We should have two methyl groups that are chemically equivalent, that come as one peak, and that are just attached to a regular carbon (so they should not be downfield any great amount).7010

And so, that makes sense with this; it's coming in 22 ppm, so this is e.7021

Here we have a CH signal that is attached to an oxygen; so that should be kind of a higher number, so that looks like a.7031

And this is at 76 ppm.7041

There is not much you can predict--not a lot of specifics you can predict--for your C-13 NMR, other than the number of signals and approximately where it might be.7044

OK, and then we have a CH2; we have three CH2s; two of them are attached to oxygen, and one is not.7056

So, this middle one is the unique one, and who would that be?7064

That would be this CH2, the one that is furthest upfield; and then, the others, b and c--we is attached to an OH; one is attached to an O-R.7070

Those are going to have different effects in their deshielding; there are tables to calculate your C-13 chemical shifts, as well, so if you had access to a table, you could more precisely determine exactly which one is b and which one is c.7084

But, without having access to those tables right now, we wouldn't be able to distinguish between those two and decide exactly which is which.7099

OK, let's try one last C-13 problem: C5 H8O2.7110

If that were saturated, what would our formula be?--C5H...2n+2, 10, 12: C5H12 would be our saturated formula.7121

We have C5H8, so we have 4 hydrogens missing.7131

So, we have 2 DU.7140

OK, what pieces do we have?--we have a carbon with no hydrogens attached; we have a CH2...I'm sorry, we have four CH2s; those are all the rest of our pieces.7145

And then, we have two oxygens.7162

And we have 2 DU; so there are a lot of missing things here.7166

Now, the one thing I will bring your attention to is this: this one carbon is very far downfield--so far, it is pretty close to 200.7170

It is 172; so what kind of carbons show up that far downfield in the C-13?7179

It must be a carbonyl; so I can take care of some of these pieces by having a carbonyl.7185

And in fact, having it closer to 170, rather than closer to 200, probably means I have an in other words, I think that other oxygen might be attached here, rather than having a ketone; but we don't really have to jump there quite yet.7195

OK, so where is our other DU?7210

It could be a double bond, or it could be a ring.7217

Do we have evidence of either?--if we had a carbon-carbon double bond, what would we see in the C-13 NMR?7222

We would see peaks somewhere in here (100 to 150); that is where our alkene or aromatic carbons show up.7230

There is nothing here; there are no other π bonds, besides this one carbonyl; so there is no carbon-carbon double bond.7238

So, what does this DU have to be?--it has to be a ring.7247

It has to be a ring; so how can we arrange these atoms?7253

Do we have any evidence of a carbon attached to an oxygen?7258

We do: we have this CH2; it is the furthest downfield, so we have one CH2 that is attached to an oxygen; the other CH2s are not.7265

So, what structure do we come up with?--the only way to put these together is by making a ring; and have one end reach around and come up to the other end.7277

Rings are pretty tricky in the NMR, because a lot of times we are kind of building our structure.7292

If you ever find yourself with two end pieces, but you have run out of parts to put in, see if it's big enough to come together.7296

5-membered rings or 6-membered rings are very common in organic molecules, and so a lot of times that is going to solve a problem for you; and this is one of those cases.7305

OK, so we could double-check: do we have C5H8O2?--we forgot to do that before we started putting them together.7313

1, 2, 3, 4, 5; yes, we have 5 carbons; we have 2, 4, 6, 8 hydrogens; and we have our 2 oxygens, so our formula is correct.7321

Our chemical shifts are good; we would expect this--just like we thought, having it be an ester brings it more in the range of 170; and it's a carbon with no hydrogens on it, so this is a carbonyl, somewhere around 170.7332

And then, we have one CH2 that is α to an oxygen, attached to an oxygen; so that is at about 70 ppm; and then all of our other random CH2s are all ordinary alkyl CH2s coming in this range, from 20 to 30.7349

OK, so do get some practice with C-13: you will find some different exercises that you have to do with those.7365

It is not so common to have to solve a C-13 with no other data, but again, if you are lucky enough to have a C-13 provided to you, along with an IR, along with a proton NMR, make sure you use all of that information when you are coming up with your pieces.7372

And most importantly, in all of these NMR problems, once you have come up with a structure and you think you have that answer, stop; put the spectrum away; look at your structure with fresh eyes; and, one by one, predict what your spectra would look like--what your IR spectrum, your proton NMR, your C-13 NMR...7386

Make sure that all the parts you have in your structure are accounted for in the spectrum, and then the good news is: you know you have the right answer before you can wrap up and move on to the next problem.7407

That wraps it up for NMR; so thanks very much for coming to

I hope to see you again soon.7425