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Lecture Comments (49)

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.

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

I. Introduction to Organic Molecules
  Introduction and Drawing Structures 49:51
   Intro 0:00 
   Organic Chemistry 0:07 
    Organic 0:08 
    Inorganic 0:26 
    Examples of Organic Compounds 1:16 
   Review Some Chemistry Basics 5:23 
    Electrons 5:42 
    Orbitals (s,p,d,f) 6:12 
   Review Some Chemistry Basics 7:35 
    Elements & Noble Gases 7:36 
    Atom & Valance Shell 8:47 
   Review Some Chemistry Basics 11:33 
    Electronegative Elements 11:34 
    Which Is More Electronegative, C or N? 13:45 
   Ionic & Covalent Bonds 14:07 
    Ionic Bonds 14:08 
    Covalent Bonds 16:17 
   Polar Covalent Bonds 19:35 
    Polar Covalent Bonds & Electronegativities 19:37 
   Polarity of Molecules 22:56 
    Linear molecule 23:07 
    Bent Molecule 23:53 
    No Polar Bonds 24:21 
    Ionic 24:52 
   Line Drawings 26:36 
    Line Drawing Overview 26:37 
    Line Drawing: Example 1 27:12 
    Line Drawing: Example 2 29:14 
    Line Drawing: Example 3 29:51 
    Line Drawing: Example 4 30:34 
    Line Drawing: Example 5 31:21 
    Line Drawing: Example 6 32:41 
   Diversity of Organic Compounds 33:57 
    Diversity of Organic Compounds 33:58 
   Diversity of Organic Compounds, cont. 39:16 
    Diversity of Organic Compounds, cont. 39:17 
   Examples of Polymers 45:26 
    Examples of Polymers 45:27 
  Lewis Structures & Resonance 44:25
   Intro 0:00 
   Lewis Structures 0:08 
    How to Draw a Lewis Structure 0:09 
    Examples 2:20 
   Lewis Structures 6:25 
    Examples: Lewis Structure 6:27 
    Determining Formal Charges 8:48 
    Example: Determining Formal Charges for Carbon 10:11 
    Example: Determining Formal Charges for Oxygen 11:02 
   Lewis Structures 12:08 
    Typical, Stable Bonding Patterns: Hydrogen 12:11 
    Typical, Stable Bonding Patterns: Carbon 12:58 
    Typical, Stable Bonding Patterns: Nitrogen 13:25 
    Typical, Stable Bonding Patterns: Oxygen 13:54 
    Typical, Stable Bonding Patterns: Halogen 14:16 
   Lewis Structure Example 15:17 
    Drawing a Lewis Structure for Nitric Acid 15:18 
   Resonance 21:58 
    Definition of Resonance 22:00 
    Delocalization 22:07 
    Hybrid Structure 22:38 
   Rules for Estimating Stability of Resonance Structures 26:04 
    Rule Number 1: Complete Octets 26:10 
    Rule Number 2: Separation of Charge 28:13 
    Rule Number 3: Negative and Positive Charges 30:02 
    Rule Number 4: Equivalent 31:06 
   Looking for Resonance 32:09 
    Lone Pair Next to a p Bond 32:10 
    Vacancy Next to a p Bond 33:53 
    p Bond Between Two Different Elements 35:00 
    Other Type of Resonance: Benzene 36:06 
   Resonance Example 37:29 
    Draw and Rank Resonance Forms 37:30 
  Acid-Base Reactions 1:07:46
   Intro 0:00 
   Acid-Base Reactions 0:07 
    Overview 0:08 
    Lewis Acid and Lewis Base 0:30 
    Example 1: Lewis Acid and Lewis Base 1:53 
    Example 2: Lewis Acid and Lewis Base 3:04 
   Acid-base Reactions 4:54 
    Bonsted-Lowry Acid and Bonsted-Lowry Base 4:56 
    Proton Transfer Reaction 5:36 
   Acid-Base Equilibrium 8:14 
    Two Acids in Competition = Equilibrium 8:15 
    Example: Which is the Stronger Acid? 8:40 
   Periodic Trends for Acidity 12:40 
    Across Row 12:41 
   Periodic Trends for Acidity 19:48 
    Energy Diagram 19:50 
   Periodic Trends for Acidity 21:28 
    Down a Family 21:29 
   Inductive Effects on Acidity 25:52 
    Example: Which is the Stronger Acid? 25:54 
    Other Electron-Withdrawing Group (EWG) 30:37 
   Inductive Effects on Acidity 32:55 
    Inductive Effects Decrease with Distance 32:56 
   Resonance Effects on Acidity 36:35 
    Examples of Resonance Effects on Acidity 36:36 
   Resonance Effects on Acidity 41:15 
    Small and Large Amount of Resonance 41:17 
   Acid-Base Example 43:10 
    Which is Most Acidic? Which is the Least Acidic? 43:12 
   Acid-Base Example 49:26 
    Which is the Stronger Base? 49:27 
   Acid-Base Example 53:58 
    Which is the Strongest Base? 53:59 
   Common Acids/Bases 60:45 
    Common Acids/Bases 60:46 
    Example: Determine the Direction of Equilibrium 64:51 
  Structures and Properties of Organic Molecules 1:23:35
   Intro 0:00 
   Orbitals and Bonding 0:20 
    Atomic Orbitals (AO) 0:21 
   Molecular Orbitals (MO) 1:46 
    Definition of Molecular Orbitals 1:47 
    Example 1: Formation of Sigma Bond and Molecular Orbitals 2:20 
   Molecular Orbitals (MO) 5:25 
    Example 2: Formation of Pi Bond 5:26 
   Overlapping E Levels of MO's 7:28 
    Energy Diagram 7:29 
   Electronic Transitions 9:18 
    Electronic Transitions 9:23 
   Hybrid Orbitals 12:04 
    Carbon AO 12:06 
    Hybridization 13:51 
   Hybrid Orbitals 15:02 
    Examples of Hybrid Orbitals 15:05 
    Example: Assign Hybridization 20:31 
   3-D Sketches 24:05 
    sp3 24:24 
    sp2 25:28 
    sp 27:41 
   3-D Sketches of Molecules 29:07 
    3-D Sketches of Molecules 1 29:08 
    3-D Sketches of Molecules 2 32:29 
    3-D Sketches of Molecules 3 35:36 
   3D Sketch 37:20 
    How to Draw 3D Sketch 37:22 
    Example 1: Drawing 3D Sketch 37:50 
    Example 2: Drawing 3D Sketch 43:04 
   Hybridization and Resonance 46:06 
    Example: Hybridization and Resonance 46:08 
   Physical Properties 49:55 
    Water Solubility, Boiling Points, and Intermolecular Forces 49:56 
    Types of 'Nonbonding' Interactions 51:47 
   Dipole-Dipole 52:37 
    Definition of Dipole-Dipole 52:39 
    Example: Dipole-Dipole Bonding 53:27 
   Hydrogen Bonding 57:14 
    Definition of Hydrogen Bonding 57:15 
    Example: Hydrogen Bonding 58:05 
   Van Der Waals/ London Forces 63:11 
    Van Der Waals/ London Forces 63:12 
    Example: Van Der Waals/ London Forces 64:59 
   Water Solubility 68:32 
    Water Solubility 68:34 
    Example: Water Solubility 69:05 
    Example: Acetone 71:29 
   Isomerism 73:51 
    Definition of Isomers 73:53 
    Constitutional Isomers and Example 74:17 
    Stereoisomers and Example 75:34 
   Introduction to Functional Groups 77:06 
    Functional Groups: Example, Abbreviation, and Name 77:07 
   Introduction to Functional Groups 80:48 
    Functional Groups: Example, Abbreviation, and Name 80:49 
  Alkane Structures 1:13:38
   Intro 0:00 
   Nomenclature of Alkanes 0:12 
    Nomenclature of Alkanes and IUPAC Rules 0:13 
    Examples: Nomenclature of Alkanes 4:38 
   Molecular Formula and Degrees of Unsaturation (DU) 17:24 
    Alkane Formula 17:25 
    Example: Heptane 17:58 
    Why '2n+2' Hydrogens? 18:35 
    Adding a Ring 19:20 
    Adding a p Bond 19:42 
    Example 1: Determine Degrees of Unsaturation (DU) 20:17 
    Example 2: Determine Degrees of Unsaturation (DU) 21:35 
    Example 3: Determine DU of Benzene 23:30 
   Molecular Formula and Degrees of Unsaturation (DU) 24:41 
    Example 4: Draw Isomers 24:42 
   Physical properties of Alkanes 29:17 
    Physical properties of Alkanes 29:18 
   Conformations of Alkanes 33:40 
    Conformational Isomers 33:42 
    Conformations of Ethane: Eclipsed and Staggered 34:40 
    Newman Projection of Ethane 36:15 
   Conformations of Ethane 40:38 
    Energy and Degrees Rotated Diagram 40:41 
   Conformations of Butane 42:28 
    Butane 42:29 
    Newman Projection of Butane 43:35 
   Conformations of Butane 44:25 
    Energy and Degrees Rotated Diagram 44:30 
   Cycloalkanes 51:26 
    Cyclopropane and Cyclobutane 51:27 
    Cyclopentane 53:56 
   Cycloalkanes 54:56 
    Cyclohexane: Chair, Boat, and Twist Boat Conformations 54:57 
   Drawing a Cyclohexane Chair 57:58 
    Drawing a Cyclohexane Chair 57:59 
    Newman Projection of Cyclohexane 62:14 
   Cyclohexane Chair Flips 64:06 
    Axial and Equatorial Groups 64:10 
    Example: Chair Flip on Methylcyclohexane 66:44 
   Cyclohexane Conformations Example 69:01 
    Chair Conformations of cis-1-t-butyl-4-methylcyclohexane 69:02 
  Stereochemistry 1:40:54
   Intro 0:00 
   Stereochemistry 0:10 
    Isomers 0:11 
   Stereoisomer Examples 1:30 
    Alkenes 1:31 
    Cycloalkanes 2:35 
   Stereoisomer Examples 4:00 
    Tetrahedral Carbon: Superimposable (Identical) 4:01 
    Tetrahedral Carbon: Non-Superimposable (Stereoisomers) 5:18 
   Chirality 7:18 
    Stereoisomers 7:19 
    Chiral 8:05 
    Achiral 8:29 
    Example: Achiral and Chiral 8:45 
   Chirality 20:11 
    Superimposable, Non-Superimposable, Chiral, and Achiral 20:12 
   Nomenclature 23:00 
    Cahn-Ingold-Prelog Rules 23:01 
   Nomenclature 29:39 
    Example 1: Nomenclature 29:40 
    Example 2: Nomenclature 31:49 
    Example 3: Nomenclature 33:24 
    Example 4: Nomenclature 35:39 
   Drawing Stereoisomers 36:58 
    Drawing (S)-2-bromopentane 36:59 
    Drawing the Enantiomer of (S)-2-bromopentane: Method 1 38:47 
    Drawing the Enantiomer of (S)-2-bromopentane: Method 2 39:35 
   Fischer Projections 41:47 
    Definition of Fischer Projections 41:49 
    Drawing Fischer Projection 43:43 
    Use of Fisher Projection: Assigning Configuration 49:13 
   Molecules with Two Chiral Carbons 51:49 
    Example A 51:42 
    Drawing Enantiomer of Example A 53:26 
    Fischer Projection of A 54:25 
   Drawing Stereoisomers, cont. 59:40 
    Drawing Stereoisomers Examples 59:41 
    Diastereomers 61:48 
   Drawing Stereoisomers 66:37 
    Draw All Stereoisomers of 2,3-dichlorobutane 66:38 
   Molecules with Two Chiral Centers 70:22 
    Draw All Stereoisomers of 2,3-dichlorobutane, cont. 70:23 
   Optical Activity 74:10 
    Chiral Molecules 74:11 
    Angle of Rotation 74:51 
    Achiral Species 76:46 
   Physical Properties of Stereoisomers 77:11 
    Enantiomers 77:12 
    Diastereomers 78:01 
    Example 78:26 
   Physical Properties of Stereoisomers 83:05 
    When Do Enantiomers Behave Differently? 83:06 
   Racemic Mixtures 88:18 
    Racemic Mixtures 88:21 
    Resolution 89:52 
   Unequal Mixtures of Enantiomers 92:54 
    Enantiomeric Excess (ee) 92:55 
   Unequal Mixture of Enantiomers 94:43 
    Unequal Mixture of Enantiomers 94:44 
    Example: Finding ee 96:38 
    Example: Percent of Composition 99:46 
II. Understanding Organic Reactions
  Nomenclature 1:53:47
   Intro 0:00 
   Cycloalkane Nomenclature 0:17 
    Cycloalkane Nomenclature and Examples 0:18 
   Alkene Nomenclature 6:28 
    Alkene Nomenclature and Examples 6:29 
   Alkene Nomenclature: Stereochemistry 15:07 
    Alkenes With Two Groups: Cis & Trans 15:08 
    Alkenes With Greater Than Two Groups: E & Z 18:26 
   Alkyne Nomenclature 24:46 
    Alkyne Nomenclature and Examples 24:47 
    Alkane Has a Higher Priority Than Alkyne 28:25 
   Alcohol Nomenclature 29:24 
    Alcohol Nomenclature and Examples 29:25 
    Alcohol FG Has Priority Over Alkene/yne 33:41 
   Ether Nomenclature 36:32 
    Ether Nomenclature and Examples 36:33 
   Amine Nomenclature 42:59 
    Amine Nomenclature and Examples 43:00 
   Amine Nomenclature 49:45 
    Primary, Secondary, Tertiary, Quaternary Salt 49:46 
   Aldehyde Nomenclature 51:37 
    Aldehyde Nomenclature and Examples 51:38 
   Ketone Nomenclature 58:43 
    Ketone Nomenclature and Examples 58:44 
   Aromatic Nomenclature 65:02 
    Aromatic Nomenclature and Examples 65:03 
   Aromatic Nomenclature, cont. 69:09 
    Ortho, Meta, and Para 69:10 
   Aromatic Nomenclature, cont. 73:27 
    Common Names for Simple Substituted Aromatic Compounds 73:28 
   Carboxylic Acid Nomenclature 76:35 
    Carboxylic Acid Nomenclature and Examples 76:36 
   Carboxylic Acid Derivatives 82:28 
    Carboxylic Acid Derivatives 82:42 
    General Structure 83:10 
   Acid Halide Nomenclature 84:48 
    Acid Halide Nomenclature and Examples 84:49 
   Anhydride Nomenclature 88:10 
    Anhydride Nomenclature and Examples 88:11 
   Ester Nomenclature 92:50 
    Ester Nomenclature 92:51 
    Carboxylate Salts 98:51 
   Amide Nomenclature 100:02 
    Amide Nomenclature and Examples 100:03 
   Nitrile Nomenclature 105:22 
    Nitrile Nomenclature and Examples 105:23 
  Chemical Reactions 51:01
   Intro 0:00 
   Chemical Reactions 0:06 
    Reactants and Products 0:07 
    Thermodynamics 0:50 
    Equilibrium Constant 1:06 
    Equation 2:35 
    Organic Reaction 3:05 
   Energy vs. Progress of Rxn Diagrams 3:48 
    Exothermic Reaction 4:02 
    Endothermic Reaction 6:54 
   Estimating ΔH rxn 9:15 
    Bond Breaking 10:03 
    Bond Formation 10:25 
    Bond Strength 11:35 
    Homolytic Cleavage 11:59 
    Bond Dissociation Energy (BDE) Table 12:29 
    BDE for Multiple Bonds 14:32 
    Examples 17:35 
   Kinetics 20:35 
    Kinetics 20:36 
    Examples 21:49 
   Reaction Rate Variables 23:15 
    Reaction Rate Variables 23:16 
    Increasing Temperature, Increasing Rate 24:08 
    Increasing Concentration, Increasing Rate 25:39 
    Decreasing Energy of Activation, Increasing Rate 27:49 
   Two-Step Mechanisms 30:06 
    E vs. POR Diagram (2-step Mechanism) 30:07 
   Reactive Intermediates 33:03 
    Reactive Intermediates 33:04 
    Example: A Carbocation 35:20 
   Carbocation Stability 37:24 
    Relative Stability of Carbocation 37:25 
    Alkyl groups and Hyperconjugation 38:45 
   Carbocation Stability 41:57 
    Carbocation Stabilized by Resonance: Allylic 41:58 
    Carbocation Stabilized by Resonance: Benzylic 42:59 
    Overall Carbocation Stability 44:05 
   Free Radicals 45:05 
    Definition and Examples of Free Radicals 45:06 
   Radical Mechanisms 49:40 
    Example: Regular Arrow 49:41 
    Example: Fish-Hook Arrow 50:17 
  Free Radical Halogenation 26:23
   Intro 0:00 
   Free Radical Halogenation 0:06 
    Free Radical Halogenation 0:07 
    Mechanism: Initiation 1:27 
    Mechanism: Propagation Steps 2:21 
   Free Radical Halogenation 5:33 
    Termination Steps 5:36 
    Example 1: Terminations Steps 6:00 
    Example 2: Terminations Steps 6:18 
    Example 3: Terminations Steps 7:43 
    Example 4: Terminations Steps 8:04 
   Regiochemistry of Free Radical Halogenation 9:32 
    Which Site/Region Reacts and Why? 9:34 
    Bromination and Rate of Reaction 14:03 
   Regiochemistry of Free Radical Halogenation 14:30 
    Chlorination 14:31 
    Why the Difference in Selectivity? 19:58 
   Allylic Halogenation 20:53 
    Examples of Allylic Halogenation 20:55 
  Substitution Reactions 1:48:05
   Intro 0:00 
   Substitution Reactions 0:06 
    Substitution Reactions Example 0:07 
    Nucleophile 0:39 
    Electrophile 1:20 
    Leaving Group 2:56 
    General Reaction 4:13 
   Substitution Reactions 4:43 
    General Reaction 4:46 
    Substitution Reaction Mechanisms: Simultaneous 5:08 
    Substitution Reaction Mechanisms: Stepwise 5:34 
   SN2 Substitution 6:21 
    Example of SN2 Mechanism 6:22 
    SN2 Kinetics 7:58 
   Rate of SN2 9:10 
    Sterics Affect Rate of SN2 9:12 
    Rate of SN2 (By Type of RX) 14:13 
   SN2: E vs. POR Diagram 17:26 
    E vs. POR Diagram 17:27 
    Transition State (TS) 18:24 
   SN2 Transition State, Kinetics 20:58 
    SN2 Transition State, Kinetics 20:59 
    Hybridization of TS Carbon 21:57 
    Example: Allylic LG 23:34 
   Stereochemistry of SN2 25:46 
    Backside Attack and Inversion of Stereochemistry 25:48 
   SN2 Summary 29:56 
    Summary of SN2 29:58 
   Predict Products (SN2) 31:42 
    Example 1: Predict Products 31:50 
    Example 2: Predict Products 33:38 
    Example 3: Predict Products 35:11 
    Example 4: Predict Products 36:11 
    Example 5: Predict Products 37:32 
   SN1 Substitution Mechanism 41:52 
    Is This Substitution? Could This Be an SN2 Mechanism? 41:54 
   SN1 Mechanism 43:50 
    Two Key Steps: 1. Loss of LG 43:53 
    Two Key Steps: 2. Addition of nu 45:11 
   SN1 Kinetics 47:17 
    Kinetics of SN1 47:18 
    Rate of SN1 (By RX type) 48:44 
   SN1 E vs. POR Diagram 49:49 
    E vs. POR Diagram 49:51 
    First Transition Stage (TS-1) 51:48 
    Second Transition Stage (TS-2) 52:56 
   Stereochemistry of SN1 53:44 
    Racemization of SN1 and Achiral Carbocation Intermediate 53:46 
    Example 54:29 
   SN1 Summary 58:25 
    Summary of SN1 58:26 
   SN1 or SN2 Mechanisms? 60:40 
    Example 1: SN1 or SN2 Mechanisms 60:42 
    Example 2: SN1 or SN2 Mechanisms 63:00 
    Example 3: SN1 or SN2 Mechanisms 64:06 
    Example 4: SN1 or SN2 Mechanisms 66:17 
   SN1 Mechanism 69:12 
    Three Steps of SN1 Mechanism 69:13 
   SN1 Carbocation Rearrangements 74:50 
    Carbocation Rearrangements Example 74:51 
   SN1 Carbocation Rearrangements 80:46 
    Alkyl Groups Can Also Shift 80:48 
   Leaving Groups 84:26 
    Leaving Groups 84:27 
    Forward or Reverse Reaction Favored? 86:00 
   Leaving Groups 89:59 
    Making poor LG Better: Method 1 90:00 
   Leaving Groups 94:18 
    Making poor LG Better: Tosylate (Method 2) 94:19 
   Synthesis Problem 98:15 
    Example: Provide the Necessary Reagents 98:16 
   Nucleophilicity 101:10 
    What Makes a Good Nucleophile? 101:11 
   Nucleophilicity 104:45 
    Periodic Trends: Across Row 104:47 
    Periodic Trends: Down a Family 106:46 
  Elimination Reactions 1:11:43
   Intro 0:00 
   Elimination Reactions: E2 Mechanism 0:06 
    E2 Mechanism 0:08 
    Example of E2 Mechanism 1:01 
   Stereochemistry of E2 4:48 
    Anti-Coplanar & Anti-Elimination 4:50 
    Example 1: Stereochemistry of E2 5:34 
    Example 2: Stereochemistry of E2 10:39 
   Regiochemistry of E2 13:04 
    Refiochemistry of E2 and Zaitsev's Rule 13:05 
    Alkene Stability 17:39 
   Alkene Stability 19:20 
    Alkene Stability Examples 19:22 
    Example 1: Draw Both E2 Products and Select Major 21:57 
    Example 2: Draw Both E2 Products and Select Major 25:02 
   SN2 Vs. E2 Mechanisms 29:06 
    SN2 Vs. E2 Mechanisms 29:07 
    When Do They Compete? 30:34 
   SN2 Vs. E2 Mechanisms 31:23 
    Compare Rates 31:24 
   SN2 Vs. E2 Mechanisms 36:34 
    t-BuBr: What If Vary Base? 36:35 
    Preference for E2 Over SN2 (By RX Type) 40:42 
   E1 Elimination Mechanism 41:51 
    E1 - Elimination Unimolecular 41:52 
    E1 Mechanism: Step 1 44:14 
    E1 Mechanism: Step 2 44:48 
   E1 Kinetics 46:58 
    Rate = k[RCI] 47:00 
    E1 Rate (By Type of Carbon Bearing LG) 48:31 
   E1 Stereochemistry 49:49 
    Example 1: E1 Stereochemistry 49:51 
    Example 2: E1 Stereochemistry 52:31 
   Carbocation Rearrangements 55:57 
    Carbocation Rearrangements 56:01 
    Product Mixtures 57:20 
   Predict the Product: SN2 vs. E2 59:58 
    Example 1: Predict the Product 60:00 
    Example 2: Predict the Product 62:10 
    Example 3: Predict the Product 64:07 
   Predict the Product: SN2 vs. E2 66:06 
    Example 4: Predict the Product 66:07 
    Example 5: Predict the Product 67:29 
    Example 6: Predict the Product 67:51 
    Example 7: Predict the Product 69:18 
III. Alkanes, Alkenes, & Alkynes
  Alkenes 36:39
   Intro 0:00 
   Alkenes 0:12 
    Definition and Structure of Alkenes 0:13 
    3D Sketch of Alkenes 1:53 
    Pi Bonds 3:48 
   Alkene Stability 4:57 
    Alkyl Groups Attached 4:58 
    Trans & Cis 6:20 
   Alkene Stability 8:42 
    Pi Bonds & Conjugation 8:43 
    Bridgehead Carbons & Bredt's Rule 10:22 
    Measuring Stability: Hydrogenation Reaction 11:40 
   Alkene Synthesis 12:01 
    Method 1: E2 on Alkyl Halides 12:02 
    Review: Stereochemistry 16:17 
    Review: Regiochemistry 16:50 
    Review: SN2 vs. E2 17:34 
   Alkene Synthesis 18:57 
    Method 2: Dehydration of Alcohols 18:58 
    Mechanism 20:08 
   Alkene Synthesis 23:26 
    Alcohol Dehydration 23:27 
   Example 1: Comparing Strong Acids 26:59 
   Example 2: Mechanism for Dehydration Reaction 29:00 
   Example 3: Transform 32:50 
  Reactions of Alkenes 2:08:44
   Intro 0:00 
   Reactions of Alkenes 0:05 
    Electrophilic Addition Reaction 0:06 
   Addition of HX 2:02 
    Example: Regioselectivity & 2 Steps Mechanism 2:03 
   Markovnikov Addition 5:30 
    Markovnikov Addition is Favored 5:31 
    Graph: E vs. POR 6:33 
   Example 8:29 
    Example: Predict and Consider the Stereochemistry 8:30 
   Hydration of Alkenes 12:31 
    Acid-catalyzed Addition of Water 12:32 
    Strong Acid 14:20 
   Hydration of Alkenes 15:20 
    Acid-catalyzed Addition of Water: Mechanism 15:21 
   Hydration vs. Dehydration 19:51 
    Hydration Mechanism is Exact Reverse of Dehydration 19:52 
   Example 21:28 
    Example: Hydration Reaction 21:29 
   Alternative 'Hydration' Methods 25:26 
    Oxymercuration-Demercuration 25:27 
   Oxymercuration Mechanism 28:55 
    Mechanism of Oxymercuration 28:56 
   Alternative 'Hydration' Methods 30:51 
    Hydroboration-Oxidation 30:52 
   Hydroboration Mechanism 33:22 
    1-step (concerted) 33:23 
    Regioselective 34:45 
    Stereoselective 35:30 
   Example 35:58 
    Example: Hydroboration-Oxidation 35:59 
   Example 40:42 
    Example: Predict the Major Product 40:43 
   Synthetic Utility of 'Alternate' Hydration Methods 44:36 
    Example: Synthetic Utility of 'Alternate' Hydration Methods 44:37 
   Flashcards 47:28 
    Tips On Using Flashcards 47:29 
   Bromination of Alkenes 49:51 
    Anti-Addition of Br₂ 49:52 
   Bromination Mechanism 53:16 
    Mechanism of Bromination 53:17 
   Bromination Mechanism 55:42 
    Mechanism of Bromination 55:43 
   Bromination: Halohydrin Formation 58:54 
    Addition of other Nu: to Bromonium Ion 58:55 
    Mechanism 60:08 
   Halohydrin: Regiochemistry 63:55 
    Halohydrin: Regiochemistry 63:56 
    Bromonium Ion Intermediate 64:26 
   Example 69:28 
    Example: Predict Major Product 69:29 
   Example Cont. 70:59 
    Example: Predict Major Product Cont. 71:00 
   Catalytic Hydrogenation of Alkenes 73:19 
    Features of Catalytic Hydrogenation 73:20 
   Catalytic Hydrogenation of Alkenes 74:48 
    Metal Surface 74:49 
    Heterogeneous Catalysts 75:29 
    Homogeneous Catalysts 76:08 
   Catalytic Hydrogenation of Alkenes 77:44 
    Hydrogenation & Pi Bond Stability 77:45 
    Energy Diagram 79:22 
   Catalytic Hydrogenation of Dienes 80:40 
    Hydrogenation & Pi Bond Stability 80:41 
    Energy Diagram 83:31 
   Example 84:14 
    Example: Predict Product 84:15 
   Oxidation of Alkenes 87:21 
    Redox Review 87:22 
    Epoxide 90:26 
    Diol (Glycol) 90:54 
    Ketone/ Aldehyde 91:13 
   Epoxidation 92:08 
    Epoxidation 92:09 
    General Mechanism 96:32 
   Alternate Epoxide Synthesis 97:38 
    Alternate Epoxide Synthesis 97:39 
   Dihydroxylation 101:10 
    Dihydroxylation 101:12 
    General Mechanism (Concerted Via Cycle Intermediate) 102:38 
   Ozonolysis 104:22 
    Ozonolysis: Introduction 104:23 
    Ozonolysis: Is It Good or Bad? 105:05 
    Ozonolysis Reaction 108:54 
   Examples 111:10 
    Example 1: Ozonolysis 111:11 
    Example 113:25 
   Radical Addition to Alkenes 115:05 
    Recall: Free-Radical Halogenation 115:15 
    Radical Mechanism 115:45 
    Propagation Steps 118:01 
    Atom Abstraction 118:30 
    Addition to Alkene 119:11 
   Radical Addition to Alkenes 119:54 
    Markovnivok (Electrophilic Addition) & anti-Mark. (Radical Addition) 119:55 
    Mechanism 121:03 
   Alkene Polymerization 125:35 
    Example: Alkene Polymerization 125:36 
  Alkynes 1:13:19
   Intro 0:00 
   Structure of Alkynes 0:04 
    Structure of Alkynes 0:05 
    3D Sketch 2:30 
    Internal and Terminal 4:03 
   Reductions of Alkynes 4:36 
    Catalytic Hydrogenation 4:37 
    Lindlar Catalyst 5:25 
   Reductions of Alkynes 7:24 
    Dissolving Metal Reduction 7:25 
   Oxidation of Alkynes 9:24 
    Ozonolysis 9:25 
   Reactions of Alkynes 10:56 
    Addition Reactions: Bromination 10:57 
   Addition of HX 12:24 
    Addition of HX 12:25 
   Addition of HX 13:36 
    Addition of HX: Mechanism 13:37 
   Example 17:38 
    Example: Transform 17:39 
   Hydration of Alkynes 23:35 
    Hydration of Alkynes 23:36 
   Hydration of Alkynes 26:47 
    Hydration of Alkynes: Mechanism 26:49 
   'Hydration' via Hydroboration-Oxidation 32:57 
    'Hydration' via Hydroboration-Oxidation 32:58 
    Disiamylborane 33:28 
    Hydroboration-Oxidation Cont. 34:25 
   Alkyne Synthesis 36:17 
    Method 1: Alkyne Synthesis By Dehydrohalogenation 36:19 
   Alkyne Synthesis 39:06 
    Example: Transform 39:07 
   Alkyne Synthesis 41:21 
    Method 2 & Acidity of Alkynes 41:22 
    Conjugate Bases 43:06 
   Preparation of Acetylide Anions 49:55 
    Preparation of Acetylide Anions 49:57 
   Alkyne Synthesis 53:40 
    Synthesis Using Acetylide Anions 53:41 
   Example 1: Transform 57:04 
   Example 2: Transform 61:07 
   Example 3: Transform 66:22 
IV. Alcohols
  Alcohols, Part I 59:52
   Intro 0:00 
   Alcohols 0:11 
    Attributes of Alcohols 0:12 
    Boiling Points 2:00 
   Water Solubility 5:00 
    Water Solubility (Like Dissolves Like) 5:01 
   Acidity of Alcohols 9:39 
    Comparison of Alcohols Acidity 9:41 
   Preparation of Alkoxides 13:03 
    Using Strong Base Like Sodium Hydride 13:04 
    Using Redox Reaction 15:36 
   Preparation of Alkoxides 17:41 
    Using K° 17:42 
    Phenols Are More Acidic Than Other Alcohols 19:51 
   Synthesis of Alcohols, ROH 21:43 
    Synthesis of Alcohols from Alkyl Halides, RX (SN2 or SN1) 21:44 
   Synthesis of Alcohols, ROH 25:08 
    Unlikely on 2° RX (E2 Favored) 25:09 
    Impossible on 3° RX (E2) and Phenyl/Vinyl RX (N/R) 25:47 
   Synthesis of Alcohols, ROH 26:26 
    SN1 with H₂O 'Solvolysis' or 'Hydrolysis' 26:27 
    Carbocation Can Rearrange 29:00 
   Synthesis of Alcohols, ROH 30:08 
    Synthesis of Alcohols From Alkenes: Hydration 30:09 
    Synthesis of Alcohols From Alkenes: Oxidation/Diol 32:20 
   Synthesis of Alcohols, ROH 33:14 
    Synthesis of Alcohols From Ketones and Aldehydes 33:15 
   Organometallic Reagents: Preparation 37:03 
    Grignard (RMgX) 37:04 
    Organolithium (Rli) 40:03 
   Organometallic Reagents: Reactions 41:45 
    Reactions of Organometallic Reagents 41:46 
   Organometallic Reagents: Reactions as Strong Nu: 46:40 
    Example 1: Reactions as Strong Nu: 46:41 
    Example 2: Reactions as Strong Nu: 48:57 
   Hydride Nu: 50:52 
    Hydride Nu: 50:53 
   Examples 53:34 
    Predict 1 53:35 
    Predict 2 54:45 
   Examples 56:43 
    Transform 56:44 
    Provide Starting Material 58:18 
  Alcohols, Part II 45:35
   Intro 0:00 
   Oxidation Reactions 0:08 
    Oxidizing Agents: Jones, PCC, Swern 0:09 
    'Jones' Oxidation 0:43 
    Example 1: Predict Oxidation Reactions 2:29 
    Example 2: Predict Oxidation Reactions 3:00 
   Oxidation Reactions 4:11 
    Selective Oxidizing Agents (PCC and Swern) 4:12 
    PCC (Pyridiniym Chlorochromate) 5:10 
    Swern Oxidation 6:05 
   General [ox] Mechanism 8:32 
    General [ox] Mechanism 8:33 
   Oxidation of Alcohols 10:11 
    Example 1: Oxidation of Alcohols 10:12 
    Example 2: Oxidation of Alcohols 11:20 
    Example 3: Oxidation of Alcohols 11:46 
   Example 13:09 
    Predict: PCC Oxidation Reactions 13:10 
   Tosylation of Alcohols 15:22 
    Introduction to Tosylation of Alcohols 15:23 
   Example 21:08 
    Example: Tosylation of Alcohols 21:09 
   Reductions of Alcohols 23:39 
    Reductions of Alcohols via SN2 with Hydride 24:22 
    Reductions of Alcohols via Dehydration 27:12 
   Conversion of Alcohols to Alkyl Halides 30:12 
    Conversion of Alcohols to Alkyl Halides via Tosylate 30:13 
   Conversion of Alcohols to Alkyl Halides 31:17 
    Using HX 31:18 
    Mechanism 32:09 
   Conversion of Alcohols to Alkyl Halides 35:43 
    Reagents that Provide LG and Nu: in One 'Pot' 35:44 
   General Mechanisms 37:44 
    Example 1: General Mechanisms 37:45 
    Example 2: General Mechanisms 39:25 
   Example 41:04 
    Transformation of Alcohols 41:05 
V. Ethers, Thiols, Thioethers, & Ketones
  Ethers 1:34:45
   Intro 0:00 
   Ethers 0:11 
    Overview of Ethers 0:12 
    Boiling Points 1:37 
   Ethers 4:34 
    Water Solubility (Grams per 100mL H₂O) 4:35 
   Synthesis of Ethers 7:53 
    Williamson Ether Synthesis 7:54 
    Example: Synthesis of Ethers 9:23 
   Synthesis of Ethers 10:27 
    Example: Synthesis of Ethers 10:28 
    Intramolecular SN2 13:04 
   Planning an Ether Synthesis 14:45 
    Example 1: Planning an Ether Synthesis 14:46 
   Planning an Ether Synthesis 16:16 
    Example 2: Planning an Ether Synthesis 16:17 
   Planning an Ether Synthesis 22:04 
    Example 3: Synthesize Dipropyl Ether 22:05 
   Planning an Ether Synthesis 26:01 
    Example 4: Transform 26:02 
   Synthesis of Epoxides 30:05 
    Synthesis of Epoxides Via Williamson Ether Synthesis 30:06 
    Synthesis of Epoxides Via Oxidation 32:42 
   Reaction of Ethers 33:35 
    Reaction of Ethers 33:36 
   Reactions of Ethers with HBr or HI 34:44 
    Reactions of Ethers with HBr or HI 34:45 
     Mechanism 35:25 
   Epoxide Ring-Opening Reaction 39:25 
    Epoxide Ring-Opening Reaction 39:26 
    Example: Epoxide Ring-Opening Reaction 42:42 
   Acid-Catalyzed Epoxide Ring Opening 44:16 
    Acid-Catalyzed Epoxide Ring Opening Mechanism 44:17 
   Acid-Catalyzed Epoxide Ring Opening 50:13 
    Acid-Catalyzed Epoxide Ring Opening Mechanism 50:14 
   Catalyst Needed for Ring Opening 53:34 
    Catalyst Needed for Ring Opening 53:35 
   Stereochemistry of Epoxide Ring Opening 55:56 
    Stereochemistry: SN2 Mechanism 55:57 
    Acid or Base Mechanism? 58:30 
   Example 61:03 
    Transformation 61:04 
   Regiochemistry of Epoxide Ring Openings 65:29 
    Regiochemistry of Epoxide Ring Openings in Base 65:30 
    Regiochemistry of Epoxide Ring Openings in Acid 67:34 
   Example 70:26 
    Example 1: Epoxide Ring Openings in Base 70:27 
    Example 2: Epoxide Ring Openings in Acid 72:50 
   Reactions of Epoxides with Grignard and Hydride 75:35 
    Reactions of Epoxides with Grignard and Hydride 75:36 
   Example 81:47 
    Example: Ethers 81:50 
   Example 87:01 
    Example: Synthesize 87:02 
  Thiols and Thioethers 16:50
   Intro 0:00 
   Thiols and Thioethers 0:10 
    Physical Properties 0:11 
    Reactions Can Be Oxidized 2:16 
   Acidity of Thiols 3:11 
    Thiols Are More Acidic Than Alcohols 3:12 
   Synthesis of Thioethers 6:44 
    Synthesis of Thioethers 6:45 
   Example 8:43 
    Example: Synthesize the Following Target Molecule 8:44 
   Example 14:18 
    Example: Predict 14:19 
  Ketones 2:18:12
   Intro 0:00 
   Aldehydes & Ketones 0:11 
    The Carbonyl: Resonance & Inductive 0:12 
    Reactivity 0:50 
   The Carbonyl 2:35 
    The Carbonyl 2:36 
    Carbonyl FG's 4:10 
   Preparation/Synthesis of Aldehydes & Ketones 6:18 
    Oxidation of Alcohols 6:19 
    Ozonolysis of Alkenes 7:16 
    Hydration of Alkynes 8:01 
   Reaction with Hydride Nu: 9:00 
    Reaction with Hydride Nu: 9:01 
   Reaction with Carbon Nu: 11:29 
    Carbanions: Acetylide 11:30 
    Carbanions: Cyanide 14:23 
   Reaction with Carbon Nu: 15:32 
    Organometallic Reagents (RMgX, Rli) 15:33 
   Retrosynthesis of Alcohols 17:04 
    Retrosynthesis of Alcohols 17:05 
   Example 19:30 
    Example: Transform 19:31 
   Example 22:57 
    Example: Transform 22:58 
   Example 28:19 
    Example: Transform 28:20 
   Example 33:36 
    Example: Transform 33:37 
   Wittig Reaction 37:39 
    Wittig Reaction: A Resonance-Stabilized Carbanion (Nu:) 37:40 
    Wittig Reaction: Mechanism 39:51 
   Preparation of Wittig Reagent 41:58 
    Two Steps From RX 41:59 
    Example: Predict 45:02 
   Wittig Retrosynthesis 46:19 
    Wittig Retrosynthesis 46:20 
    Synthesis 48:09 
   Reaction with Oxygen Nu: 51:21 
    Addition of H₂O 51:22 
    Exception: Formaldehyde is 99% Hydrate in H₂O Solution 54:10 
    Exception: Hydrate is Favored if Partial Positive Near Carbonyl 55:26 
   Reaction with Oxygen Nu: 57:45 
    Addition of ROH 57:46 
    TsOH: Tosic Acid 58:28 
    Addition of ROH Cont. 59:09 
   Example 61:43 
    Predict 61:44 
    Mechanism 63:08 
   Mechanism for Acetal Formation 64:10 
    Mechanism for Acetal Formation 64:11 
   What is a CTI? 75:04 
    Tetrahedral Intermediate 75:05 
    Charged Tetrahedral Intermediate 75:45 
    CTI: Acid-cat 76:10 
    CTI: Base-cat 77:01 
   Acetals & Cyclic Acetals 77:49 
    Overall 77:50 
    Cyclic Acetals 78:46 
   Hydrolysis of Acetals: Regenerates Carbonyl 80:01 
    Hydrolysis of Acetals: Regenerates Carbonyl 80:02 
    Mechanism 82:08 
   Reaction with Nitrogen Nu: 90:11 
    Reaction with Nitrogen Nu: 90:12 
    Example 92:18 
   Mechanism of Imine Formation 93:24 
    Mechanism of Imine Formation 93:25 
   Oxidation of Aldehydes 98:12 
    Oxidation of Aldehydes 1 98:13 
    Oxidation of Aldehydes 2 99:52 
    Oxidation of Aldehydes 3 100:10 
   Reductions of Ketones and Aldehydes 100:54 
    Reductions of Ketones and Aldehydes 100:55 
    Hydride/ Workup 101:22 
    Raney Nickel 102:07 
   Reductions of Ketones and Aldehydes 103:24 
    Clemmensen Reduction & Wolff-Kishner Reduction 103:40 
   Acetals as Protective Groups 106:50 
    Acetals as Protective Groups 106:51 
   Example 110:39 
    Example: Consider the Following Synthesis 110:40 
   Protective Groups 114:47 
    Protective Groups 114:48 
   Example 119:02 
    Example: Transform 119:03 
   Example: Another Route 124:54 
    Example: Transform 128:49 
   Example 128:50 
    Transform 128:51 
   Example 131:05 
    Transform 131:06 
   Example 133:45 
    Transform 133:46 
   Example 135:43 
    Provide the Missing Starting Material 135:44 
VI. Organic Transformation Practice
  Transformation Practice Problems 38:58
   Intro 0:00 
   Practice Problems 0:33 
    Practice Problem 1: Transform 0:34 
    Practice Problem 2: Transform 3:57 
   Practice Problems 7:49 
    Practice Problem 3: Transform 7:50 
   Practice Problems 15:32 
    Practice Problem 4: Transform 15:34 
    Practice Problem 5: Transform 20:15 
   Practice Problems 24:08 
    Practice Problem 6: Transform 24:09 
    Practice Problem 7: Transform 29:27 
   Practice Problems 33:08 
    Practice Problem 8: Transform 33:09 
    Practice Problem 9: Transform 35:23 
VII. Carboxylic Acids
  Carboxylic Acids 1:17:51
   Intro 0:00 
   Review Reactions of Ketone/Aldehyde 0:06 
    Carbonyl Reactivity 0:07 
    Nu: = Hydride (Reduction) 1:37 
    Nu: = Grignard 2:08 
   Review Reactions of Ketone/Aldehyde 2:53 
    Nu: = Alcohol 2:54 
    Nu: = Amine 3:46 
   Carboxylic Acids and Their Derivatives 4:37 
    Carboxylic Acids and Their Derivatives 4:38 
   Ketone vs. Ester Reactivity 6:33 
    Ketone Reactivity 6:34 
    Ester Reactivity 6:55 
   Carboxylic Acids and Their Derivatives 7:30 
    Acid Halide, Anhydride, Ester, Amide, and Nitrile 7:43 
   General Reactions of Acarboxylic Acid Derivatives 9:22 
    General Reactions of Acarboxylic Acid Derivatives 9:23 
   Physical Properties of Carboxylic Acids 12:16 
    Acetic Acid 12:17 
    Carboxylic Acids 15:46 
   Aciditiy of Carboxylic Acids, RCO₂H 17:45 
    Alcohol 17:46 
    Carboxylic Acid 19:21 
   Aciditiy of Carboxylic Acids, RCO₂H 21:31 
    Aciditiy of Carboxylic Acids, RCO₂H 21:32 
   Aciditiy of Carboxylic Acids, RCO₂H 24:48 
    Example: Which is the Stronger Acid? 24:49 
   Aciditiy of Carboxylic Acids, RCO₂H 30:06 
    Inductive Effects Decrease with Distance 30:07 
   Preparation of Carboxylic Acids, RCO₂H 31:55 
    A) By Oxidation 31:56 
   Preparation of Carboxylic Acids, RCO₂H 34:37 
    Oxidation of Alkenes/Alkynes - Ozonolysis 34:38 
   Preparation of Carboxylic Acids, RCO₂H 36:17 
    B) Preparation of RCO₂H from Organometallic Reagents 36:18 
   Preparation of Carboxylic Acids, RCO₂H 38:02 
    Example: Preparation of Carboxylic Acids 38:03 
   Preparation of Carboxylic Acids, RCO₂H 40:38 
    C) Preparation of RCO₂H by Hydrolysis of Carboxylic Acid Derivatives 40:39 
   Hydrolysis Mechanism 42:19 
    Hydrolysis Mechanism 42:20 
    Mechanism: Acyl Substitution (Addition/Elimination) 43:05 
   Hydrolysis Mechanism 47:27 
    Substitution Reaction 47:28 
    RO is Bad LG for SN1/SN2 47:39 
    RO is okay LG for Collapse of CTI 48:31 
   Hydrolysis Mechanism 50:07 
    Base-promoted Ester Hydrolysis (Saponification) 50:08 
   Applications of Carboxylic Acid Derivatives: 53:10 
    Saponification Reaction 53:11 
   Ester Hydrolysis 57:15 
    Acid-Catalyzed Mechanism 57:16 
   Ester Hydrolysis Requires Acide or Base 63:06 
    Ester Hydrolysis Requires Acide or Base 63:07 
   Nitrile Hydrolysis 65:22 
    Nitrile Hydrolysis 65:23 
   Nitrile Hydrolysis Mechanism 66:53 
    Nitrile Hydrolysis Mechanism 66:54 
   Use of Nitriles in Synthesis 72:39 
    Example: Nitirles in Synthesis 72:40 
  Carboxylic Acid Derivatives 1:21:04
   Intro 0:00 
   Carboxylic Acid Derivatives 0:05 
    Carboxylic Acid Derivatives 0:06 
    General Structure 1:00 
   Preparation of Carboxylic Acid Derivatives 1:19 
    Which Carbonyl is the Better E+? 1:20 
    Inductive Effects 1:54 
    Resonance 3:23 
   Preparation of Carboxylic Acid Derivatives 6:52 
    Which is Better E+, Ester or Acid Chloride? 6:53 
    Inductive Effects 7:02 
    Resonance 7:20 
   Preparation of Carboxylic Acid Derivatives 10:45 
    Which is Better E+, Carboxylic Acid or Anhydride? 10:46 
    Inductive Effects & Resonance 11:00 
   Overall: Order of Electrophilicity and Leaving Group 14:49 
    Order of Electrophilicity and Leaving Group 14:50 
    Example: Acid Chloride 16:26 
    Example: Carboxylate 19:17 
   Carboxylic Acid Derivative Interconversion 20:53 
    Carboxylic Acid Derivative Interconversion 20:54 
   Preparation of Acid Halides 24:31 
    Preparation of Acid Halides 24:32 
   Preparation of Anhydrides 25:45 
    A) Dehydration of Acids (For Symmetrical Anhydride) 25:46 
   Preparation of Anhydrides 27:29 
    Example: Dehydration of Acids 27:30 
   Preparation of Anhydrides 29:16 
    B) From an Acid Chloride (To Make Mixed Anhydride) 29:17 
    Mechanism 30:03 
   Preparation of Esters 31:53 
    A) From Acid Chloride or Anhydride 31:54 
   Preparation of Esters 33:48 
    B) From Carboxylic Acids (Fischer Esterification) 33:49 
    Mechanism 36:55 
   Preparations of Esters 41:38 
    Example: Predict the Product 41:39 
   Preparation of Esters 43:17 
    C) Transesterification 43:18 
    Mechanism 45:17 
   Preparation of Esters 47:58 
    D) SN2 with Carboxylate 47:59 
    Mechanism: Diazomethane 49:28 
   Preparation of Esters 51:01 
    Example: Transform 51:02 
   Preparation of Amides 52:27 
    A) From an Acid Cl or Anhydride 52:28 
   Preparations of Amides 54:47 
    B) Partial Hydrolysis of Nitriles 54:48 
   Preparation of Amides 56:11 
    Preparation of Amides: Find Alternate Path 56:12 
   Preparation of Amides 59:04 
    C) Can't be Easily Prepared from RCO₂H Directly 59:05 
   Reactions of Carboxylic Acid Derivatives with Nucleophiles 61:41 
    A) Hydride Nu: Review 61:42 
    A) Hydride Nu: Sodium Borohydride + Ester 62:43 
   Reactions of Carboxylic Acid Derivatives with Nucleophiles 63:57 
    Lithium Aluminum Hydride (LAH) 63:58 
    Mechanism 64:29 
   Summary of Hydride Reductions 67:09 
    Summary of Hydride Reductions 1 67:10 
    Summary of Hydride Reductions 2 67:36 
   Hydride Reduction of Amides 68:12 
    Hydride Reduction of Amides Mechanism 68:13 
   Reaction of Carboxylic Acid Derivatives with Organometallics 72:04 
    Review 1 72:05 
    Review 2 72:50 
   Reaction of Carboxylic Acid Derivatives with Organometallics 74:22 
    Example: Lactone 74:23 
   Special Hydride Nu: Reagents 76:34 
    Diisobutylaluminum Hydride 76:35 
    Example 77:25 
    Other Special Hydride 78:41 
   Addition of Organocuprates to Acid Chlorides 79:07 
    Addition of Organocuprates to Acid Chlorides 79:08 
VIII. Enols & Enolates
  Enols and Enolates, Part 1 1:26:22
   Intro 0:00 
   Enols and Enolates 0:09 
    The Carbonyl 0:10 
    Keto-Enol Tautomerization 1:17 
   Keto-Enol Tautomerization Mechanism 2:28 
    Tautomerization Mechanism (2 Steps) 2:29 
   Keto-Enol Tautomerization Mechanism 5:15 
    Reverse Reaction 5:16 
    Mechanism 6:07 
   Formation of Enolates 7:27 
    Why is a Ketone's α H's Acidic? 7:28 
   Formation of Other Carbanions 10:05 
    Alkyne 10:06 
    Alkane and Alkene 10:53 
   Formation of an Enolate: Choice of Base 11:27 
    Example: Choice of Base 11:28 
   Formation of an Enolate: Choice of Base 13:56 
    Deprotonate, Stronger Base, and Lithium Diisopropyl Amide (LDA) 13:57 
   Formation of an Enolate: Choice of Base 15:48 
    Weaker Base & 'Active' Methylenes 15:49 
    Why Use NaOEt instead of NaOH? 19:01 
   Other Acidic 'α' Protons 20:30 
    Other Acidic 'α' Protons 20:31 
    Why is an Ester Less Acidic than a Ketone? 24:10 
   Other Acidic 'α' Protons 25:19 
    Other Acidic 'α' Protons Continue 25:20 
   How are Enolates Used 25:54 
    Enolates 25:55 
    Possible Electrophiles 26:21 
   Alkylation of Enolates 27:56 
    Alkylation of Enolates 27:57 
    Resonance Form 30:03 
   α-Halogenation 32:17 
    α-Halogenation 32:18 
    Iodoform Test for Methyl Ketones 33:47 
   α-Halogenation 35:55 
    Acid-Catalyzed 35:57 
    Mechanism: 1st Make Enol (2 Steps) 36:14 
    Whate Other Eloctrophiles ? 39:17 
   Aldol Condensation 39:38 
    Aldol Condensation 39:39 
   Aldol Mechanism 41:26 
    Aldol Mechanism: In Base, Deprotonate First 41:27 
   Aldol Mechanism 45:28 
    Mechanism for Loss of H₂O 45:29 
    Collapse of CTI and β-elimination Mechanism 47:51 
    Loss of H₂0 is not E2! 48:39 
   Aldol Summary 49:53 
    Aldol Summary 49:54 
    Base-Catalyzed Mechanism 52:34 
    Acid-Catalyzed Mechansim 53:01 
   Acid-Catalyzed Aldol Mechanism 54:01 
    First Step: Make Enol 54:02 
   Acid-Catalyzed Aldol Mechanism 56:54 
    Loss of H₂0 (β elimination) 56:55 
   Crossed/Mixed Aldol 60:55 
    Crossed/Mixed Aldol & Compound with α H's 60:56 
    Ketone vs. Aldehyde 62:30 
    Crossed/Mixed Aldol & Compound with α H's Continue 63:10 
   Crossed/Mixed Aldol 65:21 
    Mixed Aldol: control Using LDA 65:22 
   Crossed/Mixed Aldol Retrosynthesis 68:53 
    Example: Predic Aldol Starting Material (Aldol Retrosyntheiss) 68:54 
   Claisen Condensation 72:54 
    Claisen Condensation (Aldol on Esters) 72:55 
   Claisen Condensation 79:52 
    Example 1: Claisen Condensation 79:53 
   Claisen Condensation 82:48 
    Example 2: Claisen Condensation 82:49 
  Enols and Enolates, Part 2 50:57
   Intro 0:00 
   Conjugate Additions 0:06 
    α, β-unsaturated Carbonyls 0:07 
   Conjugate Additions 1:50 
    '1,2-addition' 1:51 
    '1,-4-addition' or 'Conjugate Addition' 2:24 
   Conjugate Additions 4:53 
    Why can a Nu: Add to this Alkene? 4:54 
    Typical Alkene 5:09 
    α, β-unsaturated Alkene 5:39 
   Electrophilic Alkenes: Michael Acceptors 6:35 
    Other 'Electrophilic' Alkenes (Called 'Michael Acceptors) 6:36 
   1,4-Addition of Cuprates (R2CuLi) 8:29 
    1,4-Addition of Cuprates (R2CuLi) 8:30 
   1,4-Addition of Cuprates (R2CuLi) 11:23 
    Use Cuprates in Synthesis 11:24 
   Preparation of Cuprates 12:25 
    Prepare Organocuprate From Organolithium 12:26 
    Cuprates Also Do SN2 with RX E+ (Not True for RMgX, RLi) 13:06 
   1,4-Addition of Enolates: Michael Reaction 13:50 
    1,4-Addition of Enolates: Michael Reaction 13:51 
    Mechanism 15:57 
   1,4-Addition of Enolates: Michael Reaction 18:47 
    Example: 1,4-Addition of Enolates 18:48 
   1,4-Addition of Enolates: Michael Reaction 21:02 
    Michael Reaction, Followed by Intramolecular Aldol 21:03 
   Mechanism of the Robinson Annulation 24:26 
    Mechanism of the Robinson Annulation 24:27 
   Enols and Enolates: Advanced Synthesis Topics 31:10 
    Stablized Enolates and the Decarboxylation Reaction 31:11 
    Mechanism: A Pericyclic Reaction 32:08 
   Enols and Enolates: Advanced Synthesis Topics 33:32 
    Example: Advance Synthesis 33:33 
   Enols and Enolates: Advanced Synthesis Topics 36:10 
    Common Reagents: Diethyl Malonate 36:11 
    Common Reagents: Ethyl Acetoacetate 37:27 
   Enols and Enolates: Advanced Synthesis Topics 38:06 
    Example: Transform 38:07 
   Advanced Synthesis Topics: Enamines 41:52 
    Enamines 41:53 
   Advanced Synthesis Topics: Enamines 43:06 
    Reaction with Ketone/Aldehyde 43:07 
    Example 44:08 
   Advanced Synthesis Topics: Enamines 45:31 
    Example: Use Enamines as Nu: (Like Enolate) 45:32 
   Advanced Synthesis Topics: Enamines 47:56 
    Example 47:58 
IX. Aromatic Compounds
  Aromatic Compounds: Structure 1:00:59
   Intro 0:00 
   Aromatic Compounds 0:05 
    Benzene 0:06 
    3D Sketch 1:33 
   Features of Benzene 4:41 
    Features of Benzene 4:42 
   Aromatic Stability 6:41 
    Resonance Stabilization of Benzene 6:42 
    Cyclohexatriene 7:24 
    Benzene (Actual, Experimental) 8:11 
   Aromatic Stability 9:03 
    Energy Graph 9:04 
   Aromaticity Requirements 9:55 
    1) Cyclic and Planar 9:56 
    2) Contiguous p Orbitals 10:49 
    3) Satisfy Huckel's Rule 11:20 
    Example: Benzene 12:32 
   Common Aromatic Compounds 13:28 
    Example: Pyridine 13:29 
   Common Aromatic Compounds 16:25 
    Example: Furan 16:26 
   Common Aromatic Compounds 19:42 
    Example: Thiophene 19:43 
    Example: Pyrrole 20:18 
   Common Aromatic Compounds 21:09 
    Cyclopentadienyl Anion 21:10 
    Cycloheptatrienyl Cation 23:48 
    Naphthalene 26:04 
   Determining Aromaticity 27:28 
    Example: Which of the Following are Aromatic? 27:29 
   Molecular Orbital (MO) Theory 32:26 
    What's So Special About '4n + 2' Electrons? 32:27 
    π bond & Overlapping p Orbitals 32:53 
   Molecular Orbital (MO) Diagrams 36:56 
    MO Diagram: Benzene 36:58 
   Drawing MO Diagrams 44:26 
    Example: 3-Membered Ring 44:27 
    Example: 4-Membered Ring 46:04 
   Drawing MO Diagrams 47:51 
    Example: 5-Membered Ring 47:52 
    Example: 8-Membered Ring 49:32 
   Aromaticity and Reactivity 51:03 
    Example: Which is More Acidic? 51:04 
   Aromaticity and Reactivity 56:03 
    Example: Which has More Basic Nitrogen, Pyrrole or Pyridine? 56:04 
  Aromatic Compounds: Reactions, Part 1 1:24:04
   Intro 0:00 
   Reactions of Benzene 0:07 
    N/R as Alkenes 0:08 
    Substitution Reactions 0:50 
   Electrophilic Aromatic Substitution 1:24 
    Electrophilic Aromatic Substitution 1:25 
    Mechanism Step 1: Addition of Electrophile 2:08 
    Mechanism Step 2: Loss of H+ 4:14 
   Electrophilic Aromatic Substitution on Substituted Benzenes 5:21 
    Electron Donating Group 5:22 
    Electron Withdrawing Group 8:02 
    Halogen 9:23 
   Effects of Electron-Donating Groups (EDG) 10:23 
    Effects of Electron-Donating Groups (EDG) 10:24 
    What Effect Does EDG (OH) Have? 11:40 
    Reactivity 13:03 
    Regioselectivity 14:07 
   Regioselectivity: EDG is o/p Director 14:57 
    Prove It! Add E+ and Look at Possible Intermediates 14:58 
    Is OH Good or Bad? 17:38 
   Effects of Electron-Withdrawing Groups (EWG) 20:20 
    What Effect Does EWG Have? 20:21 
    Reactivity 21:28 
    Regioselectivity 22:24 
   Regioselectivity: EWG is a Meta Director 23:23 
    Prove It! Add E+ and Look at Competing Intermediates 23:24 
    Carbocation: Good or Bad? 26:01 
   Effects of Halogens on EAS 28:33 
    Inductive Withdrawal of e- Density vs. Resonance Donation 28:34 
   Summary of Substituent Effects on EAS 32:33 
    Electron Donating Group 32:34 
    Electron Withdrawing Group 33:37 
   Directing Power of Substituents 34:35 
    Directing Power of Substituents 34:36 
    Example 36:41 
   Electrophiles for Electrophilic Aromatic Substitution 38:43 
    Reaction: Halogenation 38:44 
   Electrophiles for Electrophilic Aromatic Substitution 40:27 
    Reaction: Nitration 40:28 
   Electrophiles for Electrophilic Aromatic Substitution 41:45 
    Reaction: Sulfonation 41:46 
   Electrophiles for Electrophilic Aromatic Substitution 43:19 
    Reaction: Friedel-Crafts Alkylation 43:20 
   Electrophiles for Electrophilic Aromatic Substitution 45:43 
    Reaction: Friedel-Crafts Acylation 45:44 
   Electrophilic Aromatic Substitution: Nitration 46:52 
    Electrophilic Aromatic Substitution: Nitration 46:53 
    Mechanism 48:56 
   Nitration of Aniline 52:40 
    Nitration of Aniline Part 1 52:41 
    Nitration of Aniline Part 2: Why? 54:12 
   Nitration of Aniline 56:10 
    Workaround: Protect Amino Group as an Amide 56:11 
   Electrophilic Aromatic Substitution: Sulfonation 58:16 
    Electrophilic Aromatic Substitution: Sulfonation 58:17 
    Example: Transform 59:25 
   Electrophilic Aromatic Substitution: Friedel-Crafts Alkylation 62:24 
    Electrophilic Aromatic Substitution: Friedel-Crafts Alkylation 62:25 
    Example & Mechanism 63:37 
   Friedel-Crafts Alkylation Drawbacks 65:48 
    A) Can Over-React (Dialkylation) 65:49 
   Friedel-Crafts Alkylation Drawbacks 68:21 
    B) Carbocation Can Rearrange 68:22 
    Mechanism 69:33 
   Friedel-Crafts Alkylation Drawbacks 73:35 
    Want n-Propyl? Use Friedel-Crafts Acylation 73:36 
    Reducing Agents 76:45 
   Synthesis with Electrophilic Aromatic Substitution 78:45 
    Example: Transform 78:46 
   Synthesis with Electrophilic Aromatic Substitution 80:59 
    Example: Transform 81:00 
  Aromatic Compounds: Reactions, Part 2 59:10
   Intro 0:00 
   Reagents for Electrophilic Aromatic Substitution 0:07 
    Reagents for Electrophilic Aromatic Substitution 0:08 
   Preparation of Diazonium Salt 2:12 
    Preparation of Diazonium Salt 2:13 
   Reagents for Sandmeyer Reactions 4:14 
    Reagents for Sandmeyer Reactions 4:15 
   Apply Diazonium Salt in Synthesis 6:20 
    Example: Transform 6:21 
   Apply Diazonium Salt in Synthesis 9:14 
    Example: Synthesize Following Target Molecule from Benzene or Toluene 9:15 
   Apply Diazonium Salt in Synthesis 14:56 
    Example: Transform 14:57 
   Reactions of Aromatic Substituents 21:56 
    A) Reduction Reactions 21:57 
   Reactions of Aromatic Substituents 23:24 
    B) Oxidations of Arenes 23:25 
    Benzylic [ox] Even Breaks C-C Bonds! 25:05 
    Benzylic Carbon Can't Be Quaternary 25:55 
   Reactions of Aromatic Substituents 26:21 
    Example 26:22 
   Review of Benzoic Acid Synthesis 27:34 
    Via Hydrolysis 27:35 
    Via Grignard 28:20 
   Reactions of Aromatic Substituents 29:15 
    C) Benzylic Halogenation 29:16 
    Radical Stabilities 31:55 
    N-bromosuccinimide (NBS) 32:23 
   Reactions of Aromatic Substituents 33:08 
    D) Benzylic Substitutions 33:09 
   Reactions of Aromatic Side Chains 37:08 
    Example: Transform 37:09 
   Nucleophilic Aromatic Substitution 43:13 
    Nucleophilic Aromatic Substitution 43:14 
   Nucleophilic Aromatic Substitution 47:08 
    Example 47:09 
    Mechanism 48:00 
   Nucleophilic Aromatic Substitution 50:43 
    Example 50:44 
   Nucleophilic Substitution: Benzyne Mechanism 52:46 
    Nucleophilic Substitution: Benzyne Mechanism 52:47 
   Nucleophilic Substitution: Benzyne Mechanism 57:31 
    Example: Predict Product 57:32 
X. Dienes & Amines
  Conjugated Dienes 1:09:12
   Intro 0:00 
   Conjugated Dienes 0:08 
    Conjugated π Bonds 0:09 
   Diene Stability 2:00 
    Diene Stability: Cumulated 2:01 
    Diene Stability: Isolated 2:37 
    Diene Stability: Conjugated 2:51 
    Heat of Hydrogenation 3:00 
   Allylic Carbocations and Radicals 5:15 
    Allylic Carbocations and Radicals 5:16 
   Electrophilic Additions to Dienes 7:00 
    Alkenes 7:01 
    Unsaturated Ketone 7:47 
   Electrophilic Additions to Dienes 8:28 
    Conjugated Dienes 8:29 
   Electrophilic Additions to Dienes 9:46 
    Mechanism (2-Steps): Alkene 9:47 
   Electrophilic Additions to Dienes 11:40 
    Mechanism (2-Steps): Diene 11:41 
    1,2 'Kinetic' Product 13:08 
    1,4 'Thermodynamic' Product 14:47 
   E vs. POR Diagram 15:50 
    E vs. POR Diagram 15:51 
   Kinetic vs. Thermodynamic Control 21:56 
    Kinetic vs. Thermodynamic Control 21:57 
   How? Reaction is Reversible! 23:51 
    1,2 (Less Stable product) 23:52 
    1,4 (More Stable Product) 25:16 
   Diels Alder Reaction 26:34 
    Diels Alder Reaction 26:35 
   Dienophiles (E+) 29:23 
    Dienophiles (E+) 29:24 
   Alkyne Diels-Alder Example 30:48 
    Example: Alkyne Diels-Alder 30:49 
   Diels-Alder Reaction: Dienes (Nu:) 32:22 
    Diels-Alder ReactionL Dienes (Nu:) 32:23 
   Diels-Alder Reaction: Dienes 33:51 
    Dienes Must Have 's-cis' Conformation 33:52 
    Example 35:25 
   Diels-Alder Reaction with Cyclic Dienes 36:08 
    Cyclic Dienes are Great for Diels-Alder Reaction 36:09 
     Cyclopentadiene 37:10 
   Diels-Alder Reaction: Bicyclic Products 40:50 
    Endo vs. Exo Terminology: Norbornane & Bicyclo Heptane 40:51 
    Example: Bicyclo Heptane 42:29 
   Diels-Alder Reaction with Cyclic Dienes 44:15 
    Example 44:16 
   Stereochemistry of the Diels-Alder Reaction 47:39 
    Stereochemistry of the Diels-Alder Reaction 47:40 
    Example 48:08 
   Stereochemistry of the Diels-Alder Reaction 50:21 
    Example 50:22 
   Regiochemistry of the Diels-Alder Reaction 52:42 
    Rule: 1,2-Product Preferred Over 1,3-Product 52:43 
   Regiochemistry of the Diels-Alder Reaction 54:18 
    Rule: 1,4-Product Preferred Over 1,3-Product 54:19 
   Regiochemistry of the Diels-Alder Reaction 55:02 
    Why 1,2-Product or 1,4-Product Favored? 55:03 
    Example 56:11 
   Diels-Alder Reaction 58:06 
    Example: Predict 58:07 
   Diels-Alder Reaction 61:27 
    Explain Why No Diels-Alder Reaction Takes Place in This Case 61:28 
   Diels-Alder Reaction 63:09 
    Example: Predict 63:10 
   Diels-Alder Reaction: Synthesis Problem 65:39 
    Diels-Alder Reaction: Synthesis Problem 65:40 
  Amines 34:58
   Intro 0:00 
   Amines: Properties and Reactivity 0:04 
    Compare Amines to Alcohols 0:05 
   Amines: Lower Boiling Point than ROH 0:55 
    1) RNH₂ Has Lower Boiling Point than ROH 0:56 
   Amines: Better Nu: Than ROH 2:22 
    2) RNH₂ is a Better Nucleophile than ROH Example 1 2:23 
    RNH₂ is a Better Nucleophile than ROH Example 2 3:08 
   Amines: Better Nu: than ROH 3:47 
    Example 3:48 
   Amines are Good Bases 5:41 
    3) RNH₂ is a Good Base 5:42 
   Amines are Good Bases 7:06 
    Example 1 7:07 
    Example 2: Amino Acid 8:27 
   Alkyl vs. Aryl Amines 9:56 
    Example: Which is Strongest Base? 9:57 
   Alkyl vs. Aryl Amines 14:55 
    Verify by Comparing Conjugate Acids 14:56 
   Reaction of Amines 17:42 
    Reaction with Ketone/Aldehyde: 1° Amine (RNH₂) 17:43 
   Reaction of Amines 18:48 
    Reaction with Ketone/Aldehyde: 2° Amine (R2NH) 18:49 
   Use of Enamine: Synthetic Equivalent of Enolate 20:08 
    Use of Enamine: Synthetic Equivalent of Enolate 20:09 
   Reaction of Amines 24:10 
    Hofmann Elimination 24:11 
   Hofmann Elimination 26:16 
    Kinetic Product 26:17 
   Structure Analysis Using Hofmann Elimination 28:22 
    Structure Analysis Using Hofmann Elimination 28:23 
   Biological Activity of Amines 30:30 
    Adrenaline 31:07 
    Mescaline (Peyote Alkaloid) 31:22 
    Amino Acids, Amide, and Protein 32:14 
   Biological Activity of Amines 32:50 
    Morphine (Opium Alkaloid) 32:51 
    Epibatidine (Poison Dart Frog) 33:28 
    Nicotine 33:48 
    Choline (Nerve Impulse) 34:03 
XI. Biomolecules & Polymers
  Biomolecules 1:53:20
   Intro 0:00 
   Carbohydrates 1:11 
    D-glucose Overview 1:12 
    D-glucose: Cyclic Form (6-membered ring) 4:31 
   Cyclic Forms of Glucose: 6-membered Ring 8:24 
    α-D-glucopyranose & β-D-glucopyranose 8:25 
   Formation of a 5-Membered Ring 11:05 
    D-glucose: Formation of a 5-Membered Ring 11:06 
   Cyclic Forms of Glucose: 5-membered Ring 12:37 
    α-D-glucofuranose & β-D-glucofuranose 12:38 
   Carbohydrate Mechanism 14:03 
    Carbohydrate Mechanism 14:04 
   Reactions of Glucose: Acetal Formation 21:35 
    Acetal Formation: Methyl-α-D-glucoside 21:36 
    Hemiacetal to Acetal: Overview 24:58 
   Mechanism for Formation of Glycosidic Bond 25:51 
    Hemiacetal to Acetal: Mechanism 25:52 
   Formation of Disaccharides 29:34 
    Formation of Disaccharides 29:35 
   Some Polysaccharides: Starch 31:33 
    Amylose & Amylopectin 31:34 
    Starch: α-1,4-glycosidic Bonds 32:22 
    Properties of Starch Molecule 33:21 
   Some Polysaccharides: Cellulose 33:59 
    Cellulose: β-1,4-glycosidic bonds 34:00 
    Properties of Cellulose 34:59 
   Other Sugar-Containing Biomolecules 35:50 
    Ribonucleoside (RNA) 35:51 
    Deoxyribonucleoside (DMA) 36:59 
   Amino Acids & Proteins 37:32 
    α-amino Acids: Structure & Stereochemistry 37:33 
   Making a Protein (Condensation) 42:46 
    Making a Protein (Condensation) 42:47 
   Peptide Bond is Planar (Amide Resonance) 44:55 
    Peptide Bond is Planar (Amide Resonance) 44:56 
   Protein Functions 47:49 
    Muscle, Skin, Bones, Hair Nails 47:50 
    Enzymes 49:10 
    Antibodies 49:44 
    Hormones, Hemoglobin 49:58 
    Gene Regulation 50:20 
   Various Amino Acid Side Chains 50:51 
    Nonpolar 50:52 
    Polar 51:15 
    Acidic 51:24 
    Basic 51:55 
   Amino Acid Table 52:22 
    Amino Acid Table 52:23 
   Isoelectric Point (pI) 53:43 
    Isoelectric Point (pI) of Glycine 53:44 
    Isoelectric Point (pI) of Glycine: pH 11 56:42 
    Isoelectric Point (pI) of Glycine: pH 1 57:20 
   Isoelectric Point (pI), cont. 58:05 
    Asparatic Acid 58:06 
    Histidine 60:28 
   Isoelectric Point (pI), cont. 62:54 
    Example: What is the Net Charge of This Tetrapeptide at pH 6.0? 62:55 
   Nucleic Acids: Ribonucleosides 70:32 
    Nucleic Acids: Ribonucleosides 70:33 
   Nucleic Acids: Ribonucleotides 71:48 
    Ribonucleotides: 5' Phosphorylated Ribonucleosides 71:49 
   Ribonucleic Acid (RNA) Structure 72:35 
    Ribonucleic Acid (RNA) Structure 72:36 
   Nucleic Acids: Deoxyribonucleosides 74:08 
    Nucleic Acids: Deoxyribonucleosides 74:09 
    Deoxythymidine (T) 74:36 
   Nucleic Acids: Base-Pairing 75:17 
    Nucleic Acids: Base-Pairing 75:18 
   Double-Stranded Structure of DNA 78:16 
    Double-Stranded Structure of DNA 78:17 
   Model of DNA 79:40 
    Model of DNA 79:41 
   Space-Filling Model of DNA 80:46 
    Space-Filling Model of DNA 80:47 
   Function of RNA and DNA 83:06 
    DNA & Transcription 83:07 
    RNA & Translation 84:22 
   Genetic Code 85:09 
    Genetic Code 85:10 
   Lipids/Fats/Triglycerides 87:10 
    Structure of Glycerol 87:43 
    Saturated & Unsaturated Fatty Acids 87:51 
    Triglyceride 88:43 
   Unsaturated Fats: Lower Melting Points (Liquids/Oils) 89:15 
    Saturated Fat 89:16 
    Unsaturated Fat 90:10 
    Partial Hydrogenation 92:05 
   Saponification of Fats 95:11 
    Saponification of Fats 95:12 
    History of Soap 96:50 
   Carboxylate Salts form Micelles in Water 101:02 
    Carboxylate Salts form Micelles in Water 101:03 
   Cleaning Power of Micelles 102:21 
    Cleaning Power of Micelles 102:22 
   3-D Image of a Micelle 102:58 
    3-D Image of a Micelle 102:59 
   Synthesis of Biodiesel 104:04 
    Synthesis of Biodiesel 104:05 
   Phosphoglycerides 107:54 
    Phosphoglycerides 107:55 
   Cell Membranes Contain Lipid Bilayers 108:41 
    Cell Membranes Contain Lipid Bilayers 108:42 
   Bilayer Acts as Barrier to Movement In/Out of Cell 110:24 
    Bilayer Acts as Barrier to Movement In/Out of Cell 110:25 
   Organic Chemistry Meets Biology… Biochemistry! 111:12 
    Organic Chemistry Meets Biology… Biochemistry! 111:13 
  Polymers 45:47
   Intro 0:00 
   Polymers 0:05 
    Monomer to Polymer: Vinyl Chloride to Polyvinyl Chloride 0:06 
   Polymer Properties 1:32 
    Polymer Properties 1:33 
   Natural Polymers: Rubber 2:30 
    Vulcanization 2:31 
   Natural Polymers: Polysaccharides 4:55 
    Example: Starch 4:56 
    Example: Cellulose 5:45 
   Natural Polymers: Proteins 6:07 
    Example: Keratin 6:08 
   DNA Strands 7:15 
    DNA Strands 7:16 
   Synthetic Polymers 8:30 
    Ethylene & Polyethylene: Lightweight Insulator & Airtight Plastic 8:31 
   Synthetic Organic Polymers 12:22 
    Polyethylene 12:28 
    Polyvinyl Chloride (PVC) 12:54 
    Polystyrene 13:28 
    Polyamide 14:34 
    Polymethyl Methacrylate 14:57 
    Kevlar 15:25 
    Synthetic Material Examples 16:30 
    How are Polymers Made? 21:00 
    Chain-growth Polymers Additions to Alkenes can be Radical, Cationic or Anionic 21:01 
   Chain Branching 22:34 
    Chain Branching 22:35 
   Special Reaction Conditions Prevent Branching 24:28 
    Ziegler-Natta Catalyst 24:29 
   Chain-Growth by Cationic Polymerization 27:35 
    Chain-Growth by Cationic Polymerization 27:36 
   Chain-Growth by Anionic Polymerization 29:35 
    Chain-Growth by Anionic Polymerization 29:36 
   Step-Growth Polymerization: Polyamides 32:16 
    Step-Growth Polymerization: Polyamides 32:17 
   Step-Growth Polymerization: Polyesters 34:23 
    Step-Growth Polymerization: Polyesters 34:24 
   Step-Growth Polymerization: Polycarbonates 35:56 
    Step-Growth Polymerization: Polycarbonates 35:57 
   Step-Growth Polymerization: Polyurethanes 37:18 
    Step-Growth Polymerization: Polyurethanes 37:19 
   Modifying Polymer Properties 39:35 
    Glass Transition Temperature 40:04 
    Crosslinking 40:42 
    Copolymers 40:58 
    Additives: Stabilizers 42:08 
    Additives: Flame Retardants 43:03 
    Additives: Plasticizers 43:41 
    Additives: Colorants 44:54 
XII. Organic Synthesis
  Organic Synthesis Strategies 2:20:24
   Intro 0:00 
   Organic Synthesis Strategies 0:15 
    Goal 0:16 
    Strategy 0:29 
   Example of a RetroSynthesis 1:30 
    Finding Starting Materials for Target Molecule 1:31 
    Synthesis Using Starting Materials 4:56 
   Synthesis of Alcohols by Functional Group Interconversion (FGI) 6:00 
    Synthesis of Alcohols by Functional Group Interconversion Overview 6:01 
   Alcohols by Reduction 7:43 
    Ketone to Alcohols 7:45 
    Aldehyde to Alcohols 8:26 
    Carboxylic Acid Derivative to Alcohols 8:36 
   Alcohols by Hydration of Alkenes 9:28 
    Hydration of Alkenes Using H₃O⁺ 9:29 
    Oxymercuration-Demercuration 10:35 
    Hydroboration Oxidation 11:02 
   Alcohols by Substitution 11:42 
    Primary Alkyl Halide to Alcohols Using NaOH 11:43 
    Secondary Alkyl Halide to Alcohols Using Sodium Acetate 13:07 
    Tertiary Alkyl Halide to Alcohols Using H₂O 15:08 
   Synthesis of Alcohols by Forming a New C-C Bond 15:47 
    Recall: Alcohol & RMgBr 15:48 
    Retrosynthesis 17:28 
   Other Alcohol Disconnections 19:46 
    Synthesis Using PhMGgBr: Example 2 23:05 
   Synthesis of Alkyl Halides 26:06 
    Synthesis of Alkyl Halides Overview 26:07 
   Synthesis of Alkyl Halides by Free Radical Halogenation 27:04 
    Synthesis of Alkyl Halides by Free Radical Halogenation 27:05 
   Synthesis of Alkyl Halides by Substitution 29:06 
    Alcohol to Alkyl Halides Using HBr or HCl 29:07 
    Alcohol to Alkyl Halides Using SOCl₂ 30:57 
    Alcohol to Alkyl Halides Using PBr₃ and Using P, I₂ 31:03 
   Synthesis of Alkyl Halides by Addition 32:02 
    Alkene to Alkyl Halides Using HBr 32:03 
    Alkene to Alkyl Halides Using HBr & ROOR (Peroxides) 32:35 
   Example: Synthesis of Alkyl Halide 34:18 
    Example: Synthesis of Alkyl Halide 34:19 
   Synthesis of Ethers 39:25 
    Synthesis of Ethers 39:26 
   Example: Synthesis of an Ether 41:12 
    Synthesize TBME (t-butyl methyl ether) from Alcohol Starting Materials 41:13 
   Synthesis of Amines 46:05 
    Synthesis of Amines 46:06 
   Gabriel Synthesis of Amines 47:57 
    Gabriel Synthesis of Amines 47:58 
   Amines by SN2 with Azide Nu: 49:50 
    Amines by SN2 with Azide Nu: 49:51 
   Amines by SN2 with Cyanide Nu: 50:31 
    Amines by SN2 with Cyanide Nu: 50:32 
   Amines by Reduction of Amides 51:30 
    Amines by Reduction of Amides 51:31 
   Reductive Amination of Ketones/Aldehydes 52:42 
    Reductive Amination of Ketones/Aldehydes 52:43 
   Example : Synthesis of an Amine 53:47 
    Example 1: Synthesis of an Amine 53:48 
    Example 2: Synthesis of an Amine 56:16 
   Synthesis of Alkenes 58:20 
    Synthesis of Alkenes Overview 58:21 
   Synthesis of Alkenes by Elimination 59:04 
    Synthesis of Alkenes by Elimination Using NaOH & Heat 59:05 
    Synthesis of Alkenes by Elimination Using H₂SO₄ & Heat 59:57 
   Synthesis of Alkenes by Reduction 62:05 
    Alkyne to Cis Alkene 62:06 
    Alkyne to Trans Alkene 62:56 
   Synthesis of Alkenes by Wittig Reaction 63:46 
    Synthesis of Alkenes by Wittig Reaction 63:47 
    Retrosynthesis of an Alkene 65:35 
   Example: Synthesis of an Alkene 66:57 
    Example: Synthesis of an Alkene 66:58 
    Making a Wittig Reagent 70:31 
   Synthesis of Alkynes 73:09 
    Synthesis of Alkynes 73:10 
   Synthesis of Alkynes by Elimination (FGI) 73:42 
    First Step: Bromination of Alkene 73:43 
    Second Step: KOH Heat 74:22 
   Synthesis of Alkynes by Alkylation 75:02 
    Synthesis of Alkynes by Alkylation 75:03 
    Retrosynthesis of an Alkyne 76:18 
   Example: Synthesis of an Alkyne 77:40 
    Example: Synthesis of an Alkyne 77:41 
   Synthesis of Alkanes 80:52 
    Synthesis of Alkanes 80:53 
   Synthesis of Aldehydes & Ketones 81:38 
    Oxidation of Alcohol Using PCC or Swern 81:39 
    Oxidation of Alkene Using 1) O₃, 2)Zn 82:42 
    Reduction of Acid Chloride & Nitrile Using DiBAL-H 83:25 
    Hydration of Alkynes 84:55 
    Synthesis of Ketones by Acyl Substitution 86:12 
    Reaction with R'₂CuLi 86:13 
    Reaction with R'MgBr 87:13 
   Synthesis of Aldehydes & Ketones by α-Alkylation 88:00 
    Synthesis of Aldehydes & Ketones by α-Alkylation 88:01 
    Retrosynthesis of a Ketone 90:10 
   Acetoacetate Ester Synthesis of Ketones 91:05 
    Acetoacetate Ester Synthesis of Ketones: Step 1 91:06 
    Acetoacetate Ester Synthesis of Ketones: Step 2 92:13 
    Acetoacetate Ester Synthesis of Ketones: Step 3 92:50 
   Example: Synthesis of a Ketone 94:11 
    Example: Synthesis of a Ketone 94:12 
   Synthesis of Carboxylic Acids 97:15 
    Synthesis of Carboxylic Acids 97:16 
   Example: Synthesis of a Carboxylic Acid 97:59 
    Example: Synthesis of a Carboxylic Acid (Option 1) 98:00 
    Example: Synthesis of a Carboxylic Acid (Option 2) 100:51 
   Malonic Ester Synthesis of Carboxylic Acid 102:34 
    Malonic Ester Synthesis of Carboxylic Acid: Step 1 102:35 
    Malonic Ester Synthesis of Carboxylic Acid: Step 2 103:36 
    Malonic Ester Synthesis of Carboxylic Acid: Step 3 104:01 
   Example: Synthesis of a Carboxylic Acid 104:53 
    Example: Synthesis of a Carboxylic Acid 104:54 
   Synthesis of Carboxylic Acid Derivatives 108:05 
    Synthesis of Carboxylic Acid Derivatives 108:06 
   Alternate Ester Synthesis 108:58 
    Using Fischer Esterification 108:59 
    Using SN2 Reaction 110:18 
    Using Diazomethane 110:56 
    Using 1) LDA, 2) R'-X 112:15 
   Practice: Synthesis of an Alkyl Chloride 113:11 
    Practice: Synthesis of an Alkyl Chloride 113:12 
   Patterns of Functional Groups in Target Molecules 119:53 
    Recall: Aldol Reaction 119:54 
    β-hydroxy Ketone Target Molecule 121:12 
    α,β-unsaturated Ketone Target Molecule 122:20 
   Patterns of Functional Groups in Target Molecules 123:15 
    Recall: Michael Reaction 123:16 
    Retrosynthesis: 1,5-dicarbonyl Target Molecule 124:07 
   Patterns of Functional Groups in Target Molecules 126:38 
    Recall: Claisen Condensation 126:39 
    Retrosynthesis: β-ketoester Target Molecule 127:30 
   2-Group Target Molecule Summary 129:03 
    2-Group Target Molecule Summary 129:04 
   Example: Synthesis of Epoxy Ketone 131:19 
    Synthesize the Following Target Molecule from Cyclohexanone: Part 1 - Retrosynthesis 131:20 
    Synthesize the Following Target Molecule from Cyclohexanone: Part 2 - Synthesis 134:10 
   Example: Synthesis of a Diketone 136:57 
    Synthesis of a Diketone: Step 1 - Retrosynthesis 136:58 
    Synthesis of a Diketone: Step 2 - Synthesis 138:51 
XIII. Spectroscopy
  Infrared Spectroscopy, Part I 1:04:00
   Intro 0:00 
   Infrared (IR) Spectroscopy 0:09 
    Introduction to Infrared (IR) Spectroscopy 0:10 
    Intensity of Absorption Is Proportional to Change in Dipole 3:08 
   IR Spectrum of an Alkane 6:08 
    Pentane 6:09 
   IR Spectrum of an Alkene 13:12 
    1-Pentene 13:13 
   IR Spectrum of an Alkyne 15:49 
    1-Pentyne 15:50 
   IR Spectrum of an Aromatic Compound 18:2 
    Methylbenzene 18:24 
   IR of Substituted Aromatic Compounds 24:04 
    IR of Substituted Aromatic Compounds 24:05 
   IR Spectrum of 1,2-Disubstituted Aromatic 25:30 
    1,2-dimethylbenzene 25:31 
   IR Spectrum of 1,3-Disubstituted Aromatic 27:15 
    1,3-dimethylbenzene 27:16 
   IR Spectrum of 1,4-Disubstituted Aromatic 28:41 
    1,4-dimethylbenzene 28:42 
   IR Spectrum of an Alcohol 29:34 
    1-pentanol 29:35 
   IR Spectrum of an Amine 32:39 
    1-butanamine 32:40 
   IR Spectrum of a 2° Amine 34:50 
    Diethylamine 34:51 
   IR Spectrum of a 3° Amine 35:47 
    Triethylamine 35:48 
   IR Spectrum of a Ketone 36:41 
    2-butanone 36:42 
   IR Spectrum of an Aldehyde 40:10 
    Pentanal 40:11 
   IR Spectrum of an Ester 42:38 
    Butyl Propanoate 42:39 
   IR Spectrum of a Carboxylic Acid 44:26 
    Butanoic Acid 44:27 
   Sample IR Correlation Chart 47:36 
    Sample IR Correlation Chart: Wavenumber and Functional Group 47:37 
   Predicting IR Spectra: Sample Structures 52:06 
    Example 1 52:07 
    Example 2 53:29 
    Example 3 54:40 
    Example 4 57:08 
    Example 5 58:31 
    Example 6 59:07 
    Example 7 60:52 
    Example 8 62:20 
  Infrared Spectroscopy, Part II 48:34
   Intro 0:00 
   Interpretation of IR Spectra: a Basic Approach 0:05 
    Interpretation of IR Spectra: a Basic Approach 0:06 
    Other Peaks to Look for 3:39 
   Examples 5:17 
    Example 1 5:18 
    Example 2 9:09 
    Example 3 11:52 
    Example 4 14:03 
    Example 5 16:31 
    Example 6 19:31 
    Example 7 22:32 
    Example 8 24:39 
   IR Problems Part 1 28:11 
    IR Problem 1 28:12 
    IR Problem 2 31:14 
    IR Problem 3 32:59 
    IR Problem 4 34:23 
    IR Problem 5 35:49 
    IR Problem 6 38:20 
   IR Problems Part 2 42:36 
    IR Problem 7 42:37 
    IR Problem 8 44:02 
    IR Problem 9 45:07 
    IR Problems10 46:10 
  Nuclear Magnetic Resonance (NMR) Spectroscopy, Part I 1:32:14
   Intro 0:00 
   Purpose of NMR 0:14 
    Purpose of NMR 0:15 
   How NMR Works 2:17 
    How NMR Works 2:18 
   Information Obtained From a ¹H NMR Spectrum 5:51 
    No. of Signals, Integration, Chemical Shifts, and Splitting Patterns 5:52 
   Number of Signals in NMR (Chemical Equivalence) 7:52 
    Example 1: How Many Signals in ¹H NMR? 7:53 
    Example 2: How Many Signals in ¹H NMR? 9:36 
    Example 3: How Many Signals in ¹H NMR? 12:15 
    Example 4: How Many Signals in ¹H NMR? 13:47 
    Example 5: How Many Signals in ¹H NMR? 16:12 
   Size of Signals in NMR (Peak Area or Integration) 21:23 
    Size of Signals in NMR (Peak Area or Integration) 21:24 
   Using Integral Trails 25:15 
    Example 1: C₈H₁₈O 25:16 
    Example 2: C₃H₈O 27:17 
    Example 3: C₇H₈ 28:21 
   Location of NMR Signal (Chemical Shift) 29:05 
    Location of NMR Signal (Chemical Shift) 29:06 
   ¹H NMR Chemical Shifts 33:20 
    ¹H NMR Chemical Shifts 33:21 
   ¹H NMR Chemical Shifts (Protons on Carbon) 37:03 
    ¹H NMR Chemical Shifts (Protons on Carbon) 37:04 
   Chemical Shifts of H's on N or O 39:01 
    Chemical Shifts of H's on N or O 39:02 
   Estimating Chemical Shifts 41:13 
    Example 1: Estimating Chemical Shifts 41:14 
    Example 2: Estimating Chemical Shifts 43:22 
    Functional Group Effects are Additive 45:28 
   Calculating Chemical Shifts 47:38 
    Methylene Calculation 47:39 
    Methine Calculation 48:20 
    Protons on sp³ Carbons: Chemical Shift Calculation Table 48:50 
    Example: Estimate the Chemical Shift of the Selected H 50:29 
   Effects of Resonance on Chemical Shifts 53:11 
    Example 1: Effects of Resonance on Chemical Shifts 53:12 
    Example 2: Effects of Resonance on Chemical Shifts 55:09 
    Example 3: Effects of Resonance on Chemical Shifts 57:08 
   Shape of NMR Signal (Splitting Patterns) 59:17 
    Shape of NMR Signal (Splitting Patterns) 59:18 
   Understanding Splitting Patterns: The 'n+1 Rule' 61:24 
    Understanding Splitting Patterns: The 'n+1 Rule' 61:25 
   Explanation of n+1 Rule 62:42 
    Explanation of n+1 Rule: One Neighbor 62:43 
    Explanation of n+1 Rule: Two Neighbors 66:23 
   Summary of Splitting Patterns 66:24 
    Summary of Splitting Patterns 70:45 
   Predicting ¹H NMR Spectra 70:46 
    Example 1: Predicting ¹H NMR Spectra 73:30 
    Example 2: Predicting ¹H NMR Spectra 79:07 
    Example 3: Predicting ¹H NMR Spectra 83:50 
    Example 4: Predicting ¹H NMR Spectra 89:27 
  Nuclear Magnetic Resonance (NMR) Spectroscopy, Part II 2:03:48
   Intro 0:00 
   ¹H NMR Problem-Solving Strategies 0:18 
    Step 1: Analyze IR Spectrum (If Provided) 0:19 
    Step 2: Analyze Molecular Formula (If Provided) 2:06 
    Step 3: Draw Pieces of Molecule 3:49 
    Step 4: Confirm Pieces 6:30 
    Step 5: Put the Pieces Together! 7:23 
    Step 6: Check Your Answer! 8:21 
   Examples 9:17 
    Example 1: Determine the Structure of a C₉H₁₀O₂ Compound with the Following ¹H NMR Data 9:18 
    Example 2: Determine the Structure of a C₉H₁₀O₂ Compound with the Following ¹H NMR Data 17:27 
   ¹H NMR Practice 20:57 
    ¹H NMR Practice 1: C₁₀H₁₄ 20:58 
    ¹H NMR Practice 2: C₄H₈O₂ 29:50 
    ¹H NMR Practice 3: C₆H₁₂O₃ 39:19 
    ¹H NMR Practice 4: C₈H₁₈ 50:19 
   More About Coupling Constants (J Values) 57:11 
    Vicinal (3-bond) and Geminal (2-bond) 57:12 
    Cyclohexane (ax-ax) and Cyclohexane (ax-eq) or (eq-eq) 59:50 
    Geminal (Alkene), Cis (Alkene), and Trans (Alkene) 62:40 
    Allylic (4-bond) and W-coupling (4-bond) (Rigid Structures Only) 64:05 
   ¹H NMR Advanced Splitting Patterns 65:39 
    Example 1: ¹H NMR Advanced Splitting Patterns 65:40 
    Example 2: ¹H NMR Advanced Splitting Patterns 70:01 
    Example 3: ¹H NMR Advanced Splitting Patterns 73:45 
   ¹H NMR Practice 82:53 
    ¹H NMR Practice 5: C₁₁H₁₇N 82:54 
    ¹H NMR Practice 6: C₉H₁₀O 94:04 
   ¹³C NMR Spectroscopy 104:49 
    ¹³C NMR Spectroscopy 104:50 
   ¹³C NMR Chemical Shifts 107:24 
    ¹³C NMR Chemical Shifts Part 1 107:25 
    ¹³C NMR Chemical Shifts Part 2 108:59 
   ¹³C NMR Practice 110:16 
    ¹³C NMR Practice 1 110:17 
    ¹³C NMR Practice 2 118:30 
  Mass Spectrometry 1:28:35
   Intro 0:00 
   Introduction to Mass Spectrometry 0:37 
    Uses of Mass Spectrometry: Molecular Mass 0:38 
    Uses of Mass Spectrometry: Molecular Formula 1:04 
    Uses of Mass Spectrometry: Structural Information 1:21 
    Uses of Mass Spectrometry: In Conjunction with Gas Chromatography 2:03 
   Obtaining a Mass Spectrum 2:59 
    Obtaining a Mass Spectrum 3:00 
   The Components of a Mass Spectrum 6:44 
    The Components of a Mass Spectrum 6:45 
   What is the Mass of a Single Molecule 12:13 
    Example: CH₄ 12:14 
    Example: ¹³CH₄ 12:51 
    What Ratio is Expected for the Molecular Ion Peaks of C₂H₆? 14:20 
   Other Isotopes of High Abundance 16:30 
    Example: Cl Atoms 16:31 
    Example: Br Atoms 18:33 
    Mass Spectrometry of Chloroethane 19:22 
    Mass Spectrometry of Bromobutane 21:23 
   Isotopic Abundance can be Calculated 22:48 
    What Ratios are Expected for the Molecular Ion Peaks of CH₂Br₂? 22:49 
   Determining Molecular Formula from High-resolution Mass Spectrometry 26:53 
    Exact Masses of Various Elements 26:54 
   Fragmentation of various Functional Groups 28:42 
    What is More Stable, a Carbocation C⁺ or a Radical R? 28:43 
    Fragmentation is More Likely If It Gives Relatively Stable Carbocations and Radicals 31:37 
   Mass Spectra of Alkanes 33:15 
    Example: Hexane 33:16 
    Fragmentation Method 1 34:19 
    Fragmentation Method 2 35:46 
    Fragmentation Method 3 36:15 
   Mass of Common Fragments 37:07 
    Mass of Common Fragments 37:08 
   Mass Spectra of Alkanes 39:28 
    Mass Spectra of Alkanes 39:29 
    What are the Peaks at m/z 15 and 71 So Small? 41:01 
   Branched Alkanes 43:12 
    Explain Why the Base Peak of 2-methylhexane is at m/z 43 (M-57) 43:13 
   Mass Spectra of Alkenes 45:42 
    Mass Spectra of Alkenes: Remove 1 e⁻ 45:43 
    Mass Spectra of Alkenes: Fragment 46:14 
    High-Energy Pi Electron is Most Likely Removed 47:59 
   Mass Spectra of Aromatic Compounds 49:01 
    Mass Spectra of Aromatic Compounds 49:02 
   Mass Spectra of Alcohols 51:32 
    Mass Spectra of Alcohols 51:33 
   Mass Spectra of Ethers 54:53 
    Mass Spectra of Ethers 54:54 
   Mass Spectra of Amines 56:49 
    Mass Spectra of Amines 56:50 
   Mass Spectra of Aldehydes & Ketones 59:23 
    Mass Spectra of Aldehydes & Ketones 59:24 
   McLafferty Rearrangement 61:29 
    McLafferty Rearrangement 61:30 
   Mass Spectra of Esters 64:15 
    Mass Spectra of Esters 61:16 
   Mass Spectrometry Discussion I 65:01 
    For the Given Molecule (M=58), Do You Expect the More Abundant Peak to Be m/z 15 or m/z 43? 65:02 
   Mass Spectrometry Discussion II 68:13 
    For the Given Molecule (M=74), Do You Expect the More Abundant Peak to Be m/z 31, m/z 45, or m/z 59? 68:14 
   Mass Spectrometry Discussion III 71:42 
    Explain Why the Mass Spectra of Methyl Ketones Typically have a Peak at m/z 43 71:43 
   Mass Spectrometry Discussion IV 74:46 
    In the Mass Spectrum of the Given Molecule (M=88), Account for the Peaks at m/z 45 and m/z 57 74:47 
   Mass Spectrometry Discussion V 78:25 
    How Could You Use Mass Spectrometry to Distinguish Between the Following Two Compounds (M=73)? 78:26 
   Mass Spectrometry Discussion VI 82:45 
    What Would be the m/z Ratio for the Fragment for the Fragment Resulting from a McLafferty Rearrangement for the Following Molecule (M=114)? 82:46 
XIV. Organic Chemistry Lab
  Completing the Reagent Table for Prelab 21:09
   Intro 0:00 
   Sample Reagent Table 0:11 
    Reagent Table Overview 0:12 
    Calculate Moles of 2-bromoaniline 6:44 
   Calculate Molar Amounts of Each Reagent 9:20 
    Calculate Mole of NaNO₂ 9:21 
    Calculate Moles of KI 10:33 
   Identify the Limiting Reagent 11:17 
    Which Reagent is the Limiting Reagent? 11:18 
   Calculate Molar Equivalents 13:37 
    Molar Equivalents 13:38 
   Calculate Theoretical Yield 16:40 
    Theoretical Yield 16:41 
   Calculate Actual Yield (%Yield) 18:30 
    Actual Yield (%Yield) 18:31 
  Introduction to Melting Points 16:10
   Intro 0:00 
   Definition of a Melting Point (mp) 0:04 
    Definition of a Melting Point (mp) 0:05 
    Solid Samples Melt Gradually 1:49 
    Recording Range of Melting Temperature 2:04 
   Melting Point Theory 3:14 
    Melting Point Theory 3:15 
   Effects of Impurities on a Melting Point 3:57 
    Effects of Impurities on a Melting Point 3:58 
    Special Exception: Eutectic Mixtures 5:09 
    Freezing Point Depression by Solutes 5:39 
   Melting Point Uses 6:19 
    Solid Compound 6:20 
    Determine Purity of a Sample 6:42 
    Identify an Unknown Solid 7:06 
   Recording a Melting Point 9:03 
    Pack 1-3 mm of Dry Powder in MP Tube 9:04 
    Slowly Heat Sample 9:55 
    Record Temperature at First Sign of Melting 10:33 
    Record Temperature When Last Crystal Disappears 11:26 
    Discard MP Tube in Glass Waste 11:32 
    Determine Approximate MP 11:42 
   Tips, Tricks and Warnings 12:28 
    Use Small, Tightly Packed Sample 12:29 
    Be Sure MP Apparatus is Cool 12:45 
    Never Reuse a MP Tube 13:16 
    Sample May Decompose 13:30 
    If Pure Melting Point (MP) Doesn't Match Literature 14:20 
  Melting Point Lab 8:17
   Intro 0:00 
   Melting Point Tubes 0:40 
   Melting Point Apparatus 3:42 
   Recording a melting Point 5:50 
  Introduction to Recrystallization 22:00
   Intro 0:00 
   Crystallization to Purify a Solid 0:10 
    Crude Solid 0:11 
    Hot Solution 0:20 
    Crystals 1:09 
    Supernatant Liquid 1:20 
   Theory of Crystallization 2:34 
    Theory of Crystallization 2:35 
   Analysis and Obtaining a Second Crop 3:40 
    Crystals → Melting Point, TLC 3:41 
    Supernatant Liquid → Crude Solid → Pure Solid 4:18 
    Crystallize Again → Pure Solid (2nd Crop) 4:32 
   Choosing a Solvent 5:19 
    1. Product is Very Soluble at High Temperatures 5:20 
    2. Product has Low Solubility at Low Temperatures 6:00 
    3. Impurities are Soluble at All Temperatures 6:16 
    Check Handbooks for Suitable Solvents 7:33 
   Why Isn't This Dissolving?! 8:46 
    If Solid Remains When Solution is Hot 8:47 
    Still Not Dissolved in Hot Solvent? 10:18 
   Where Are My Crystals?! 12:23 
    If No Crystals Form When Solution is Cooled 12:24 
    Still No Crystals? 14:59 
   Tips, Tricks and Warnings 16:26 
    Always Use a Boiling Chip or Stick! 16:27 
    Use Charcoal to Remove Colored Impurities 16:52 
    Solvent Pairs May Be Used 18:23 
    Product May 'Oil Out' 20:11 
  Recrystallization Lab 19:07
   Intro 0:00 
   Step 1: Dissolving the Solute in the Solvent 0:12 
   Hot Filtration 6:33 
   Step 2: Cooling the Solution 8:01 
   Step 3: Filtering the Crystals 12:08 
   Step 4: Removing & Drying the Crystals 16:10 
  Introduction to Distillation 25:54
   Intro 0:00 
   Distillation: Purify a Liquid 0:04 
    Simple Distillation 0:05 
    Fractional Distillation 0:55 
   Theory of Distillation 1:04 
    Theory of Distillation 1:05 
   Vapor Pressure and Volatility 1:52 
    Vapor Pressure 1:53 
    Volatile Liquid 2:28 
    Less Volatile Liquid 3:09 
   Vapor Pressure vs. Boiling Point 4:03 
    Vapor Pressure vs. Boiling Point 4:04 
    Increasing Vapor Pressure 4:38 
   The Purpose of Boiling Chips 6:46 
    The Purpose of Boiling Chips 6:47 
   Homogeneous Mixtures of Liquids 9:24 
    Dalton's Law 9:25 
    Raoult's Law 10:27 
   Distilling a Mixture of Two Liquids 11:41 
    Distilling a Mixture of Two Liquids 11:42 
   Simple Distillation: Changing Vapor Composition 12:06 
    Vapor & Liquid 12:07 
    Simple Distillation: Changing Vapor Composition 14:47 
    Azeotrope 18:41 
   Fractional Distillation: Constant Vapor Composition 19:42 
    Fractional Distillation: Constant Vapor Composition 19:43 
  Distillation Lab 24:13
   Intro 0:00 
   Glassware Overview 0:04 
   Heating a Sample 3:09 
    Bunsen Burner 3:10 
    Heating Mantle 1 4:45 
    Heating Mantle 2 6:18 
    Hot Plate 7:10 
   Simple Distillation Lab 8:37 
   Fractional Distillation Lab 17:13 
   Removing the Distillation Set-Up 22:41 
  Introduction to TLC (Thin-Layer Chromatography) 28:51
   Intro 0:00 
   Chromatography 0:06 
    Purification & Analysis 0:07 
    Types of Chromatography: Thin-layer, Column, Gas, & High Performance Liquid 0:24 
   Theory of Chromatography 0:44 
    Theory of Chromatography 0:45 
   Performing a Thin-layer Chromatography (TLC) Analysis 2:30 
    Overview: Thin-layer Chromatography (TLC) Analysis 2:31 
   Step 1: 'Spot' the TLC Plate 4:11 
   Step 2: Prepare the Developing Chamber 5:54 
   Step 3: Develop the TLC Plate 7:30 
   Step 4: Visualize the Spots 9:02 
   Step 5: Calculate the Rf for Each Spot 12:00 
   Compound Polarity: Effect on Rf 16:50 
    Compound Polarity: Effect on Rf 16:51 
   Solvent Polarity: Effect on Rf 18:47 
    Solvent Polarity: Effect on Rf 18:48 
    Example: EtOAc & Hexane 19:35 
   Other Types of Chromatography 22:27 
    Thin-layer Chromatography (TLC) 22:28 
    Column Chromatography 22:56 
    High Performance Liquid (HPLC) 23:59 
    Gas Chromatography (GC) 24:38 
    Preparative 'prep' Scale Possible 28:05 
  TLC Analysis Lab 20:50
   Intro 0:00 
   Step 1: 'Spot' the TLC Plate 0:06 
   Step 2: Prepare the Developing Chamber 4:06 
   Step 3: Develop the TLC Plate 6:26 
   Step 4: Visualize the Spots 7:45 
   Step 5: Calculate the Rf for Each Spot 11:48 
   How to Make Spotters 12:58 
   TLC Plate 16:04 
   Flash Column Chromatography 17:11 
  Introduction to Extractions 34:25
   Intro 0:00 
   Extraction Purify, Separate Mixtures 0:07 
    Adding a Second Solvent 0:28 
    Mixing Two Layers 0:38 
    Layers Settle 0:54 
    Separate Layers 1:05 
   Extraction Uses 1:20 
    To Separate Based on Difference in Solubility/Polarity 1:21 
    To Separate Based on Differences in Reactivity 2:11 
    Separate & Isolate 2:20 
   Theory of Extraction 3:03 
    Aqueous & Organic Phases 3:04 
    Solubility: 'Like Dissolves Like' 3:25 
    Separation of Layers 4:06 
    Partitioning 4:14 
   Distribution Coefficient, K 5:03 
    Solutes Partition Between Phases 5:04 
    Distribution Coefficient, K at Equilibrium 6:27 
   Acid-Base Extractions 8:09 
    Organic Layer 8:10 
    Adding Aqueous HCl & Mixing Two Layers 8:46 
    Neutralize (Adding Aqueous NaOH) 10:05 
    Adding Organic Solvent Mix Two Layers 'Back Extract' 10:24 
    Final Results 10:43 
   Planning an Acid-Base Extraction, Part 1 11:01 
    Solute Type: Neutral 11:02 
    Aqueous Solution: Water 13:40 
    Solute Type: Basic 14:43 
    Solute Type: Weakly Acidic 15:23 
    Solute Type: Acidic 16:12 
   Planning an Acid-Base Extraction, Part 2 17:34 
    Planning an Acid-Base Extraction 17:35 
   Performing an Extraction 19:34 
    Pour Solution into Sep Funnel 19:35 
    Add Second Liquid 20:07 
    Add Stopper, Cover with Hand, Remove from Ring 20:48 
    Tip Upside Down, Open Stopcock to Vent Pressure 21:00 
    Shake to Mix Two Layers 21:30 
    Remove Stopper & Drain Bottom Layer 21:40 
   Reaction Work-up: Purify, Isolate Product 22:03 
    Typical Reaction is Run in Organic Solvent 22:04 
    Starting a Reaction Work-up 22:33 
    Extracting the Product with Organic Solvent 23:17 
    Combined Extracts are Washed 23:40 
    Organic Layer is 'Dried' 24:23 
   Finding the Product 26:38 
    Which Layer is Which? 26:39 
    Where is My Product? 28:00 
   Tips, Tricks and Warnings 29:29 
    Leaking Sep Funnel 29:30 
    Caution When Mixing Layers & Using Ether 30:17 
    If an Emulsion Forms 31:51 
  Extraction Lab 14:49
   Intro 0:00 
   Step 1: Preparing the Separatory Funnel 0:03 
   Step 2: Adding Sample 1:18 
   Step 3: Mixing the Two Layers 2:59 
   Step 4: Draining the Bottom Layers 4:59 
   Step 5: Performing a Second Extraction 5:50 
   Step 6: Drying the Organic Layer 7:21 
   Step 7: Gravity Filtration 9:35 
   Possible Extraction Challenges 12:55