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Lecture Comments (27)

1 answer

Last reply by: Professor Starkey
Thu Oct 6, 2016 1:07 PM

Post by Tane Boghozian on October 4 at 06:47:10 PM

Hi Dr. Starkey, I have been using educator. com for almost four months and its great!
Thank you very much for the great lectures. I have learned a lot. I have a question about your availability. Would be possible to talk to you about my experiments and research whenever I have questions? will you answer my questions?
Thanks a lot for your help.

1 answer

Last reply by: Professor Starkey
Thu Aug 13, 2015 9:59 AM

Post by sandi imayeguahi on August 13, 2015

That was your best lecture by far. Your really great.

1 answer

Last reply by: Professor Starkey
Sun Apr 26, 2015 9:49 PM

Post by Nagasrinivas Tripuraneni on April 26, 2015

Professor, thank you for the informative lecture.

1 answer

Last reply by: Professor Starkey
Fri Oct 18, 2013 2:10 PM

Post by brandon oneal on October 18, 2013

The only question I have is how can you tell which compound it is?

1 answer

Last reply by: Professor Starkey
Sun Jul 21, 2013 10:44 PM

Post by Parabjit Kaur on July 21, 2013

when will you post lectures on the mass spectrometry ?

1 answer

Last reply by: Professor Starkey
Thu May 9, 2013 10:19 AM

Post by Nawaphan Jedjomnongkit on May 6, 2013

If we have internal and symmetrical alkyne, can we differentiate from normal alkane? Because what we can get from IR spectrum is only SP3 CH no SP CH and CC triple bond.

1 answer

Last reply by: Professor Starkey
Sun Feb 3, 2013 11:03 PM

Post by Ramin Sadat on February 3, 2013

Will you be posting anything on Mass spectrometry ?

1 answer

Last reply by: Professor Starkey
Tue Jan 29, 2013 9:47 PM

Post by Edi William Yapi on January 29, 2013

Good Lord.... You are just a great instructor !

1 answer

Last reply by: Professor Starkey
Sun Jan 13, 2013 11:54 AM

Post by Janaki Dharmarpandi on January 12, 2013

you are awesome!!!!

1 answer

Last reply by: Professor Starkey
Sun Dec 2, 2012 11:43 AM

Post by ali aden on December 1, 2012

you are amazing, i did understand this material, excepts i have a little confussion o the finger print area.

1 answer

Last reply by: Professor Starkey
Mon Aug 13, 2012 9:05 PM

Post by Daniel Ugsang on August 10, 2012

Honestly, you are a LIFE SAVER. My O-chem prof is good but goes off on tangents. You get right to the point. Succinct. Good explanations. I really hope I pass Ochem and if I do, it'll be because of you, Professor Starkey!

1 answer

Last reply by: Professor Starkey
Tue Jul 17, 2012 9:55 AM

Post by nouf alkusayer on July 7, 2012

U r amazing !!!!

1 answer

Last reply by: Professor Starkey
Fri Feb 3, 2012 11:33 PM

Post by Lukasz Skora on February 3, 2012

Hi Dr. Starkey, awesome lectures! They are life savers!. I have a question, does it depend on the finger print region if you wanted to deduce how long the compounds is. How for example can one tell if it's a butane or a decane, or like pentanol, hexanol, or octanol. They are all sp3 CH's with an alcohol at the end. Or a benzene with a longer substituent compared to a shorter one. Thank you.

0 answers

Post by Thomas Notto on November 23, 2011

Great Job on the basics of Infrared Spectroscopy... I think I got now!!!

Infrared Spectroscopy, Part II

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

  • Intro 0:00
  • Interpretation of IR Spectra: a Basic Approach 0:05
    • Interpretation of IR Spectra: a Basic Approach
    • Other Peaks to Look for
  • Examples 5:17
    • Example 1
    • Example 2
    • Example 3
    • Example 4
    • Example 5
    • Example 6
    • Example 7
    • Example 8
  • IR Problems Part 1 28:11
    • IR Problem 1
    • IR Problem 2
    • IR Problem 3
    • IR Problem 4
    • IR Problem 5
    • IR Problem 6
  • IR Problems Part 2 42:36
    • IR Problem 7
    • IR Problem 8
    • IR Problem 9
    • IR Problems10

Transcription: Infrared Spectroscopy, Part II

Hi, and welcome back to Educator.0000

We are going to continue the topic of IR spectroscopy next.0002

We are going to think about how to approach a spectrum if we need to interpret them.0006

What I have done is: I have written out a kind of a basic approach that is a good place to start; it's nice to have a systematic approach, so that you are not kind of floundering around and missing things.0011

If you have a systematic approach, and you do the same thing every time, you are going to be more consistently getting things right.0023

OK, so what I suggest is that you start by just reading the spectrum from right to left, like you would read a sentence.0029

The first thing you should do is look for the following obvious peaks: these are things that you should not miss, anytime they are in an IR spectrum.0036

OK, as we read from right to left, the first thing we would encounter is an OH or an NH signal; that comes somewhere around 3300.0042

Remember, that is a very broad signal; OK, we want to be careful of this, though, because a lot of times, we see an OH signal in our spectrum; we are not expecting an OH--the molecule that we are analyzing shouldn't have an OH.0050

OK, and that is because there is a very common contaminant that we might have in our sample that does have an OH, and that is water.0064

So, sometimes, if our sample is just wet (meaning we have done, let's say, an extractive workup, and we have isolated our product, but we haven't properly dried the sample), we might be taking an IR spectrum of water along with our sample.0074

And so, we will see a peak around 3,300, and we will learn to disregard that as water (or maybe a protic solvent, or something like that--an alcohol).0087

OK, but we should be able to pick out whether or not there is an OH present in the sample.0097

OK, then we are going to come right to that 3,000 cutoff, because we know that is where our CH bonds are going to appear, and we know that if they are just above 3,000--just to the left of it--that is where our sp2 CH's come from.0102

And, if they are just below, just to the right of it, that is where our sp3 CH is; so that is going to tell us whether we have maybe an aromatic or an alkene, or maybe some alkyl carbons.0120

OK, if it's aromatic or if it's alkene, we can also look for those substitution patterns--those strong peaks below 1,000.0129

They are in the fingerprint region, so it's not easy to find right away, but we might look for those later.0140

OK, and like we saw in the previous spectra that we looked at: most spectra are going to have the sp3 hybridized CH's in them, and so we are usually going to see these peaks just below 3,000.0147

So, we can kind of get used to seeing that and labeling that.0160

OK, we are going to continue reading along; we are going to look at the region around 2,200, and that is where we find our triple bonds (either a nitrile or an alkyne).0163

OK, now, these are often very weak: they are usually very, very small signals; but because that area around 2,200 is normally completely flat, if you see a peak at 2,200, that is something that you should recognize and pick out.0173

OK, and finally, around 1,700 is: we have our carbonyl; this is a very strong peak; this is very obvious; all of these peaks are things that you should never miss if they are present in your IR spectrum.0188

OK, and somewhere around 1,700 is where you say it is lowered by conjugation; we are going to expect the lower number if it's conjugated to a benzene ring or another double bond, and so on.0200

So, like I said, where exactly around 1,700 it is depends on what kind of a functional group: is it a ketone?--is it an ester?--and so on.0210

OK, now, besides those obvious bands, some things that we can look for for additional evidence are the other things we will find in our correlation tables.0219

We might have an sp hybridized CH; that is also around 3,300, but that is now a sharp peak, compared to the alcohol's OH broad peak.0228

An aldehyde CH is very small peaks, hard to find, hard to see initially; but we can look for them.0238

Remember, those are our little vampire fangs: 2,850 and 2,750; we can look for those.0245

Carbon-carbon double bonds: we may or may not find those; they are often variable intensity; sometimes they disappear altogether.0250

If it's a symmetrical, symmetrically substituted alkene, we won't see the peaks at all.0258

OK, C-O stretches: again, they are typically strong, somewhere around 1,050; but sometimes it is hard to find those, too, because they are in the fingerprint region.0264

OK, carboxylic acids are pretty unique, because that OH stretches so spread out, and we have that in combination with the carbonyl (that is something we can look for).0272

And then, if we do have an aromatic ring--a benzene ring--in our structure, then depending on the substitution pattern (so I have shown one group here, and then, kind of attached at the middle here, I have another group meeting--it could maybe be ortho or meta or para), we can look for those out-of-plane bending peaks around 1,000 and those overtone peaks at the aromatic ripple.0282

OK, and there are also characteristic patterns if it's trisubstituted and so on; I just wanted to show you some examples of that ortho, meta, and para, and the monosubstituted.0308

OK, so let's get started with our first spectrum; and what I have shown here are the 8 compounds that we have already analyzed, and predicted the peaks for, because it turns out that the spectra we are going to be looking at from this point forward actually match one of these.0318

Let's see if we can interpret the spectrum, and then find out which structure it matches.0332

Here is what we are going to do: we are going to read it from left to right, just like we would a sentence, and we are going to start and look for our obvious peaks.0339

We start, and we look: around 3,300 is where we find an alcohol, an OH: do we have an alcohol here--do we have an OH?0347

We sure do--this big, broad peak around 3,300 is exactly what an OH looks like.0354

We can point to it; we can circle it; we can do something to label it.0361

Very clearly, though, we want to get in that habit of labeling our spectra.0365

OK, then we continue reading left to right; we move to this 3,000 mark, and we look just above 3,000 for peaks; and sure enough, we see some peaks here.0371

We also see some peaks just below 3,000, so what does that tell us?--just above 3,000 is where we find sp2 CH's, and just below 3,000 is where we find sp3 CH's.0380

OK, we continue reading: we go up to 2,200; here is 2,200; we look for peaks there; do we have a triple bond?0394

There is no triple bond there; we continue to 1,700; here is 1,700--is there a carbonyl?--could that be a carbonyl--this little tiny peak--could that be a carbonyl?0401

No way--remember, a carbonyl is the biggest, strongest peak in your whole spectrum; so these are the obvious bands that you can't miss, so there is no carbonyl; there is no triple bond.0411

OK, so just based on these functional groups that we know are present, let's take a look at our structures 1 through 8 and decide which spectrum we have--which compound we have.0422

Well, we know there is an OH; and actually, there are only two choices that have OH's; how would we distinguish between 6 and 7?0432

Well, there are two ways: one of them is: 7 must have a carbonyl, and we know we don't have a carbonyl (right?--we could just kind of make that note here: no carbonyl--just for your information; I wouldn't really put that on my spectrum).0442

OK, but it can't be 7, so it must be 6; it has the OH; it has the sp2 CH; it has the sp3 CH.0454

OK, now what else can we look for?--if we know it's 6, and we are thinking that it's 6, now let's go back and find some of those needles in the haystack that will more completely label this spectrum and give us further evidence.0462

For example, because it has an aromatic ring, how would you describe that aromatic ring?0476

It is a monosubstituted ring; in fact, I see this aromatic ripple up here; that is consistent with the monosubstituted: 1, 2, 3, 4.0481

And we also look to find...we should find two peaks around 700 and 750; and there they are, so this is monosubstituted aromatic; we call this out-of-plane bending.0491

This gives us more evidence for that structure.0506

OK, is there anything else?--maybe we could look for that C-O stretch, and maybe one of these is a C-O; I'll put a little question mark here; there are two peaks here; I don't know for sure which is the C-O.0509

So, that is nice to attempt to label, but some things we are not going to be able to identify with certainty.0524

OK, also, with the benzene ring, maybe this peak around here at 1,600 is the carbon-carbon double bond; maybe we could put a little question mark there--I don't know for sure.0531

That is, we would expect to find something around that region, so that is probably a good guess.0542

OK, let's try another one.0548

We will start by reading the spectrum from left to right: the first thing we should be looking for around 3,300 is an alcohol--do we have any OH's?0552

Well, we have this tiny little blip, but that sure doesn't look like anything significant; so we have no OH here; we continue to 3,000; here is 3,000.0559

We look just above--anything just above 3,000?--that is empty, so there is no sp2 CH (we'll just put a little note up here on what we are seeing).0570

How about just below 3,000?--yes, here we have our first significant peak, so we will label those: we have sp3 CH; we continue down, reading left to right.0579

At 2,200, do we have any triple bonds?--no triple bonds.0590

And then, we move down to 1,700; at 1,700, what do we find?--oh, we do have a carbonyl.0599

OK, again, a systematic approach means we are going to be looking for our obvious peaks, and we are not going to missing them any time they show up.0606

Let's take a look at structures 1 through 8: what compound has a carbonyl, an sp3 CH, and nothing else?0613

OK, so it can't be the carboxylic acid, because it had an OH; and we actually have three compounds with carbonyls.0624

All right, so we can cross these ones off as not being possible; these ones are our three contenders--how would we distinguish between them?0632

These two are ketones--1 and 4 are ketones, and 3 is an aldehyde; and there is something else that distinguishes them--how about our CH's?0642

We said that there is no sp2 CH, but 3 has a benzene ring and 4 has a benzene ring, so those cannot be our structures.0653

So, we actually have just one possibility here: this has to be this ketone, because all we have are sp3 CH's and a carbonyl.0660

OK, is there anything else I can look for to label?--really, there is nothing else.0672

Look at all of those peaks in the fingerprint region that we are just leaving alone; OK, sometimes that is going to be frustrating--you might find that frustrating--but the key to solving IR problems is knowing which peaks are significant, being able to pick those out.0678

We have to resist the temptation to label every single squiggle and wiggle, because these are not significant peaks--it is simply your molecule bending and wobbling all around, and that doesn't really tell us anything diagnostic.0694

OK, so just those two peaks are all we need to label to positively identify its structure.0706

OK, how about our next one?--again, systematic approach: what is the first thing we should look for?0714

Reading left to right, we look at 3,300, and we see that there is no OH; we just make a little note to ourselves here, and we move to 3,000.0719

We move to 3,000, and what do we find?--anything above 3,000?--sure; we have these three peaks up here.0729

It doesn't matter if they are big or small or anything; any peaks up here are significant--right about 3,000 but just above it means we have sp2 CH.0736

It could be a benzene ring; it could be an alkene; it doesn't matter right now--all we know is: we have sp2 CH; and then just below 3,000, we have sp3 CH.0746

OK, systematic approach--what else do we look for?0757

2,200: there is no triple bond; 1,700: we look for a carbonyl--no carbonyl.0759

OK, so those are our obvious peaks; let's see if that is enough to identify the compound.0773

We need a compound without any carbonyl, without any OH; all we have are sp2 CH's and sp3 CH's.0779

So, what do you think?--it looks like compound 8 is the only structure that has no other functional group in it, that has both sp2 CH and sp3 CH.0788

That looks good; now, is there anything else I can find if I know 8 is my structure?0802

What about that benzene ring?--we would expect...oh, there it is: there is our aromatic ripple.0808

And this is a little harder to read, very often, about the substitution pattern, but we can look over here to find our substitution pattern (this is another example of a monosubstituted): here it is--750, 690--monosubstituted aromatic is the way we describe that out-of-plane bending motion.0818

OK, very good.0842

OK, let's try the next one: let's read it from left to right.0845

Do we have any OH at 3,300?--well, we have a peak at 3,300; does that look like an OH to you?--remember, an OH, because of its hydrogen bonding, is going to be a very broad signal; this is not consistent with an OH.0848

It is obviously something, but we will come back to that; we have no OH.0863

We move down to 3,000; do we have anything above 3,000?--we have something way above 3,000, but remember, we are looking immediately at 3,000.0867

There is nothing here, so there is no sp2 CH; but just below 3,000, we do have sp3 CH.0877

We continue down to 2,200; do we have anything at 2,200?--oh, we sure do, and what is it that we find at 2,200?0887

That is where we find carbon-carbon triple bonds or carbon-nitrogen triple bonds--we don't know which one it is yet.0895

They both come around 2,200; so we have a triple bond; and how about a carbonyl--any carbonyls?0901

That would be at 1,700 (remember, feel free to use your correlation table: you don't have to memorize these numbers--feel free to use that to confirm), but there is no carbonyl.0908

OK, so we look at our structures: we are lucky enough to have just a single structure that has a triple bond, but let's see if this is consistent.0921

What else would we need to have if this were our structure?0930

This hydrogen is very special--how would you describe that hydrogen?0935

It is attached to an sp hybridized carbon, so this is called an sp CH bond, and where does that stretch show up?--it shows up right around 3,300.0940

So, why is this peak so skinny--so narrow and sharp?--it is because this is an sp CH stretch, rather than an OH.0951

OK, and so, if this is a CH, that tells us we don't have a nitrile; it's not a C-N triple bond, because we need a hydrogen on a carbon; so this has a hydrogen on one side, and then the rest of this alkyl chain on the other side.0962

Now, we know exactly what we have here.0977

Anything else we can pick out from that structure?--nothing else of interest; so we'll ignore this fingerprint region and just label those three peaks.0980

OK, let's continue: let's read this left to right.0994

If you want to pause any time and try the rest of the spectrum (or spectra) on your own, that is a very good idea--a very good exercise.0997

OK, the same systematic approach: any alcohols (3,300)?--we have this tiny little blip, but that is not a nice, strong peak, so we have no OH.1004

What do we see at 3,000?--at 3,000, we have these peaks above--very tiny, but that is OK--it is still significant: sp2 CH is just above 3,000; sp3 CH is just below 3,000.1018

OK, 2,200--anything at 2,200?--it is nice and flat, as usual.1034

1,700--where is 1,700?--sometimes it is hard to read these numbers, but look: this is 1,500; this is 2,000, so this must be 1,700; use the axis labels down here to help you pick out the thing.1040

The big peak at 1,700--what does that tell you?--carbonyl, for sure.1056

So, what we need is a compound that has a carbonyl; it also has both sp2 and sp3 CH's.1062

We come back to our carbonyl compounds: there is no OH, so it can't be that one; it can't be #1--what is wrong with #1?--#1 has no sp2 hybridized carbon with a hydrogen on it.1071

It has this sp2 hybridized carbon as the carbonyl, but it has no sp2 CH.1088

OK, but 3 and 4 both fit the bill here: they both have a carbonyl; they both have an alkane-type carbon; and they both have an sp2 CH.1093

So, how would we distinguish between 3 and 4?1106

What sets them apart?1110

3 has a hydrogen attached to the carbonyl (we call that an aldehyde); 4 does not (it's a ketone).1111

How would we find the presence of an aldehyde?--we would look for our vampire fangs at 2,850 and 2,750; and they are not there.1119

OK, so there is no aldehyde; it must be the ketone.1128

We can make a little note here on how we made that decision: there is no aldehyde CH.1131

And so, that is what makes it compound 4 instead of compound 3.1139

You want to make sure that, after you have made a should go ahead and draw the structure right on the spectrum, because then there is no mistaking: every functional group that is on this structure needs to be in the spectrum and vice versa.1144

If you said you had a carbonyl, your structure has to have a carbonyl; so that is a very rewarding part about solving spectroscopy problems: you can see the evidence when you have made the right choice.1159

OK, how about our next one?--let's look for an OH; at 3,300, do we have an OH?1174

Well, not really, but we have kind of a big mess here; that looks a little interesting; we will kind of leave that for now.1180

But what do you see when we move to 3,000?--just above 3,000...1190

We could make a little question mark here; there is no obvious OH peak like we are used to seeing; so we will leave that as an unknown for now.1195

Just above 3,000, though, I see some peaks here, so there is an sp2 CH; and just below 3,000, I have sp3 CH.1202

And you know what, I just realized that we just did compound 4 with the aromatic; we forgot to circle the aromatic ripple on the monosubstituted aromatic; so we can go back to that slide and do that.1213

That would be good, to finally label those.1224

OK, but back to this one: we have sp2; we have sp3; anything at 2,200--any triple bond?1228

Nothing at 2,200; how about 1,700--anything at 1,7000?--for sure, there it is--our biggest, strongest, most obvious peak is our carbonyl, C-O double bond--a carbonyl.1234

OK, so we need something with a carbonyl and an sp2 and an sp3, and something that looks a little strange with this big, broad peak here.1248

Does this maybe look like a carboxylic acid?1262

Remember how a carboxylic acid OH, because it dimerizes, is such a strong hydrogen bonder that it is really, really, really spread out?--that is exactly what we have here.1265

So, in fact, we do have an OH; this is an RCO2H carboxylic acid OH, and it is this whole peak here.1273

It's the OH; so this big, broad peak is our carboxylic acid OH; you could label it as a carboxylic acid OH, if you would like; or you could abbreviate it, RCO2H.1286

It is the OH bond that is stretching, so we label it as an OH peak.1299

And so, this is compound 7; we also must have the carbonyl, if it is a carboxylic acid; we have our sp2 CH; we have our sp3 CH; and what else should we find?1303

That is what I just noticed--we forgot to do that for compound 4: what else should we find?--we have another one of our monosubstituted aromatics; that is getting kind of common here.1315

So, what out-of-plane bends do we expect to see there?--750, 700; there they are...monosubstituted aromatic.1323

OK, and we also get that aromatic ripple, but because we have this carbonyl, it kind of obscured--you didn't see all four peaks here--but there is our little aromatic ripple, too; that pattern is consistent with a monosubstituted aromatic.1336

Very good; OK, how about our next one?1352

We read from left to right: do we have an alcohol (3,300)--do we have an OH?--no OH.1357

What do you see at 3,000?--anything above 3,000, just above 3,000?--nothing there, so there is no sp2 CH.1366

But, just below 3,000, I see this nice, big set of peaks; those are sp3 CH's.1375

Move along; anything at 2,200?--no triple bond; anything at 1,700?--no carbonyl; so we have no triple bond, no carbonyl.1384

Just a little reminder of our systematic approach: what decisions did we just make?--the only significant functional group we have here is an sp3 CH.1397

What does our compound look like?--well, we only have a single alkane with no other interesting functional group here, so it must be compound 5.1406

And remember, every alkane has a very similar IR, because all we are doing is functional group analysis; I can't tell you exactly which alkane this is.1416

All I know is that the only thing present in my sample are sp3 CH's--no other functional groups.1425

That is why this type of a problem, where you are matching it to a set of possible structures, is ideal--because, if I said just "What is the structure of this IR?" you wouldn't be able to answer that definitively.1432

I could ask, "What is a possible structure for this IR?"; but again, it would have hundreds of possible answers--thousands of possible answers; and so, it is a little vague that way.1447

This kind of matching problem is pretty commonly encountered, but make sure that not only do you match it with your structure and draw your structure on here, but that you label your spectrum to show why this is the compound that matches.1457

Oops, I missed a methyl group there.1473

OK, we don't have to worry about anything in this fingerprint region; that is the only thing to label on this spectrum.1475

OK, and last but not least, this is our eighth spectrum, so let's see what we have at 3,300; do we have an OH?--well, we have a broad peak here, but that is kind of a weird shape.1481

An OH is usually a nicely shaped, broad peak; so you know we have something here--let's put an OH with a question mark.1495

Maybe it's water or something like that; it's something weird, so put a little question mark here.1502

OK, now we move to 3,000; what do we have at 3,000?--it looks like we have some peaks above 3,000 (sp2 CH); it looks like we have some peaks below 3,000 (those are sp3 CH).1508

So, we have both of those.1522

Let's look at 2,200--any triple bonds?--no triple bonds.1524

Let's look at 1,700--at 1,700, it sure is...right there, a big, big peak; we must have a carbonyl in this compound.1528

Let's take a look at our structures and decide what possibilities we have.1539

OK, we need an sp2 CH and an sp3 CH; do we have an aromatic?--we don't have any other alkenes in this example, right?--so the sp2 CH is coming from our aromatic.1547

All of these happen to be monosubstituted aromatic, so we have our 700 and 750.1561

Let me go ahead and find that, because that is the only way I can get my sp2 CH.1571

It can't be 1, because 1 only has the sp3 CH; so we're coming back down to these; those are our only two possibilities; they both have carbonyls; they both have sp3 CH; they both have sp2 CH.1577

How do we distinguish between these two?--we have an aldehyde and a ketone; what is the determining factor?1590

Here it is: that aldehyde CH is very unique; when you have an aldehyde, how are you going to find that?--you are going to look for little vampire fangs at 2,850 and 2,750.1597

And they are going to be hard to find typically (we overlook them a lot), but here they are; here is one at 2,750, and here is one at 2,830 1, 2.1611

These two peaks are my aldehyde CH: so actually, my sp3 CH--I labeled it incorrectly here at first; it only goes out to here, because these last two are the aldehyde CH.1620

And so, it is not obvious at first; but when we look for it, we can pick out an aldehyde pretty well--especially the one at 2,750; it usually stands alone; the 2,850 is usually touching the sp3, so it's a little harder to see.1636

OK, so this, in fact, is structure #3.1651

Very good; but 3 doesn't have an alcohol--it doesn't have an OH; so what is this big, messy peak here--why do we have this?1657

I mean, it could be water from our sample, but what else do we have?1664

Remember, when we have a carbonyl, very often we see this overtone around 3,400, and sometimes it is really big overtone; so it's that, plus maybe a combination of some water; so we can ignore it.1668

It is not an OH; it is just our carbonyl overtone.1680

This is consistent with the aldehyde structure given.1686

Now, another type of problem we can have, besides a matching sort of problem--I mean, there are all sorts of problems you can have; but here is another example.1692

Let's say we used IR to analyze our product after we ran a reaction.1700

Now again, this is a very common scenario, because IR spectrometers are very affordable; and so, it is one of the major pieces of equipment that you will have in almost every organic lab.1704

So, it is a routine part of analysis, after you do a reaction and isolate your product, to take an IR of that product as a way of identifying what compound you have.1717

Now, if that IR spectrum is going to tell you whether or not your reaction worked, you have to know what to expect from the IR; so you need to be able to predict it.1729

Let's take a look at some sample reactions and see how you could use IR to tell whether or not this reaction had gone to completion, or was successful.1739

So, to be complete--that means all of your starting material is gone and all of your product is present.1753

The only thing present is your product; so that assumes that the reaction has run its course; you have isolated your product; you have purified your product somewhat; and we are looking for this as our product.1760

What we can ask is, "Well, what peaks are going to be new in the IR spectrum--what is going to show up? What peaks are going to disappear that used to be in the starting material?"1771

So, if you compare the IR of the starting material to the IR of the product, what changes would there be?--this is, again, a very common analysis that we are going to be doing after running a reaction in an organic lab.1781

What do you expect to be different--do you find different in the starting material and the product?1792

And again, this is maybe a good point to pause it and try some of these problems on your own, and then we can work through it together.1798

Well, clearly, we have a carbonyl here that is missing in our product--that is gone in our product; so our carbonyl, which is around 1,700, should disappear.1805

If our reaction was successful, there should be no carbonyl peak in our product.1819

What else can we look for?--well, we should now have an OH that we didn't have before.1823

So, our OH around 3,300 appears; that is something new.1829

Perhaps we can look for a C-O single bond; there is no C-O single bond in our structure before, so we can maybe look for that around 1,050.1837

It's a little harder to find in that fingerprint region, but we can maybe find that as well.1846

OK, but remember, if our sample is just wet, we might find that we have the OH peaks in there, even though we have not made an alcohol.1852

That is why it is important to also check that the carbonyl peak is absent; and that way, we know if we have an organic product, we must have converted our starting material to something; and that gives us evidence that the OH is actually part of our structure.1860

OK, how about the next one--what changes have taken place, and what changes that have taken place are significant in the IR?1875

How would it present itself in the IR spectrum?1884

We have gotten rid of this double bond; we used to have a double bond, and now it's gone.1888

So, how does that affect the IR?--well, the IR of the starting material--how do you see evidence of that double bond?1892

True, the carbon-carbon double bond might be there, but that is kind of a weak signal; what is a much more significant signal is the sp2 CH that is here and here.1899

Those bonds stretching are going to be very obvious in the IR; and so we want those to disappear, because there is no sp2 hybridized CH in the product.1909

OK, that is going to be the biggest one; we could put also the carbon-carbon double bond disappearing, but that is not going to be nearly as significant, because that might not have been a significant peak to start off with.1921

OK, anything else?--we had a carbonyl; we still have a carbonyl; but because this carbonyl was conjugated, the new carbonyl in the product, or the carbonyl that is now in the product, is going to be at a different wavenumber.1935

So, let's just say the carbonyl moves; it is going to be at a different peak, and it is going to...when you have conjugation, that is something that lowers the number, so this is actually going to be a higher reciprocal centimeter.1952

It is going to move to the left, because we have gotten rid of that double bond; so a subtle difference, but it also will be a difference that we should observe, if this reaction was successful.1970

What if we did this substitution reaction--we had a bromine here, and now it is a nitrile?1983

OK, is there anything in the starting material that is unique--do we know how to identify a bromine in an IR?1990

We have not talked about halides in the IR, and that is because they are not very diagnostically useful; this carbon-bromine bond will stretch, but it is something that either is weak in intensity or comes in the fingerprint region or so on.1998

So, there is nothing significant here; we have sp3 hybridized carbons; we still have sp3 hybridized carbons; so there is nothing significant to note in our starting material that is going to disappear.2016

But how would we tell if we had this nitrile--what does this structure look like, when you have a C-N group?2028

What does that look like?--to have four bonds, it must be a C, triple bond, N, right?--and that triple bond is something that we can pick out in the IR.2038

So, what is going to show up?--the C, triple bond, N--the nitrile--is going to appear; it appears at around 2,200.2045

So, if we have that peak showing up after we have isolated our organic product, then that confirms that we must have synthesized the nitrile.2055

OK, and how about our next one?--we had an alkene, and now we have an ether.2065

We haven't really talked about ethers specifically; we haven't seen any examples of ethers; but what is interesting in the ether?2070

What peaks would you predict for this product--what kinds of CH's?2078

This CH3 and this CH3 and this CH, CH2, CH2...all of those are the same by IR: they are sp3 CH's, just below 3,000.2083

OK, but we already had those--we already had sp3 CH's--so that is not going to be something new.2097

We can just mention they are still there, so that is not too big of a deal; but what is going to be new--what is going to be different?2103

Well, perhaps we can find our C-O: we have a C-O bond here; we have a C-O bond here.2111

So, we might actually see those separately--so maybe a C-O around 1,050.2117

Sometimes those are obvious.2123

But what is the really significant change that is going to happen--what did we have in our starting material that is now gone?2126

We had an sp2 hybridized carbon; so we have sp2 CH's just above 3,000 that are going to disappear.2131

There should be no more of those in our product; so that is going to be the big change that is going to be happening.2141

OK, let's look at this multi-step transformation as we go from an alcohol to an aldehyde to a carboxylic acid; these are oxidation reactions.2152

How would our IR change as we move along--what are the most significant changes?2161

Well, in our alcohol, we have an OH peak around 3,300; that is always an obvious peak, and that is going to disappear.2167

If we oxidize this to an aldehyde, what would our aldehyde product show?2179

It is going to show the carbonyl, which is always significant--a large peak at 1,700--a very strong signal.2185

And what else should be new?--this aldehyde CH is a significant peak that will be unique: that is our little double set of peaks, 2,850 and 2,750--small, but they are there, and you can look for those.2192

And then, if we were to oxidize this further to carboxylic acid, what is going to happen?2213

We are still going to have a carbonyl; our carbonyl is going to shift to a different wavenumber--you can take a look at a table of carbonyl shifts to see how an aldehyde compares to a carboxylic acid.2218

So, the carbonyl will still be there, but its wavenumber is likely to be different.2231

Of course, we are going to have this OH; it is going to be a carboxylic acid OH, which means it is going to be from 3,400 all the way down to 2,400--really, really broad.2236

So, instead of having a nice little OH, you are going to have a big, spread-out OH.2249

That is how we know we have a carboxylic acid.2253

Anything else significant here?--well, our aldehyde C-H should be gone; the aldehyde C-H disappears.2256

That is how you can tell, maybe, that not only do I have some of my product--sometimes a reaction doesn't go to completion; so you have some product, but you also still have some starting material left.2268

The IR could maybe tell you if you have a significant amount of starting material: you would see both the carboxylic acid peaks and the aldehyde peaks, so that you could maybe see that you have a mixture there.2278

So, it is not always just one or the other; but to have this product (carboxylic acid) pure means there should no longer be an aldehyde CH, and we should have the OH on the carbonyl.2288

OK, here is another multistep transformation: we go from this alkyne to this alkyne, and then an alkyne to an alkene; how can we follow that by IR and verify that these reactions have worked?2303

Well, there is a big difference between these two alkynes; what is the big difference there?2321

They both have a carbon-carbon triple bond; we will see that in this first structure (this is around 2,200); but in this compound, because we have symmetrically substituted triple bond (we have the same group on either side), what happens to a bond when it is totally symmetrical?2327

And there is no change in dipole when you stretch that bond; what happens to the absorption of the IR?2353

It doesn't happen: there is no absorption of the IR, which means there is no signal in the IR spectrum.2358

We would expect this carbon-carbon triple bond peak to disappear.2364

We'll put a little question mark there, because we don't know for sure that is going to happen; but that could be very likely, and we would predict that to happen, and that is because it is symmetrically substituted.2370

OK, so I wouldn't be surprised if our triple bond went away; but what other change is taking place?2386

There is something else that is special about this triple bond; when you have a terminal alkyne, it is called "terminal" because it's at the end of a chain, and when you are at the end of the carbon chain, what does this carbon have on it?2392

It has a hydrogen on it.2405

This is an internal alkyne, which means it has a carbon group on the other side; there is no CH.2407

So, would that CH show up in the IR spectrum?--it absolutely would: it's an sp hybridized CH, and that comes somewhere around 3,300; that is the sharp peak we sometimes see at around 3,300.2417

And it would be in our starting material; it would not be in our product; so our sp CH, for sure, disappears, because it is no longer there.2430

They both have sp3 CH's, so this methyl group in each of them means we have our normal peak below 3,000; so that is not going to be useful to us--it's going to appear in both cases.2444

OK, but then, if we took that alkyne and we converted this to an alkene, this is a trans-1,2-disubstituted alkene.2456

I'm explaining, and I'm looking at the detail of this substitution pattern, because, just like in a benzene ring, when you have monosubstituted, or ortho, or meta, or para, you can see a difference in those substitution patterns; you can pick that out for a benzene ring.2467

It turns out, you can also do the same thing for an alkene; I'm not going to...again, this is kind of more of an advanced IR topic; it is not often covered at the introductory level.2487

But just to let you know, you can still have those interesting bending motions; we can see that substitution pattern, kind of in that same region around 1,000 reciprocal centimeters.2495

So, it's in the fingerprint region; it's not something we would notice right away; but we can look for it, and trans is going to be different from a cis, is going to be different from a monosubstituted alkene, and so on.2511

OK, but if we didn't know to look for those bending patterns, what differences would we expect to find here, in going from the alkyne to the alkene?2524

Well, it is these hydrogens that are significant; and those show up--it is the stretching of those C-H bonds that is significant.2534

We are going to see these sp2 CH's appear, and where does that come?--that is the one that is just above 3,000.2541

OK, so we would be able to follow this multistep transformation by IR.2552

OK, and finally, just a little more challenging problem here: let's look at all of these alcohols and see if we can predict how to distinguish them by their IR's--because, clearly, they all have OH's, so they are all going to have that peak around 3,300; so they all have the OH stretch around 3,300, so that is not going to be something that is going to be useful to us.2558

We should be able to identify when we have an alcohol, but what if we had different types of alcohols--how can we look for ways to distinguish them?2586

OK, and it's going to have to come from the rest of the molecule--what else is in the molecule?2597

We have a benzene ring here; so what significant peaks do we see for the benzene ring?2600

We have...what type of CH's do we have?--we have sp2 CH's, and those are just above 3,000; and because it's monosubstituted, we expect to have our peaks at...a monosubstituted aromatic is around 750 and 690.2607

We have those out-of-plane bending, so we would expect to find those; and that is going to be different from this one, because this next one also has sp2 CH just above 3,000, but what else does it have--any other CH's?2637

We have this CH3 and this CH2; we also have the sp3 CH's that occur just below 3,000.2655

That would be a way to distinguish between these two.2665

What would be another difference that you would find in their spectra?--is this a monosubstituted aromatic?2667

No, this is a disubstituted aromatic; and what do we call it when they are 1,4 like this (1, 2, 3, 4--when they are opposite)?--we call that para.2675

We have a para substituted aromatic; and where do those peaks show up?--there is just a single peak somewhere around 815; so that pattern is going to be different than the monosubstituted; that little aromatic ripple region around 2,000 is going to look at little different, as well.2685

OK, so we can distinguish, just based on substitution patterns, too.2702

And how does this one differ from this one?--we have this (this is phenol, actually, when you have a benzene ring with an OH) phenol; how is that different--why is this a different structure?2708

We are missing the double bond; so this is not benzene any longer--this is just cyclohexane.2718

And so, what kind of CH's do we have here?--we have just sp3 CH's--just below 3,000.2725

So, in this case, we have only sp2; here we have only sp3; and in this case, we have both sp2 and sp3.2733

OK, is this a monosubstituted aromatic?--this is not a monosubstituted aromatic; this is...just a little note here; just a little reminder--this is not aromatic, because there are no π bonds.2742

You need to have those three π bonds (1, 2, 3) to make a benzene.2755

So, every time we talked about "aromatic" (the aromatic ripple, the substitution patterns, ortho/meta/para, and so on), we were always talking about benzene in particular with those three π bonds.2759

OK, and how about this last one--how is this going to be distinguished; what kind of CH's do we have?2771

We have the double bonds; so we see sp2 CH's just above 3,000; and we have a regular alkyl, sp3 CH, just below 3,000.2777

So, it is similar to this second compound in that it has both sp2 and sp3; so how could we tell the difference?2791

This is where we are really going to have to dig into that fingerprint region and pick out those out-of-plane bendings.2801

Here, you would have the para substituted at 815; here you would not--there is no para; OK.2807

But what we do have is an alkene; I would describe this alkene as a monosubstituted alkene; it has just one carbon group on it.2816

And a monosubstituted alkene has peaks that we can look for; we can look for those bending patterns.2829

OK, so to do this problem, it really relies on some of the advanced knowledge of IR, in order to distinguish between these two.2838

In order to distinguish between the three cyclic compounds, it can really just be done by looking at the sp2 versus sp3 CH's and the type of CH's you have in your structure.2847

OK, so I think we have had a pretty good introduction into the theory of IR and the application of IR.2859

This is an incredible important technique that you are going to be using throughout organic chemistry laboratory; and in fact, it is typically taught as one of the first spectroscopies.2864

And on this, you will be building some additional ones, such as NMR spectroscopy; and a lot of times, spectra problems are going to be combinations.2875

You will get some IR data, and then you will also get some NMR data, and you have to use them together to come up with a structure.2887

Any time you are given an IR spectrum, what you are going to be getting from that is what functional groups are present: do we have an OH?--do we have a carbonyl?--do we have an aromatic ring?--etc.2893

And if given a correlation table, you should be able to pick up those functional groups with confidence.2904

Thanks very much, and I hope to see you again soon.2911