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  • A couple of people are filing in, I’ll get started.

  • So last time I tried to give an overview of the course and also an overview to IR spectroscopy.

  • And I guess one of the take home messages I was trying to give was that IR spectroscopy

  • really should be something that you [inaudible]

  • And you want to make it easy to do the experiments and I suggested techniques like placing a

  • drop on a thin film or [inaudible] form a solution in forming a calcium fluoride cell;

  • anything to make it easy to do.

  • There are thingsand I hope the example that I started class with really brings it

  • homethere are things that IR just excels at that, I’ll add, NMR doesn’t excel at.

  • NMR is going to talk to you about some things; if youve got an aromatic ring or an alkene in your molecule

  • with a hydrogen on it and itll probably talk to you pretty well.

  • But if you have an unusual ring size like we saw in the beta lactone or a 5 membered

  • ring with a carbonylIR can talk to you in a way that NMR just can’t talk to you,

  • that mass spec isn’t going to talk to you, that UV, which were not going to cover

  • in this course, isn’t going to talk to you.

  • Other things like a terminal alkynewell have a problem later on, I guarantee, everyone

  • is going to look at this and say oh wow, afterward, you know, only with NMR data and not realize,

  • and then something really talksthere are probably about 2 dozen pieces of information

  • that easily IR can talk to you and this is what I’m going to put in.

  • So today I want to talk about CH containing functional groups; in other words I want us

  • to be able to know how to identify alkane groups and alkene groups, or fragments within

  • molecules, arene, in other words aromatic, alkyne.

  • And one of the pieces of philosophy that I’m going to try to take particularly as we get

  • to certain functional groups is to start to look for patterns here.

  • Because it’s not so much about tabulating peaksit’s really about being able to

  • read spectra.

  • I want to talk today about oxygen containing functional groups, and well go through:

  • alcohols, aldehydes, ketones, esters, acidscarboxylic acids of courseacid chlorides, and let’s

  • say acid anhydrides.

  • obviously there are other functional groups and I’m just picking ones that I think are

  • particularly important.

  • Some of the features I hope we get a chance to talk about today, if not today tomorrow,

  • more about ring size which I think is really cool in the case of cyclic ketones and esters,

  • conjugation and understanding the principles involved, inductive effects, andprobably

  • not this time, as a matter of fact, certainly not this timewell talk about nitrogen

  • containing functional groups.

  • well talk about amides, amines, ammonium salts, maybe nitrilesnitriles are another

  • one that IR I think really shines for identifyingand maybe nitro compounds.

  • So one of the things I’m going to urge you to do when you look at an IR spectra is really

  • to learn how to read the IR spectra.

  • So let me give you my take because part of reading is not just sitting and tabulating

  • every peak, it’s knowing what’s important.

  • So when I look at an IR spectrum, generally in my mind’s eye I think from about 4000

  • to about 600 and probablyand of course these are centimeters^(-1) , wavenumbers,

  • [inaudible] waves per centimeterand probably also in my mind’s eye I end up drawing a

  • line at 3000 wavenumbers.

  • And I think this really ends up dividing the spectrum into some important regions. So if

  • you have an IR spectrum maybe you see something like this. Youre reading in—I’m going

  • to label my axis, of course youre going to have %T, and of course your spectrum is

  • going to look something like this.

  • So you see some peaks, and some stuff over here. Now, the region over here from 1600

  • to about 3600 is the functional group region. And that’s generally what I look at.

  • There’s information to be had in this region from 1400 to about 600 wavenumbers; that’s

  • the fingerprint region.

  • If we have a chance maybe on Wednesday well take a look at how you can find aromatic substitution

  • patterns from here; I think in one of the homework problems there’s a—the textbook,

  • by the way, take the time to flip through it. It’s got very nice appendices that are

  • extremely useful.

  • One of the appendices, for example, will give you some correlating peaks to be able to figure

  • out whether an alkene iswhat the substitution pattern is on an alkene; whether it’s terminal,

  • whether it’s cis, or whether it’s trans.

  • Anywayso generally my eye will sort of be drawn to this region; above 1600 I’ll

  • look for carbonyls, I’ll generally look about this region, just above 2000, because

  • normally you don’t expect anything there.

  • If I see anything there it’s going to send up a red flag that there may be something

  • interesting; we already talked about alkynes.

  • Generally I won’t pay too much attention to this region right below 3000. But there

  • are a few things that can show up here at about 2820 and 2720 can be indicative of an

  • aldehyde if you have the right sort of carbonyl peak.

  • Things down from 3000 of course might clue you in to alkenes and aromatics. And then

  • over here, and particularly over here, you get clued in to carboxylic acids.

  • So it’s really this issue of being able to read the spectrum that I think is really

  • so key in being able to get some useful information out of IR.

  • One of the thingsit’s not just the position of peaks that counts it’s their relative

  • intensities, and that can clue you in a lot.

  • So, for example carbonyl and alkenes, C double bond C, both show up in the same region but

  • they have different appearances to them. So for example, a carbonyl C=O is generally strong,

  • and that’s going to mean usually strong relative to other peaks, but again part of

  • that is what fraction of the molecule is the carbonyl occupying.

  • If you have a ketone with 6 carbons in it, youve got a lot more carbonyl groups in

  • that path length than if you had a ketone with 60 carbons. So your ketone peak, your

  • carbonyl peak, is going to be stronger in smaller molecules than in bigger molecules.

  • Carbon-oxygen single bonds, anything with a big dipole moment in generalnot always,

  • but in generalhas strong peaks. So anything where you have substantial differences in

  • electronegativity between 2 groups.

  • Bonds that are often weak or moderate are compounds like alkynes, carbon-carbon triple

  • bond stretches, alkenes, carbon-carbon double bond stretches of various sorts, say, terminal

  • alkenes and internal alkenesand we already talked about if there’s no change in dipole

  • moments, so if you have, say, an alkyne with two alkyl groups on the end, youre probably

  • not going to see-or most certainly won’t seethe carbon-carbon triple bond stretch.

  • If you have an alkene that’s tetra-substituted youre probably not going to see the carbon-carbon

  • double bond. So let’s say usually not seeninternal alkenes, internal alkynes.

  • We talked a little bit when I was talking about methylene groups and I talked about

  • the asymmetric stretch and symmetric stretch, we talked a little bit about CH groups.

  • CH stretches are generally very sharp; so things that might clue you inas I said,

  • alkyl generally not that informative, but usually they fall between about 3000 and about

  • 2840 wavenumbers. And if youre good, you really should be able to draw a line at about

  • 2840 in the spectrum because, as I was saying, aldehydes can have little CH stretches and

  • a Fermi resonance that can clue you in.

  • Alkenes, arenesaromatics if you prefergenerally were talking about 3100 to about 3000.

  • And again youre going to get corroboratory peaksfirst of all, NMR will also be useful

  • but youre going to see patterns.

  • For example, aromatics will have a series of bands from about 1650 to about 2000 wavenumbers

  • that can clue you in.

  • Alkenes, generally you can see the carbon-carbon double bond stretch. And that’s generally

  • pretty sharp but not so strong at about 1640 to 1670 wavenumbers.

  • Alkynes, if they are terminal alkynes of course you have a CH group, that tends to stand out.

  • So itll be at about 3300 and as I was saying theyre generally very sharp, so alcohols

  • also show up at about 3300, but the pattern recognition is going to be completely different

  • because an alcohol is going to be a broad band and an alkyne is going to be the sharp

  • band.

  • And, as I said, if you are able to see the C – C triple bond stretch it’s somewhere

  • just below 2000 depending on the exact substitution, about 2100 to 2260.

  • And again, if you just see something in that general region of about 2000 to 2500 it should

  • clue you in that there’s something unusual.

  • Aldehydes, as I’ve said, the CH stretch of aldehydes shows up and you get 2 bands.

  • What happens is there’s an inactive band, one of these 2 bands I believe it’s the

  • 2820 is the CH and the other one is an inactive band that essentially gets pumped by it’s

  • called a Fermi resonance. And then thatll go along with your carbonyls so youll be

  • looking in that region of around 1700 is going to clue you in; let’s say about 1740 to

  • 1720 would be typical for an aldehyde carbonyl, it could a little bit lower if there is some

  • conjugation there.

  • Let me put up some examples here just so we see what these guys look like.

  • Let me get this handed out. [inaudible] with this particular class the best way is to get

  • these handed out, but we will figuretherell be lots of hand outs in the class.

  • My recommendation, by the way, is to get a loose-leaf binder or else get good at sort

  • of organizing the handouts.

  • Let’s see, I have a few extras, let me send some more down here.

  • Well be using transparencies a lot for discussion sections, where there’s really

  • no substitute for looking at stuff which of course you can do with an LCD projector but

  • also marking on stuff.

  • So when youre up hereand everyone, I hope, is going to have a chance to be up here

  • in our discussion sectionit takes only minor skill but you get in the habit of not

  • standing there with your shoulder in the beam.

  • Alright, let’s just take a look at the first one. What is this first one? well just

  • look at the first 2 right now, what iswhat class of compounds is the first one?

  • The first one here. Alkeneok great, actually I’m hearing exactly the sort of stuff that

  • I like to be hearing here. So, I’ve heard carbonyl, aromatic, and alkene. And we can

  • actually take a very, very good guess.

  • This peak here obviously stands out, remember I said draw a line in your mind at 1600

  • So it doesn’t look like a carbonyl; typical carbonyls are going to be a little stronger

  • and a little bit fatter. It looks like an alkene. This should clue us in, so I’ll

  • just pick some numbers here. I’ll just write 1642 and here were at about 3080.

  • Now, if I had to guess, and fortunately youll probably get other data but if I had to guess

  • I’d say what were seeing hereusually with an aromatic youll have a series of

  • small, very weak bands.

  • For, like, a phenyl group there will be like 4 of them between 2000 and 1600. And this

  • doesn’t look anything like that. Youll usually see more CH’s for a phenyl, belowfrom

  • about 3000 to 3100.

  • So this is indeed an alkene. Now, one of the reasons I’m not putting as much emphasis

  • on the fingerprint regionsso here, for example, you get some CH bandsis as you

  • get to bigger molecules the fingerprint region is going to become more and more complicated

  • and less and less easy to read.

  • And, part of the problem of the pedagogy of IR spectroscopy is it’s a relatively mature

  • area; it’s been around largely since the 1950s and used as a tool by chemists but the

  • molecules that we study have gotten bigger and more complex and people for pedagogical

  • purposes still tend to look at small molecule examples.

  • This is a small alkene, maybe octene or something. But as you get to larger and more multi-functional

  • region molecules, this region gets harder to read.

  • Alright, so what else do I want to do right now?

  • What about the second example here? Alkyne, yeah. Terminal alkyne.

  • And so, some of the things that we see here is this band at about 3310. There’s another

  • band at about 2119 and the band at 3310 is quite sharp and so that’s very characteristic

  • of an alkyne.

  • And so indeed we have an alkyne. This region at 2000 jumps out at me and says at about

  • 2000-2100 you have something jumping out.

  • In general alcohols will be broad; if you have an alcohol that’s not hydrogen bondedin

  • pure form alcohols tend to hydrogen bond to each other, if you have them very dilute they

  • tend not to hydrogen bond but then they tend to be about here over at about 3400-3500.

  • Silanols could even be a little bit further.

  • If the alcohol is very sterically hindered, like a tertiary alcohol, youll have less

  • hydrogen bonding than if it’s more sterically hindered, like a primary alcohol and so you

  • may see a [inaudible] band at about 3400-3500.

  • Alright I want to go back to scratching out more examples on the blackboard.

  • Alright, so alcohols, as I said your OH if youre neatneat is just another way of

  • saying not dissolved in a solvent for example on a film or a solid plate or a KBr pellet

  • because on a KBr pellet if you are dealing with a solid or a [inaudible] youre going

  • to have particles of your alcohol that are hydrogen bonded together and [inaudible] molecules

  • that hydrogen bond together.

  • But usually youll see a band at about 3300, itll be broad; and on the pattern recognition,

  • again, if I’m drawing these two points in my mind’s eye say the end of the spectrum

  • at about 4000—I guess the examples I have here may run to 4600—what youre going

  • to be seeing is a band that sort of picks up at about 3600 and comes down maybe somewhere

  • around 3000, and then of course youll have some alkyl stuff over here.

  • So you might say well it’s 3300 but you look and in your eye youre seeing this

  • thing pick up at about 3600, so youre going to see this pattern.

  • And this will be different than the carboxylic acid that may pick up around here but is going

  • to go down around here; it's going to be really, really ugly.

  • As I’ve said, I’m not putting so much emphasis on the fingerprint region, you can

  • sometimes look for corroboration. C – O single bond, we talked about it last time

  • in comparison to a double bond where we were talking about a harmonic oscillator and frequencies

  • and the bond strength for single bond versus double bond.

  • So usually somewhere between about 1300 and 1000, but I’m going to say often buried

  • in other stuff, and again even more so in bigger molecules.

  • Alright, let me take an example of an alcohol just so you can see something apart from my

  • little drawing over here.

  • And this is again, this is a small—a small alcohol. This is a small alcohol so we have

  • this band at about 3300. We have, you know, here this small alcohol you can see your C

  • – O single bonds stretch. And again the main thing I’m looking at is that this is

  • sort of picking up and coming down.

  • Alright, any thoughts or questions on this one?

  • Ah, C is our number of, I believecertainly in this region you have your CH bands, remember

  • we talked about methylene groups and we said there’s one at about 1380 and another at

  • about 1460.

  • So certainly a big part of what youre seeing here isbecause this is going to be a small

  • chain linear alcohol—a big part of what youre seeing is CH bands there.

  • Yeah, so the stretchingthe CH stretching modes are much stronger, much higher frequency

  • those are at about 3000 and CH bending modes are at about 14—or about 1360 and 14—you

  • know, the 13s and the 14s you have the asymmetric, the in-plane and out-of-plane bending.

  • There are actually things you can discern at this region and again there are better

  • ways to do it in general. But in this region right here in the CH bending region you can

  • actually determine the difference, for example between a single methyl group and an isopropyl

  • group and again some of them is coupled vibrations where an isopropyl group gives you a special

  • coupled vibration that you can pick out.

  • But there are probably better ways to do that. And in big molecules things are just awfully

  • crowded.

  • The carbonyl region from about 1650 to about 1850 really should talk to you. The carbonyls

  • are going to be strong. I’ll take ester, aldehydes, ketone, carboxylic acid, and amides

  • as 5 groups that are really important. and I’ve written them in general in order of

  • decreasing frequencies.

  • So were talking, for this particular group, with sort of normal, non-unusual carbonyls

  • let’s say 1750 to 1650 and were usually talking about strong.

  • So esters generally were talking about 1750 to 1735 if there’s nothing to perturb

  • it like conjugation or strained rings.

  • Aldehydes in general were talking about 1740 to 1720. And as I’ve said you can often

  • pick out the CH stretch in the Fermi resonance if you look at right on the edge of the region

  • at about 2800—about 2820 and 2720.

  • A plain old vanilla, normal ketone were typically talking about 1725 to 1705 wavenumbers.

  • With carboxylic acid, again without any sort of conjugation or anything special were

  • talking about say 1725 to 1700. Carboxylic acids love to hydrogen bond. And you get these

  • very nice hydrogen bonded dimer stretches that generally produce broadbroad non-specific

  • CH stretches that as I said will cover that whole region from about 3500 to about 2500.

  • Amides typically were talking a little bit lower frequency and I’ll talk about

  • why in just a second, but let’s say about 1690 to about 1750. If were talking about

  • primary amides that have 2 NHssecondary amides that have 1NH and tertiary amides that

  • have no NHs.

  • So as I said carboxylic acids are one of the ugliest things that youre going to see

  • in the IR and theyre ugliness really makes them stand out it’s kind of like a pug dog.

  • You look at a pug and you say that’s so ugly it’s almost cute [inaudible].

  • So you look at a carboxylic acid and you just sort of have this misformed, misshapen band

  • maybe with some lumps and some CHsmaybe I over exaggerated a little bit. But suffice

  • it to say it gets a little bit lumpy and then youre going to see your carbonyl standing

  • out probably a little stronger for this example, let’s say at 1725 to 1700 unless it’s

  • conjugated, in which case itll be lower.

  • And then some stuff in the fingerprint region, and generally my eye kind of catches the shift

  • in the baseline at about 3500 and it catches it coming down at about 2500. If youve

  • got a big molecule it’s going to be a lot smaller band.

  • This is one of the reasons I’m not a huge fan of KBr pellets; with KBr pellets it’s

  • really hard to get your KBr dry and when you get to bigger molecules being able to figure

  • out if that’s just a little bit of [inaudible] water in my KBr or is that an alcohol, is

  • that a carboxylic acid? It can be tough.

  • As I’ve said aldehydesso I’ll just write; let me write some text here I’ll

  • say 3500 to 2500 and I’m going to say broad and ugly. For an aldehyde, just to show some

  • corroboratory stuff let me draw out the same sort of spectrum here from about 400 to about

  • 3000 to about 600 to about 1600.

  • I’m going to make this a little bit bigger so you guys in the back can see it. So usually

  • whatll happen is right on the edge of thingsthat’s a really lousy carbonyl. Right on the edge

  • of things youll probably be able to, if you look hard, pick out a band at about 2820

  • and a band at about 2720.

  • And then this band at about 1740 to about 1720 again unless it’s conjugated. So that

  • could help you clue you in.

  • Now, if I had an IR spectra like that the next thing I’d do is look at my NMR spectrum

  • and see if there’s a peak at about 9 to 10 parts per million that I expect associated

  • with an aldehyde.

  • I’ll give you a moment to finish up sketching this.

  • Alright, so let’s take one moment to read a couple more spectra here.

  • What is this third one? So you have aldehydealdehyde, everyone agree?

  • So, if I’m looking here I really don’t see much of anything over here, so I’m suspicious

  • on an aldehyde. Our peak hereand you should get good at reading analog, I know everyone

  • is in a digital mindset but learn to take this scale and go ok, 1700 1800 these tick

  • marks are 20, so I go 20, 40, 60, 80.

  • So that’s right at about 1715. The number digitally was 1717. So it’s a little low for an aldehyde; were shy on

  • anything over here; and again you really can draw inwith a pencil draw and you notice

  • here the tick marks are closer together but you can draw a line right here as the line

  • at 2840 and there’s just nothing that looks distinct, there are just little, teeny, teeny

  • resonance.

  • So there’s really nothing to speak of here at 2820 and 2720. So, data for common carbonyl

  • compounds it’s low to be an ester, we don’t see a corroboratory C – O single bond peak

  • but, you know, it’s impossible to tell anyway, but it’s low to be a normal ester, it’s

  • a little low to be an aldehyde, we don’t see that sort of big, ugly associated with

  • a carboxylic acid, it’s high to be an amide, we don’t see any substantial NHthis type

  • of thing is not uncommon, that can just be a little bit of water in your sample nut this

  • little band is not uncommon.

  • And that really puts this at a ketone. Just sort of a simple unstrained ketone. Last one?

  • Carboxylic acid.

  • That onethat one screams carboxylic acid. Now youve got this carbonyl at about 1710,

  • weve got this big, ugly here at about 2520 to 3500.

  • And so if youre on top of thingsif youre on top of things like that youre in pretty,

  • pretty good shape.

  • Alright, I want to talk about somesome other effects here.

  • So, let’s talk about effects of ring size and then were going to talk about some

  • conjugation and some other effects.

  • So, okso remember a normal ketone is about 1715 wavenumbers. A small ring is going to

  • bring it up to higher wavenumbers; so a cylcopropanone is at about 1825 wavenumbers. Cyclopropanones

  • generally aren’t stable, they generally undergo ring opening, but on the other hand

  • cyclobutanones are at about 1780.

  • Cyclopentanone still have a little bit of ring strain, it’s at about 1745. And as

  • I said by the time youre at cyclohexanone youre right back where you’d expect to

  • be, at about 1715.

  • So what’s going on? What does it mean that the frequency is higher for small rings? Stretches

  • faster, which means the bond is stronger or weaker? Stronger, right?

  • A spring with a—a very stiff spring vibrates quickly, a very slack spring, a very weak

  • spring, vibrates slower. We already saw that trend, a C – O single bond has a stretch

  • at about 1100, a C-O double bond has a stretch at about 1700. Remember it’s that root k

  • over u term; that root force constant over [inaudible] mass.

  • So if you increase from 1715 to 1825, that’s a good bit stiffer spring. So that means the

  • carbonyl bond is stronger or weaker? Stronger. Is that surprising? It is surprising. How

  • can that be?

  • Carbon-carbon bond could be weakerit’s a good way to start thinking about it.

  • Do you have some specifics? We are on track here. P—yes, so let’s think about p-character.

  • So, what sort of orbitals, if it were a perfect world what sort of orbitals would you use

  • towell what sort of orbitals would you first use to make up say a regular, normal,

  • 109.5 degree carbon?

  • sp3. If youre going to go to a strained ring, do you need more s-character or more

  • p-character for carbonyls?

  • in the carbon-carbon bonds? Which way are our p-orbitals orientated? So the p-orbitals

  • would be at right angles, right? So if you just use p-character, you could make this.

  • Now, in fact you don’t but you put in more p-character in here, so that leaves what?

  • More s-character.

  • And which is lower in energy, which forms stronger bonds? s.

  • So it’s sort of counter intuitive because you think strain, oh that should be bad and

  • yet it goes in the opposite direction and when you think it through it actually makes

  • sense.

  • And were going to see this coming up as a trend in couple of times. As I was also

  • saying, this should talk to you. If you haveNMR is not great for telling a cylcopentanone

  • from a cyclohexanone. But if you get these types of wavelengths it’s going to be telling

  • something’s going on. That’s why when I got my beta lactone that I was talking about

  • last time at 1820, even just a perfunctory glance at the spectrum said that there’s

  • something going on in a way that no other technique was going to say.

  • And just as in the case of the ketone in the case of the esters, your numbers go down with

  • less strain, and by the time youre atso this is a beta lactone, gamma lactone, and

  • delta lactoneby the time youre at a delta lactone youre right at the typical

  • ester position, right at 1735 wavenumbers.

  • And remember, when youre carrying out reactions to make molecules, youre not just corroborating

  • what you think youve done youre actually asking a question.

  • I had a hypothesis: if I mix this stuff I’m going to form this, and sometimes you get

  • a surprise, and it teaches you something.

  • So, let’s talk aboutso that’s ring size, and that’s something that IR really

  • shines at, let’s talk about something else.

  • Let’s talk about conjugation of carbonyls.

  • So remember, a normal ketone is about 1715, if we go to cyclohexenone, conjugated, were

  • at about 1690 wavenumbers. So does that surprise us, or does that make sense?

  • Makes sense, why does it make sense?

  • Conjugation makes the carbonyl bond have more single-bond character and it should be lower

  • in wavelength, lower in frequency rather.

  • So you can write a resonance structure and it dominates the reactivity for alpha-beta

  • unsaturated compoundsremember theyre Michael acceptors, nucleophiles like to add

  • into the beta position.

  • Beta position has a delta positive on it. Of course we know, at the graduate level,

  • it’s not that it’s one structure and the other and resonating between them, it is simply

  • both structures at once; this is a representation of the molecular orbital character of the

  • molecule and so the molecular orbital character of the molecule says it’s not quite a full

  • double bond, it’s a little bit weaker.

  • It’s, you know, 90% double bond, 10% single bond characters.

  • Alright, so, just so I can play with your minds, then why is it if you look atat

  • least certain compoundslike this, well see, let’s say, a band atmaybe a little

  • bit weaker—a band at 1690 and a band at 1640 for an alpha-beta unsaturated compound

  • like this.

  • Not exactly, but kind ofsort of.

  • What other stretch do we have? We have the alkene stretch. And in factso you have

  • your C-C stretch and your C-O stretch and in fact because your alkene is really polarized,

  • the polarization leads to stronger stretchesso because it’s really polarized you have a

  • bigger dipole moment or specifically a bigger change in dipole moment as you vibrate it,

  • itll be kind of strong.

  • Alright, one lastone last category to play with, and that’s electron withdrawing groups.

  • And this is another one where IR really shines, because let’s say youre making an acid

  • chloride. Telling an acid chloride from a carboxylic acid by NMR often isn’t that

  • easy. But in IR, it’s , you know, like falling off a log.

  • So an acid chloride shows up at about 1820 wavenumbers. And just in general if you have

  • any sort of electron withdrawing groupremember, normal ketone, say, would be about 1715, an

  • ester would be about 1735, so this is often a region that should be making you stand up

  • and pay attention.

  • So what does that say about the carbon-oxygen bond? Stronger bond, again counterintuitive

  • until you think about it for a second.

  • If the electron withdrawing group is pulling electrons away, you can think of a second

  • resonance, one you could call a non-bond resonance structure. Like so.

  • And that non-bond resonance structure is going to have carbon-oxygen triple bond character.

  • There’s not much of this contributing, this is just a representation of the molecular

  • orbitals of the molecules saying that youre pulling electron density onto the electron

  • withdrawing group.

  • But it’s enough to shift the carbonyl group a little bit. If you go to an acid anhydride,

  • an acyl group is of course electron withdrawing, any group that is an acylating agent, any

  • group that is prone to attack by a nucleophile, especially prone, is going to be a—is going

  • to be a carbonyl with an electron withdrawing group.

  • So a nucleophile—a weak nucleophile like water or an amine is going to react very rapidly

  • with an acid anhydride or with the acid chloride.

  • And we see a band again at about 1820, and we see another band at about 1750 wavenumbers,

  • what gives here?

  • 2 carbonyls, and that means? Asymmetric stretch. You get coupled vibrations. Even if the carbonyls

  • aren’t in the same molecule in certain protein structures where you have carbonyls near each

  • other, you get vibrational coupling that splits your carbonyl stretches into two.

  • Ok, last example for today.

  • Take an amide. Generic amide, and I already gave you the number: 1650 to 1690 is sort

  • of typical.

  • What does that say about our carbonyl stretch? Weaker, and? So you can write this resonance

  • structure, and that’s—so that’s an interesting conundrum of various groups here.

  • In that all of these groups with lone pairs can be donating by resonance but they can

  • be inductively electron withdrawing, and in some cases one wins out and in some cases

  • the other wins out.

  • Nitrogen and chlorine are both electronegative but the big difference is that nitrogen is

  • in the same row of the periodic table as carbon. So you get good pi donation, you get good

  • overlap and nitrogen the donation wins out.

  • Whereas chlorine, although about equal in electronegativity to nitrogen is down a row

  • and so you get a big sigma electron withdrawing effect and not so much pi donation effects.

  • And in this case you simply have, say, compared to an ester where in an ester this oxygen

  • can donate into just one, in the case of an anhydride, this oxygen has to donate into

  • two. So you end up with more electron withdrawing.

  • Anyway, this is probably a—oh yeah, one last thing and then I’ll wrap this up. Alright,

  • last thing.

  • Alright, so we said a normal, unstrained ester or lactone is at about 1735; if I go ahead

  • and take an unsaturated lactone where now were not conjugated with the carbonyl,

  • but conjugated in the other way, conjugated with the oxygen, we move to 1760.

  • And again you can think of this as a resonance effect.

  • You can write a resonance structure like so; to put it another way this is a lactone of

  • an enol ether. Enol ethers are electron rich at the position here, basically if you have

  • an electrophile it’s going to attack here and you see that in the IR spectrum.

  • Alright, so that promptly sums up everything I want to say about C-O containing functional

  • groups, well pick up next time with nitrogen containing functional groups.

A couple of people are filing in, I’ll get started.

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