Chapter 2: IR Spectroscopy: Reading IR Spectra

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Welcome back to the Deep Dive.

Today we're cutting through some of the real complexity of organic chemistry.

We're diving deep into the tools that chemists actually use, you know, every single day to check their work.

You spend months in the classroom learning substitution, elimination, all these addition reactions, you learn how to make something compound X.

But the real question, and this is the one that separates, you know, textbook chemistry from what happens in the lab, is all about verification.

In the real world, how do you actually know you made compound X?

And not some other junk, right?

And not some unwanted byproduct, exactly.

It's amazing when you think about it.

If you look back at the history of this science, just figuring out the structure of one new compound,

it used to take months, sometimes years.

Oh, it was a brutal process.

It involved all this complex chemical degradation.

Painful is an understatement.

But now, that whole process, it's down to minutes.

It is.

And that's all thanks to modern analytical techniques, specifically spectroscopy.

I think it's fair to say that spectroscopy is, without question, the most important tool we have for figuring out molecular structures in OCAM.

So before we get completely lost in terms like frequency and wave number, let's just ground this whole thing in a really simple idea.

I love the analogy from our source material here.

The bakery one.

Yeah, it's a good one.

The bakery one.

So imagine you run a bakery, okay, and you've got 10 friends, and you know exactly what each of them orders every single morning.

John always gets a brownie, Mary always gets a muffin, Peter gets the French roll.

So if you walk into that bakery one morning, and you see that the brownie is missing,

you don't even need to see John's face.

You just know he was there.

The missing item, the absence of that brownie, is all the evidence you need.

And we apply that exact same logic to the molecular world.

We replace the bakery items with different frequencies of light, and the friends are specific molecular bonds.

Right.

So when we shine a whole spectrum of light through our compound,

certain very specific frequencies get absorbed.

They go missing.

Just like the brownie.

Just like the brownie.

And by analyzing exactly which frequencies disappeared, we can piece together the structure of the compound.

I think the challenge for any student, though, is figuring out where this specific tool, infrared or IR spectroscopy, fits into the whole toolkit.

It's a big toolkit.

Organic chemists really rely on a few key types, and each one gives you a different piece of the puzzle.

You've got NMR, nuclear magnetic resonance spectroscopy.

That uses radio waves.

That's sort of the big

gun.

It's the tool that gives you the specific arrangement of pretty much all the carbon and hydrogen atoms.

It builds the complete molecular skeleton for you.

Then you have something called UVV spectroscopy, which uses visible and ultraviolet light.

And that's much more of a niche tool.

It's really focused on these things called conjugated pi systems, long chains of alternating single and double bonds.

We're not really going to get into that one today.

No, that's a deep dive for another day, because our focus is entirely on infrared spectroscopy.

I do.

So we're using infrared radiation.

What's the specific piece of information that IR gives us that the others don't?

IR spectroscopy reveals the functional groups present in the molecule.

Okay, so it's like the molecule's fingerprint, but maybe not its full identity.

That's a great way to put it.

It tells you the type of compound you have.

Is it an alcohol, a ketone, an amyloine, an alkene?

It's that essential, really fast first step.

It lets you know if the basic building blocks you were trying to make are actually there.

That really sets the stage for our mission today, then.

We are trying to translate a pattern of missing light into a functional group.

But before we can read the charts, we have to understand what's actually happening on a physical level.

Right.

We need to go from the big picture down the quantum mechanics of it all.

So molecules can store energy in a lot of different ways, but the one that matters for IR is the vibration of the chemical bonds.

The vibrate?

The vibration.

The best way to visualize this is to think of every single chemical bond as a tiny, perfectly behaving spring that's connecting two atoms.

And this energy, the energy that's stored in that vibrating spring, is governed by quantum mechanics, which means?

Which means it's quantized.

Quantized.

Okay, what does that mean in simple terms?

It means a bond can't just vibrate at any old energy level it wants.

It's restricted to specific, discrete, allowed vibrational energy levels.

Think of it like a ladder.

The bond can be on the first rung, which is its low energy state, or the second rung, the high energy state.

But it can't just float in between the rungs.

It cannot float in between.

It has to be on one or the other.

So if the bond is just sitting there, vibrating in its low energy state, how does it jump up to that higher state?

I'm guessing that's where the light comes in.

That's exactly where the light comes in.

That jump only happens if a photon of incoming IR light has exactly the right amount of energy.

Not close to the right amount.

Exactly the right amount.

It has to be a perfect match.

We call that energy difference delta E, or delta E.

If the photon's energy perfectly matches that delta E between the two allowed states, the bond absorbs the photon.

When it absorbs the photon, that energy gets stored, and it causes what you call a vibrational excitation.

The bond just starts vibrating much more energetically.

It's like you've plucked a guitar string.

But that energy is temporary, right?

It doesn't stay there forever.

No, it's very temporary.

Eventually,

that excited vibration just kind of fizzles out, and it releases the energy, usually as heat, back into the environment.

But the important thing is that for a split second, it did absorb that specific photon.

And that's the event we record.

Now, I've heard chemists talk about different kinds of vibrations.

There's stretching, but there's also bending.

That's right.

A bond can store energy by stretching.

That's moving the atoms apart and then back together like a spring, or by bending, which is a lot more complicated.

It's about changing the angle between bonds.

So which one do we care about?

For our purposes, for identifying functional groups, we overwhelmingly focus on the stretching vibrations.

They give us the clearest, the most reliable, and just the most useful information.

And this is where the specificity of IR really comes from, isn't it?

Because that energy gap, that delta EE, is completely unique for every different type of bond.

It's the entire foundation of the technique.

The energy required is determined by the stiffness of that bond spring and the masses of the atoms it's connecting.

So the energy gap for a carbon -hydrogen bond, for example, is way bigger than the gap for a carbon -oxygen bond.

So the CH bond needs a more energetic photon to make it jump.

A much higher energy photon.

And that uniqueness means every functional group absorbs light at its own characteristic frequency.

And let's just say with confidence, okay, that absorption right there, that's an OH bond.

Or, yes, we successfully made a carbon -oxygen double bond.

Okay, so we've got the physics down.

We shine IR light on a sample.

Some photons get absorbed.

Now let's talk about the data we get back.

How do you actually, you know, read the output?

The output is called an IR spectrum.

And it's basically just a graph.

The instrument measures the full range of IR frequencies, and it plots how much light was absorbed at each one.

So the vertical axis is measuring absorption.

Well, technically, it's measuring how much light passes through the sample.

We call percent transmittance, or proceed.

And this is a really important visual point.

All the signals we care about, they all point down.

All point down, exactly.

So a big dip in the graph means a lot of absorption happened there.

Less light was transmitted at that specific frequency.

The deeper the dip, the stronger the absorption.

Okay, and the horizontal axis, that's where we measure the frequency of the light that got absorbed.

But you don't use normal frequency units like Hertz.

Chemists use this thing called wave number.

Wave number, yeah.

The symbol is a little new with a tilt over it.

New dollars, just the frequency divided by the speed of light.

So the units are weird.

Inverse centimeters?

The units are inverse centimeters.

Centimeters to the minus one.

And for most organic molecules, we're looking at a range from about 400 inverse centimeters up to 4 ,000.

So why?

Why complicate things with inverse centimeters?

Why not just stick with frequency?

It's mostly a historical convenience.

But the most important thing, the only thing you really need to remember, is that wave number is directly proportional to frequency.

And therefore, it's directly proportional to energy.

Okay, so that's the key takeaway.

That's the key.

So if you see a big wave number, like 4 ,000, that means a high energy photon was needed to cause that vibration.

So signals on the far left of the chart are high energy.

Signals on the right are low energy.

Exactly.

Reading the spectrum is really just about translating that location, that wave number, into the energy required for a specific bond to vibrate.

Now, before we jump into the specific numbers, you said earlier that you can't just look at one thing.

When we see a peak, we have to analyze three different characteristics.

Yes, this is the strategic approach.

You have to look at all three, because functional groups can sometimes overlap on one characteristic, but be totally different on the others.

These three traits are one, wave number.

So where does the signal appear?

The location.

The location.

That's our main tool for figuring out the bond type.

Two is intensity.

How deep is the signal?

Is it strong or is it weak?

This is vital for telling apart, say, a polar bond like CO, from a non -polar one like CO.

And the third.

And the third is shape.

Is the signal broad and smooth or is it sharp and narrow?

This one is a dead giveaway for certain interactions, especially hydrogen bonding.

Okay, so it's like a three -digit code for every functional group.

You can't just rely on the first digit.

You need all three to unlock the identity.

That's a perfect way to think about it.

Let's start with that first and most critical characteristic,

then.

Wave number.

What are the physical properties of the bond that decide exactly where it's going to absorb light on that 400 to 4000 scale?

Well, we can go right back to our spring and mass analogy.

The two things that matter most are the stiffness of the spring, which is the bond strength, and the weight of the atoms attached to it, their atomic mass.

Okay, let's take the stiffness first.

The bond strength or bond order.

A skipper spring is harder to stretch, right?

It takes more energy.

So stronger bonds require more energy to vibrate and they absorb at higher frequencies, which means higher wave numbers.

And we see this really clearly when we compare single, double, and triple bonds between the same two atoms.

Oh, absolutely.

The classic example is the carbon -nitrogen bond.

A single C -N bond is the weakest of the three.

And it shows up pretty far to the right, around 1100 inverse centimeters.

Okay.

You step up to a double C -N bond that's stiffer and stronger, so the absorption jumps way up to about 1600.

And the triple bond?

The triple bond C -cripple bond N is the strongest and stiffest of all.

It requires the most energy to stretch, and that pushes its absorption all the way up to around 2200 inverse centimeters.

So that relationship is fundamental.

Stronger bond, higher wave number.

Stronger bond, higher centimeter, minus one.

Got it.

Okay, now for the second factor.

The atomic mass of the atoms in the bond.

This one's even more straightforward.

Bonds that involve smaller, lighter atoms vibrate at higher frequencies, which means they absorb at higher wave numbers.

And this is why bonds to hydrogen are always on the far left of the spectrum.

Always.

Because hydrogen is the lightest atom there is.

Right.

I mean, look at the numbers.

A typical C -H bond absorbs around 3000.

But if you swap that hydrogen for duperium, which is twice as heavy.

Oh, so a C -D bond.

A C -D bond.

The signal immediately drops down to around 2200.

And if you keep going to heavier atoms, the signal just plummets.

A C -O bond is around 1100.

And a C -C -L bond is way down around 700.

The weight of the atoms really is a primary factor that defines the major zones of the whole spectrum.

And this logic, this combination of mass and strength, is what lets chemists map out the spectrum into these two really important zones.

The diagnostic region and the fingerprint region.

Yes.

The diagnostic region is from 4000 down to 1500 inverse centimeters.

This is where all the useful information is.

So this is where we live.

This is where you live.

It's much cleaner, it has fewer peaks, and it contains all the important stretching signals.

Double bonds, triple bonds, and all the bonds to hydrogen, C -H, N -H, O -H.

If you want to know what functional group you have, you look here.

And then there's the other part.

The fingerprint region.

That's from 1500 down to 400.

Why?

Why is it called the fingerprint region?

It's called that because it is just a chaotic mess.

It is packed with a huge number of signals.

Mostly from those complex bending vibrations and a ton of low energy single bond stretches.

So it's just too crowded to be useful for identifying one specific group.

It's almost impossible.

There's so much overlap.

However, that complex pattern that's created by all that chaos is absolutely unique to every single molecule.

It's literally its fingerprint.

So while it's great for matching a sample to a known compound, for our main goal today, identifying the type of functional group, we just ignore it.

Okay, so to summarize the map, then it's pretty simple.

Bonds to hydrogen are highest energy.

So they're on the far left because of hydrogen's light mass.

Triple bonds are next because they're super strong.

Then double bonds.

Exactly.

You've got the whole layout right there.

That general map is a great start, but we need to fine tune it.

For any student, I think one of the most important distinctions is trying to interpret the CH signal.

How does IR tell the difference between, say, an alkene -CH bond and an alkene -CH bond?

This is a classic question.

And it gives us probably the single most important checkpoint on the entire spectrum.

The absorption frequency for a CH bond depends almost entirely on the hybridization state of the carbon atom it's bonded to.

So whether it's B3, Sb2, or a C.

Right.

And the trend is incredibly consistent.

The CH bond on a Sb hybridized carbon, so from an alkene, is the highest energy.

It's around 3 ,300.

OK.

The CH bond on an Sb2 carbon from an alkene is next around 3 ,100.

And then the most common one, the CH bond on an Sb3 carbon from an alkene, is the lowest energy down around 2 ,900.

And this trend is so reliable that it gives us what's called the 3 ,000 wave number dividing line.

The 3 ,000 dividing line, yes.

So for the student, the practical instruction is to just immediately draw an imaginary line right at 3 ,000.

What's the benefit of that?

The benefit is huge.

If you see signals that appear to the right of that 3 ,000 line, those are your standard Csp3H bonds.

Almost every organic molecule has them, so they're usually strong, but they don't tell you anything special.

They're just background noise almost.

They're background noise.

The excitement, the real information, comes when you see a signal to the left of that 3 ,000 line.

Because if you see a signal there, you know for a fact that your compound has either a Csp2H bond, which means an alkene or aromatic ring, or a CspH bond, which means a terminal alkene.

It's an immediate confirmation of unsaturation.

It's fantastic.

So let's talk about the why.

Why does the bond strength change with hybridization?

I know this has to do with the shape of the orbitals, right?

The percentage of his character.

You got it.

That is the key mechanism.

Nsp orbital is 50 % character.

Nsp2 is 33%.

Nsp3 is only 25 % character.

And because s orbitals are spherical, their electron density is held closer to the nucleus.

So the more character a carbon orbital has, the more tightly it holds its electrons, and so the shorter and stronger the bond it forms.

Precisely.

So the CspH bond with 50 % character is the shortest strongest Ch bond there is.

And strongest bond means highest stretching energy, which puts it way up at 3 ,300.

Exactly.

And the Csp3H bond is the longest and weakest, so it's at the lowest wave number, around 2 ,900.

It's a beautiful direct link between orbital theory and the spectrum you see on the screen.

Now there's a really important caveat here, and I think this is a classic student trap.

The absence of a signal above 3 ,000 doesn't automatically mean there are no double or triple bonds in the molecule.

No, absolutely not.

That is a huge trap.

The 3 ,000 line only looks for Ch bonds that are on unsaturated carbons.

But what about something like a tetrasubstituted double bond?

Right.

A Cc bond where all four positions are attached to other carbons, not hydrogens.

It's clearly a double bond, but there's no Csp2H bond there to stretch.

So you won't see anything above 3 ,000.

But you'd still see the Csp bond stretch itself right, down in the double bond region.

Yes.

You'd still expect to see that around 1 ,600 to 1 ,680.

So the lesson is, always check both regions.

Check above 3 ,000 for the special Ch bonds, but also check the 1 ,600 to 2 ,300 region for the double and triple bonds themselves.

So we've established that the main location of a signal is set by these big factors, like mass and hybridization.

But chemistry always has nuance.

There are secondary factors that can kind of tweak these locations.

And I think the most common one is the effect of resonance and conjugation.

Yes.

This is where we see electron delocalization actually changing the physical stiffness of a bond.

So just to refresh, conjugation means you have pi bonds, double or triple bonds, separated by exactly one single bond.

Right, like CcCO.

That lets the electrons spread out over multiple atoms.

So let's use the standard example to see how this works.

Comparing a regular saturated ketone to a conjugated unsaturated ketone, a standard ketone usually gives that really strong CO signal around 1720.

Right.

But when we introduce conjugation, we put a Cc next to it, that CO signal shifts to a lower wave number.

It moves down to about 1 ,680.

OK.

So the question is, why?

Why does linking it to a double bond make the carbonyl absorb a lower energy photon?

It feels counterintuitive.

It does feel a bit counterintuitive.

It happens because the conjugation actually weakens the CO bond a little bit.

How does it weaken it?

It all comes down to resonance theory.

In a normal non -conjugated ketone, you can draw two main resonance structures.

The big one, the major contributor, shows a standard neutral CO double bond.

And the other one is a minor contributor with charge separation.

Right, a C plus O minus single bond.

Yeah.

But that one contributes very little.

So the bond has almost full double bond character.

But when that ketone is conjugated, those pi electrons have somewhere else they can go.

Exactly.

The conjugation allows you to draw one extra resonance structure.

And in this new structure, the carbonyl is shown as a single bond.

Now, even though this third structure is also a minor contributor, the fact that it exists at all means that overall, the true hybrid bond has slightly more single bond character than it did before.

And more single bond character means a slightly weaker bond.

A slightly weaker, slightly longer bond.

And a weaker bond requires less energy to stretch.

So it absorbs a lower energy photon.

And the signal shifts to a lower wave number from 1720 down to 1680.

That's the whole mechanism.

And this resonance shift is a critical piece of diagnostic information.

If you see a strong CO signal below 1700, you can be pretty sure that molecule is conjugated.

And our sources say this applies to more than just ketones, right?

Esters do the same thing.

Yes, it's a universal principle.

And ester will shift from around 1740 down to 1710 when it's conjugated.

It's all about that electron delocalization.

We've spent a lot of time on location, on wave number.

Now we have to talk about that second characteristic.

Intensity.

How strong or weak is the signal?

And what does that tell us?

Intensity is all about the magnitude of the bond's dipole moment.

To absorb IR radiation really efficiently, a bond has to have a large dipole moment that changes as it vibrates.

Bonds with big fluctuating dipoles give you really strong signals.

They give you huge signals, yes.

Okay, so let's use this to make the classic distinction between a CO carbonyl group and a CEC alken group.

They both show up in that same double bond region between 1600 and 1850.

But their intensities are night and day.

The carbonyl group is the poster child for a strong signal.

Oxygen is super electron negative, so it's pulling a ton of electron density away from the carbon.

Which creates a huge partial negative on the oxygen and a huge partial positive on the carbon.

Right, a massive charge separation, which means a very large dipole moment.

And the result is?

The result is that the CO stretch absorbs IR light incredibly efficiently.

It produces a very strong signal.

It's often the deepest, most obvious peak in the entire spectrum.

You just can't miss it.

But in contrast, the CEC bond, it usually has a tiny dipole moment, if it has one at all.

That's right.

And that's why CEC bonds usually give you a pretty weak signal.

The atoms are similar in electronegativity, so unless the double bond is substituted really asymmetrically, the electronic balance is pretty even.

So practically, if I see a peak in that double bond region and it's absolutely massive, I can pretty much say right away it's a carbonyl and not an alkene.

That is a highly effective strategy.

Intensity is a great first screen.

Now this brings up the extreme case.

What happens if the CEC or a C triple bond C is perfectly symmetrical?

You mean like in 2 part and 3 dimethyldebutene, where both sides of the double bond are identical?

Exactly.

If that bond is perfectly symmetrical, it has zero net dipole moment.

And if there's no dipole moment to change during the vibration, the bond is completely transparent to IR radiation.

So you see no signal at all?

No signal is observed.

And that is a major, major trap for students.

Just because you don't see a CCC signal does not mean a double bond isn't there.

It might just be symmetrical.

OK, one final point on intensity.

The CH signals, the ones just below 3000, they're often really strong.

But a single CH bond isn't all that polar, so why are they so intense?

That's a great question.

It's really just a numbers game.

We call it the multiplier effect.

That signal doesn't represent one bond stretching.

It represents the stretching of many Csp3H bonds all over the molecule.

And when you add up the absorption from every single one of those protons, the combined effect is usually strong enough to create a pretty deep signal in that 2850 to 3000 range.

We have wave number for location and intensity for strength.

Now we get to the third characteristic, shape.

This is where we look at whether a signal is broad or narrow, and this can be an instinct drive away for some of the most important functional groups.

Shape is almost always about intermolecular forces, and specifically hydrogen bonding.

This is how you spot alcohols and carboxylic acids immediately.

Let's start with the alcohol -OH bond.

Okay, so in a normal lab sample, all the alcohol molecules are bumping into each other.

They're interacting.

The hydrogen from one OH group is attracted to the oxygen on a neighbor.

And that interaction, that hydrogen bond,

actually weakens the OH bond itself.

And the key part of the logic here is that this weakening isn't uniform across the whole sample?

Exactly.

At any given instant, if you could take a snapshot of all the molecules in your sample, some of those OH bonds would be really strongly hydrogen bonded, making them quite weak.

Others would be only weakly bonded, and a few might be totally free, not bonded at all.

So you get a whole distribution of bond strengths.

Whole distribution.

And since a range of bond strengths requires a range of vibrational energies to excite them, the absorption doesn't happen at one sharp point.

It gets smeared out across a range of frequencies.

And that gives us the characteristic broad signal for the OH stretch.

That's it.

That broad, smooth curve you see sweeping across the region from about 3200 to 3600, that is the signature of hydrogen bonding in an alcohol.

Now, the source material points out you can actually manipulate this.

What happens if you take that same alcohol and you dilute it way, way down in a solvent that doesn't hydrogen bond?

If you do that, the alcohol molecules get isolated from each other.

They can't find a partner to hydrogen bond with.

So now all the OH bonds are free OH bonds.

So they're all uniform in strength.

They're all uniform.

And so they all absorb at a single narrow frequency, right around 3600.

Sometimes you'll actually see both in a real spectrum if the concentration is just right.

A sharp little spike for the free ones sitting on top of the big broad hump for the H bonded ones.

Now let's go to carboxylic acids, where this broadening effect just goes into overdrive.

It really does.

Carboxylic acids don't just do casual H bonding.

They can form this really stable cyclic dimer, where two molecules are locked together by two very strong hydrogen bonds.

So the interaction is just way stronger than in an alcohol.

Much stronger and much more consistent across the sample.

And the result is a signal that is just massive and unmistakable.

Yes, the OH signal in a carboxylic acid is extremely broad.

It often starts up near 3600 and stretches all the way down to 2200.

It's this giant hazy cloud that literally covers up the normal CH signals around 2900.

It looks like a big hairy beard sometimes.

If you see that feature, it is a guaranteed carboxylic acid.

And this is important.

That giant broad OH signal must always come with the strong CO signal just above 1700.

Absolutely.

It's a two -part signature.

If one of them is missing, it's not a carboxylic acid.

So hydrogen bonding explains broad versus narrow.

But shape can also be about the number of peaks, which comes from the molecule's internal geometry.

This is the case for amines.

This is a great example.

You can use this to tell apart primary and secondary amines.

A secondary amine, which is R2NH, has only one NH bond.

And it gives you one single sharp signal between about 3350 and 3500.

But a primary amine, RNH2, has two NH bonds.

And it produces two sharp signals.

It looks like a little fork or a pair of bunny ears.

Right.

And the reason for the two peaks isn't that each NH bond is stretching on its own.

The whole NH2 group acts as one unit.

What happens is half the molecules are undergoing symmetric stretching, where both bonds stretch and phase at the same time.

The other half are doing asymmetric stretching, where one stretches in while the other stretches out.

And because those two different motions require slightly different amounts of energy, they absorb at two different sharp frequencies.

And that's why you get the fork.

So if you see two sharp peaks up there, it's the primary amine.

One sharp peak, it's a secondary.

It's a very reliable tell.

We've covered the physics, the data, and the three key characteristics.

Wave number, intensity, and shape.

Now let's put it all together.

How do we attack an unknown spectrum strategically?

Okay, the analysis has to be systematic.

And the first step is to just manage the clutter.

You start by drawing a line at 1500.

Just ignore everything to the right of that.

That's the fingerprint region.

Right.

We stay in the diagnostic region.

And we target those three major zones, going from high energy to low energy.

That's right.

You check for signals in the bonds to H region, which is 2700 to 4000.

You check the triple bond region, 2100 to 2300.

And you check the double bond region, 1600 to 1850.

And for every single signal you find, you have to analyze all three things.

Location, intensity, and shape.

And within that bonds to H region, the very next step is to draw that line at 3000.

That's your unsaturation alert.

You see signals above it.

You've got CSP2H or CSPH.

You see a broad signal.

You've got an OH.

You see that twin fork.

You've got an NH.

It immediately tells you about alcohols, amines, carboxylic acids.

Now you mentioned a classic trap earlier, this little artifact that can trick students into thinking they have an alcohol when they really don't.

Yes, the overtone.

I remember getting caught by this one myself years ago.

It usually shows up as this small kind of insignificant little bump, often between 3400 and 3500.

And it's especially common in compounds that have a really strong CO signal.

So what is an overtone exactly?

Think of it like a harmonic in music.

When a bond vibrates at its main fundamental frequency, let's say the CO frequency, it can sometimes also produce a very faint signal at exactly twice that frequency.

So CO is around 1720.

Twice that is 3440.

Right, which puts it smack in the middle of the OH zone.

So if you see a tiny weak little bump around 3400, and you also have a giant CO signal, chances are it's just the overtone, it's not an alcohol.

You need that classic broad strong shape to confirm an alcohol.

That's a great summary of the strategy.

Let's put it into practice with that complex example from the source.

Identifying a compound with the formula C6H10O.

This is a perfect exercise.

We have to use that formula first.

For C6H10O, we can calculate the degrees of unsaturation, the DOU.

Right, a saturated 6 -carbon alkane would be C6H14, so we're missing four hydrogens.

Missing four hydrogens means two pairs, which means a DOU of two.

Okay, so a DOU of two means the molecule must have some combination that adds up to two, like a ring and a double bond or two double bonds or a triple bond.

Two elements of unsaturation, now we go to the spectrum.

Step one, check the triple bond region, 2100, 2300.

And looking at the spectrum, there's nothing there, it's completely flat.

So that rules out triple bonds.

Our two elements of unsaturation must be coming from double bonds or rings.

Step two, check the double bond region, 1600, 1850.

Here we see two very distinct signals.

There's a weak, narrow one at 1650 and then a really strong, intense one at 1720.

The intensity is everything here.

The weak signal at 1650 is a perfect match for a CKC double bond.

The strong, intense signal at 1720 is the unmistakable signature of a CO carbonyl group.

So that's it.

That accounts for both our degrees of unsaturation, one CC and one CO.

Exactly.

Now step three, check the bonds to H region, 2700, 4000.

We draw our 3000 line.

We see the expected strong CC3H signals just below 3000.

But crucially, there's also a small signal just above 3000.

And that signal above the line confirms we have a vanillic CSSP2H bond, a hydrogen that's attached directly to that CC double bond.

Right.

And we also look for other shapes.

No broad signals, no fork shapes, so we can immediately rule out alcohols, carboxylic acids, and amines.

OK, so let's put it all together.

We have a CC bond.

We have a CO bond.

And the CC has at least one hydrogen on it.

This leaves us with two possible structures that fit the C6H10O formula, one where the CC and CO were right next to each other, so they're conjugated.

And one where they're separated by an SP3 carbon, so they're non -conjugated.

This is the final step.

How do we tell them apart?

We go back to our wave number logic, specifically that resonance effect.

We said that conjugation weakens the CO bond and shifts its signal down to around 1680.

A non -conjugated ketone, on the other hand, should absorb at the normal spot around 1720.

And our observed CO signal is at 1720, the higher unshifted location.

So we can conclude that the compound must be the non -conjugated structure.

That's our answer.

If the molecule had been conjugated, that 1720 peak just wouldn't be there.

The signal would be lower.

It's a fantastic example of how a subtle shift in wave number can let you distinguish between two very similar isomers.

So before we wrap up, let's just quickly run through the absolute essential patterns, the things that should just jump out at you from a spectrum.

Sure, the unassailable signs.

The OH of an alcohol is defined by its broad shape from 3200 to 3600.

The OH of a carboxylic acid is that very broad, massive cloud from 2200 to 3600, plus the CO signal.

The NH of a primary amine is those two sharp signals, the fork, up around 3350 to 3500.

The CSPH of a terminal alkan is that sharp signal way up around 3300.

And finally, the CO signal is always very strong and located between 1700 and 1750, unless conjugation pulls a little lower, towards 1680.

That pattern recognition really is what makes learning IR so satisfying.

It is.

So if you're a student and you're focused on synthesis, what does mastering IR really mean for you?

It means efficiency.

It means confidence in the lab.

IR is the chemist's first check.

It's fast, you don't need much sample, and it confirms the big picture.

It lets you know if your reaction fundamentally worked before you spent hours on purification or running a much more complex analysis like NMR.

Right, so if your goal was to oxidize an alcohol to a ketone, the first thing you do is run an IR.

If you see that broad OH signal is gone, and this new, sharp, strong CO signal has appeared at 1720.

You have very high confidence that your functional group transformation was a success.

It's that vital confirmation.

It's a molecular fingerprint that validates all the chemical intuition you've been building up.

As we move forward in this topic of structure determination,

there is one last provocative thought that builds directly on what we just talked about with resonance.

Okay, lay it on me.

We established that when you have a conjugated system, the resonance weakens the CO bond, which lowers its vibrational energy, and its signal shifts down from 1720 to 1680.

But what's happening to the other bond in that system?

What about the neighboring CC bond?

Does the delocalization of electrons weaken the CC bond as well, shifting its absorption to a lower energy?

Or does energy conservation mean that something has to compensate, and maybe it absorbs at a higher wave number?

That is a great question.

That forces you to think about how electron distribution affects the entire molecule, not just one isolated bond.

It's moving from functional groups to system -wide electronic effects.

Exactly.

Something for you, the learner, to noodle on as you're looking at those spectrum tables.

That was the deep dive into the structure and utility of infrared spectroscopy.

Thanks for joining us.

We really hope this helps make sense of this spectroscoping puzzle.

We'll catch you on the next deep dive.