Chapter 10: Nucleophilic Substitution at the Carbonyl Group

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Welcome learners to another deep dive.

Today we're embarking on a fascinating journey into the heart of organic chemistry.

We are indeed.

We're pulling back the curtain on a really foundational topic, nucleophilic substitution at the carbonyl group.

That's right.

If you've felt the pull of the carbonyl's reactivity and aldehydes and ketones, well prepare to see it in a whole new light today.

Absolutely.

We're diving deep into chapter 10 of Playden, Greaves, and Warren's organic chemistry second edition.

Great chapter.

Yep.

Our mission really for this deep dive is to unravel the intricate mechanistic reasoning behind these reactions.

Okay.

Explore a whole bunch of pathways for functional group transformations and even look strategically at how we can forge new carbon bonds.

Right.

You'll gain hopefully a crystal clear understanding of why these reactions are so crucial in synthesis, all while, you know, keeping the theory solid.

No oversimplifying.

Okay.

Let's unpack this then.

We've touched on simple nucleophilic additions before, right?

Carbonyls, green yards, organolithiums in earlier chapters.

Chapter six and nine, yeah.

But this chapter, chapter 10, introduces a critical difference, the initial addition step.

It's followed by other steps.

Exactly.

Meaning the overall reaction isn't just addition,

it's

substitution.

Can you elaborate on what makes this different?

That's the absolute core of it.

We're swapping one group attached to the carbonyl for another.

It's not just adding across the CO bond and staying there.

Okay.

And well, at the heart of pretty much all these transformations is a key intermediate and maybe even more importantly, a critical concept,

the leaving group.

Ah, the leaving group.

So that concept becomes central here.

What exactly defines a good leaving group and how does that tell us if we get addition or substitution?

Well, the key lies in what happens after the nucleophile attacks the carbonyl.

Think about, say, an alcohol adding to a ketone.

Right.

Forming a hemiacetal.

Exactly.

Chapter six stuff.

We said that was often reversible, right?

Especially if it wasn't cyclic.

And why?

Because the alkoxide, the RO, could easily get kicked back out.

It can leave.

Now, apply that to a situation where the starting material already has a potential leaving group.

Like an ester reacting with the Grignard.

Right.

The Grignard attacks, you form this negatively charged intermediate.

The tetrahedral one.

The tetrahedral intermediate, yes.

But now, instead of just getting protonated, it can collapse.

You can reform the stable CO double bond by kicking out the RO group from the original ester.

Ah, so that's a substitution.

The RO group is swapped for the Grignard's alkyl group.

Precisely.

And the defining feature of a good leaving group is its ability to depart as a stable, relatively unreactive anion.

Think Cl, acetate, even alkoxide sometimes.

That's fascinating.

So this tetrahedral intermediate you mentioned, it seems to be the absolute linchpin for understanding all of this.

What makes it so crucial?

It absolutely is the linchpin.

You have a trigonal planar as P2 carbonyl carbon, right?

Yeah.

The nucleophile attacks it and bang, it becomes tetrahedral C3.

Yeah.

This intermediate is formed in all these substitutions, its formation, and crucially, its collapse.

That's the whole mechanistic story.

And this intermediate isn't always stable, you said.

We've seen, I think, how adding a Grignard to an ester,

it doesn't stop at the ketone usually.

That's a perfect example.

Exactly.

You add one Grignard, the tetrahedral intermediate forms, it collapses, kicks at the RO, and you get a ketone.

Right.

But the ketone is also reactive to the Grignard, often more reactive.

So a second Grignard attacks, forms another tetrahedral intermediate, which then gets protonated in the workup to give a tertiary alcohol.

So the stability of that first intermediate and what it does next dictates everything.

Precisely.

And here's your core predictive tool.

The intermediate will collapse by expelling the best leaving group, the group that forms the most stable anion when it leaves.

Okay.

So comparing, say, chloride, Cl, ethoxide, Eto, or I don't know, a methyl anion, I mean, chloride is an excellent leaving group.

Why?

HCl is a strong acid, so Cl is a stable, happy anion.

Makes sense.

Methyl anion, forget it.

Super unstable, very strong base.

Ethoxide, it's okay, but it's still a pretty strong base.

So Cl is the best leaving group of those three.

Master that principle stability of the departing anion and you're way ahead.

And how do we know this intermediate actually exists?

You can't just, like, put it in a bottle.

That is an excellent question.

And it really highlights the cleverness of early mechanistic chemists.

The definitive proof came way back in 1951 from Myron Bender.

Brilliant work.

What did he do?

Used oxygen 18 labeling, isotope labeling.

He took carboxylic acid derivatives, like esters, labeled with 18O in the group,

and reacted them with normal water, H2 -16O.

Okay.

Now, if it was just a direct swap, a simple SN2 type thing, which it isn't, but if it were, the product alcohol or acid would get one 18O.

Right.

But what he found is really cool.

If he used insufficient water, so some starting material was left over, that leftover starting material had actually lost some of its 18O label.

Well, wait, the label got washed out of the starting material.

Exactly.

That was the killer evidence.

The only way that can happen is if there's an intermediate.

The tetrahedral one?

The tetrahedral intermediate, where both the original 18O from the carbonyl and the 16O from the water are attached to the same carbon.

And if there are protons around, they can rapidly swap between those oxygens.

So when the intermediate collapses back to the starting ester, sometimes the 16O leaves and sometimes the original 18O leaves.

So the 18O gets exchanged for 16O from the water via that intermediate.

Precisely.

This addition elimination mechanism, as it's called, isn't just accepted now.

It's fundamental.

It was a triumph of experimental design, revealing this fleeting, invisible intermediate.

Gave chemists a powerful lens.

Okay.

So that mechanism is solid.

Nucleophile attacks, tetrahedral intermediate forms, best -leaving group leaves.

Now let's meet the family of these carboxylic acid derivatives.

You've got RCOX, acid chlorides, esters and hydrides, amides.

Each has a different X group that changes its reactivity, right?

Absolutely.

That X group is everything.

You've got acyl chlorides, XEL, acid anhydrides, exosubture, esters and amides, X and R2.

Their reactivity varies hugely.

For instance, making esters.

Reacting alcohols with acid chlorides or anhydrides is bread and butter chemistry.

Acetyl chloride plus cyclohexanol.

Needs a base, right?

Like pyridine.

Usually, yes.

Pyridine mops up the HCl produced.

You get cyclohexyl acetate.

Same idea with acetic anhydride reacting with an alcohol.

Gives you the ester and acetic acid.

And the mechanism is that same two -step dance we just talked about.

Alcohol attacks, tetrahedral intermediate forms, collapses, kicks out Cl or acetate.

That's the core pathway, yes.

And pyridine, as you said, acts as a base to deprotonate the alcohol after it's added, or maybe the intermediate.

Okay.

But pyridine can be even cleverer.

It's actually a nucleophile itself.

It can attack the SCl first to form a highly reactive acyl pyridinium ion.

That then gets attacked by the alcohol, which is often faster than the direct reaction.

Pyridine acts as a nucleophilic catalyst.

Wow.

Okay.

So it speeds things up by changing the mechanism slightly.

Exactly.

Neat trick.

What about making amides?

Do acyl chlorides react with amines too?

Oh, absolutely.

Very readily.

The mechanism is very similar to ester formation.

The amine, being a good nucleophile, attacks the SCl.

A second molecule of the amine then acts as a base to pull a proton off the nitrogen in the intermediate.

Got it.

And then the intermediate collapses, kicking out the chloride ion.

Gives you the amide and an ammonium salt.

Is that the Schott and Bellman thing?

The Schott and Bellman synthesis is a practical way to do this.

Yeah.

Yeah.

Often uses aqueous sodium hydroxide as the base in a two -phase system.

The base stays in the water layer, neutralizes the HCl,

and the organic reactants stay in the organic layer.

Very useful.

So with all these options, Cl, Ocr, or R, Nr2 leaving,

what's the best way to predict which one will leave?

Is there a simple guide?

There is a really useful guide, yeah.

It comes back to acid strength.

Think about the conjugate acid Hx of the potential leaving group X.

The lower the pKa of Hx, meaning the stronger the

better X is as a leaving group.

Ah, right.

Because a stable anion comes from a strong acid.

Exactly.

So HCl, pKa, way down below zero, strong acid.

So Cl is an excellent leaving group.

Makes sense.

Acetic acid, pKa, around 4 .8.

So acetate, ACO, is a decent leaving group.

Ethanol, pKa, around 16.

So ethoxide, ETO is,

okay, but getting worse.

Ammonia, NH3, pKa, around 35, 40 maybe.

Very weak acid.

So NH2 is a terrible leaving group.

Terrible leaving group.

That pKa compass is invaluable.

So is it generally true then that good nucleophiles make bad leaving groups and the other way around?

This is a very good rule of thumb, yes.

Generally, the stronger the base X is, higher pKa of Hx, the better it is as a nucleophile towards the carbonyl.

And the weaker the base X is, lower pKa of Hx, the better it is as a leaving group.

Okay.

Andions are usually better nucleophiles than neutral compounds too.

This whole picture explains why, for example, you can react an amine with an ester to make an amide.

Because the amamine is a decent nucleophile and the OR group from the ester is a decent leaving group.

Right.

But trying to react an alcohol with an amide to make an ester, much harder.

Right.

The NR2 group is a terrible leaving group.

You usually can't just swap them like that.

Okay.

That clarifies a lot.

It sounds like there's a definite

pecking order for reactivity among these derivatives then.

There is.

A very clear hierarchy.

Acylchlorides are the undisputed kings of reactivity.

Super electrophilic.

Right.

Then acid anhydrides, then esters, and at the bottom, the least reactive are the amides.

And why are amides so much less reactive?

It's all about resonance or delocalization.

The nitrogen atom in an amide has a lone pair of electrons.

And that lone pair can be delocalized into the carbonyl orbital.

Yeah.

You can draw a resonance structure where there's a double bond between carbon and nitrogen and a negative charge on the oxygen.

Right.

I remember that.

That delocalization does two things.

It stabilizes the amide molecule itself, making it less keen to react, and it puts more electron density towards the carbonyl group, making the carbonyl carbon less electrophilic, less attracted to nucleophiles.

And you can actually see that difference.

You can.

In infrared spectroscopy, the CO stretching frequency tells you about the bond strength and electron density.

Acylchlorides, highly electrophilic, have a CO stretch way up around 1 ,850 centimeters one.

Their CO stretch is much lower, typically around 1 ,650, 1 ,690 centimeters one.

That lower frequency directly reflects the weaker, less polarized CO bond due to that nitrogen lone pair donation.

It's concrete evidence.

That's really cool.

So acylchlorides are reactive because chlorine pulls electrons away, making the carbonyl carbon very positive.

Exactly.

Inductive withdrawal by the chlorine wins out.

Okay.

What about carboxylic acids themselves?

They have an HOH group.

Can they undergo substitution?

Good question.

Under basic conditions, generally no.

If you add an NAMEN, say it just acts as a base, deprotonates the acid.

Forms the carboxylate salt, RCOO.

Right.

And that carboxylate anion is negatively charged, has resonance stabilization.

It's very unreactive towards nucleophiles.

Dead end.

But there must be a way.

There is.

Acid catalysis.

This is where acid steps in and performs some, well, not magic, but very clever chemistry.

Ah, okay.

What does the acid do?

Two crucial things.

First, it protonates the carbonyl oxygen.

That makes it R -C -O -H -O -H plus set.

Sort of, yeah.

That positive charge makes the carbonyl carbon much more electrophilic.

Suddenly, even weak nucleophiles like alcohols can attack it effectively.

Okay.

That's step one.

More reactive carbonyl.

What's step two?

Step two is fixing the leaving group problem.

The OH group in a carboxylic acid is, like we saw with NH2, a terrible leaving group.

Hydroxide, HO, is a strong base.

But an acid, when that OH group gets protonated after the nucleophile attacks, it becomes and HOH2 plus water.

And water is a great leaving group.

Excellent leaving group.

Stable neutral molecule.

So the acid catalyst activates a carbonyl and turns a bad leaving group into a good one.

That's the secret behind Fischer esterification, reacting a carboxylic acid with an alcohol, using an acid catalyst to make an ester.

And Fischer esterification, that's an equilibrium, isn't it?

All the steps are reversible.

They are, indeed.

Which means you need to push it towards the products.

How do you do that?

Le Chatelier's principle.

Use a large excess of the alcohol, for instance.

Or, more commonly, remove the water as it's formed, maybe by using a Dean -Stark trap or drying agents.

Shift the equilibrium.

Got it.

And this applies to reversing it too, like acid -catalyzed ester hydrolysis.

Exactly.

Add lots of water and acid, you drive it back to the carboxylic acid and alcohol.

Same principle applies to transesterification, swapping one alcohol group on an ester for another.

Maybe you distill off the lower boiling alcohol product to drive it.

Okay.

So we can make esters, we can break them with acid.

What about breaking them with base?

Suponification.

Ah, suponification.

Base -catalyzed ester hydrolysis.

This one is different because it's irreversible.

Why irreversible?

The hydroxide ion, OH, is a strong nucleophile.

It attacks the ester carbonyl, forms the tetrahedral intermediate.

That collapses, kicks out the alkoxide, RO.

But now you've formed a carboxylic acid, RCOOH.

Right.

And you're in strong base.

That carboxylic acid is immediately deprotonated by the hydroxide or the alkoxide.

To form the carboxylate salt, RCOOH.

Exactly.

And that carboxylate anion, as we said, is resonance stabilized and unreactive towards nucleophiles.

It won't react back with the alcohol.

The final deprotonation step pulls the whole equilibrium over, making it irreversible.

And that's literally soap making.

Pretty much.

Suponification of fats and oils, which are tracers of glycerol with strong base like sodium hydroxide, gives you glycerol and the sodium salt to the fatty acids.

That's soap.

Bender used 18 -O labeling here, too, by the way, confirming OH attacks the carbonyl, not replaces the OR group directly.

Fascinating.

Okay, what about the stubborn ones?

Amides.

The least reactive.

How do you hydrolyze those?

They definitely need a bigger push.

Vigorous conditions.

Either strong acid in heat or strong base in heat.

How does strong acid work?

Similar logic to a sterification.

The acid protonates the amide oxygen.

Amide is usually protonate on oxygen, not nitrogen, because of resonance making the carbonyl more electrophilic for water to attack.

Okay.

Then the leaving group, the amide, R2NH, gets protonated as it leaves, forming R2NH2 plus gri.

That prevents it from reacting back.

And strong base.

Again, hydroxide is a strong nucleophile.

It attacks the amide carbonyl.

It requires high temperatures.

But eventually the tetrahedral intermediate collapses, kicking out the amide anion R2N, which is a very strong base.

But like saponification,

the carboxylic acid formed is immediately deprotonated to the carboxylate salt, driving the reaction forward.

Got it.

Tough, but doable.

What about nitriles, RCN?

Are they related to this family?

They are very closely.

You can think of a nitrile as a sort of dehydrated primary amide R2NH2 minus H2O.

And they hydrolyze similarly.

Under acidic or basic conditions, water adds across the CN triple bond, first forming a primary amide intermediate.

And then that primary amide hydrolyzes further, just like we discussed, to give the carboxylic acid an ammonia or an ammonium salt.

So you can get carboxylic acids from nitriles too.

Absolutely.

It's a common route.

For example, you can make mandelic acid, which is found in almonds from benzaldehyde.

You react benzaldehyde with HCN to make a and then hydrolyze the nitrile group to a carboxylic acid.

Cool example.

Okay.

We've talked a lot about going down the reactivity ladder from acyl chloride to ester to amide.

How do you climb up?

How do you make, say, an acyl chloride from a carboxylic acid?

You need to replace that bad -leaving group OH.

Excellent question.

Because that's often the crucial first step to access all the other derivatives.

You need special regions for that.

Thinochloride SOCl2 is a classic.

Phosphorus pentachloride PCL5 is another.

What do they do?

Their job is to convert that awful hydroxyl leaving group into something much better.

Take cyanochloride.

The carboxylic acid oxygen attacks the sulfur atom.

Okay.

There's a rearrangement, and essentially the oxygen gets incorporated into a highly unstable intermediate that wants to fall apart.

It collapses, kicks out chloride ion, which can then attack the carbonyl.

But importantly, it also expels sulfur dioxide, SO2, and HCl gas.

Gases.

So they leave the reaction mixture.

Exactly.

Driving the reaction irreversibly, according to Louis Chatelier.

The key is transforming that OH into something that can leave easily attached to the sulfur or phosphorus atom.

So once you've made the acyl chloride from the acid,

you've reached the summit.

You've reached the summit, the reactivity ladder.

From the acyl chloride, you have the power.

You can react it with alcohols to get esters, amines to get amines, even carboxylate salts to get anhydrides.

It gives you amazing synthetic flexibility.

That's a really powerful concept.

Okay.

Let's switch gears slightly.

We've talked about interconverting derivatives.

What if the goal is different?

What if we want to make, say, a ketone from an ester using a Grignard reagent?

You mentioned earlier that usually overshoots.

Right.

That's the classic problem.

The Grignard adds, the intermediate collapses, you get the ketone.

But the ketone is more reactive than the ester you started with.

So as soon as that ketone forms, boom, a secondary Grignard molecule attacks it, leading eventually to the tertiary alcohol after you add water or acid in the workup.

You can't easily stop it at the ketone stage.

Same problem with reducing esters with strong reducing agents like LiOH4.

Same problem.

LiOH4 reduces the ester to an aldehyde intermediate, but the aldehyde is more reactive than the ester, so it immediately gets reduced further to the primary alcohol.

You go all the So how do chemists get around this overshooting and actually stop at the ketone or aldehyde stage?

There are two main strategies, really clever ones.

Strategy one, make the starting material way more reactive than the products.

Strategy two, make the intermediate stable so it doesn't collapse until you tell it to.

Okay, tell me about strategy one, making the starting material more reactive.

Well, instead of an ester, start with an acyl chloride.

Super reactive, remember.

Right.

You can use a less reactive organometallic reagent like an organocuprit, often called a Gilman reagent, like lithium dimethylcuprit, B2 -q -lay.

These cuprates are softer, less reactive nucleophiles than Grignard's or organolithiums.

They're reactive enough to attack the highly electrophilic acyl chloride to form the ketone, but they're not reactive enough to attack the ketone product itself, so the reaction stops cleanly at the ketone stage.

Nice control.

Clever.

Using a less reactive nucleophile with a more reactive electrophile.

What about the other strategy, the stable intermediate one?

That sounds intriguing.

It really is elegant chemistry.

The idea is to form that tetrahedral intermediate, but trap it somehow, prevent it from collapsing to the ketone until you're ready.

Then at the end you add acid, the acid quench.

What does the acid do?

The acid does two things.

It protonates the intermediate, triggering its collapse to the ketone, and it simultaneously destroys any leftover organometallic reagent, preventing it from attacking the newly formed ketone.

Okay, how do you trap the intermediate, like with lithium carboxylates?

You mentioned those earlier.

Yes, that's one way.

Carboxylates salts themselves are pretty unreactive electrophiles, but if you hit them with a really powerful nucleophile, like an organolithium region, not usually Grignard's here, it can add.

You form a two negative charges on oxygen.

That thing is surprisingly stable because there's no good leaving group, just O2.

It just sits there.

Until you add acid.

Until you add aqueous acid at the end.

Prognation, collapse, ketone forms, leftover organolithium is quenched.

Bingo.

Neat.

And the wine -rub amides.

Are they famous for this?

The wine -rub amides.

N -methoxy -N -methyl amides.

Absolutely brilliant design by Steven Weinreb.

They're probably the gold standard for this controlled addition.

What makes them special?

When a Grignard or an organolithium attacks a wine -rub amide, the tetrahedral intermediate formed is stabilized by chelation.

The lithium or magnesium ion coordinates to both the oxygen that just formed and the methoxy oxygen.

Like a little claw holding it together.

Exactly like a claw.

That chelation makes the intermediate stable and stops it from collapsing prematurely.

It also makes it unreactive towards a second equivalent of the organometallic region.

So it just waits?

It just waits patiently until you add dilute acid.

The acid breaks up the tail, protonates it, it collapses, and out pops your ketone.

Nice and clean.

That's incredibly elegant.

It is.

And it's so versatile.

You can even make aldehydes this way if you start with the wine -rub amide of formic acid or just use N -N -dimethylformamide DMF with an organolithium followed by quench.

Wow.

And nitriles can do this too.

Stop the reaction part way.

Yes.

Nitriles offer another route.

When an organometallic adds to a nitrile, RCN, it forms an iminanian intermediate RCN.

That iminanian is not very electrophilic.

It's stable enough under the reaction conditions.

It just sits there until you add aqueous acid.

And the acid?

The acid first protonates the nitrogen, giving it an iminium ion, and then water adds, and through a series of steps, it hydrolyzes completely to the ketone.

Again, the ketone is only formed during the acidic workup, safely away from the organometallic region.

So when you boil it all down,

all these reactions making esters for perfumes, amides for polymers like nylon, synthesizing complex drug molecules, they all hinge on this same basic theme.

Pretty much.

Nucleophilic attack on a carbonyl, formation of that tetrahedral intermediate, and then its fate usually collapse by kicking out the best leaving group.

It's all guided by these fundamental principles of leaving groupability, nucleophilicity, and intermediate stability.

It's amazing how understanding those core ideas lets chemists design such precise synthetic routes.

It really is.

It's about mastering that subtle dance of electrons, understanding stability, predicting which bonds will form and break.

You choose your reagents, your conditions, to gently persuade the molecules to do exactly what you want, stopping at the right place, like with the wine -reb amides.

It connects right back to Bender's work, figuring out those invisible steps through clever experiments.

It really shows how a few powerful principles can explain so much complexity in organic chemistry.

It's quite beautiful, actually.

Indeed it is.

And maybe that raises a question for you, our listener, to think about.

As you reflect on these mechanisms, the attack, the intermediate, the leaving group,

how might really understanding that interplay of electrophilicity, nucleophilicity, stability, how might that inform your approach if you were designing a synthesis?

Maybe for functional groups we haven't even discussed today.

Something to mull over.

Definitely something to think about.

Well, we hope you enjoyed this deep dive into the fascinating world of carbonyl substitution chemistry.

Thank you as always for being part of the Last Minute Lecture family.