Chapter 6: Ketones & Aldehydes: Nucleophilic Addition

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Okay, let's unpack this.

Welcome to the deep dive.

The shortcut to fluency in, well, the very complex landscape of organic chemistry.

Today, we are undertaking a deep dive into what I think is really the heart of synthesis, the world of ketones and aldehydes.

Oh, absolutely.

These aren't just molecules.

They are, I mean, they're the workhorses of the synthetic laboratory.

If you are learning how to solve multi -step synthesis problems, the reactions in this area, chapter six,

are just, they're essential.

They're almost all about forming new carbon bonds or interconverting major functional groups.

And our mission today, really, is to give you a rigorous step -by -step roadmap that's built on cause and effect logic.

We want to take what looks like this mountain of memorization and turn it into a logical flow chart.

And the genus of this topic and how it's organized is that nearly every reaction revolves around that central carbonyl carbon.

We can categorize every single reaction based on what kind of nucleophile attacks it.

It is the ultimate exercise in recognition.

Once you understand the fundamental electronic nature of the carbonyl group and the two major guiding rules that follow, you can predict the outcome of almost any reaction we discussed today.

From reductions to creating new rings.

Exactly.

From reductions to the creation of these new complex cyclic structures, it all comes back to the same core principles.

So before we can start reacting these molecules, we have to know how to create them.

Of course.

Section 6 .1 preparation is, it feels absolutely foundational, especially when we start thinking about working backwards in retrosynthesis problems.

Indeed.

Yeah, when a chemist looks at an alcohol and a ketone, the most powerful and common interconversion is using oxidation to create that C double bondo.

The carbonyl.

The carbonyl group from an alcohol.

This is just a crucial starting point.

And the rules for alcohol oxidation are, they're really rigid, aren't they?

It's all based on substitution.

Completely.

A primary alcohol, which has the OH group on a carbon bonded to only one other carbon that oxidizes to form an aldehyde.

A secondary alcohol bonded to two other carbons that oxidizes to a ketone.

And that's usually where the oxidation stops for a secondary alcohol.

But the tertiary alcohol, the one bonded to three other carbons, that's the great exception, it absolutely refuses to react under standard oxidation conditions.

Why is that?

It seems like it should be able to.

Well, it comes down to basic valency rules in organic chemistry.

I mean, think about what oxidation is in this context.

You're forming a double bond between the carbon and the oxygen.

Exactly.

You fundamentally need to form that double bond.

So if that central carbon is already bonded to three other carbon atoms,

creating that fourth bond, the C double bondo, would require that carbon to temporarily form five bonds in the mechanism.

And that's the inviolable rule, carbon never forms five bonds.

Never.

So since there's no way to safely remove an attached carbon group or hydrogen atom during that initial step for a tertiary alcohol, the reaction is simply impossible.

It's a guaranteed no reaction result.

Which is actually a blessing in a way.

It's a huge blessing when you're designing selective synthesis steps.

This brings us to what's often called the Goldilocks problem of reagent selection.

Because when we oxidize alcohols, the choice of reagent is just critical, especially for primary alcohols.

Right.

If you're dealing with a secondary alcohol, life is easy.

It's really straightforward.

You can use strong oxidizing agents, the chromium -based heavy hitters.

Like sodium dichromate.

Sodium dichromate.

So that's Na2Cr2O7 with sulfuric acid in water, or the Jones regent, which is CrO3 in aqueous acetone.

These agents efficiently convert the secondary alcohol into the ketone you want.

And it stops there.

It stops there because the ketone has no remaining hydrogen on that carbonyl carbon, so it cannot be oxidized any further.

It's a dead end, which is great.

But the primary alcohol is the tricky one.

It's the problem child.

It really is.

If we use one of those same strong oxidizers on a primary alcohol, the oxidation,

it doesn't stop.

We hit our target, the aldehyde, but then we blow right past it.

That's because that intermediate aldehyde is highly, highly susceptible to further oxidation, especially into the strong, acidic conditions.

Water, which is present in the solution, adds across the C double bond O bond to form what's called a hydrate, or a geminal dial.

It's a carbon with two OH groups on it.

And that new OH group can be oxidized.

That hydrate intermediate now has a new alcohol functional group, which is readily oxidized by the chromic acid into a carboxylic acid.

So if your goal is the carboxylic acid, strong oxidation is fantastic.

But if your goal is the aldehyde, you need a way to pump the brakes.

You need to stop the process at that first stage.

And the solution is the Goldilocks agent,

pyridinium chlorochromate, or PCC.

This is a mild, non -aqueous oxidizing agent.

It's a complex of CRO3Cl with a protonated pyridine ring, and you usually dissolve it in an organic solvent like dichloromethane.

Because it avoids water and strong acid, it prevents the aldehyde from forming that oxidizable hydrate intermediate.

So PCC is the control knob.

It is.

It's the control.

It converts primary alcohols to aldehydes, and it can also convert secondary alcohols to ketones.

It's the versatile, gentle option.

The strategic insight here is, well, it's the PCC offers control.

If you have a primary alcohol and you absolutely must stop at the aldehyde, you choose PCC.

Okay, that makes sense.

And the other major preparation method we need to revisit is ozonolysis.

This is a functional group interchange that allows us to create ketones and aldehydes not from alcohols, but by breaking a carbon double bond, an alkene.

Ozonolysis is conceptually so simple, but it's incredibly powerful for breaking large molecules down into smaller pieces.

It is.

You use step one, ozone, O3, followed by step two, a reductive workup,

most commonly dimethyl sulfide or DMS.

And what that does is it performs a clean, almost surgical cut right across that CC double bond.

Every carbon that was part of the alkene is replaced by a carbonyl group.

If the original alkene carbon was connected to two alkyl groups, the result is a ketone.

And if it was connected to an alkyl group and a hydrogen.

The result is an aldehyde.

It's a direct translation of the alkene substitution pattern into the products.

So this reaction gives us a really powerful tool for analyzing unknown alkenes.

You can break them down into recognizable fragments and also for synthesis, right?

Absolutely, for synthesis.

It lets you generate specific carbonyl starting materials from cheaper, more readily available alkene precursors.

It's a staple of the toolkit.

Now that we have the starting materials, we get to the core question.

Why do they react the way they do?

Yes.

This next section contains the philosophical, the electronic underpinning of every single mechanism that follows.

We're focusing squarely on the carbonyl group, that C double bond O.

And we should be careful to distinguish this from the SL group, right?

That's a very good point.

The SL group is a carbonyl attached to an alkyl group, a term we reserve mostly for the next chapter on carboxylic acid derivatives.

Here, our focus is the C double bond O itself.

And all of the chemistry flows from two simple electronic effects, induction and resonance.

And they're completely complementary.

They point in the exact same direction and dictate the fate of the molecule.

Let's look at induction first.

Oxygen is one of the most electronegative elements on the periodic table, much more so than carbon.

And that huge disparity causes a powerful inductive effect.

Oxygen relentlessly pulls electron density away from the carbon atom through both the sigma bond and the pi bond.

Which means the oxygen ends up with a significant partial negative charge.

We call that delta minus.

Right.

And the carbon atom is left highly electron poor, bearing a partial positive charge, a delta plus.

And resonance doesn't just confirm this, it takes it to an extreme, doesn't it?

It really does.

If we draw the major resonance structure, we show the pi bond breaking entirely, and that electron pair is shifting completely onto the oxygen atom.

So you end up with a formal negative charge on the oxygen.

And, critically, a full positive charge on the carbon atom.

Now, even though this charged resonance structure is a minor contributor to the overall hybrid, the fact that we can even draw it demonstrates the carbon atom of the carbonyl group is fundamentally electrophilic.

It's always ready to accept an electron pair.

Always.

It's electron hungry.

So the central insight, the big takeaway, is that the carbonyl carbon is the primary site for nucleophilic attack.

It is the target.

And its geometry makes it an incredibly easy target.

That carbonyl carbon is B2 hybridized.

Which gives it a trigonal planar geometry.

It's flat.

It's flat.

And what's the practical implication of that flat geometry?

Well, there's not much getting in the way.

Minimal steric hindrance.

Exactly.

The incoming nucleophile doesn't have to navigate crowded alkyl groups to reach that carbon.

It can easily attack the plane from either the top face or the bottom face.

High electrophilicity combined with low steric hindrance translates directly into high kinetic reactivity.

So carbonyls react fast.

Very fast.

Okay, so kinetically they're reactive.

What about the thermodynamics?

What's the driving force?

That double bond, the C double bond -O bond, is exceptionally stable.

It is thermodynamically favored.

The core principle you have to internalize is that the formation of a carbonyl group is often the driving force for a reaction.

Its stability dictates the fate of the entire mechanism.

It does.

And this stability leads us to the two guiding rules.

The golden rules for all the carbonyl chemistry we're going to talk about.

Okay, what are they?

Rule one is simple.

A carbonyl group is easily attacked by a nucleophile.

That's a kinetic rule.

We've established that.

Rule two is a challenging one.

The one that requires thought.

After a carbonyl group is attacked, it will try to reform the double bond, if possible.

If possible.

That's the key phrase.

That's the million dollar phrase.

Yeah.

The rest of this deep dive is simply applying rule two, and especially its crucial caveat, to all the different kinds of nucleophiles.

Let's start with H -nucleophiles, or reducing agents.

These are reagents that deliver a hydride ion, H-, to that electrophilic carbonyl carbon.

Now let's go back to a fundamental question.

If we need H-, why can't we just use something like sodium hydride, Na?

This is such a classic point of confusion, and it forces us to distinguish between two really important concepts.

Bacicity and nucleophilicity.

They're not the same thing.

They're absolutely not the same thing.

H - from NaH is a phenomenal base.

Have a highly concentrated,

unstable negative charge.

So it will rip off any available proton it can find.

But it is a miserable nucleophile.

It has a negative charge.

Shouldn't it be attracted to that positive carbon?

It's about size and the ability to interact effectively.

It's a property chemists call polarizability.

Nucleophilicity relies on an atom's ability to sort of distort its electron cloud, to reach out and overlap with the electrophile's orbital to form a new bind.

Hydrogen is the smallest atom on the table.

Right.

It is fundamentally the least polarizable.

It cannot effectively share its electron density with the carbonyl carbon.

It prefers to just grab a free proton instead.

It's a much easier, faster reaction for it.

So we can't use free H-, we need specialized reagents.

These hydride delivery agents where the hydride is tethered but ready to launch.

Precisely.

And that brings us to sodium borohydride, NaH4, and lithium aluminum hydride, LiOH4.

The two big ones.

The two big ones.

NaH4 is the mild selective option.

Boron is relatively small, which makes the BH bonds less reactive and less able to sustain that negative charge.

Consequently, NaBH4 is highly selective.

It will only react with the most electrophilic carbonals.

Meaning ketones and aldehydes.

Meaning ketones and aldehydes.

And crucially, it leaves things like esters and carboxylic acid derivatives

completely untouched.

And LiOH4 is the sledgehammer.

It is the sledgehammer.

It's vastly more reactive.

Why the huge jump in power?

What's the difference?

Aluminum is directly below boron in the periodic table, which makes it a significantly larger atom.

Okay, so it's bigger.

And larger size means greater polarizability.

The aluminum hydride complex can distort its electron cloud much more effectively, making it a superior, highly energetic nucleophile.

This power allows it to reduce almost any functional group containing a carbonyl.

Including those less reactive esters and even carboxylic acids.

Everything.

So this distinction is absolutely critical for synthesis.

If you have a molecule with both a ketone and an ester, and you only want to touch the ketone.

You have to choose NABH4 for that selective reduction.

But if you need to reduce both, you bring out the big guns.

You choose LiOH4.

Absolutely.

Now for the mechanism, which is conceptually identical for both of them, the deliberate hydride attacks the electrophilic carbonyl carbon.

The C double bond O pi bond breaks, placing a negative charge on the oxygen atom.

And then we formed the tetrahedral intermediate,

an alkoxide ion.

An alkoxide.

And now we apply our golden rule number two.

Does the carbonyl reform?

Well, we have to look at what's attached.

We do.

We must recall the crucial caveat for ketones and aldehydes.

They are only attached to carbon groups, which would leave as C minus, or hydrogen, which would leave as H minus.

And the caveat states, never expel H minus or C minus.

Because they're incredibly unstable, highly basic ions, it would be energetically disastrous to just spit them out into solution.

Exactly.

So the alkoxide intermediate is trapped.

The reaction is, for all intents and purposes, stuck.

It cannot proceed any further on its own.

And this is why the crucial second step, the workup, is required.

Yes.

We have to introduce a source of protons, usually water or aqueous acid, to quench the reaction by protonating that negatively charged oxygen.

And the final product is an alcohol.

This is a reduction.

Ketones become secondary alcohols.

Aldehydes become primary alcohols.

And this brings us to a critical detail about notation.

And, well, frankly, lab safety.

For the super aggressive Lyle H4, you must write the reaction as two separate steps.

Step one, Lyle H4.

Step two, H2O.

Yes.

And that two -step notation reflects the reality that Lyle H4 is violently reactive with water.

It instantly produces highly flammable hydrogen gas.

You cannot mix them.

So you have to let the first step, the attack, finish completely before you add the water in step two.

You have to.

Conversely, the milder NABH4 is tame enough that it can tolerate a proton source in the solvent like methanol MeOH.

So for that reason, it's typically written as a single step.

Pay attention to those numbered steps on an exam.

They convey critical information.

OK.

Let's move to the O -nucleophiles, specifically alcohols, which lead to the formation of acetyls.

This is one of the most complex mechanisms we cover, I think.

It requires a really deep dive into the choreography of proton transfers.

It does.

And before we even start, we have to tackle the initial challenge.

An alcohol is a relatively weak nucleophile.

Right.

So if it attacks an unprotonated carbonyl, the resulting intermediate is just too unstable.

It immediately reverts back, expelling the alcohol to reform that highly stable C double bondo.

No net reaction occurs.

So if our nucleophile is weak, we have to employ the classic strategy.

Activate the electrophile.

Precisely.

We use a catalytic amount of a strong acid, like sulfuric acid, to protonate the ketone's oxygen atom first.

So you're making the carbonyl an even better target.

Much better.

By protonating that oxygen, we now force the carbon atom to carry a full, or at least a significant positive charge.

It's been transformed from a partially positive delta plus to a fully activated starving electrophile.

And now the weak alcohol nucleophile is strong enough to successfully attack this new, better target.

Yes.

That activation step is the key that unlocks the entire mechanism.

So let's walk through the seven -step journey, focusing on what bonds are forming and breaking.

All right.

So the mechanism begins with a weak ROH, attacking the highly activated protonated ketone.

This yields the first tetrahedral intermediate.

This molecule now contains a newly attached alkoxy group, the OR group, and a hydroxyl group, an OH.

And the molecule wants to reform a double bond, but we have two different oxygen atoms attached.

We need a good leaving group.

And this is where we encounter the second inviolable rule, this time for acidic chemistry.

Never expel HO minus hydroxide in acidic conditions.

Right, because hydroxide is a terrible leaving group, and it's far too basic to exist freely under these conditions.

It is.

And this forces this elaborate what's often called the proton transfer dance.

The proton shuffle.

The proton shuffle.

The purpose of these transfers is purely functional.

To move protons around until the poorer leaving group, OH, is converted into the excellent leaving group, water,

H2O.

So we first deprotonate the attached OR group, and then we protonate the OH group.

Yes, and that halfway point, before the water is expelled, is the hemiacetal.

This neutral intermediate now contains the excellent leaving group, H2O, all primed and ready to go.

So the molecule is now ready for the second critical event.

The lone pair electrons from the other attached oxygen atom push down to reform a carbon -oxygen double bond, and that simultaneously ejects the H2O.

This leads to a second, highly resonance -stabilized, caesonic intermediate.

And that cation is extremely electrophilic.

Extremely.

Which sets up the third critical event.

A second molecule of alcohol attacks the cation.

Remember, you need two moles of alcohol for every one mole of ketone.

And a final deprotonation step gives us the neutral product, the acetyl, the ROCOR linkage.

So the whole cycle is protonation, attack, proton shuffle, loss of water, second attack, and final deprotonation.

The complexity is entirely necessary to get around that poor leaving group problem.

Okay, now for the thermodynamics.

You mentioned the C double bondo is really stable.

It is.

Which means acetyl formation is generally uphill in energy.

The equilibrium usually strongly favors the starting materials, the ketone and alcohol.

So how do we force this unfavorable reaction to actually happen in the lab?

We exploit Le Chatelier's principle.

Since water is a product, if we constantly remove the water from the reaction mixture as it's formed, we shift the equilibrium to the right.

We force the formation of the acetyl.

And that's done using specialized equipment, right?

Like a Dean -Stark trap.

Exactly.

A Dean -Stark trap or azeotropic distillation.

When you see H2O written over the reaction arrow, it means the reaction is being actively dried to force it to completion.

This reversibility is key though, because the reverse reaction, acetyl hydrolysis, is favored just by drowning the mixture in excess water and catalytic acid.

That's right.

The excess water reverses the equilibrium, driving the molecule all the way back to the stable ketone.

And this ability to easily turn a ketone off and then turn it back on is the origin of its most important synthetic application.

The protecting group.

Let's talk strategy.

Imagine you have a complex molecule with both a ketone and an ester.

You need to use LiOH4 to reduce the ester to an alcohol, but you need the ketone to remain untouched.

We already know LiOH4 will attack both of them.

So we need a chemical shield.

We need to hide the ketone.

By reacting the molecule with a diol, like ethylene glycol, under acidic conditions with water removal, we form a highly stable non -reactive cyclic acetyl from the ketone.

And critically, esters do not react under these conditions.

So step one, protection.

The ketone is now temporarily deactivated.

It's inert to strong nucleophiles and bases, including our LiOH4.

Step two, the chemistry you actually want to perform.

You now introduce the LiOH4.

It will selectively reduce that ester group to an alcohol, leaving the cyclic acetyl completely untouched.

It doesn't even see it.

And then step three, deprotection.

You simply introduce aqueous acid, H2O, with an H plus catalyst to hydrolyze the acetyl back into the original ketone.

You've achieved a selective reduction on a polyfunctional molecule, all thanks to the clever use of the acetyl as a temporary chemical mask.

OK, moving down the periodic table from oxygen, we encounter sulfur nucleophiles.

And since sulfur is directly below oxygen, its chemistry is highly analogous, but with one unique, really powerful application.

So we treat a ketone with a diphyll,

specifically something like ethylene thioglycol, under catalytic acid.

And instead of an acetyl, we form a thioacetyl.

Right, with the S -S linkage.

The mechanism is identical to acetyl formation, just using sulfur atoms, which are generally even better nucleophiles than oxygen.

The result is a cyclic S -S analog of the cyclic acetyl.

But where this becomes strategically superior is that the thioacetyl intermediate can be used for a total reduction of the carbonyl group.

Correct.

Thioacetyls can be reduced completely to an alkane, an HH, on that carbon by treating them with rainy nickel.

Rainy nickel, what exactly is that?

It's just finely divided, highly porous nickel metal that's typically loaded with adsorbed hydrogen.

It acts as a powerful hydrogenation catalyst.

So the nickel breaks the carbon -sulfur bonds and replaces them with carbon -hydrogen bonds, a process called desulphurization.

Exactly.

And the net synthetic result is the conversion of that original C double bond O into a simple methylene group, a CH2.

This is a total reduction.

This is significant because it provides us with our second method for achieving that CHO to CH2 transformation.

We already saw the Clemson reduction, which needs really strongly acidic conditions.

Right.

And this rainy nickel reduction only requires catalytic acid for the initial thioacetyl formation.

And the desulphurization itself is neutral.

It's much gentler.

So if your target molecule has some base -sensitive functional group that would be destroyed by strong acid, like in the Clemson, you now have an alternative strategic pathway.

Synthesis is all about having options.

This gives you another tool.

Our next category is nitrogen nucleophiles amines.

And these reactions require us to distinguish sharply between primary and secondary amines because the end products are structurally very different.

Very different, yes.

A primary amine, RNH2, reacts with a ketone or aldehyde, again under catalytic acid and then removal of water, to form an inamine.

That's a C double bond N, often called a shift base.

OK, C double bond N.

And a secondary amine, R2NH.

A secondary amine reacts similarly to form an enamine.

That's a carbon -carbon double bond adjacent to the nitrogen atom.

NC double bond C.

So let's compare the endoming formation mechanism to the acetyl one.

We noted two key differences.

The first is the order of the initial steps.

For acetyls, we had to protonate the ketone first.

Right.

But here, with amamines, the reaction often starts with a nucleophilic attack first, followed by the proton transfers.

And this is a subtle but important detail that's governed by relative acidities by pKa values.

The catalytic acid we use is often a protonated amine, an ammonium ion, which has a pKa around 10.

That's just too weak an acid to efficiently protonate the ketone's oxygen.

It is.

However, the amine itself is a much stronger nucleophile than the neutral alcohol we use for acetyl formation.

Right.

So the amine is strong enough to attack the unprotonated ketone directly.

The resulting intermediate then undergoes the proton transfers, leading to a neutral analogous intermediate called the carbonyl amine.

Which is structurally similar to the hemiacetyl.

Very similar.

After the carbonyl amine is protonated and loses water, we get a highly charged intermediate.

And this leads to the second key difference, the final step.

With a primary amine, that charged intermediate still has a proton attached to the nitrogen atom.

So the mechanism concludes by a base coming in and removing that proton from the nitrogen.

This creates the final, stable, neutral amine.

But with a secondary amine, the initial steps are identical, leading to that same kind of positively charged intermediate after water leaves.

But that secondary amine has no hydrogen atom attached to the nitrogen to remove.

So the base has to look elsewhere.

It removes a proton from the adjacent alpha carbon atom.

This deprotonation simultaneously forms the new carbon -carbon double bond, yielding the final product, the enamine.

So the double bond is essentially forced onto the backbone of the molecule.

It has no other choice.

No.

And there's a note on regioselectivity for enamine formation.

If you start with an unsymmetrical ketone, where two different alpha carbons have protons to remove, the formation of the enamine surprisingly favors the product with the less substituted double bond.

That's counterintuitive.

It runs counter to general alkene stability rules, Zaitsev's rule, and it's a key exception to memorize for this reaction.

And this ability of these mechanisms to proceed through the loss of water is exploited by using special N nucleophiles for identification and further synthesis.

Yes, for instance, if the R group on the primary amine is a hydroxyl group, you're using hydroxyl amine and the product is an oxime, a C double bond NOH.

Or if R is another amine group, NH2, the regent is hydrazine.

And the product is a hydrozone, a C double bond NNH2.

And the hydrozone leads us to our third CO to CH2 method, the Wolff -Kishner reduction.

Yes.

Here, that hydrozone intermediate is reduced to an alkane using powerful basic conditions, typically potassium hydroxide, KOH, and heat.

The Wolff -Kishner mechanism is a really beautiful illustration of thermodynamics just triumphing over kinetics.

It is.

It proceeds through a very high energy carbanion intermediate, which would normally make the reaction incredibly unfavorable and slow.

You'd think it would never work.

So why does the reaction go to completion?

What's the trick?

The trick is the step immediately following the carbanion formation.

It involves the irreversible and immensely favorable loss of N2 gas molecular nitrogen.

The N triple bond N is one of the strongest, most stable bonds known.

Ah, so the relentless bubbling of that N2 gas out of the solution constantly removes a product from the equilibrium.

It's Le Shapelier's principle in action again.

It completely drives the entire process forward, pushing the reaction past that unstable intermediate all the way to completion.

So to summarize the synthesis toolbox,

you now have three ways to convert a ketone to an alkane.

Clemson, which is strong acid,

Rainy Nickel, which is catalytic acid, and Wolff -Kishner, strong base.

And your strategic choice depends entirely on which other sensitive functional groups are present in your molecule.

We've arrived at, I think, the most strategically important section,

the carbon nucleophiles.

These reactions are the building blocks of larger molecules, they give us the power to form new carbon bonds.

And we start with the giants.

The Grignard reagents, RMGX.

They're created simply by inserting magnesium metal into an alkyl halide bond, often in diethyl ether solvent.

And that insertion of magnesium radically changes the reactivity of the attached carbon.

Carbon is way more electronegative than magnesium.

Much more.

So the CMG bond is highly polarized, effectively creating a species where the carbon acts as a highly reactive, strongly basic, and strongly nucleophilic C - source.

It's a carbanion equivalent.

This extreme reactivity leads to what's often called the reaction condition cruelty of the Grignard reagent.

Yes.

Because it's essentially a powerful base, it is completely incompatible with any acidic proton.

If you introduce water and alcohol, a carboxylic acid, even a terminal alkene.

The Grignard will act as a base first.

Always.

That proton transfer, the Grignard ripping off a proton, is instantaneously faster than its nucleosilic attack on a carbonyl.

The region is irreversibly destroyed.

It's quenched.

So because the Grignard is such a potent nucleophile, it attacks the carbonyl group directly.

No acid catalysis needed.

In fact, acid would just destroy it.

It would.

And the mechanism mirrors the hydride reduction we saw earlier.

The R - attacks, forming the tetrahedral alkoxide intermediate.

And critically, we apply the golden rule.

Never expel C -.

Since the newly attached group is a carbon atom, that intermediate is trapped.

It cannot reform the carbonyl.

Consequently, just like with LiOH4, the Grignard reaction must be written in two distinct steps.

Step one, RMGX in ether.

And then step two, a proton source, water, to protonate the alkoxide.

And the synthetic utility is huge.

You form a new alcohol, and you simultaneously install a brand new alkyl group on what used to be the carbonyl carbon.

Our second C -nucleophile system is the phosphorus yeliides, which power the Wittig reaction.

Okay, first, what exactly defines an allylide?

An allylide is a neutral molecule that contains two adjacent atoms that are oppositely charged.

In this case, it's a formal positive charge on the phosphorus P +, and a formal negative charge on the adjacent carbon C -.

And they are prepared in two steps.

First, an SN2 reaction between triphenylphosphine PTH3 and an alkyl halide to form a salt.

And then you deprotonate that salt using a super strong base like emboli.

That powerful base pulls off an alpha proton, generating the nucleophilic C -.

The Wittig reaction mechanism is entirely different from the tetrahedral attack mechanisms we've seen so far.

Completely different.

The C - attacks the ketone or aldehyde, and the resulting intermediate immediately collapses in a four -atom cyclotation reaction.

This forms a highly strained four -membered ring intermediate called an oxyphosphatane.

Right, and this strained ring then spontaneously breaks apart or fragments.

Why does it do that?

The fragmentation is the key.

It separates the phosphorus group from the rest of the molecule.

And the reason this is so favorable and irreversible is the tremendous thermodynamic stability gained by forming the new phosphorus -oxygen double bond.

That P -double bondo formation is the massive driving force that makes the entire Wittig reaction go.

And the net result.

The C -double bond O is converted directly into a C -double bond C.

We form an alkene.

Exactly.

This gives us a direct and really elegant way to synthesize alkenes from ketones.

And when you combine it with ozonolysis, which goes from alkene back to ketone, the Wittig reaction provides complete interconvertibility between these two crucial functional groups.

It's a hallmark of synthetic mastery.

Finally, we examine the sulfur yellowides.

They're prepared similarly to the Wittig region, starting from dimethyl sulfide, DMS, followed by reaction with an alkyl halide, and then deprotonation with a strong base.

But despite the similar preparation, the sulfur halide yields a totally unique product when it reacts with a ketone or aldehyde.

It gives you an epoxide, a three -membered cyclic ether.

So why does the presence of sulfur instead of phosphorus change the product from an alkene to a ring?

This is a great question.

The initial attack is the same, yielding a tetrahedral intermediate with a negatively charged oxygen, O-, and a positively charged sulfur, S +, nearby.

But unlike the Wittig region, where the PO bond formation drives fragmentation, the S -plus intermediate is highly susceptible to an internal reaction.

The O -intermediate immediately performs an intramolecular SN2 -type process.

The oxygen attacked the adjacent carbon atom, which is next to the positive sulfur, forming the three -membered ring and simultaneously expelling the stable, neutral dimethyl sulfide molecule.

So the expulsion of that neutral sulfur compound is the driving force for the ring formation.

It is.

So sulfur halides are an alternative method for making epoxides, which we've previously only seen via the oxidation of alkenes.

Again, multiple strategic pathways to the same functional group.

We have to now address the most famous exception to our golden rule.

Never expel H - or C -.

We're talking about reactions where the molecule finds a clever way to avoid generating those unstable ions.

Right.

We can briefly mention the Kanazaro reaction, which, on the surface, appears to expel a hydride ion.

But a mechanistic study shows it's actually a direct transfer of a hydride from one molecule to another.

It avoids having a free high -energy ion in solution.

But the crucial exception, the one you absolutely need to master for synthesis problems, is the Bayer -Villager oxidation.

Oh, this one is strategically vital for changing the carbon backbone of a molecule.

The regions are a peroxy acid, RCO3H, most commonly MCPBA.

And the reaction inserts an oxygen atom right next to the carbonyl group.

Right.

This converts a ketone into an ester or an aldehyde into a carboxylic acid.

So how does it get around the rule?

The mechanism starts conventionally.

Protonation of the ketone, then nucleophilic attack by the peroxy acid, creating a highly charged tetrahedral intermediate.

Here's the crucial point.

To reform the carbonyl, the molecule needs to expel the RCO2 - group.

But that expulsion is coupled with an internal change.

So instead of just kicking out an alkyl group as an unstable C-, a unique rearrangement occurs.

Exactly.

An alkyl group attached to the carbonyl carbon migrates to the neighboring oxygen atom at the same time as the carbonyl double bond reforms and the peroxy group leaves.

This concerted migration avoids the unstable intermediate entirely.

It's a beautiful workaround.

Because it's a migration, we need to know how to predict which group moves.

And that's defined by the migratory aptitude order.

This dictates which group is most likely to move on to the oxygen atom, thus determining which side the new oxygen inserts on.

And that order is hydrogen is greater than tertiary, which is greater than secondary or phenol, which is greater than primary, which is greater than methyl.

Yes.

H3 degrees, 2 degrees, fill 2, 1 degrees, methyl.

Hydrogen migrates most easily.

The predictive power here is enormous.

For an unsymmetrical ketone, the oxygen atom will be inserted on the side that has the group with the higher migratory aptitude.

So if you have a ketone with a tertiary carbon on one side and a secondary on the other, the oxygen inserts next to the tertiary carbon.

And if you apply this to a cyclic ketone, the oxygen inserts into the ring.

Resulting in a lactone, which is a cyclic ester, via ring expansion.

This predictable expansion makes Bayer -Villager a beautiful tool for building complex ring systems.

This leads us naturally into our final section on strategy.

Section 6 .9 really emphasizes that knowing the reaction mechanisms is only half the battle.

You have to transition to synthetic thinking.

You do.

And the most powerful tool here is retrosynthetic analysis, working backwards.

You work backwards from the product to the starting material.

You look at the final functional group and you ask, what was the single reaction that created it?

Exactly.

For example, if your target molecule contains an alkene, and you know you have a ketone starting material, your last step should immediately scream, Wittig reaction.

Or if your product is an alcohol that has a new carbon chain attached.

Your final step must have involved a Grignard region.

There's no other way we've learned to do that.

This strategic thinking is so essential because synthesis problems in introductory courses are constrained, right?

They're usually only three to five steps.

They are.

So you focus on the functional group conversion, identify the reaction that performs it, and then find the necessary precursor.

Synthesis is a puzzle, but you have the playbook.

Every new functional group we cover today is both a target and a potential starting material for the next step.

The goal for you, the learner, is to internalize these patterns until they just become second nature.

This has been a truly comprehensive deep dive into ketones and aldehydes, which we found are really unified by two simple powerful concepts.

The high electrophilicity of the carbonyl carbon and the subsequent fate of that resulting tetrahedral intermediate.

And that fate is dictated by the thermodynamic stability of the C double bond O, causing it to attempt to reform, governed by the golden rule.

Only suitable leaving groups can be expelled, which usually excludes the unstable H minus or C minus.

Right, forcing the reaction to halt or, in rare cases like Bay or Villager, leading to an elegant rearrangement.

We map the whole landscape, covering preparation via oxidation and ozonolysis, and then we meticulously explore the five categories of nucleophiles.

H NUX for selective reduction.

O NUX for acetyl formation and that critical protecting group strategal.

S NUX for the total rainy nickel reduction.

And NUX for imines, imines, and the powerful Wolff -Kishner reduction.

And finally, C NUX for forming new carbon bonds via Grignard, Wittig, and sulfur ethylide reactions.

To leave you with a final thought on the value of pattern recognition and strategic choice, remember our three routes to reduce a ketone to an alkane,

Clemson, rainy nickel, and Wolff -Kishner.

The true insight there is that they all yield the same product.

But their conditions, acidic, catalytic acid, and basic, are vastly different.

Synthesis mastery isn't just about knowing how to perform the reduction.

It's about understanding which condition must be chosen to ensure compatibility with every other functional group present on the molecule.

That ability to choose the optimal least destructive condition, that is the very definition of chemical control.

Thank you for joining us for the deep dive.

We wish you the best of luck as you apply these patterns to your synthesis problems.