Chapter 12: Alkynes

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Welcome to the Deep Dive.

We're the show that cuts through the textbook jargon and gets straight to the core concepts.

That's the idea.

And today we are diving deep, really deep into alkynes.

Chapter 12 from organic chemistry as a second language, if you're following along.

Yep, alkynes.

They're kind of a cornerstone in orgochem with those carbon triple bonds.

So our mission today, take this chapter and really break it down.

We want you to get the essential insights, the stuff that actually matters for understanding, not just memorizing.

Exactly.

We'll look at their structure, how they're put together, how they behave, how you actually make them, how you can change them into other things.

And the common mistakes, right.

We always need those.

Oh yeah.

Definitely cover the common pitfalls to watch out for.

Okay, so stick with us.

And we promise to clear up the key reactions, those sometimes tricky mechanisms, and give you some solid problem -seeming strategies.

You'll be ready to tackle alkynes.

Let's unpack this.

Sounds good.

All right, first things first.

What exactly is an alkyne?

At its heart, it's pretty simple.

Any molecule that has a carbon triple bond, that's C triple bond C unit.

C, got it.

And that triple bond, well, it dictates the shape.

Each of those carbons involved in the triple bond is what we call spout hybridized.

Okay, spout hybridized.

What does that mean in practice?

It means it's linear, straight line.

Think 180 degrees bond angles around those carbons.

So when you draw them, seriously, draw them straight.

Don't put a bend to the triple bond part.

It looks wrong, and it actually affects how you think about their reactions.

Right, drawing it correctly reflects the actual geometry.

Good tip.

Now, I know there are two main types,

terminal and internal.

Yeah, exactly.

A terminal alkyne has a hydrogen directly attached to one of the triple bond carbons.

It's like at the end of the chain, HCCR.

An internal alkyne has carbon groups attached to both sides of the triple bond, RCCR.

No hydrogen directly on the CECR.

And that hydrogen on the terminal alkyne, that's important, isn't it?

Oh, it's super important.

It leads to probably the most surprising property of terminal alkenes.

They're weakly acidic.

Acidic, a CH bond.

That seems wrong.

Carbon isn't very electronegative.

I know, right?

But it's true.

That specific proton, the one on the SPiM hybridized carbon, can actually be removed by a strong enough base.

You form what's called an alkanide ion, which is a carbon anion, C minus.

A carbanion.

But aren't those usually incredibly unstable?

They usually are.

But this is where that SPiP hybridization comes back in.

And SPiP orbital has more a Marys character than SPiP2 or SPiP3 orbitals.

That means the electrons in a CYP orbital are held closer on average to the positively charged nucleus.

Ah, so that stabilizes the negative charge on the carbon better than you'd expect?

Precisely.

It's stable enough to form.

Now you still need a very strong base.

Remember the ARO rules for comparing stability.

Atom resonance induction orbital.

Yeah, vaguely.

Well, usually atom is the most important factor.

Oxygen anions are more stable than nitrogen anions, which are more stable than carbon anions.

But here the orbital factor, Cidid versus SPiP2 versus SPiP3, is so significant that an Amherst hybridized carbon anion is actually more stable than an ANTB3 nitrogen anion, like in an amide ion.

Factor four beats factor one here.

Wow, okay, that is surprising.

So practically what bases work and what bases don't.

Can I use sodium hydroxide?

Nope, hydroxide isn't strong enough.

Neither are alkoxides like sodium methoxide, NaOE, or sodium methoxide, NaOMe.

Their conjugate acids, water and alcohols, are stronger acids than the terminal alkyne.

So the equilibrium lies the wrong way.

Okay, so no common lab bases like hydroxide or alkoxides.

What do we need then?

You need something whose conjugate acid is a weaker acid than the alkyne.

So you need a stronger base.

Think carbanions, or more practically, amide ions.

Sodium amide NNH2 is the classic choice.

Sodium amide, NNH2.

Yep, or another good one is sodium hydride, NaH.

It's a source of H-, which is a very strong base.

A neat thing about using NaH is that the byproduct is hydrogen gas, H2.

And that bubbles out of the solution.

Exactly, Le Chatelier's principle in action.

Removing a product drives the reaction forward to completion.

So NaH is often very effective.

So the bottom line is you need a super strong base, like NaNNH2 or NaH, to rip off that terminal proton and make the alkanide ion, understanding that base strength is crucial.

Absolutely critical for predicting if a reaction will even happen.

Okay, makes sense.

Now that we know about their structure and this cool acidity, how do we actually make alkynes in the lab?

Well, the most common route starts from dihalides, compounds that have two halogen atoms.

Dihalides, like two chlorines or two bromines.

Yeah, exactly.

And they can be either geminal dihalides, meaning both halogens are on the same carbon atom.

Geminal, like twins on the same carbon.

Right, or they can be vicinal dihalides, where the halogens are on adjacent carbons.

Vicinal -like neighbors.

Perfect.

Either way, you can treat these dihalides with a very strong base to make an alkyn.

Let me guess, sodium amide.

You got it.

NaNNH2 is the usual suspect here.

What happens is a double elimination reaction, two successive E2 eliminations.

Okay, so E2 happens twice.

Pull off H and X, then pull off another H and X.

That's the gist of it.

You form a vinyl haliety intermediate after the first E2, and then the second E2 reaction on that vinyl halide gives you the alkyne triple bond.

And stoichiometry matters here, I bet.

How much base do we need?

It does.

For making an internal alkyne, you need two equivalents of the strong base, like NaNNH2, one for each E2 step.

Okay, two equivalents for internal.

What about terminal?

There's always a catch with terminal alkyne.

Always.

For a terminal alkyne, you need three equivalents of NaNNH2, or sometimes just excess is written.

Three.

Why the extra equivalent?

We only have two elimination steps.

But remember that terminal alkyne proton, it's acidic.

As soon as the terminal alkyne forms in the presence of that super strong NaNNH2 base, what happens?

Ooh.

The base rips off the terminal proton immediately.

Exactly.

So the third equivalent of base is used up just deprotonating the product you just made.

You end up with the alkanite salt.

So you don't get the neutral alkyne straight away.

Correct.

To get the neutral terminal alkyne, you need a second separate step.

After the reaction with NaNNH2 is done, you add a mild proton source, usually just water, H2O.

Water.

Is that acidic enough?

It is.

Because water's conjugate base is hydroxide OH, which is much more stable and therefore a weaker base than the alkanite ion.

So the alkanite happily takes the protons from water to become the neutral alkyne.

Okay, so the full procedure for a terminal alkyne is, step one, NaNNH2, three equivalents or excess.

Step two, add H2O.

That's the sequence.

Nail that down.

It's a standard synthetic playbook move.

Knowing how to make them is fundamental.

Great.

So we've made our Alpens.

We know terminal ones can form these alkanite ions.

What can we do with these ions?

Now for the fun part,

making new bonds.

These alkanite ions, besides being formed from that weak acid, are actually fantastic nucleophiles, really strong nucleophiles.

Meaning they like to attack electron core centers.

Exactly.

They're great for SN2 reactions.

This is a key way to make new carbon bonds, extending the carbon chain.

You react the alkanite ion with an alkaniide.

Like methyl iodide or ethyl bromide.

Perfect examples.

But here's a really important restriction.

This SN2 reaction works well only with methyl halides like CH3I or primary alkaniides like RCH2Br.

Only methyl and primary.

Why not secondary or tertiary halides?

Because alkanite ions are also strong bases.

If you try to react them with a secondary or tertiary alkyl halide, which are more sterically hindered.

Ah, elimination takes over.

E2 will happen instead of SN2.

You got it.

You'll get an alkane product from elimination instead of the longer alkane you wanted from substitution.

So stick to methyl or primary halides for this alkylation reaction.

Good rule to remember.

So the process is always two steps then.

Step one, deprotonate the terminal alkyne with nan -NH2 to make the alkaniide nucleophile.

Step two, add the methyl or primary alkyl halide for the SN2 reaction.

Okay, step one, nan -NH2.

Step two, RCH2X.

Makes sense.

What about acetylene itself?

HACAH has two acetic protons.

Right, acetylene is the simplest alkane.

You can alkylate both sides, putting two different R groups on or maybe the same one twice.

Can you do it all at once, like two equivalents of nan -NH2 and two equivalents of ethyl iodide?

Ah, that's a super common mistake people try.

No, you cannot do that effectively.

Trying to form the dianion of acetylene, CAC, by ripping off both protons at the same time is just too energetically unfavorable.

It doesn't really happen.

If you add two equivalents of NNH2 and then two equivalents of alkyl halide, the excess nan -H2 will just react with the alkyl halide itself, leading to side products and a mess.

Okay, so no dianion formation.

How do you alkylate both sides then?

You have to do it sequentially, one side at a time.

Ah, so four steps total.

Exactly.

Step one, NNH2, one equiv.

Step two, first alkyl halide, R1X.

Step three, nan -H2, one equiv again.

Step four, second alkyl halide, R2X.

Each ethylation is its own two -step sequence.

That makes sense.

Do one side completely, then do the other side.

Understanding that prevents a major synthetic headache.

Totally.

It's about controlling the stoichiometry and the order of events.

Okay, let's shift gears slightly.

We've made alkynes, we've added things to them.

What about reducing them, adding hydrogen?

Right, reduction, adding H2 across the triple bond.

There are a few key ways to do this, depending on what product you want.

What's the simplest way?

Just blast it with hydrogen.

Pretty much.

If you take an alkin and treat it with hydrogen gas, H2, and a standard metal catalyst like platinum PT,

palladium PD, or nickel -9.

The usual hydrogenation catalyst.

Yeah, it goes all the way.

You get complete reduction to the alkene.

The triple bond becomes a single bond.

C goes to CH2CH2.

Does it stop at the alkene, the double bond, along the way?

It forms, yes.

But the alkene intermediate is actually more reactive towards hydrogenation than the starting alkene under these conditions.

So as soon as it forms, it reacts again immediately.

You generally can't isolate the alkene this way.

Okay, so H2 with PDT, PD, or Ni means full reduction to the alkene.

But what if I want to stop at the alkene?

Good question.

For that, you need a special catalyst,

a poisoned catalyst or partially deactivated one.

Poisoned.

Sounds dangerous.

Not quite like that.

It just means its activity's been reduced.

The most famous one is Lindlar's catalyst.

Lindlar's catalyst, I've heard of that.

What is it?

It's basically palladium metal deposited on calcium carbonate, KCO3, and then treated with lead acetate or lead oxide, and sometimes quinoline.

These additives poison the catalyst just enough so that it reacts with the alkene but stops at the alkene stage.

Okay, so H2 with Lindlar's catalyst gives an alkene.

But what kind of alkene?

Cis or trans?

Cis alkene.

Always cis.

This is because the reaction happens on the surface of the catalyst, and both hydrogen atoms add to the same face of the alkene molecule as it sits on the surface.

We call this syn addition.

Syn addition gives cis alkene.

Got it.

Lindlar's catalyst only makes cis alkenes.

Correct, it cannot make a trans alkene.

So,

how do we make a trans alkene from an alkene?

There must be a way.

There is.

It's a completely different reaction called the dissolving metal reduction.

Dissolving metal?

Sounds interesting.

What are the reagents?

You use elemental sodium metal, NOAA, dissolved in liquid ammonia, NH3 as the solvent, usually at low temperature, like negative 70 degrees Celsius.

Okay, sodium metal and liquid ammonia.

Now, wait, didn't we just talk about sodium amide and NH2?

Are these related?

That's a critical point of confusion.

They're not the same and do totally different things.

NaNNH2, sodium amide, is a strong base.

NaMetal in liquid NH3 is a reducing agent system.

Sodium metal provides electrons and the ammonia acts as the solvent and eventually as a proton source.

Don't mix them up.

Okay, Na is the electron source, NH3 is the solvent proton source.

Different from NaNNH2, the base, how does this give a trans alkene?

The mechanism is pretty cool and different from hydrogenation.

It involves single electron transfers, usually drawn with Fischer -Garros.

Sodium donates an electron to the alkyne, forming a radical anion.

This intermediate is key.

Radical anion.

Okay, now, in this radical and an intermediate, the two R groups attached to what was the triple bond want to be as far apart as possible to minimize repulsion between the radical electron and the electron lone pair on the other carbon, so they adopt a trans configuration.

Ah, the intermediate prefers to be trans.

Exactly.

Then it picks up a proton from ammonia, gets another electron from sodium, picks up another proton,

and the trans geometry of that key intermediate dictates the final stereochemistry.

It's an anti -addition overall, giving the trans alkenes.

Wow.

So three reduction methods.

HDPT gives alken, H2 -Lindlar gives cis -alkene via syn -addition, and NaNNH3 gives trans -alkene via anti -addition through a radical anion.

You nailed it.

Choosing the right reduction method depends entirely on whether you want an alkane, a cis -alkene, or a trans -alkene.

It's beautiful synthetic control.

It really is.

Okay, moving on from reduction, what about adding water across the triple bond?

Hydration.

Right, hydration adding H and OH.

There are two main ways to do this, and just like reduction, they give different products.

Let's hear them.

First method, acid catalyzed hydration.

The reagents are usually dilute sulfuric acid, H2SO4, in water, but it needs a catalyst to work well with alkanes.

Mercuric sulfate, HgSO4.

Mercury, sounds a bit nasty.

It is, yeah.

Mercury compounds are toxic, so chemists try to avoid this if possible, but it's the classic way.

This reaction follows Markovnikov's rule.

Markovnikov.

The hydrogen adds to the carbon that already has more hydrogens, or in this case, the OH, adds to the more substituted carbon of the alkyne.

Exactly.

The OH group ends up on the more substituted side.

Now, the initial product you form looks like an alcohol on a double bond.

It's called an enol.

Enol.

En plus ol.

Makes sense.

But here's the thing.

Enols are almost always unstable relative to their ketone or aldehyde cousins.

They rapidly rearrange.

Rearrange.

Through a process called keto -enol tautomerization.

The enol quickly flips into a ketone.

Tautomerization.

Okay, we need to unpack that.

We do.

Basically, tautomers are constitutional isomers that are in rapid equilibrium with each other.

They differ in the location of a proton and a double bond.

In this case, the enol, CCOH, inter converts with the ketone, HCCO.

And the ketone is usually much more stable.

Way more stable, usually.

The equilibrium lies heavily towards the ketone side.

This tautomerization is catalyzed by either acid or base.

In this case, the sulfuric acid we added.

So the mechanism involves protonating the enol's double bond, making a resin -stabilized cutation, and then water pulls off the proton from the oxygen.

That's the acid -catalyzed pathway, spot on.

And the end result is the ketone.

So, big warning.

When you do acid -catalyzed hydration of an alkene, do not draw the enol as the final product.

It immediately becomes the ketone.

Got it.

Final product is the ketone.

What about regiochemistry, if I start with a terminal alkyne?

Good question.

With a terminal alkyne, following Markovnikov's rule, the OH adds to the internal carbon, C2, not the terminal one, C1.

So after tautomerization, you always get a methyl ketone.

The CO is at C2.

Always a methyl ketone from a terminal alkyne with HgSO4, H2SO4, H2O.

What if the alkyne is internal and unsymmetrical?

Then you typically get a mixture of two different ketones because the OH could add to either side of the triple bond.

Unless, of course, the internal alkyne is symmetrical, then you only get one ketone product.

Okay, that covers acid -catalyzed hydration.

What's the alternative?

The alternative achieves anti -Markovnikov hydration.

It's hydroboration oxidation.

Ah, like with alkenes.

But maybe slightly different reagents.

Similar idea, but yes, slightly different.

For alkynes, you typically use a bulkier borane reagent first, like Dissimilbrane or 9 -BBN.

This is to prevent the borane from adding twice across the triple bond.

Okay, so step one, R2BH, a hindered borane.

Then step two.

Step two is the oxidation, just like with alkenes, hydrogen peroxide, H2O2, and sodium hydroxide, NaOH, so basic conditions.

And this gives anti -Markovnikov addition.

Yes, the boron adds to the less substituted carbon, and then when it's replaced by OH in the oxidation step, the OH group ends up on the less substituted carbon.

So for a terminal alkyne, the OH adds to C1, the end carbon.

Precisely, and just like before, this initially forms an enol, but again, enol's tautomerize.

But this time, the conditions are basic NaOH.

Right, so the tautomerization happens via a base -catalyzed mechanism.

Hydroxide pulls off the proton from the enol's OH group, forming a resonance -stabilized anion, an enolate, and then water protonates the carbon atom.

And what's the final product after tautomerization in this case?

If you started with a terminal alkyne and did anti -Markovnikov addition of OH to C1, the resulting tautomer is an aldehyde.

Aldehyde, okay, this is huge, let me summarize.

Acid -catalyzed hydration, HgSO4, of a terminal alkyne gives a metal ketone, CO at C2.

Hydroboration oxidation, R2BH, then H2O2 and NaOH, of a terminal alkyne gives an aldehyde, CO at C1.

That is the absolute key takeaway for hydration.

You have regiochemical control.

You can choose whether to make a ketone or an aldehyde from the same terminal alkyne just by picking the right regions.

Powerful stuff.

Very powerful.

Now, you mentioned tautomerization a few times.

Let's just circle back and make sure we're crystal clear on that.

It seems important.

It is, it pops up everywhere.

The crucial thing is not to confuse tautomers with resonance structures.

Okay, what's the difference again?

Tautomers are distinct, actual compounds that are in equilibrium with each other, like the enol and the ketone.

Atoms have moved, specifically a proton has moved, and pi bonds have shifted.

They are constitutional isomers.

Different compounds in equilibrium, got it, and resonance structures.

Resonance structures are not different compounds.

They're different ways of drawing the same single species, usually an intermediate, like a carbocation or an anion.

Only electrons move in resonance structures.

All the atoms stay in the exact same place.

They're just different representations of electron distribution within one molecule or ion.

That's a critical distinction.

Tautomers is different molecules swapping back and forth.

It's just different drawings of the same thing.

Exactly, and remember for keto -enol tautomerism, the equilibrium almost always heavily favors the keto, ketone or aldehyde form, because the CO double bond is generally stronger and more stable than the CC double bond.

Almost always.

Is there a time when the enol is favored?

Yes.

The big exception is when the enol form is aromatic.

The classic example is phenol.

Phenol looks like an enol attached to a benzene ring, right?

C6H5OH.

Yeah, it does.

If it were to tautomerize to its keto form, it would lose the aromaticity of the benzene ring.

Aromaticity provides a huge amount of stabilization.

So in this case, the enol form, phenol, is way more stable and is the form that exists.

The equilibrium lies almost entirely on the enol side.

Because becoming non -aromatic is too high an energy cost, makes sense.

Any tips for drawing these tautomerization mechanisms?

They seem tricky.

Yeah, a couple of key things students mess up.

First, be consistent with conditions.

If you're an acid, like H3O +, the first step should always be protonation of the molecule.

And any base you use later, the mechanism should be weak, like water, not hydroxide.

Okay, acid starts with protonation, uses weak bases like water.

What about base?

If you're in base, like OH, the first step should always be deprotonation.

And any acid you use later should be weak, like water, not H3O plus L.

Base starts with deprotonation, uses weak acids like water, makes sense.

What are common mistakes?

Under acid conditions, a really common error is protonating the wrong spot on the enol.

Don't protonate the oxygen of the OH group.

Protonate the double bond, specifically the carbon atom, to generate that resonance stabilized carbocation intermediate.

That's the productive pathway.

Protonate the CTC pi bond, not the oxygen, got it.

And the other big one is using the wrong species.

Don't suddenly invoke hydroxide, OH, as a base in your mechanism if the overall conditions are acidic.

Use water.

Similarly, don't use H3O plus as an acid if the conditions are basic.

Use water.

Keep the species consistent with the reaction conditions.

Right, the species present in solution matter.

Water is usually your go -to weak acid base in these mechanisms.

Understanding the how the mechanism is just as important as the what, the product.

It really is, it separates memorization from true understanding.

Okay, one last major reaction type for alkenes, ozonolysis, we saw this with alkenes where it chopped the double bond.

Does it do something similar with triple bonds?

It does, it's also an oxidative cleavage reaction.

For alkenes, the regents are typically step one, ozone, O3, and step two, water, H2O as the workup.

Note,

you don't need a reductive workup like with alkenes, like DMS or zinc, just water.

Okay, O3 then H2O, what happens to the triple bond?

The triple bond gets completely cleaved, broken apart, and each carbon atom that was part of the triple bond gets oxidized all the way up to a carboxylic acid functional group, COH.

Both sides become carboxylic acids.

Generally, yes, if you have an internal alkyne, RCCR, ozonolysis gives you two carboxylic acid molecules, RCOH and RCOH.

What about terminal alkenes?

HDR, one side is just a hydrogen.

Ah, special case, good catch.

The internal carbon, the one attached to R, still becomes a carboxylic acid, RCOH, but the terminal carbon, the CHN, gets oxidized all the way to carbon dioxide, CO2.

Okay, so it just becomes gas and bubbles away.

Exactly, so ozonolysis of a terminal alkyne gives one carboxylic acid and CO2.

Like alkenozynolysis, this seems useful for figuring out the structure of an unknown alken, right?

By looking at the carboxylic acid fragments.

Absolutely, it's a classic structure determination tool.

Break the molecule into smaller, identifiable pieces.

Fantastic, okay, wow.

We have covered a lot of ground today on alkenes.

We really have, from their basic linear structure and that surprising acidity of terminal ones.

To making them via double elimination.

Alkylating them using SN2 on methyl and primary halides.

Reducing them selectively to alkanes, cisalkanes, or transalkanes.

Hydrating them to make ketones via Markovnikov addition or aldehydes via anti -Markovnikov addition.

Understanding that crucial ketoenol tautomerization.

And finally, chopping them up with ozone to make carboxylic acids.

It really shows how versatile these triple bonds are in synthesis.

Definitely.

And hopefully this deep dive has given you a clearer roadmap through it all.

We hit the key concepts,

tried to explain the why behind the mechanisms, and pointed out those common mistakes.

Yeah, the goal was to arm you with the understanding needed to tackle problems confidently.

And what's really cool, I think, is seeing how interconnected it all is.

You know, how that pro -hybridization affects acidity and geometry.

How choosing a catalyst completely changes the stereochemical outcome of a reduction.

It's like a puzzle.

It really is.

So our advice to you listening is, take this framework, go practice some problems.

See how these reactions are used in synthesis.

Connect the dots.

The more you apply it, the more intuitive it all becomes.

Absolutely.

Don't just memorize the reagents.

Understand why they do what they do.

Think about the electrons, the intermediates, the stability.

That's where the real mastery comes from.

Every reaction is a tool.

Learn how and when to use it effectively.

Well said.

Thank you so much for joining us on this deep dive into the fascinating world of alkenes.

We hope this helps you on your organic chemistry journey.

Yeah, keep asking questions.

Keep exploring.

It's challenging, but also really rewarding when it clicks.

Until next time, keep diving deep.