Chapter 11: Addition Reactions

Welcome to Last Minute Lecture.

This free chapter overview is designed to help students review and understand key concepts.

These summaries supplement not replaced the original textbook and may not be redistributed or resold.

For complete coverage, always consult the official text.

Welcome curious minds to the deep dive.

Today we're tackling a topic that can feel pretty daunting in organic chemistry.

We're diving into addition reactions.

That's right.

And if this has ever felt like, you know, a whole different language, this deep dive is really your shortcut to getting it.

Exactly.

We're drawing directly from chapter 11 of organic chemistry as a second language.

First semester topics, fourth edition.

It's a book famous for making things click.

And our mission today is simple.

Extract the most crucial stuff you need.

We're talking terminology,

mechanisms, and how to actually solve problems, predicting products, figuring out synthesis.

It's not just about memorizing what happens.

We want you to get the why.

That's where the confidence comes from, right?

Yeah, absolutely.

It's about building that intuition.

So you can look at a reaction and think, okay, I see what's going to happen here.

So let's jump in.

The basics first.

What exactly is an addition reaction?

Okay.

Fundamentally, you've got a carbon -carbon double bond.

An addition reaction is where two new groups, let's call them X and Y, add across that double bond.

And the double bond itself disappears.

Right.

It becomes a single bond.

The pi bond breaks to form two new sigma bonds, one to X and one to Y.

It's a core transformation.

Got it.

And you mentioned there are sort of three main skills we need for this chapter.

Yeah, exactly.

Predicting products.

What do you get?

Drawing mechanisms.

How do you get there?

Step by step.

And proposing syntheses.

How do you make a specific molecule using these reactions?

Okay.

To do any of that, we need the language.

Let's start with regiochemistry.

Sounds complicated.

It just means where the X and Y groups end up.

Specifically, on which of the two carbons that used to be in the double bond.

And it only matters if X and Y are different, right?

And the alkene isn't symmetrical.

Precisely.

If you're adding, say, two bromines or adding something to ethene, regiochemistry isn't really a question.

But add H and Br to propene.

Then you need to know where the H goes and where the Br goes.

Which brings us to Markovnikov's rule, the classic rich -get -richer idea.

Kind of, yeah.

The simple way to think about Markovnikov addition is that the non -hydrogen group, the Br, the OH, whatever X, goes to the more substituted carbon of the original double bond.

The one that already had more carbons attached to it.

Okay.

So anti -Markovnikov must be the opposite.

Exactly.

The non -hydrogen group goes to the less substituted carbon.

But it's not just some arbitrary rule we memorize.

Not at all.

And this is key.

Whether a reaction follows Markovnikov or anti -Markovnikov rules is entirely explained by its mechanism.

It usually comes down to forming the most stable intermediate, which we'll definitely get into.

Okay.

So that's where things add.

What about how they add in 3D space?

That's stereochemistry.

Right.

This describes the spatial relationship between X and Y after they've added.

Do they add to the same side or opposite sides of where the double bond used to be?

So we have syn addition.

Meaning both groups add to the same face of the double bond plane.

Imagine them both coming in from the top or both from the bottom.

And anti -addition.

They add to opposite faces.

One comes from the top, the other from the bottom.

Crucial point here.

Don't mix them up.

Anti is stereochemistry, how they add in space.

Anti -Markovnikov is regiochemistry, where they add totally different concepts.

Yes.

That's a really common mistake.

So keep those straight.

When does this syn versus anti stuff really become critical for predicting products?

It matters most when the reaction creates two new stereocenters.

Remember, those are carbons bonded to four different groups.

If you only make one stereocenter, or none, the synantin nature doesn't usually change the number of products you get, though it might affect which specific stereoisomer forms.

Ah, but there's a trap.

The meso compound trap.

Oh yeah.

Always watch out for meso compounds.

Even if you form two stereocenters via, say, syn addition, if the resulting molecule has an internal plane of symmetry, it's meso.

And that means there's only one product, not a pair of enantiomers.

It's a classic exam trick.

Okay, language down.

Let's look at some actual reactions.

Starting simple.

Adding H and H -hydrogenation.

Reagents,

H2 gas, and a metal catalyst, usually platinum, PT, palladium PD, or nickel.

Regiochemistry.

Doesn't apply.

You're adding two identical hydrogens.

Stereochemistry.

Always syn addition.

Okay.

Why always syn?

What's the mechanism doing here?

It's pretty cool, actually.

It's a surface reaction.

Think of the metal catalyst like a workbench.

The H2 molecules land on it, and the HH bond breaks.

The alkene also lands on the surface, kind of lies flat.

Ah, so it just picks up both hydrogens from that surface side.

Exactly.

They add simultaneously, or nearly so, from the same face the face is absorbed onto the metal.

That's why it has to be syn.

So, solving problems.

Watch for two new stereocenters.

If you make them, just draw the syn pair.

Unless it's meso.

Always check for symmetry.

Okay.

Next reaction.

Adding H and X, like HCl, or HBr, Markovnikov addition.

Right.

Regions are just Hx, HCl, HBr, Hi, and as you said, it's Markovnikov regiochemistry.

The X goes to the more substituted carbon.

What about stereochemistry here?

Syn or anti?

For this one, it's often a mix, or not relevant in typical scenarios where you'd form two stereocenters.

It's usually considered beyond the scope of introductory courses because the intermediate can be attacked from either face.

So focus on the regiochemistry.

So why Markovnikov?

What's the mechanism telling us?

Okay.

This is fundamental.

It's an ionic mechanism.

First, the alkenes pi electrons, they're nucleophilic, attack the H plus of the Hx.

This proton adds to the less substituted carbon.

Why the less substituted one?

Because doing that forms a more stable carbocation on the more substituted carbon.

Remember, carbocation stability.

Tertiary is way better than secondary, which is way better than primary.

The reaction takes the path of least resistance, forming the most stable intermediate possible.

Got it.

So the positive charge ends up on the more substituted carbon, then what?

Then the halate ion, X-, which is just floating around, acts as a nucleophile and attacks that positively charged carbocation, boom, product formed.

Markovnikov regiochemistry, explained by carbocation stability.

But wait, carbocations, they can do sneaky things, right?

Rearrangements.

Exactly.

This is the major pitfall.

If you form a carbocation, say a secondary one, and right next door, there's a spot where it could become a more stable tertiary carbocation.

Right.

It might rearrange.

Like a hydrogen jumps over.

Yep.

A hydride shift, H moving, or sometimes a methyl shift, C3 group moving.

This happens before the halate attacks.

So your halate might end up on a carbon you didn't initially expect.

So the takeaway is, any reaction involving a carbocation intermediate,

watch out for Always.

Check if a 12 -del -2 shift, hydride or methyl, can lead to a more stable carbocation.

If it can, it probably will.

Okay, but what if we want the B to go on the less substituted carbon?

We want anti -Markovnikov -HBr addition.

Ah, now we get clever.

You use HBr, but you add peroxides, often written as ROR.

Just adding that one extra region completely flips the outcome.

So HBr alone gives Markovnikov.

HBr with peroxides gives anti -Markovnikov.

Why?

How do peroxides do that?

Because peroxides trigger a totally different pathway.

A radical mechanism, not ionic.

Peroxides are weak.

They break easily with a bit of heat or light, splitting symmetrically to form RORadical species with an unpaired electron.

Radicals, they're highly reactive, right?

Very.

The RORadical bumps into an HBr molecule and rips off a hydrogen atom, forming ROH, and crucially, a bromine radical, pria.

Okay, so now we have a bromine radical.

What does it do?

This is what attacks the alkenes double bond.

And just like carbocations, carbon radicals also have stability trends.

Tertiary, secondary, primary.

So the Brom adds to the less substituted carbon.

To form the more stable carbon radical on the more substituted carbon.

Exactly.

That dictates the anti -Markovnikov regiochemistry.

Then that carbon radical plucks a hydrogen atom from another HBr molecule, forming the product and regenerating a Brom radical.

Which can then attack another alkene molecule, a chain reaction.

Precisely.

It's a radical chain mechanism.

Super efficient.

And the key practical point is you control the regiochemistry.

No peroxides, ionic Markovnikov.

Add peroxides, radical, anti -Markovnikov.

Very cool.

Very useful.

Okay, let's add H and OH now.

First, Markovnikov.

Acid catalyzed hydration.

Regions here look like H3O +, or maybe H2SO4, and water.

Basically water with an acid catalyst.

And the mechanism sounds familiar.

It's almost identical to the ionic HX addition we just did.

Step one, alkene attacks H +, from the acid catalyst, to form the most stable carbocation.

Markovnikov principle again.

Okay.

Carbocation formed on the more substituted carbon.

Step two, water H2O, which is the solvent in a decent nucleophile, attacks the carbocation.

That gives a positive oxygen.

Right.

So step three, another water molecule comes along and acts as a base, just plucks off one of those extra protons in the oxygen.

And you're left with your neutral alcohol product, with the OH on the more substituted carbon.

And you mentioned this one is an equilibrium, like it can go backwards.

Yeah, this is actually the reverse of an E1 elimination.

Because it's in equilibrium, you can push it.

Use lots of water -elite acid, and you favor the alcohol product, hydration.

Use very little water -concentrated acid, and you favor the alkene elimination dehydration.

Le Chatelier's principle in action.

Smart.

Okay, but what if we need the OH on the less substituted carbon, anti -Markovnikov hydration?

That calls for a different reaction entirely.

Hydroaboration -oxidation is a two -step process.

Regens.

Step one is BH3, borane, usually complex with THF.

Step two is hydrogen peroxide, H2O2, and a base like sodium hydroxide, NaOH.

Okay, so BH3 first, then peroxide base, and the outcome is anti -Markovnikov OH addition.

What about stereochemistry?

Ah, this one also gives synodish, and both the H and the eventual OH add to the same face of the double bond.

No carbocation here, then.

How does it work?

Correct, no carbocation.

That means no rearrangements to worry about, which is nice.

The first step, hydroboration is concerted.

The alkene pi bond attacks the boron atom, and BH3 boron has an empty orbital, making it electrophilic.

At the same time.

At the same time, one of the hydrazines attached to boron shifts over, and adds to the other carbon of the double bond.

It happens through a sort of four -membered ring transition state.

So how does that give anti -Markovnikov regiochemistry?

Two reasons, really.

CERICs.

The boron group, BH2, is bulkier than the hydrogen, so it prefers to add to the less crowded, less substituted carbon.

And electronics play a role, too.

The result is boron on the less substituted carbon, hydrogen on the more substituted.

And the synodition?

Because the H and the B add simultaneously in that concerted step, they must add to the same face of the double bond.

Okay, so after step one, we have boron on the less substituted carbon.

How does that become an OH?

That's step two.

Oxidation.

The H2O2 and NaH react with the boron compound in a way that essentially replaces the carbon -boron bond with the carbon -oxygen bond, with retention of stereochemistry.

So the OH ends up exactly where the boron was, preserving that syn relationship with the hydrogen that added.

Wow, that's elegant.

Concerted, regioselective, stereoselective.

It's a really powerful reaction.

Okay, let's zoom out again.

We have these individual tools.

How do we use them to build things?

Synthesis.

Right, the simplest is just a one -step synthesis.

You see the starting material, you see the target, you pick the one reaction that does the job.

But often it's more complex.

Like what if you need to move a functional group?

Say, change the position of a leaving group, like a bromine.

Good strategy.

You can do that in two steps.

Eliminate then add.

First, use a base to eliminate HBr and form an alkene.

And we can control where the double bond forms?

Use a small strong base, like sodium methoxide, NEOET.

For Zaitsev elimination, the more substituted alkene.

Use a big bulky base like potassium tert -butoxide, TBOK.

For Hoffman elimination, the less substituted alkene.

Okay, so you form the alkene you want then.

Then you add HBr back across the double bond, but you choose your conditions carefully.

Want Markovnikov addition to put bra on the more substituted spot.

Use HBr alone.

Want anti -Markovnikov to put bra on the less substituted spot.

Use HBr with peroxides.

Clever.

You control elimination, then you control addition.

What if you start with an alcohol though?

OH isn't a good leaving group for elimination.

Great point.

You have to make it a good leaving group first.

The standard way is to convert the alcohol, ROH, into a tosylate, using tosyl chloride, DSCl, and pyridine.

Tosylate is an excellent leaving group.

Then you can proceed with your chosen elimination.

Okay, another strategy.

What if we want to move the double bond itself?

You could do that too.

Add, then eliminate.

Add a group across the existing double bond, maybe HBr, maybe something else.

Choosing Markovnikov or anti -Markovnikov to set up the next step.

Then eliminate again.

Right.

Now you have a leading group, like Br, in a new position.

Choose your elimination conditions, like Sev or Hoffman, to form the double bond where you want it.

It lets you shuffle that pi bond around.

That gives you a lot of control over the molecule's structure.

It really does.

Careful choice of reagents is everything in multi -step synthesis.

One more synthesis trick.

What if you start with just an alkane?

No double bonds, no leaving groups.

Seems stuck.

Ah, the unreactive alkane problem.

Your entry point here is usually radical bromination.

Use N -bromosepkinamide, NBs, and light -daitich V.

This selectively puts a bromine atom onto the most substituted carbon atom available, usually a tertiary position, if one exists.

Why the most substituted?

Because the reaction goes via a carbon radical intermediate, and tertiary radicals are the most stable.

So, bromination happens there preferentially.

So NBsV puts a BrR on the most substituted carbon.

Now you have a handle.

Exactly.

Now you have a leaving group.

From there, you can eliminate to form a double bond, and then do all the other addition chemistry we've talked about.

It's how you introduce functionality into a plain alkane.

That's super useful.

Okay, let's add a few more key addition reactions to our toolkit.

How about adding BrR and BrR, just bromination?

Reagents.

Just Br2.

Maybe in an inert solvent like CCl4, radiochemistry isn't relevant again.

Stereochemistry.

Always anti -addition.

Always anti.

Why?

Does it have a special intermediate like hydrogenation had the surface?

It does.

When the alkene attacks Br2, it doesn't form a simple carbocation.

Instead, it forms a cyclic, three -membered bromonium ion.

One bromine atom bridges across the two carbons, carrying a positive charge.

A triangle with bromine at one corner?

Kind of, yeah.

It's strained.

Now the other bromide ion, Br, that was kicked off, acts as a nucleophile.

But it can't attack from the same side as the bulky bridged bromine.

It must attack from the backside.

Like an SN2 reaction.

Exactly like an SN2 backside attack.

It attacks one of the carbons in the ring, popping the ring open, and forcing that carbon's bond to the bridged bromine to break.

The result.

One BrO ends up on one face, the other BrO on the opposite face.

Anti -addition.

Very neat.

Now what happens if you do this reaction not in CCl4 but in water, adding Br2 and H2O?

Now things get interesting.

This is halohydrin formation.

You had BrR and OH.

Where do they go?

Regiochemistry?

The OH group ends up on the more substituted carbon, and the BrR on the less substituted carbon.

And stereochemistry.

Still anti.

Still anti -addition.

The mechanism starts the same way, formation of that bridged bromonium ion.

But now water attacks instead of Br.

Exactly.

Water is the solvent, so there's way more of it around the Br.

Water attacks the bromonium ion.

Now which carbon does it attack?

It preferentially attacks the more substituted carbon of the bromonium ion.

By that one.

Because that carbon bears more partial positive charge.

It has more carbocationic character, even within the bridged ion.

So the nucleophile water is more attracted there.

And it still attacks from the backside.

Still backside attack for anti -addition.

Water attacks, pops open the ring, then another water molecule deprotonates the added water to give the neutral OH group.

So you get Br and OH added anti, with OH on the more substituted carbon.

The solvent makes a huge difference.

It absolutely can.

Competing nucleophiles.

Okay, two more ways to add OH groups.

What if we want to add two OHs, anti to each other?

Anti -dihydroxylation.

This is also a two -step process.

First, you react the alkene with a peroxy acid, like MCPBA, metachloroperoxid and zoic acid.

This forms an epoxide.

An epoxide?

That's a three -membered ring with an oxygen.

Yep.

A cyclic ether.

Then step two is adding aqueous acid, H3O+.

The acid protonates the epoxide oxygen, making it more reactive.

Then water attacks.

Backside attack again.

Backside attack.

Water attacks one of the carbons of the protonated epoxide from the side opposite the CO bond, popping the ring open.

This ensures the two OH groups, one from the epoxide oxygen, one from the attacking water, end up anti to each other.

So epoxide formation, then acid -catalyzed opening, gives anti -dials.

What about syn addition of two OHs, syn -dihydroxylation?

Two main ways to do this.

One uses osmium tetroxide, O4, followed by a workup with something like H2O2 or NMO.

The other uses cold -dilute potassium permanganate, KMnO4, under basic conditions, NaOH.

Osophore or cold KMnO4?

How do these give syn addition?

Both involve a concerted mechanism.

The metal region, osophore or MnO4, adds across the double bond in a single step, forming a cyclic intermediate where both oxygens from the region attach to the carbons from the same face of the alkanes simultaneously.

Ah, like the hydrogenation on the catalyst surface, adding from one side.

Similar idea, yes.

The geometry of the addition forces both oxygens onto the same face.

Then the intermediate is broken down, hydrolyzed, in the workup step to get the dial, but the syn -stereochemistry established in that first concerted step is maintained.

Concerted addition leads to syn -stereochemistry.

Got it.

Last one, ozonolysis.

Sounds destructive.

It is.

It completely cleaves the C -C's double bond.

Regions are one, ozone, O3, and two, a reducing agent, typically dimethyl sulfide, DMS, sometimes zincacetic acid.

So you literally cut the molecule in half at the double bond.

Pretty much.

You break both the sigma and the pi bond of the C -C unit.

Each carbon that was part of the double bond ends up double bonded to an oxygen atom instead.

So you form aldehydes or ketones.

Exactly.

Depending on what was attached to the original double bond carbons.

If a carbon had two R groups, it becomes a ketone, R -C -O -R.

If it had one R group and one H, it becomes an aldehyde, R -C -O -H.

If it had two hydrogens, it becomes formaldehyde.

Regiochemistry and stereochemistry are kind of irrelevant here since you're breaking the bond apart.

Right.

The main thing is identifying the two carbonyl compounds that result from splitting the alkene.

And the prediction trick.

Yeah.

It's super helpful.

Just visually erase the double bond in the starting alkene and draw two oxygen atoms double bonded where the C -C -FET bond used to be.

That directly gives you your products.

Nice shortcut.

The mechanism involves ozone forming a weird intermediate called a melosinide.

Yeah.

It forms an initial cyclic adduct, melosinide, which rearranges to an ozonide.

And then the reducing agent, DMS, cleaves that ozonide to give the final aldehydectin products and DMSO.

The mechanism's complex, but the outcome cleavage to carbonols is the key.

Wow.

Okay.

That's a whirlwind tour of Chapter 11.

We went from basic definitions of regio and stereochemistry all the way through 10 major addition reactions, their mechanisms, and even synthesis strategies.

It's a dense chapter, for sure.

But hopefully breaking it down like this, focusing on the why behind the rules.

The carbocation stability, the radical stability, the concerted steps, the intermediates like ammonium ions and epoxides helps them make more sense.

Absolutely.

It's about seeing the patterns, understanding the driving forces.

That's how you really learn organic chemistry, not just memorizing reagents.

Right.

And seeing how a tiny change like adding peroxides to HBR or just changing the solvent from to water during bromination can completely change the product.

It really highlights the subtlety and honestly, the elegance of these reactions.

It really does.

It makes you wonder what other seemingly small changes in reaction conditions, maybe things we encounter every day without thinking about it, could be causing dramatically different chemical transformations right under our noses.

That's a fascinating thought.

Chemistry is happening everywhere, often dictated by these subtle factors.

That's definitely something for you all to ponder.

Thank you so much for joining us on this deep dive into organic chemistry's addition reactions.

Keep exploring, keep asking questions and keep learning.

We look forward to our next deep dive with you.