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Hi! I’m Deboki Chakravarti and welcome to Crash Course Organic Chemistry!
The Portuguese explorer Vasco da Gama sought a sea route to India in 1497 to bring black pepper and other spices back to Europe,
which helped kick off the Age of Discovery.
So, in some way, the pursuit of black pepper, and the organic chemical piperine, changed the world.
Named for the tropical vine, Piper nigrum, piperine triggers the "spicy" receptors in our mouths that lead to a little bit of pain.
But flavorful pain, because it makes spicy food taste good!
The E-double bonds in piperine can be converted to cis bonds in a rare isomerization reaction with sunlight.
And these isomers of piperine don't have the same heat to them.
It's unclear if this isomerization is why old, ground pepper isn't super hot, and if you want to become an organic chemist to investigate that mystery, go for it!
With this spicy pepper inspiration, though, let’s focus on more chemical reactions involving alkenes.
[Theme Music]
Over the past few episodes, we've been doing reactions with alkenes that follow the same general pattern.
Alkenes are nucleophilic and can attack electrophiles, so the pi bond gets broken and two sigma bonds form.
Two groups get added to the former alkene, one to each carbon on either side of the double bond.
But that’s a mouthful, so organic chemists say those two groups get added across the double bond, and we call these addition reactions.
Before we dive into these puzzles, which can feel huge and messy, let's touch base.
Organic chemistry is full of patterns, and we can use those patterns to predict, instead of memorize, reaction products.
For alkene addition reactions, there are three key questions we can ask to help us figure out what's going on.
Think of these questions as three moves that combine to make up a dance, a cool fighting combo, or a secret handshake.
Except... that dance or combo or handshake is just a specific kind of addition reaction with a special name.
Question Number 1:
What are we adding across the double bond?
Maybe we're adding hydrogen bromide, like we have in previous episodes,
so there's a proton on one carbon of the former double-bond and there's a bromine on the other carbon.
Or maybe we're adding water across a double bond, H on one side and OH on the other.
And if it is water… we call it hydration.
Question Number 2:
Where will the group add on an asymmetrical molecule?
For example, the left-hand carbon of the double bond in this molecule is bonded to two carbons.
And the right-hand carbon of the double bond in this molecule is bonded to two hydrogens.
As a refresher from episode 14, we learned Markovnikov's rule,
which tells us that a proton adds to the carbon of the double bond that has the most hydrogens.
We call that Markovnikov addition.
With some special reactants, these groups can add in the opposite orientation.
And if a proton adds to the carbon with less hydrogens, it's anti-Markovnikov addition.
In general, this question is defining a reaction’s tendency to bond groups to one atom over another in an asymmetrical molecule,
a property called regioselectivity.
And Question Number 3:
What is the expected stereochemistry of the added groups?
Here, we need to pay attention to how the groups add in 3D space.
If the groups we’re adding add to the same face of the double bond, it's called syn addition.
Or if they add on opposite faces, it's called anti addition.
That's it!
These three questions are the key to deciphering the secret handshakes of addition reactions.
Let's start with an addition reaction called halogenation.
We're adding chlorine or bromine across the double bond, both of which are in the halogen group of the periodic table.
That answers Question Number 1!
By the way, we typically perform this reaction in a non-nucleophilic solvent like carbon tetrachloride.
For Question Number 2, we're adding the same group, a bromine, on both sides so regioselectivity doesn't matter here.
And for Question Number 3, see that one bromine adds to the top of the ring and one bromine adds to the bottom of the ring, so this is anti addition.
We've got to step through the reaction to understand why it happens this way though.
Let's look at the mechanism of cycloheptene with molecular bromine.
Remember HBr is a strong acid and completely dissociates in water.
This means that bromide, Br minus, is pretty stable on its own.
So, the nucleophilic alkene can attack bromine, pushing off Br minus and forming a cation.
But here's a possible dilemma: thinking about stereochemistry, the bromide ion could attack a planar carbocation on either face of the ring.
That would give us a mixture of syn and anti addition, but experimentally that doesn’t happen here, we only get the anti product.
So there's something tricky going on that prevents the syn addition: we don’t form a simple carbocation here!
If we think about bromine donating a pair of its electrons to the nearby positive charge, we form a structure called a bromonium ion.
This high-energy bridge helps stabilize the positive charge.
To get straight to the bromonium ion, we need to adjust our arrow pushing for the first step.
The alkene attacks, Br-minus is kicked off, and we also show a lone pair from the bromine we attacked reaching back to attack the alkene to form the bridge.
The bromonium ion is a major contributor to the resonance hybrid and blocks that face of the ring.
So the remaining bromide ion is forced to attack the opposite side of the ring, and it's anti-addition.
To sum up our combo key to halogenation, we've got this notecard!
We'll make one for all these reactions.
Now let's learn a twist on halogenation: an addition reaction called halohydrin formation.
"Halo-" comes from halogen and the "hydrin" comes from water, think hydration.
So for Question Number 1, we're still adding chlorine or bromine across the double bond, just in a solvent like water that's also a nucleophile.
Because there are many solvent molecules around, we get a different final product.
To see what I mean, let's predict the products if we had an asymmetrical alkene such as Z-3-methylpent-2-ene
and react it with molecular chlorine, but this time using water as a solvent.
First, the nucleophilic alkene attacks chlorine.
Like what we saw with bromination, a chloronium ion forms to stabilize the positive charge.
Stereochemically speaking, the chloronium bridge can be sticking up or down,
which ultimately means that we'll get equal amounts of these two enantiomers reacting with water to make our final products.
If you remember all the way back to episodes 8 and 9, that's a racemic mixture!
Now it's time to think about Question Number 2.
The next step in a halohydrin reaction involves a water molecule as our nucleophilic attacker.
Oxygen’s lone pair is going to attack at the side where we can draw a carbocation that contributes more to the resonance hybrid, which is the tertiary one.
So we would expect water to add to the more substituted side of the bond, and this reaction is regioselective.
For Question Number 3, the chloronium ion blocks syn addition, so the water molecule adds anti.
And then a second water molecule deprotonates the oxonium ion… forming an alcohol group in the final products.
They're called chlorohydrins, because this is a halohydrin formation reaction.
Whew!
We can sum up halohydrin formation with this notecard.
Both of the addition reactions we just learned involve halogens and alkenes, but in episode 14 we looked at acid-catalyzed hydration.
An acid catalyst helped us add water across the double bond.
This reaction involves a 1,2 hydride shift, but not all acid-catalyzed hydrations do.
If you want a detailed reaction mechanism, you can go watch episode 14.
But here's a notecard to compare it with the other reactions we're learning in this episode.
To add a nucleophile like water without carbocation rearrangements, we can use a reaction called oxymercuration/reduction.
To help remember this, think of mercury as a tiny red stop sign of a planet, blocking shifty carbocations.
The name oxymercuration means using a mercury atom to help add oxygen across the double bond.
And the reduction part is how we get rid of the mercury atom after it's done its job.
So for Question Number 1, we're adding a nucleophile with an oxygen in it across a double bond, with the help of mercury acetate.
We’re going to use water here to be consistent with our other reactions, but we can also use alcohols in this reaction and form an ether as the product.
Let's start by looking at our reaction arrow.
Next to the "1." we have mercury (II) acetate and water.
What's interesting about metal complexes like our mercury friend here, is that the bonds aren’t as tight as organic molecules,
which means the attached groups dissociate easily.
Our mercury (II) can either be attached to one or two acetate molecules at any time in solution.
And when it's single-bonded, the mercury has a formal positive charge.
That makes our first step pretty clear: mercury (II) acetate is our cation and electrophile.
So the alkene, our nucleophile, has something to attack!
But, like bromine and chlorine, the mercury has a lone pair of electrons that can reach back and make another high-energy bridge… do you sense a pattern here?
This makes an intermediate called a mercurinium ion, which helps stabilize the positive charge,
but also blocks any sort of 1,2-shifts we’d get with a plain old carbocation!
The mercurinium can form on either face of the alkene, like we saw with the chloronium ion before.
We still haven't used up everything from the "1." above the reaction arrow.
So in this next step, water is our nucleophile and attacks the carbon that better stabilizes a positive charge.
Then, a second molecule of water swoops in and deprotonates, just like the one-two punch in other hydration reactions we’ve seen.
That's another pattern to notice as we do more reactions!
And to answer Question Number 2 about regioselectivity, water adds to the more substituted carbon.
Finally, we're ready to move onto the "2." below the reaction arrow, which is sodium borohydride.
We're actually going to skip the details here because it's a metal reaction mechanism, which gets into inorganic chemistry.
Basically, all we need to know for this series is that a hydrogen atom replaces the mercury and we get our alcohol.
And to answer Question Number 3, the mercurinium ion blocks a side of the double bond, so we have anti addition of the alcohol.
The really important piece to remember here is that we get Markovnikov addition and no rearrangement products.
So to sum up our combo key to oxymercuration, here's another notecard.
OK, we're just going to squeeze in one more alkene addition reaction here: deep breath, we got this!
Our last reaction is called hydroboration.
Its name comes from the fact that we're adding water across the double bond, which gives us "hydro-"
and we're going to use boron to help us do it, which gives us "-boration."
So the answer to Question Number 1 is: we add borane and replace it with an OH.
Overall, it’s like adding water across the double bond.
Like most of these alkene addition reactions, hydroboration has two separate steps.
To start the reaction, our nucleophile, 1-methylpent-1-ene, attacks the boron-containing reagent, which we will represent here as BH3.
And in hydroboration, this kicks off a weird chain reaction where all bonds break and form at the same time, called a concerted reaction.
Because a boron atom is chunkier it has to add to the less hindered carbon and our four-membered transition state looks like this.
We show the remaining two H's from BH3 still on the boron, but in reality, those hydrogens can also add to two other alkene molecules.
So to answer Question Number Two, this is anti-Markovnikov addition.
The boron reagent is big, so the smaller hydrogen has to be added to the more substituted carbon.
For Question Number Three, because the B and the H are still attached during this funky concerted reaction, they have to add to the same side of the ring.
So hydroboration is always syn addition.
This isn't our final product because we still have to use the chemicals in step 2 to get rid of the borane and make our alcohol, in a reaction called oxidation.
The peroxide, H2O2, and sodium hydroxide act like referees in this process, breaking off the boron and leaving behind the alcohol group.
We'll get into oxidation more next time, but for now, let's sum up our three-combo key to hydroboration.
And here's a quick comparison of the three alkene addition reactions that form alcohols:
Acid-catalyzed hydration has Markovnikov regioselectivity, is not stereoselective, and can have carbocation rearrangements.
Oxymercuration has Markovnikov regioselectivity and blocks carbocation rearrangements.
And hydroboration has anti-Markovnikov regioselectivity, syn-stereochemistry, and no carbocation rearrangements.
Depending on where we want the alcohol to end up, we can choose the perfect reagent!
So remember to practice recognizing these patterns… even though they might be tricky at first.
In this episode:
We began to create three-move combos for alkene addition reactions that specify
1) what is being added across the alkene,
2) regioselectivity, and
3) stereochemistry.
And we learned a bunch of addition reactions to alkenes
In the next episode, we'll investigate even more alkene addition reactions, focusing on oxidation and reduction.
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