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Hi! I’m Deboki Chakravarti and welcome to Crash Course Organic Chemistry!
The Australian Blue Mountains and the American Blue Ridge Mountains both get their colorful names from a blue haze that blankets them on hot summer days.
This haze comes from small molecules that scatter sunlight, specifically small-wavelength blue light.
One of these small molecules is isoprene, a volatile, biogenic organic compound…
or, in simpler terms, an organic chemical made by living things, in this case, trees, that readily evaporates into the air.
Like many alkenes, isoprene's pi electrons in double bonds make it a reactive molecule.
It reacts with ozone, nitrogen dioxides, and other atmospheric pollutants -- not always in good ways.
And isoprene polymerizes to make a major component of natural rubber.
The reactions we’ll learn over the next few episodes will let us add different things to alkenes.
But first, we have to revisit some alkene nomenclature and another familiar friend:
carbocations, those molecules with positively charged carbon atoms.
[Theme Music]
An alkene is a molecule containing carbon-carbon double bonds.
Unlike single bonds, double bonds are rigid, they can’t easily rotate because the pi bond would need to break first and that costs energy.
In episode 2, we talked briefly about cis-and trans-isomers of alkenes.
As a refresher, this naming system helps us describe two different geometric isomers around a double bond.
For example, let's look at pent-2-ene.
If the methyl and ethyl groups are on opposite sides around the double bond, we call it trans-pent-2-ene.
And if the methyl and ethyl groups are on the same side around the double bond, we call it cis-pent-2-ene.
But, in the context of alkenes, cis- and trans- is an old school naming system that only works when the double-bond carbons are attached to two hydrogens and two R-groups.
If we consider something like 2-chloropent-2-ene, the cis- and trans- system fails to help us accurately communicate where the groups are.
So we need a better way!
Thankfully, organic chemists have us covered.
Similar to assigning R and S enantiomers of molecules, we can prioritize the groups on each carbon of the double bond, using the rule that higher atomic number means higher priority.
In this first isomer of 2-chloropent-2-ene, the carbon on the left side of the double bond is attached to an ethyl group and a hydrogen atom.
The ethyl wins priority, because its carbon has a higher atomic number than hydrogen.
The carbon on the right side of the double bond is attached to a chlorine and a methyl group.
Here the chlorine wins priority.
If we mark our priority winners and losers, we can see that high priority groups on each double-bonded carbon are on the same side of the double bond.
So this is the Z isomer, which comes from the German word zusammen for together.
Or as I remember it: on ze Zame Zide.
So we have (Z)-2-chloropent-2-ene.
The other isomer of 2-chloropent-2-ene has high priority groups on opposite sides with respect to the double-bonded carbon atoms.
So this is the E isomer, derived from the German word entgegen or opposite.
This isn’t quite as clever, but I remember it as Ecross for across.
So it's (E)-2-chloropent-2-ene.
Assigning priorities isn't always so easy, so if we have a tie, we need to keep hopping along until one group wins.
For example, in this molecule, the left-hand carbon of the double bond is straightforward.
Bromine wins priority over the methyl group.
But on the right-hand carbon, the first position gives us the same thing: a carbon with two hydrogens attached.
A tie!
So we have to keep going and compare the next two atoms.
There, we can see that the triple-bonded carbon wins priority, because it's like 3 carbons at once compared to 1 carbon on the other side.
Now we can see our priority winners are on the Zame Zide, so it's a Z-isomer, and we can call this guy (Z)-5-bromo-4-ethylhex-4-en-1-yne.
Returning to our first example, trans-pent-2-ene can be more precisely called E-pent-2-ene, and the cis isomer is the Z-alkene.
And if one side of the alkene has two of the same group, like 2-methylpent-2-ene here, we don’t need to use E and Z.
So now we know how to name alkenes a little bit better, and we can start playing around with some reaction chemistry.
Many of the chemical reactions associated with alkenes are addition reactions,
which means the pi electrons are attracted to electrophiles, and groups get added to the carbons on each side of the double bond.
In episode 12, we saw what happens when we mix hydrogen bromide and cis-but-2-ene.
It's a pretty straightforward nucleophilic attack where the alkene double bond attacks a proton, and one of the previously double-bonded carbons gets a bromine.
No matter which carbon bonds to the bromine, the product is the same.
But things aren't usually so simple.
For instance, if we add hydrogen bromide to 2-methylbut-2-ene, which isn't symmetrical, we can potentially make two different products.
Here's the catch though -- when we do this reaction in a lab, we only make one of these two products.
This seemingly-mysterious observation actually has a perfectly good explanation.
It's because of the stability of different carbocations, which are positively charged carbon atoms.
To see where carbocations come in, we have to break this reaction into steps.
Remember that hydrogen bromide is a strong acid, so it's fully dissociated into H+ and Br- ions.
Most textbooks show an attack on an undissociated molecule of hydrogen bromide, so that’s how we’ll show it in this series.
So first, the alkene initiates a nucleophilic attack on the proton formed by the dissociation of hydrogen bromide --
donating a pair of electrons, snagging the proton, and leaving the remaining carbon of the double bond short two electrons.
This creates a positive charge: a carbocation.
There are two possible carbocations that can form here.
One where the positive charge ends up on a carbon surrounded by 3 other carbons, called a tertiary carbocation.
And one where the positive charge is on a carbon surrounded by 2 carbons and a hydrogen, called a secondary carbocation.
The next step is to form the product: a bromide nucleophilic attack on the carbocation, to form a bond.
Where the carbocation forms ultimately determines what product we get.
And for some reason the tertiary carbocation is what forms, because that's what leads to our observed product!
So, I don't know about you, but I'm wondering: why does this reaction happen this way?
Well, we describe carbons with more carbon-carbon bonds as more substituted and more stable.
If we were to draw four similar carbocations, as we replace carbon-carbon bonds with carbon-hydrogen bonds, the carbocation becomes less and less stable.
In other words, a tertiary carbocation is more stable than a secondary one, which is more stable than a primary one, which is more stable than a plain ol' methyl carbocation!
The more substituted carbocation is stabler than a less substituted one.
This pattern in stability is caused by two things: an inductive effect and hyperconjugation.
The inductive effect is where the electron density is sort of spread out through the sigma bonds to stabilize the positive charge.
It's like if you're trying to balance on a unicycle.
The carbon-carbon bonds are your friends with their feet firmly on the ground, who can stick out a hand (or some electrons in bonds) for support.
However, the carbon-hydrogen bonds are friends that aren't strong enough to help.
Basically, more alkyl groups surrounding the carbocation lead to a bigger stabilizing effect.
Hyperconjugation is an even more spread-out kind of stabilization, borrowing electron density from sigma bonds kinda sideways from the positive charge.
To sort of continue the unicycle metaphor, it's like instead of just a hand, your friends actually lean their body up against you to help you balance,
sort of like a big bear hug.
So here, too, additional carbon atoms have more sigma bonds that can stabilize the carbocation.
So, when you add a hydrogen halide to an alkene, the product will be formed from the most stable carbocation.
This pattern is called Markovnikov’s rule: in these reactions, the proton will add to the side of the double bond that has the most hydrogens.
That means the positively charged carbon will be connected to more sp3 carbons,
which means it's more substituted, and therefore the more stable of the two possible carbocations!
With practice, we can use Markovnikov’s rule to help us predict lots of products.
For example, let's try adding hydrogen bromide to 1-methylcyclohex-1-ene.
We know the nucleophilic attack starts at the double bond and attacks the proton.
Then, to follow Markovnikov's rule, we have to put the proton on the side of the double bond with the most hydrogens --
that's the carbon with one carbon-hydrogen bond instead of none.
The positive charge will end up on the other carbon, which gives it more stabilization!
We'll make a tertiary carbocation instead of a secondary one.
The bromide does a nucleophilic attack...
And our product is 1-bromo-1-methylcyclohexane.
Seems pretty simple, right?
Almost too simple…
So let's try another example and use Markovnikov’s rule again to predict the product.
Here we have hydrogen bromide and 3,3-dimethylbut-1-ene.
If we go through the same steps, do a nucleophilic attack, put the proton on the carbon with two carbon-hydrogen bonds instead of one, and get a positive charge on the other carbon...
our expected product would be 2-bromo-3,3-dimethylbutane.
But when we do this reaction in the lab, we get a mixture of two products.
And our expected product is just the minor one -- not even the majority!
We just learned Markovnikov's rule, and now this reaction puts the bromine in a place where we didn’t even have a double bond to start?!
Okay. Deep breath.
Organic chemistry is tricky but it does make sense a lot of sense with practice.
Are we really breaking science or is something more subtle going on here?
Let’s see where and why this reaction gets weird.
First, like we've seen before, the alkene attacks the proton from hydrogen bromide, and the positive charge ends up on the carbon that creates the more stable secondary carbocation,
instead of the less stable primary carbocation that we didn't make.
This is where I tried to do the next step of the reaction and how we get our minor product.
But for the major product to form, the positive charge ends up on a different carbon somehow.
That "somehow" is a sneaky move called a 1,2-alkyl shift.
As a reminder, alkyl groups are methyls, ethyls, and things like that.
And in this case, it's a 1,2-methyl shift where a methyl group, CH3, moves over to the positively charged carbon next door, taking its electrons with it.
This move only happens with groups on the adjacent carbon atom.
That creates a new, even more stable tertiary carbocation.
So when the bromide does its nucleophilic attack, we get the major product that we see experimentally:
2-bromo-2,3-dimethylbutane.
So… technically my earlier prediction wasn't wrong, it was just incomplete.
Depending on how we follow our reaction mechanism roadmap, and whether the 1,2-methyl shift happens, this reaction can have two different products.
Not all carbocations rearrange like this, as we saw with the first two reactions.
This multi-directional-map only happens when there's an even more stable carbocation that can form,
so we have to keep an eye out for carbons branching out next to the spot where the carbocation forms.
And not just for shifty alkyl groups.
Turns out, hydrogen atoms can also do their own sneaky move called a 1,2-hydride shift.
To see this in action, let's look at an example of hydrogen bromide and 3-methyl-pent-1-ene.
We do the same nucleophilic attack initiated by our alkene, and pick the secondary carbocation as our product
because it's more stable than the primary option.
Then, a reaction with bromide would form the product, 2-bromo-3-methylpentane.
But this time we won't be fooled!
We're keeping an eye out for more stable carbocations and, what do you know, there's another possible product.
Bumping the positive charge left one carbon would give us a more stable tertiary carbocation.
And we can make that happen with a 1,2-hydride shift.
It's a very similar dance where a hydride, or hydrogen with its pair of electrons, shifts over to the positively charged carbon next door.
By taking the electrons from its bond with it, the hydrogen leaves behind a positive charge on the carbon it abandoned!
Now, there is a methyl group here too so I guess we could try a 1,2-methyl shift again… but that actually wouldn't help us much.
See, the hydride shift creates a tertiary carbocation when we had a secondary carbocation.
But if we were to do a methyl shift, it doesn't help with the stability.
These two secondary carbocations are energetically the same, so we need to remember that rearrangements don't just happen for the heck of it --
they form a more stable intermediate.
After we do the hydride shift to gain stability and the bromide does its nucleophilic attack, we get the major product that we see in the lab:
3-bromo-3-methylpentane.
This is another reaction that has two possible roadmaps, and a major and minor product.
It just used a different kind of shift to help increase carbocation stability!
Throughout organic chemistry, energy and stability will continue to be guiding forces as we puzzle our way through reactions.
But for this episode, we learned that:
The cis/trans nomenclature for alkenes is limited, so we can use the E/Z system for more precision
Markovnikov’s rule can help us predict products of addition reactions involving alkenes
Carbocations are stabilized by the inductive effect and hyperconjugation
And we have to inspect alkene addition reactions for possible rearrangements through 1,2 shifts that make more stable carbocations.
Next episode we’ll get into thermodynamics and how to use free energy and kinetics to help us predict reaction products.
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