Leah here from leah4sci.com and in this video we're going to look at Markovnikov versus
Anti-Markovnikov when it comes to alkene addition reactions.
When working on your alkene addition reactions, if you have a perfectly symmetrical alkene
and you subjected to a reaction like hydro-halogenation reaction with HBr I have an option of putting
the hydrogen on bromine on either of the two carbons for the pi-bond, if we call this carbon
two and carbon three and draw out the two products where the hydrogen is on two and
bromine on three and then reverse it, you'll notice that this is actually the same exact
product.
Drawing the second one backwards is redundant because when I name it, i'm forced to number
it from the side that gives bromine the lower number for a product of two bromobutane, given
that we have symmetry on the molecule, it didn't matter where I put the bromine.
But what happens if I try to do the same reaction with an asymmetrical starting alkene where
in this case, the pi-bond sits between carbon two and three where carbon two is a tertiary
carbon and three is a secondary carbon.
If I carry out the same reaction, first breaking the pi-bond and then adding hydrogen on bromine
for the first product, I'll add hydrogen to carbon two and bromine to carbon three, for
the second we put bromine on two and hydrogen on three.
Remember that in skeletal structure, hydrogens are invisible so all we're left with is bromine
sitting on carbons two or three, now we have two completely different products.
The first one is 2-bromo-3-methylbutane, the second is 2-bromo-2-methylbutane, very different
molecules so the question is which is the correct product and this is where Markovnikov's
role comes in.
Markovnikov discovered that when you have hydrohalogenation of an asymmetrical starting
alkene such as this one, where the pi-bond sits between a tertiary and secondary carbon.
We're going to get two different products but one of them will be favored, the product
is going to favor the molecule that has the halogen on the more substituted carbon rather
than the halogen on the less substituted carbon, making 2-bromo-2-methylbutane a major product
and this one here the minor product.
But this rule can be expanded to all the alkene addition reactions where we say a molecule
that follows Markovnikov's rule is one where the substituent adds to the more substituted
carbon, specifically of the two carbons that used to form the alkene and the reaction that
goes in to Markovnikov is one where the substituent add to the less substituted carbon of the
two that used to form the alkene.
Anytime you learn a rule, you're looking for a quick way to remember and apply it, but
here's the warning, if you memorized without understanding, you might get confused and
mess it up later.
So let me show you the shortcuts and then lets explain why Markovnikov's rule is that
way.
So if you're faced with something complicated like a halohydrin formation, you'll still
be able to figure out how to follow Markovnikov's rule.
The first way to memorize this to recognize that the nucleophile adds to the more substituted
carbon.
In the case of hydrohalogenation, the intermediate has a bromide ion and solution, that would
be the nucleophile and if choosing between a secondary and a tertiary carbon, we know
it's going to add tertiary.
Another way to remember this even though its backwards, it does work is that hydrogen adds
to the less substituted carbon when looking at the same molecule knowing that the nucleophile
has to add to the tertiary carbon, the leftover hydrogen is going to add to the less substituted
or secondary carbon.
The reason some students prefer this option is the nucleophile changes and sometimes it’s
hard to recognize, but if you know you have a hydrogen, it doesn't matter what is attached
to you, you know where to put the hydrogen.
Shortcut 3 is actually what I learned back in undergrad and it's simply another version
of shortcut number and that is to remember the quote, that the rich get richer but specifically
we're talking about the hydrogen rich, get hydrogen richer.
If we at the same starting alkene, even though it's written in skeletal structure with the
hydrogens are invisible, you have to remember that a tertiary carbon with a pi-bond has
a total of four bonds, one, two, three, and four, that means it has no hydrogen atoms,
it's hydrogen poor but a secondary carbon with a pi-bond only has three bonds to carbon,
that means there is a fourth bond to hydrogen and so this is the only carbon `that has hydrogen
rich and therefore it gets hydrogen richer, meaning the hydrogen adds to the less substituted
carbon, this is more obvious when your asymmetrical alkene is between a secondary and a primary
carbon because the secondary carbon has one hydrogen and three bonds to carbon, the primary
only has one sigma and one pi-bond to carbon and two bonds to hydrogen.
If we look at the hydrogen rich gets hydrogen richer, the primary or less substituted carbon
has two hydrogen atoms, the secondary or more substituted carbon only has one hydrogen atom,
therefore the hydrogen rich gets hydrogen richer.
If you're finding this video helpful so far, make sure you give it a thumbs up and let’s
continue.
Now if you're thinking Leah these are all great, which one should I memorize?
This is what I recommend, not a single one.
These are great if you're looking to pull it off quickly but if you don't understand
what you're looking at, you're going to have a hard time applying this to tricky questions,
so lets sidetrack from this conversation and make sure that we understand what Markovnikov's
rule is actually all about so that no matter how it's presented you know how to tackle
the problem.
The principles from Markovnikov's rule goes back to a very simple phenomenon in organic
chemistry in any science and in life.
Something happy will be very stable and unreactive.
If it's unhappy, it will be unstable and therefore very very reactive, let’s take this concept
to step further and apply it to the mechanism of this reaction.
As I teach in the hydrohalogenation video, the first step in this reaction is the nucleophilic
pi-electrons on the alkene are going to reach out for and grab that hydrogen atom.
Hydrogen can only have one bond so when it's grabbed and a new bond forms between itself
and carbon, the electrons between hydrogen and bromine have to collapse and break away
from hydrogen and on to bromine.
We have two potential intermediates for this reaction, with the pi-bond initially between
carbons two and three, I have the option of placing hydrogen on carbon number two or on
carbon number three where the other carbon that used to hold the pi-bond is now missing
a bond, it's deficient giving us a positive charge or a carbocation.
This is the key, this right here is what helps you determine the direction for Markovnikov's
rule, carbocation stability, that's the key.
As I teach in the carbocation stability tutorial link below, the more substituted the carbocation,
the more stable it's going to be, that means a tertiary carbocation is going to be much
more stable than a secondary carbocation which is much more stable than a primary carbocation.
For this intermediate, we're faced between a secondary carbocation and a tertiary carbocation,
what is it about the more stable carbocation that tells us which product is going to form?
It's a kinetics thing, a reaction speed.
Reactions are only as fast as their slowest step, the alkene is relatively stable and
when it attacks that hydrogen, it forms an unstable intermediate, yes the tertiary carbocation
is more stable than the secondary, but that's simply the lesser of two evils and an unstable
intermediate.
Therefore, the more stable the intermediate that is going to form, the more likely it's
going to form, the faster it’s going to form.
If an intermediate has to form that is not stable, it's going to be very very slow and
it may or may not form.
So if we're looking at a reaction where we can have some secondary and some tertiary
forming, the tertiary are going to form at such a faster rate compared to secondary that
if we compare the amount of product we're going to have a lot more that resulted from
that tertiary intermediates.
The last step in this reaction is where the nucleophile attacks the carbocation, but the
nucleophile doesn't attack the more substituted carbon because it wants to, the nucleophile
is greedy, it's negative, it's looking for a positive charge and wherever that positive
charge happens to be, that's where it's going to attack.
The fact that the tertiary carbocation formed so much faster means that there are so many
more tertiary carbocations available to attack and all the bromines and solution are rushing
to attack whatever they can, that is why we get so many more of this product.
So if we have to summarize Markovnikov's rule again, then we can say Markovnikov's rule
tells us that the nucleophile adds to the carbon that would have form the most stable
carbocation.
If you need to hear this again, rewind the video, listen to the explanation, make sure
it clicks and then lets try a few problems to make sure you really get it.
So you're given a simple alkene reaction where H2SO4 react with propene, you recognize that
H2SO4 is simply a source of H+ and solution and that this reaction is going to proceed
by a carbocation intermediate.
You look at your starting molecule, you identify that the pi-bond sits between a primary carbon
and a secondary carbon and you know that Markovnikov's rule says that the nucleophile in this case
the OH is going to add to the carbon that would have form the most stable carbocation
intermediate, you know to put the alcohol at the secondary carbon, what about a situation
where you have a carbocation rearrangement?
How does that follow Markovnikov's rule?
Say we're given this molecule where a pi-bond is sitting between a primary and a secondary
carbon but we also have a tertiary carbon nearby, if we're told to react this with the
same reagent, H2SO4 in the presence of H2O, we once again look for where the most stable
carbocation would have formed.
Our first thought is to put the carbocation on the secondary carbon but here's another
trick that you need to remember, anytime you have a secondary carbocation near a tertiary
carbon, you're going to get a hydride shift, if you're not familiar with this concept,
make sure you watch the tutorial link below, that will show you how the carbocation winds
up at the tertiary carbon and then by the time the water, the nucleophile is looking
to attack, it doesn't care where it attacks, it simply goes to where the carbocation is
and that gives us an unexpected product where the alcohol sits on the most substituted carbon.
If you blindly followed the memorization from before and simply look for what's more or
less substituted without taking into account the carbocation intermediate, you would likely
have put your alcohol here and that would be the incorrect product.
What about a reaction that doesn't follow Markovnikov's rule?
How do we use what we now understand to answer products that are into Markovnikov?
Once again, I want to remind you Markovnikov's rule was discovered for reactions that have
carbocation intermediate but ultimately it was about stability with anti-Markovnikov
products, we still want to look at what is the most stable intermediate?
Two reactions that may come to mind will be the hydroboration of an alkene where we react
the pi-bond with BH3, THF, NaOH and H2O2 or radical halogenation where we react the alkene
with H(X), in this case HBr and peroxides.
These both gave us anti-Markovnikov products but not because we don't have a carbocation
intermediate, but rather because of what we do have in the intermediate.
For hydroboration, the intermediate has both a boron and a hydrogen binding to the alkene
as demonstrated in the hydroboration mechanism link below, to give us an intermediate that's
quite bulky, so when the pi-bond has to line up with the boron and the hydrogen, it's less
hindered when the less substituted carbon lines up with the bulk of the boron and everything
it's attached to and the more substituted carbon lines up with the tiny hydrogen.
This is why alcohol ultimately winds up on the less substituted carbon, again the mechanism
is shown in the video link below.
With the radical halogenation of alkenes, once again we take the intermediate into consideration.
The step that ultimately breaks the alkene is one in which a radical halogen is going
to attack the alkene, forcing the alkene to break apart in homolytic cleavage where one
electron goes to each side, giving us an intermediate that has the halogen on the less substituted
carbon and the radical sitting on the more substituted carbon and this is key.
Radicals like carbocations are more stable when they're more substituted, because the
bromine is the thing that added first, it had to add somewhere that would allow the
resulting radical to be on a more substituted and more stable carbon and in the final step
when a hydrogen is added, the hydrogen just adds as the a by the way because that is the
only spot but bromine has already added the less substituted carbon due to forming the
most stable intermediate.
In both of these situations, Anti-Markovnikov has nothing to do with a carbocation intermediate
but it has everything to do with the most stable possible intermediate, and finally
how do you figure out Markovnikov's rule?
When the two things that we're adding are different but there's no hydrogen, for example
in halohydrin formation where we'll react Cl2 in the presence of H2O, we know that this
reaction follows Markovnikov's rule but there's no carbocation intermediate and there's no
hydrogen to add to the carbon chain.
Once again, it's the intermediate that tells us what to do, as you work through the mechanism
which I teach in the alkene reactions series link below, we have an intermediate that forms
a chloronium bridge that's a chlorine atom bound to both of the carbons that used to
hold the pi-bond, only one lone pair and a positive charge.
Chlorine being a highly electronegative atom is not very happy with the positive charge
and pulls on the electron density between itself and the carbon atoms, making the carbon
atoms partially positive.
In an asymmetrical carbon, they're not equally partially positive, the less substituted carbon
that would form a less stable carbocation holds less of the positive charge.
The more substituted carbon that would have formed a more stable carbocation is going
to hold more of the positive charge so when water shows up in the next step, looking to
attack something positive, it's more attracted to the more partially positive carbon which
happens to be the more substituted carbon, telling us that in this case Markovnikov's
rule has to deal with water attacking the more substituted carbon, breaking this bond
between carbon and chlorine meaning chlorine gets stuck on the left substituted carbon
not because that's where it wants to go but rather because it was holding on to both carbons
and water kicked it out.
The final halohydrin now has a halogen on the less substituted carbon because that's
where it got bumped to and then alcohol on the more substituted carbon because it attacked
the more partially positive carbon again following Markovnikov's rule.
I realized this was a lot to take in so make sure you go through it slowly and carefully
and once you understand this, I want you to go back to the entire alkene reaction series,
mechanism by mechanism and make sure you can apply what you learned here to the individual
reactions.
Once you feel confident with alkene reactions, make sure you try the alkene reaction practice
quiz on my website where you can find this entire series and the alkene reactions cheat
sheet by visiting leah4sci.com/alkenereactions , again, that's leah4sci.com/alkenereactions.