Leah here from leah4sci.com and in this video we're going to look at the reaction
and mechanism for the Anti-Markovnikov Radical Halogenation of Alkenes. You can
find my entire alkene reaction series along with the alkene reaction cheat sheet and practice quiz
by visiting my website linked below or going to leah4sci.com/alkene-reactions.
Radical Halogenation often called the light reaction compared to the non-radical
dark reaction is a reaction where an Asymmetric Alkene for example this 2-methyl, 2-butene
reacts with Hx, here H-Br in the presence of peroxide to give us an anti-Markovnikov less
substituted product where the bromine will add to the secondary carbon and the hydrogen adds to the
tertiary more substituted carbon. Compare this to the regular hydrohalogenation reaction that does
not take place in the presence of peroxide where when the pi bond breaks, the bromine adds to the
tertiary or more substituted carbon and hydrogen adds to the secondary or less substituted carbon.
This is the Markovnikov product but the radical reaction is the opposite giving us
the anti-markovnikov product. Since radical reactions are initiated by heat or light,
this is often called the light reaction and the non-radical reaction
often called the dark reaction. Two things to remember before we jump into the mechanism.
The first is that Markovnikov's Rule is less about the product itself and more about the stability
of the intermediate that drives this specific outcome. And 2, since this is a radical reaction,
we'll be referring to the Initiation and Propagation and termination steps.
If you're not comfortable with either, pause this video and see the tutorials linked below
or go to my website leah4sci.com/radical. Ready? Let's take a look. Most alkene reactions
begin where the nucleophilic pi electrons will reach out and attack some kind of electrophile
not in this reaction. Radical reactions always begin with the initiation step
with the formation of a radical and in this case peroxides are going to set it off.
Recall from General Chemistry that peroxide or per-oxide is a molecule that has an oxygen bound
to another oxygen atom. You're probably very familiar with hydrogen peroxide H2O2 and you’ve
likely come across some alkyl peroxides written as R-O-O-R. For this reaction we'll look at a methyl
peroxide which is a CH3 bound to an Oxygen bound to an Oxygen and bound to another CH3.
Unlike the initiation for breaking apart something like Chlorine or Bromine, Oxygen does not require
too much input of energy because it's already so unstable. This is why the hydrogen peroxide in
your medicine cabinet is in a dark bottle. Because just the light in your bathroom
is enough to start breaking those bonds and that's exactly what happens here. We have a homolytic
cleavage reaction where the oxygens each grab one electron and break apart to give us 2-OR radicals
in this case CH3 bound to an Oxygen, remember each Oxygen had 2 lone pairs. And now we get that extra
single unpaired radical electron. This is very unstable and looking to attack another molecule.
And if there's something like an H-Br hanging around, Oxygen will use its electron to attack
that Hydrogen, forcing Hydrogen to provide one of its own electrons to form that bond, remember come
over to dinner but you gotta bring dinner. And that second electron will break off onto bromine.
The two products in this step include an alcohol which comes from that CH3O radical
combining with a hydrogen along with an unstable and very reactive bromine radical that has 3 lone
pairs and one lone electron from the bond between hydrogen breaking apart. Since we had one radical
in the reactants and one radical in the products, this is a propagation step. And this
unstable Bromine radical is looking to continue this chain of propagation
because it's unhappy as it is. And now finally that alkene comes into play. The alkene has two
electrons that are sitting in P-orbitals above and below the molecule and that makes it very
easy for a bromine radical to attack. When bromine attacks it's looking to form a bond between itself
and carbon and one of the radical electrons will come up to meet it so that they can form a bond
where the second electron will break on to the molecule to sit as a lone electron on one of the
carbon atoms. Question is, which one? And this step here is critical because the stability of
this intermediate is what determines the regioselectivity of the final product.
My two options are to place bromine on the more substituted carbon with a bond between itself and
that red lone electron giving me that blue lone electron as a radical on the secondary carbon. My
second option is to place bromine on the secondary carbon with a bond between itself and that blue pi
electron leaving the red electron to sit as a lone radical on the tertiary carbon. The way to
determine the final product is to understand the stability of the radical intermediate.
It's not about the bromine, bromine is looking to meet either of the two electrons, instead it's
about understanding which intermediate is more stable and that intermediate is more likely to
form. If you recall that radical stability is the same as carbocation stability and that the more
substituted radical is the more stable radical. You'll see that this is the preferred intermediate
because the radical sits on the tertiary carbon having the support of 1,2,3 other carbon atoms.
But even a stable tertiary radical is still a radical and that means it's still very reactive
and so it looks for another molecule of H-Br floating around in solution and
attacks the hydrogen to form a bond forcing hydrogen to provide one of its own electrons
and causing that second electron to break off onto bromine. This gives us our final product
with the bromine sitting on the secondary or less substituted carbon. A hydrogen atom bound at the
tertiary or more substituted carbon and another bromine radical floating around in solution.
Since we have a radical in the reactants and another radical on the products,
this is yet another propagation step where bromine is gonna look for another alkene to attack,
ultimately forming another bromine radical and another one and another one and so on. A quick
note about the stereochemistry of this reaction, notice that bromine is sitting on a secondary
carbon that now has 4 unique substituents. We have a methyl, an isopropyl, bromine,
and an invisible hydrogen atom. That means this right here is a chiral carbon, but because we're
starting with an alkene which means an Sp2 carbon atom that is trigonal planar or flat,
bromine can attack from the top or the bottom so that if the final product is chiral we get
a racemic mixture with 50% R and 50% S. For even more on radicals, covering everything from
formation, hybridization, resonance to radical reactions and more, visit my website linked below
or go to leah4sci.com/Radical. The link again is leah4sci.com/Radical.