Class 11 Exam > Class 11 Videos > Friedel-Crafts alkylation - Chemistry, Class 11
Friedel-Crafts alkylation - Chemistry, Class 11 Video Lecture
FAQs on Friedel-Crafts alkylation - Chemistry, Class 11 Video Lecture
| 1. What is Friedel-Crafts alkylation? |  |
Ans. Friedel-Crafts alkylation is a chemical reaction in organic chemistry that involves the introduction of an alkyl group onto an aromatic compound. It is a type of electrophilic substitution reaction where an alkyl halide reacts with an aromatic compound in the presence of a Lewis acid catalyst to form a new carbon-carbon bond.
| 2. What are the main steps involved in Friedel-Crafts alkylation? | |
Ans. The main steps involved in Friedel-Crafts alkylation are as follows:
1. Activation: The Lewis acid catalyst, such as aluminum chloride (AlCl3), activates the alkyl halide by forming a complex with it.
2. Formation of carbocation: The activated alkyl halide reacts with the aromatic compound, leading to the formation of a carbocation intermediate.
3. Rearrangement (if necessary): Sometimes, the carbocation may undergo rearrangement to form a more stable carbocation through a hydride or alkyl shift.
4. Deactivation: The Lewis acid catalyst deactivates the carbocation by abstracting a proton from the product, regenerating the catalyst for further reactions.
| 3. What are the limitations of Friedel-Crafts alkylation? | |
Ans. Friedel-Crafts alkylation has several limitations, including:
1. Rearrangement: The carbocation intermediate formed during the reaction can undergo rearrangement, leading to the formation of unexpected products.
2. Polyalkylation: When multiple alkyl groups are present in the reactant, polyalkylation (more than one alkyl group being added to the aromatic compound) may occur, resulting in a mixture of products.
3. Side reactions: Friedel-Crafts alkylation can also lead to side reactions, such as the formation of alkylated catalyst complexes or the formation of alkylated byproducts.
4. Reactivity of aromatic compounds: Not all aromatic compounds are suitable for Friedel-Crafts alkylation. Some aromatic compounds may not react readily or may require harsh reaction conditions.
5. Limitation to alkyl halides: Friedel-Crafts alkylation usually requires the use of alkyl halides as the alkylating agent, limiting the range of alkyl groups that can be introduced.
| 4. How can the regioselectivity of Friedel-Crafts alkylation be controlled? | |
Ans. The regioselectivity of Friedel-Crafts alkylation can be controlled through the use of different reaction conditions or by modifying the aromatic compound. Some strategies to control the regioselectivity include:
1. Use of different Lewis acids: Different Lewis acids can have varying effects on the regioselectivity of the reaction. For example, using boron trifluoride (BF3) as the catalyst can lead to para-selectivity, while using aluminum chloride (AlCl3) can lead to a mixture of regioisomers.
2. Use of protecting groups: Protecting groups can be used to temporarily block certain positions on the aromatic compound, directing the alkyl group to a specific site during alkylation.
3. Variation of reaction conditions: Changing the reaction temperature, solvent, or concentration of the reactants can also influence the regioselectivity of the reaction.
4. Substituent effects: The presence of electron-donating or electron-withdrawing substituents on the aromatic compound can affect the regioselectivity by influencing the stability of the carbocation intermediate.
| 5. Can Friedel-Crafts alkylation be used to introduce more than one alkyl group onto an aromatic compound? | |
Ans. Friedel-Crafts alkylation can lead to the introduction of more than one alkyl group onto an aromatic compound, but it often results in polyalkylation. Polyalkylation occurs when multiple alkyl groups are added to the aromatic compound, leading to a mixture of products. This can be a limitation of Friedel-Crafts alkylation, as it can be difficult to control the selectivity of the reaction and obtain a single desired product. However, by carefully selecting reaction conditions and controlling the reactivity of the alkylating agent, it may be possible to achieve monoalkylation, where only one alkyl group is added to the aromatic compound.
Text Transcript from Video
Let's look at the reaction
for Friedel-Crafts alkylation.
So we start with
our benzene ring,
and to benzene we're going
to add an alkyl chloride,
and our catalyst is
aluminum chloride.
And the end result is to
substitute an R group, the R
group that was on
the alkyl chloride,
for a proton on
the aromatic ring.
Let's look at the mechanism
for Friedel-Crafts alkylation.
We start with our
alkyl chloride and we
add our aluminum chloride,
which we've already seen
can function as a Lewis acid.
So it could be a
Lewis acid because it
can accept a pair of electrons.
And so our Lewis
base over here is
going to be the alkyl chloride.
It's going to donate
an electron pair.
So I'm going to say this
electron pair is going
to be donated to that aluminum.
So we'd go ahead and draw the
result of R-Lewis acid base
reaction here.
So now the chlorine is
bonded to the aluminum,
and the aluminum is bonded to
these other chlorines here,
and I'm going to leave off
the lone pairs of electrons
on those just to save time.
In terms of formal
charges, aluminum now
has a negative 1 formal
charge, and this chlorine now
has a plus 1 formal
charge like that.
So we know that halogens are
relatively electronegative,
and so this chlorine here
could take these electrons,
we could show these
electrons in here moving off
onto that chlorine.
And when we do that, we're
taking a bond away from R alkyl
group and so therefore,
R alkyl group
is left with a positive
1 formal charge.
And this is a carbocation.
And this is our electrophile
in our mechanism
for electrophilic
aromatic substitution.
And since we actually form a
carbocation in this mechanism,
a carbocation
intermediate, rearrangement
is actually possible so
we have to be careful
when we're looking at a
Friedel-Crafts alkylation
reaction in terms of
predicting the products.
Our other product here, right?
Well, we would have
our aluminum still
bonded to these chlorines,
and of course, it's
bonded to this
other chlorine too.
And this chlorine now has
three lone pairs of electrons
around it so these
electrons in here,
ones that were in the
bond between our R group
and our chlorine
left, and they are now
on our chlorine like that.
And so our aluminum still has
a negative 1 formal charge
in there like that.
All right, so now we have
generated our electrophile,
which is our carbocation.
And we know, of course,
that our benzene ring
is going to react
with our electrophile.
So here is our
benzene ring, and here
is our carbocation so
positively charged.
So we know that the pi
electrons in our benzene ring
are going to function
as a nucleophile
and attack the electrophiles so
negative charges are attracted
to positive charges like that.
And so we can go ahead
and draw the results
of our nucleophilic attacks so
we now have our hydrogen still
bonded here.
And once again, I could
add the alkyl group
to either of those two carbons.
Let me just highlight
those again.
So either one of
these two, but just
to be consistent with
how I've been doing it
in the previous videos, right?
I'm going to add the R group
to the top carbon here, which
means that I took a bond
away from the bottom carbons.
Let me go ahead and
highlight that one.
So this, of course,
lost a bond, which
means that is where our
positive 1, a formal charge,
would go so.
Positive 1, formal
charge, a carbocation.
And just to save
time, I'm not going
to draw the other resonance
structures for this ion.
OK?
So remember, this will
represent our sigma complex,
and you can watch the
previous videos for how
to draw resonance structures
for this carbocation.
And so in the last step
of our mechanism here,
we're going to reform
our aromatic ring,
and so we need to deprotonate
our sigma complex in order
to do that so we could
think about these electrons
in here moving out
to take that proton,
and these electrons
moving in here
to take away the
plus 1 formal charge
and reform our aromatic ring.
So now we have our
aromatic ring reformed
to R group is on
our ring like that.
Our other products
would be HCL and also we
would regenerate our catalyst,
aluminum chloride like that.
So let's go ahead and
follow some electrons.
All right?
So these electrons in
magenta right here.
When the proton leaves, those
electrons in magenta reform
this right here.
And I can say that
these electrons in here
are forming a bond between
the chlorine and the proton
like that to form HCL.
So that's our mechanism for
Friedel-Crafts alkylation.
Let's look at a few
examples of this reaction.
So we'll start with
benzene, and to benzene
we are going to add
two chloropropane.
So go ahead and draw
two chloropropane.
That's going to be
our alkyl chloride.
And once again, we need
our aluminum chloride
as our catalyst.
And so the way I do
Friedel-Crafts alkyl reactions
is I think about what sort of a
carbocation am I going to get.
And so without drawing
the entire mechanism,
I know that this
chlorine is going
to leave during the
mechanism, and if the chlorine
leaves during the
mechanism, well, that
means that we took a bond
away from this carbon
right here, and so that's where
our plus 1 formal charge is
going to be.
So we have a plus
1 formal charge
on that top carbon right there.
If we think about a
possible rearrangement,
this is a secondary carbocation
and there's no possible way
that this carbocation
could rearrange
to form a tertiary carbocation.
And so secondary carbocations
are relatively stable,
and so this will
be the carbocation
that reacts with
our benzene ring.
And so you could think
about once again,
these pi electrons forming a
bond with that carbon so let's
go ahead and draw the results
of our nucleophilic attacks.
So we have our ring, we
have our other pi electrons,
we have our hydrogen
that's already on our ring,
and now we formed a bond to
three carbons so let's go ahead
and once again highlight
some electrons here.
So these electrons
in magenta have now
formed a bond between those
two carbons right there.
And always double check
yourself and count your carbons
so we have one, two, three
carbons on that carbocation.
So we have to make sure we have
the same number over here--
one, two, and three.
So in our last step, we
know that deprotonation
of our sigma complex, right?
So this is a positive
1 formal charge here.
Deprotonation of
our sigma complex,
we get rid of that
positive 1 formal charge
and reform our
aromatic ring so we've
seen that these electrons
would move in here,
and so we have our product.
So we have our aromatic
ring like that.
And then we have this isopropyl
group coming off of our ring
so we form isopropyl benzene
as our final product.
Let's do one more
Friedel-Crafts alkylation.
This one will have a possible
rearrangement so let's go ahead
and start with benzene again.
So we have our benzene
ring, and to benzene
we are going to add butyl
chloride here so one, two,
three, four carbons.
And then a chlorine like that,
and then aluminum chloride,
once again, as our catalyst.
And so if you're
approaching this problem,
once again, I like
to think about what
sort of a carbocation is going
to form in the mechanism.
So I know that this
chlorine would kick off
during the mechanism, and so
that would take away a bond
from this carbon
right here so we're
going to get a plus 1 formal
charge on that carbon.
So if I go ahead and draw
out my four carbons here,
and I know there's
going be a plus 1
formal charge on that
carbon on the end,
which we, of course, know
this is a primary carbocation,
and primary carbocations
are not very stable
so there could be some
rearrangement here.
And if you think
about what's the best
way to rearrange our carbocation
to see if we can form a more
stable one, you know
that there are hydrogens
attached to the carbon next
door to that positive 1
formal charge.
And so we get a
hydride shift here.
So we've seen how to do
that in earlier videos.
So you could think about
the hydrogen and these two
electrons, all of them are
going to move over here.
So if we go ahead and show the
result of that hydride shift
so that hydrogen and
those two electrons
are going to move
over there, that
takes away the plus
1 formal charge
of the carbon on the end.
And we took a bond
away from this carbon
now so that's this
carbon over here so that
is where the plus 1 formal
charge is going to go.
And so now we have a
secondary carbocation.
So just to refresh your memory,
this is a secondary carbocation
because the carbon in magenta
is bonded to two other carbons.
So we have a
secondary carbocation,
which we know is more stable
than a primary carbocation.
And so if we can think about
two possible products here.
So our benzene ring could react
it with our primary carbocation
or more likely
it's going to react
with our secondary carbocation
over so let's go ahead
and draw the
results if it reacts
with a secondary carbocation.
So I could think about
my electrons in here
and my pi bonds going
all the way over here
to react with that carbocation,
that secondary carbocation,
and so we can go ahead
and show the result
of our nucleophilic attack.
So once again, we
have our pi electrons.
We had a hydrogen already
bonded to that ring there,
and we're going to
form a bond to what
used to be our carbocation
so we have four carbons,
and one of them is going
to branching like that.
So let's go ahead and analyze
what we've done so far.
So plus 1 formal charge
on our sigma complex.
So our electrons, our
pi electrons in here,
are going to be these
electrons right here so forming
our carbon-carbon
bond like that.
And so when we look at
what's attached to that--
let me go ahead and highlight
some of these carbons here.
So I'll start in red.
So this carbon right
here I'm going to say
is this carbon like that.
So in magenta this carbon
down here in magenta
is, of course, this
carbon right here
in magenta, the one
that's bonded to our ring.
And then, of course, we could
follow a few more carbons here.
So I'm going to say that
this carbon in green
that's, of course,
this one right here.
And then finally, this
carbon right here in blue
is this carbon.
So always count your
carbons and make sure
that you have the
right branching
for something like this.
So in the last step, of course,
deprotonation of our sigma
complex, these electrons would
move into here like that.
And we have one of our
final products here.
So let's go ahead
and draw this one
so we would have that
as our major product
for this reaction.
It is possible for the
primary carbocation
to react as well
so let's go ahead
and draw the result of that.
So if these pi
electrons instead went
for the primary
carbocation, they'd
form a bond with that
carbon right there.
And so we could go ahead
and draw the result of that
so we would have
our ring once again.
And once again, we
have our hydrogen,
and this time we would have four
carbons coming off of our ring.
So one, two, three,
four like that.
And once again,
positive 1 formal charge
on our sigma complex.
So let's go ahead and
once again the highlight
those electrons in magenta.
So the electrons
in magenta right
here they formed this
carbon-carbon bond.
And they're bonding to
this carbon this time.
So the one in the end,
so I'm going to say it's
that one right there.
So next we have the carbon
in magenta right here,
and so that would be this one.
And just to be consistent
with our colors,
I'll go with green next so
this carbon is green and so
that one is green right here.
And then finally, this carbon
in blue is this carbon in blue.
The reason I'm taking the time
to show all of these carbons
is I find that this is one of
the areas where students will
make the most
mistakes so you have
to be careful about
following your electrons
and counting your
carbons when you're
drawing your final
products here.
So deprotonation of
our sigma complex,
these electrons move into there
and we would form butylbenzene
as the minor product in
this reaction so let's
go ahead and draw that out.
So we have four carbons coming
off of our benzene ring,
and this would be
the minor product
since it comes from the
less stable carbocation.
The primary
carbocation is not as
stable as the
secondary carbocation.
And so this shows one
of the limitations
to the Friedel-Crafts
alkylation reaction.
You can't always
control the alkyl group
that you're putting on
to your ring because
of the possible rearrangement
because there's a carbocation
present in the mechanism.
So if your goal was
to make butylbenzene,
you wouldn't be able to make it
in extremely high yield using
a Friedel-Crafts alkylation,
and so we'll see a better way
to do that in the
next video, which
is on Friedel-Crafts acylation.
for Friedel-Crafts alkylation.
So we start with
our benzene ring,
and to benzene we're going
to add an alkyl chloride,
and our catalyst is
aluminum chloride.
And the end result is to
substitute an R group, the R
group that was on
the alkyl chloride,
for a proton on
the aromatic ring.
Let's look at the mechanism
for Friedel-Crafts alkylation.
We start with our
alkyl chloride and we
add our aluminum chloride,
which we've already seen
can function as a Lewis acid.
So it could be a
Lewis acid because it
can accept a pair of electrons.
And so our Lewis
base over here is
going to be the alkyl chloride.
It's going to donate
an electron pair.
So I'm going to say this
electron pair is going
to be donated to that aluminum.
So we'd go ahead and draw the
result of R-Lewis acid base
reaction here.
So now the chlorine is
bonded to the aluminum,
and the aluminum is bonded to
these other chlorines here,
and I'm going to leave off
the lone pairs of electrons
on those just to save time.
In terms of formal
charges, aluminum now
has a negative 1 formal
charge, and this chlorine now
has a plus 1 formal
charge like that.
So we know that halogens are
relatively electronegative,
and so this chlorine here
could take these electrons,
we could show these
electrons in here moving off
onto that chlorine.
And when we do that, we're
taking a bond away from R alkyl
group and so therefore,
R alkyl group
is left with a positive
1 formal charge.
And this is a carbocation.
And this is our electrophile
in our mechanism
for electrophilic
aromatic substitution.
And since we actually form a
carbocation in this mechanism,
a carbocation
intermediate, rearrangement
is actually possible so
we have to be careful
when we're looking at a
Friedel-Crafts alkylation
reaction in terms of
predicting the products.
Our other product here, right?
Well, we would have
our aluminum still
bonded to these chlorines,
and of course, it's
bonded to this
other chlorine too.
And this chlorine now has
three lone pairs of electrons
around it so these
electrons in here,
ones that were in the
bond between our R group
and our chlorine
left, and they are now
on our chlorine like that.
And so our aluminum still has
a negative 1 formal charge
in there like that.
All right, so now we have
generated our electrophile,
which is our carbocation.
And we know, of course,
that our benzene ring
is going to react
with our electrophile.
So here is our
benzene ring, and here
is our carbocation so
positively charged.
So we know that the pi
electrons in our benzene ring
are going to function
as a nucleophile
and attack the electrophiles so
negative charges are attracted
to positive charges like that.
And so we can go ahead
and draw the results
of our nucleophilic attacks so
we now have our hydrogen still
bonded here.
And once again, I could
add the alkyl group
to either of those two carbons.
Let me just highlight
those again.
So either one of
these two, but just
to be consistent with
how I've been doing it
in the previous videos, right?
I'm going to add the R group
to the top carbon here, which
means that I took a bond
away from the bottom carbons.
Let me go ahead and
highlight that one.
So this, of course,
lost a bond, which
means that is where our
positive 1, a formal charge,
would go so.
Positive 1, formal
charge, a carbocation.
And just to save
time, I'm not going
to draw the other resonance
structures for this ion.
OK?
So remember, this will
represent our sigma complex,
and you can watch the
previous videos for how
to draw resonance structures
for this carbocation.
And so in the last step
of our mechanism here,
we're going to reform
our aromatic ring,
and so we need to deprotonate
our sigma complex in order
to do that so we could
think about these electrons
in here moving out
to take that proton,
and these electrons
moving in here
to take away the
plus 1 formal charge
and reform our aromatic ring.
So now we have our
aromatic ring reformed
to R group is on
our ring like that.
Our other products
would be HCL and also we
would regenerate our catalyst,
aluminum chloride like that.
So let's go ahead and
follow some electrons.
All right?
So these electrons in
magenta right here.
When the proton leaves, those
electrons in magenta reform
this right here.
And I can say that
these electrons in here
are forming a bond between
the chlorine and the proton
like that to form HCL.
So that's our mechanism for
Friedel-Crafts alkylation.
Let's look at a few
examples of this reaction.
So we'll start with
benzene, and to benzene
we are going to add
two chloropropane.
So go ahead and draw
two chloropropane.
That's going to be
our alkyl chloride.
And once again, we need
our aluminum chloride
as our catalyst.
And so the way I do
Friedel-Crafts alkyl reactions
is I think about what sort of a
carbocation am I going to get.
And so without drawing
the entire mechanism,
I know that this
chlorine is going
to leave during the
mechanism, and if the chlorine
leaves during the
mechanism, well, that
means that we took a bond
away from this carbon
right here, and so that's where
our plus 1 formal charge is
going to be.
So we have a plus
1 formal charge
on that top carbon right there.
If we think about a
possible rearrangement,
this is a secondary carbocation
and there's no possible way
that this carbocation
could rearrange
to form a tertiary carbocation.
And so secondary carbocations
are relatively stable,
and so this will
be the carbocation
that reacts with
our benzene ring.
And so you could think
about once again,
these pi electrons forming a
bond with that carbon so let's
go ahead and draw the results
of our nucleophilic attacks.
So we have our ring, we
have our other pi electrons,
we have our hydrogen
that's already on our ring,
and now we formed a bond to
three carbons so let's go ahead
and once again highlight
some electrons here.
So these electrons
in magenta have now
formed a bond between those
two carbons right there.
And always double check
yourself and count your carbons
so we have one, two, three
carbons on that carbocation.
So we have to make sure we have
the same number over here--
one, two, and three.
So in our last step, we
know that deprotonation
of our sigma complex, right?
So this is a positive
1 formal charge here.
Deprotonation of
our sigma complex,
we get rid of that
positive 1 formal charge
and reform our
aromatic ring so we've
seen that these electrons
would move in here,
and so we have our product.
So we have our aromatic
ring like that.
And then we have this isopropyl
group coming off of our ring
so we form isopropyl benzene
as our final product.
Let's do one more
Friedel-Crafts alkylation.
This one will have a possible
rearrangement so let's go ahead
and start with benzene again.
So we have our benzene
ring, and to benzene
we are going to add butyl
chloride here so one, two,
three, four carbons.
And then a chlorine like that,
and then aluminum chloride,
once again, as our catalyst.
And so if you're
approaching this problem,
once again, I like
to think about what
sort of a carbocation is going
to form in the mechanism.
So I know that this
chlorine would kick off
during the mechanism, and so
that would take away a bond
from this carbon
right here so we're
going to get a plus 1 formal
charge on that carbon.
So if I go ahead and draw
out my four carbons here,
and I know there's
going be a plus 1
formal charge on that
carbon on the end,
which we, of course, know
this is a primary carbocation,
and primary carbocations
are not very stable
so there could be some
rearrangement here.
And if you think
about what's the best
way to rearrange our carbocation
to see if we can form a more
stable one, you know
that there are hydrogens
attached to the carbon next
door to that positive 1
formal charge.
And so we get a
hydride shift here.
So we've seen how to do
that in earlier videos.
So you could think about
the hydrogen and these two
electrons, all of them are
going to move over here.
So if we go ahead and show the
result of that hydride shift
so that hydrogen and
those two electrons
are going to move
over there, that
takes away the plus
1 formal charge
of the carbon on the end.
And we took a bond
away from this carbon
now so that's this
carbon over here so that
is where the plus 1 formal
charge is going to go.
And so now we have a
secondary carbocation.
So just to refresh your memory,
this is a secondary carbocation
because the carbon in magenta
is bonded to two other carbons.
So we have a
secondary carbocation,
which we know is more stable
than a primary carbocation.
And so if we can think about
two possible products here.
So our benzene ring could react
it with our primary carbocation
or more likely
it's going to react
with our secondary carbocation
over so let's go ahead
and draw the
results if it reacts
with a secondary carbocation.
So I could think about
my electrons in here
and my pi bonds going
all the way over here
to react with that carbocation,
that secondary carbocation,
and so we can go ahead
and show the result
of our nucleophilic attack.
So once again, we
have our pi electrons.
We had a hydrogen already
bonded to that ring there,
and we're going to
form a bond to what
used to be our carbocation
so we have four carbons,
and one of them is going
to branching like that.
So let's go ahead and analyze
what we've done so far.
So plus 1 formal charge
on our sigma complex.
So our electrons, our
pi electrons in here,
are going to be these
electrons right here so forming
our carbon-carbon
bond like that.
And so when we look at
what's attached to that--
let me go ahead and highlight
some of these carbons here.
So I'll start in red.
So this carbon right
here I'm going to say
is this carbon like that.
So in magenta this carbon
down here in magenta
is, of course, this
carbon right here
in magenta, the one
that's bonded to our ring.
And then, of course, we could
follow a few more carbons here.
So I'm going to say that
this carbon in green
that's, of course,
this one right here.
And then finally, this
carbon right here in blue
is this carbon.
So always count your
carbons and make sure
that you have the
right branching
for something like this.
So in the last step, of course,
deprotonation of our sigma
complex, these electrons would
move into here like that.
And we have one of our
final products here.
So let's go ahead
and draw this one
so we would have that
as our major product
for this reaction.
It is possible for the
primary carbocation
to react as well
so let's go ahead
and draw the result of that.
So if these pi
electrons instead went
for the primary
carbocation, they'd
form a bond with that
carbon right there.
And so we could go ahead
and draw the result of that
so we would have
our ring once again.
And once again, we
have our hydrogen,
and this time we would have four
carbons coming off of our ring.
So one, two, three,
four like that.
And once again,
positive 1 formal charge
on our sigma complex.
So let's go ahead and
once again the highlight
those electrons in magenta.
So the electrons
in magenta right
here they formed this
carbon-carbon bond.
And they're bonding to
this carbon this time.
So the one in the end,
so I'm going to say it's
that one right there.
So next we have the carbon
in magenta right here,
and so that would be this one.
And just to be consistent
with our colors,
I'll go with green next so
this carbon is green and so
that one is green right here.
And then finally, this carbon
in blue is this carbon in blue.
The reason I'm taking the time
to show all of these carbons
is I find that this is one of
the areas where students will
make the most
mistakes so you have
to be careful about
following your electrons
and counting your
carbons when you're
drawing your final
products here.
So deprotonation of
our sigma complex,
these electrons move into there
and we would form butylbenzene
as the minor product in
this reaction so let's
go ahead and draw that out.
So we have four carbons coming
off of our benzene ring,
and this would be
the minor product
since it comes from the
less stable carbocation.
The primary
carbocation is not as
stable as the
secondary carbocation.
And so this shows one
of the limitations
to the Friedel-Crafts
alkylation reaction.
You can't always
control the alkyl group
that you're putting on
to your ring because
of the possible rearrangement
because there's a carbocation
present in the mechanism.
So if your goal was
to make butylbenzene,
you wouldn't be able to make it
in extremely high yield using
a Friedel-Crafts alkylation,
and so we'll see a better way
to do that in the
next video, which
is on Friedel-Crafts acylation.
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