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Organic chemistry
Course: Organic chemistry > Unit 9
Lesson 4: Electrophilic aromatic substitutionFriedel-Crafts acylation
The mechanism of acylation. Created by Jay.
Want to join the conversation?
- Can you please tell about Clemmensen reduction?(5 votes)
- Clemmensen reduction is the reduction of ketones (or aldehydes) to alkanes using zinc amalgam and hydrochloric acid.
R(C=O)R' --> RCH2R'(11 votes)
- To reduce the ketone in the last example, could you also use LiAlH4? Or will that result in an alcohol?(5 votes)
- That's right. LiAlH4 reduces ketones to alcohols.(10 votes)
- Do we need to neutralize the excess of the acid by adding base after the reaction is completed? If yes, why do we need to do it (can't we just wash it or leave it)?(2 votes)
- Washing once removes most of the acid. Washing a second time removes most of what is left, and so on. Washing with base removes all the acid in one step.
Excess acid in the mixture can catalyze unwanted reactions when you try to purify the product by distillation.(7 votes)
- can we reduce alcohol by using zn hg conhcl(2 votes)
- No. HCl and Zn amalgam reduce ketones to alcohols, but the reaction stops at that stage.(3 votes)
- Could there be a resonance structure (of the acylium ion) where one of the bonds from the double bond between the carbon and the oxygen moves onto that carbon, leaving the oxygen with a positive charge?(2 votes)
- No, because the vacant orbital on the carbon is an sp² orbital, which is orthogonal to (in the nodal plane of) the π orbital.(2 votes)
- To reduce the ketone in the last example, can H2 and Pd/C be used?(2 votes)
- Yes for that example you could use H2 Pd/C because the double bonded O is so close to the ring, if it were further away then it would turn into an OH.(2 votes)
- how many electrons are shared between the C-atom and the O-atom in the propionylium cation?(2 votes)
- Basically, there are 2 Resonance Structures formed in the propionylium cation :
1) Triple bond between C and O, with a +ve Charge on O.
2) Double bond between C and O, with +ve Charge on C.
- Therefore, in 1), 6 electrons will be shared between C and O.
- Therefore, in 2), 4 electrons are shared between C and O.(2 votes)
- When the resonance occurs in acylium ion to form C0R+ with 3 bonds between carbon and oxygen I think the latter one is more stable and contributing to the resonance hybrid so finally the Oxygen should bond with the carbon in benzene and not carbon? Can this happen.........(2 votes)
- R-C⁺=O ⟷ R-C≡O⁺
The contributor with a positive charge on O is a minor contributor.
O is the second-most electronegative atom in the Periodic Table, so it will not give up its electrons easily. The carbon atom will be the electrophile.(2 votes)
- in clemmensen's method for preparation of alkanes from alkanones, I didn't understand one thing:
R - (C = O ) - R' + 2H+Cl- -------> R-(C-(OH2+))-R' + 2Cl- ------> R - C+ - R' .
R--C+--R' + Zn ------> (R--C(-)--R')Zn2+ . R--C(-)--R' + 2H+ -----> R--CH2(+)--R.
WE ARE GETTING A POSITIVE CHARGE EXTRA ON C. what goes wrong in my explanation of the mechanism?(2 votes) - is there a video on the clemmensen reduction(2 votes)
Video transcript
In the last video we looked at
our Friedel-Crafts alkylation. In this video, we're going
to an acylation, which is very similar
to the alkylation. We start off with
benzene, and to benzene we add an acyl chloride. And so this right here,
you could think about as an acyl group. We're also going to use
aluminum chloride once again as our catalyst, and you can see
the acyl group has substituted in for one of the
aromatic protons, and so that's our electrophilic
aromatic substitution reaction. The mechanism for an acylation
is similar to an alkylation, although there is an
important difference, but they start off the same in
that aluminum chloride is going to function as a Lewis acid
and accept a pair of electrons. And so this lone pair of
electrons on this chlorine, you could think about
that chlorine donating that pair of electrons
and the aluminum accepting that electron pair. And so if I go ahead and draw
the result of that Lewis acid base reaction, we
have our carbonyl, we have our chlorine attached
to our carbonyl carbon. The chlorine has two
lone pairs of electrons. It's now formed a bond
with the aluminum, and the aluminum is bonded
to three other chlorines. And I'm not going to draw in
the lone pairs of electrons on those other chlorines
just to save time, but the aluminum gets a
negative 1 formal charge, and this chlorine now has
a positive 1 formal charge. So to highlight our electrons,
these electrons right here in magenta, are forming
a bond between the chlorine and the aluminum like that. So in order to find
our electrophile, you could think
about these electrons in here kicking off
onto the chlorine. And so I'm taking a bond away
from that carbonyl carbon, and so if I take a bond away
from that carbonyl carbon, that carbon is now
positively charged. That carbon is still
double-bonded to an oxygen, and that carbon is
bonded to an R group. And so we've created
an acyl cation. And we can think
about this acyl cation as being the electrophile
in our mechanism for electrophilic
aromatic substitution. This cation is
resonance stabilized. I could take a lone pair of
electrons here on this oxygen and move them into here so
I could draw a resonance structure where now
the carbon would be triple-bonded to
this top oxygen here. This top oxygen would still
have a lone pair of electrons and have a plus 1
formal charge like that. And this carbon is still bonded
to an R group, so I am saying that this lone pair of
electrons here on this oxygen can move into here like that. And this resonance
stabilization of the acyl cation is one difference between an
acylation and an alkylation. Because our cation in an
acylation is resonance stabilized there is
no rearrangement, and that's different
from what we saw on our alkylation reaction. We formed a carbocation that
was capable of rearranging to form a more
stable carbocation in the previous
video on alkylation, and so that made it a
little bit difficult to control the types of
products that we got. And so with the acylation
there is no rearrangement, and again, it's due to this
resonance stabilization of our acyl cation here. So we would also form
this complex over here where the aluminum is
bonded to four chlorines. So we could think
about this chlorine now as having three lone pairs
of electrons around it, so I'm going to highlight
those electrons in red. So these electrons
in here kick off onto the chlorine like that,
and once again, we still have a negative 1 formal charge
on this aluminum like that. So the catalyst has
generated our electrophile, and now we can show our
electrophile, our acyl cation reacting with a benzene ring. And for that mechanism, you
could show either one of these. You could show either one of
these resonance forms reacting with your benzene ring. I'm just going to take the
one on the right and the one with the positive
charge on the carbon. So we have our
benzene ring, and we have one of the protons on
our benzene ring like that. And I'm choosing the resonance
structure on the right so I'm going to have a plus
1 formal charge on my carbon. And I'm going to have an R
group attached to that carbon as well. So we have a nucleophile
electrophile situation. So once again, negative charges
attract to positive charges. These pi electrons are going
to function as a nucleophile, attack our electrophile, and
so we can add our electrophile onto our benzene ring. And so once again, I'm going
to show our electrophile adding to the top carbon here. So the top carbon
has a hydrogen, and now it's going to form a
bond to our carbonyl carbon like that. So I put in my lone
pairs of electrons, and there's an R group attached
to that carbon as well. So follow our
electrons in magenta, functioning as a
nucleophile, forming a bond between this carbon and
this carbonyl carbon like that. We took a bond away from
this carbon down here, so that's a plus 1 formal
charge, and so, of course, this is one possible
resonance structure. And we could draw a few more
possible resonance structures. I'm not going to do
that for time reasons. I've done it in
the earlier videos, so please watch
the earlier videos if you're confused about
resonance structures. I'm going to use this
resonance structure to represent our sigma complex. And of course, to
finish our reaction, we need to deprotonate
our sigma complex and regenerate
our aromatic ring. So these electrons
in here are going to pick up this proton,
which would cause these electrons to move in
to reform our aromatic ring, and to take away the
plus 1 formal charge. And so when we do that,
we form our benzene ring with our acyl group now
attached to our benzene ring, so it substituted for
that proton there. So let's follow those
electrons as well. So I'm going to make those
electrons in here green. So these electrons in here
are going to move into here. And then also we could think
about what else is formed. So these electrons
up here in red are going to bond
to that proton, and so we would also have HCL. So let me go ahead and highlight
those electrons in here in red. And then of course, we would
also regenerate our catalyst so we would make AlCl3. So we've formed our product. We've installed an acyl
group on our benzene ring, and so that is
Friedel-Crafts acylation. Let's look at a situation where
a Friedel-Crafts acylation might be used instead of a
Friedel-Crafts alkylation. So let's say that
our goal was to go from benzene to butylbenzene. So let me go ahead and
draw butylbenzene out here. So four carbon alkyl group
coming off of our benzene ring. So we saw in the last video
that a Friedel-Crafts alkylation would make butylbenzene
as a minor product because of the rearrangement
of the carbocation, and so this would be
formed as a minor product. If you wanted to form
it in a higher yield, you could use a
Friedel-Crafts acylation. And so if I wanted to
get to butylbenzene using an acylation, I would
need to install an acyl group on my benzene ring that has
the same number of carbons. So an acyl group that
has four carbons on it. And so let me go
ahead and draw that. So we would show
an acyl chloride that has a total
of four carbons. So there is my acyl
chloride, and I can highlight the four carbons. So this carbon, two,
three, and four. So to that acyl
chloride, we would also need to add our catalyst, right? So we need some aluminum
chloride as well. So AlCl3 to catalyze this
Friedel-Crafts acylation, and we're going to put this acyl
group onto our benzene ring. So here's our acyl
group, and you could think about just
putting that onto your ring. So when I draw my ring, I know
that the carbon on my ring is going to be directly attached
to this carbonyl carbon, and there's a total
of four carbons here so one, two,
three, and then four. And so that's my
Friedel-Crafts acylation. And so now I need to go from
this compound to my target compound up here,
my butylbenzene, and so somehow I need to
get rid of that carbonyl. And a reaction that's been
historically used to do this is the Clemmensen reduction,
which involves a zinc amalgam with mercury, and also a
source of proton, so HCL. And the zinc amalgam is
going to reduce that carbonyl to our alkyl group so it
actually will form the desired alkyl group and gets
rid of that carbonyl, and so the Clemmensen reduction
is a very useful reaction and synthesis. And so this is a way to make
our butylbenzene molecule in high yields using
an acylation, which is a little bit more
reliable than our alkylation.