Here's the general reaction
for the nitration of benzene. So we start off with
benzene, and to it, you add concentrated nitric and
concentrated sulfuric acids. And that puts a nitro group
onto your benzene ring, in place of this proton. Let's look at the mechanism
for the nitration of benzene. So we start over here with the
dot structure for nitric acid. And here's the dot
structure for sulfuric acid. And sulfuric acid is actually a
stronger acid than nitric acid. So the first step
is sulfuric acid is going to function as
a Bronsted-Lowry acid and donate a proton. So I'm going to say, it's
this proton right here. And nitric acid is actually
going to function as a base and accept that proton. So we can go ahead and show this
lone pair of electrons picking up this proton and
these two electrons in here remaining
behind on that oxygen. So let's go ahead and show
that acid-base reaction. So we would have our
compound over here. So let's go ahead and
draw in these atoms. So this oxygen is
still going to have a negative 1 formal
charge on it. And then over here on
the right, this oxygen already had one
hydrogen bonded to it. It just picked up
another proton. So it still has one
lone pair of electrons, which gives that a
plus 1 formal charge. So let me go ahead and
highlight these electrons here. So I'm saying that this
lone pair of electrons right here picked up a
proton like that, which gives that oxygen a
plus 1 formal charge. And notice that forms
water as a leaving group. And so if these electrons
were to move in here, that would kick these electrons
in here off onto the oxygen. So let's go ahead and
draw the result of that. So we would have
water that left. So let's go ahead and show
H2O over here on the right. So we'll go ahead and draw
in those electrons like that. So I'm saying that these
electrons in here in magenta are the ones that came off onto
the water molecule, so water as our leaving group. And if water leaves,
that leaves what's called the nitronium ion behind. So let me go ahead and draw
the nitronium ion here, so it looks like this. Now, let me go ahead and
highlight those electrons. So I'm saying that
these electrons in here moved in to form a pi bond. So I'm saying it could be
represented by those electrons down here on my structure. Now, for formal charges,
this nitrogen has a plus 1 formal charge. And that nitrogen is
actually SP hybridized, which makes the
nitronium ion linear. And the nitronium ion
is positively charged. This is going to be the
electrophile in our mechanism. So this is our electrophile,
our positively-charged ion. So up here, if we
think about it, we would also create
the conjugate base to sulfuric acid, so HSO4
minus would be over here. All right, so let's
go ahead and show what happens now that we
formed our electrophile. So the electrophile is going
to add to the benzene ring. So we're going to have a
nucleophile-electrophile reaction. So let's go ahead and draw
our benzene ring over here. And I draw in my hydrogen
on my benzene ring. We've now formed
our nitronium ion. So the point of the catalyst
was to produce our electrophile here. So we have a positively charged
nitronium ion like that. And so now we have a
nucleophile-electrophile situation where, once
again, the pi electrons in our benzene ring are going
to function as a nucleophile and attack our electrophile. So those electrons
are going to attack that positively-charged
nitrogen, which will kick these electrons in
here off onto that oxygen. So let's go ahead
and draw the results of our nucleophilic attack. So we have our ring right here. We have a hydrogen. And once again, just
for convention sake, I'm going to show our
electrophile adding to that top carbon
double bond there, of what used to be
the double bond. And so now we have
a nitrogen there. We have a nitrogen double
bonded to our top oxygen. And then over
here, we would have an oxygen with three
lone pairs of electrons, giving that a negative
1 formal charge. And the nitrogen,
of course, is still going to have a plus 1
formal charge like that. All right, let me go ahead
and highlight those electrons. So once again,
these pi electrons are going to be attracted
to the positive charge, nucleophile-electrophile. And those pi electrons are
going to form this bond right here to our nitro group. Well, once again, as we've
seen several times before, we took away a bond
from this carbon. So that's where our
plus 1 the formal charge is going to go like that. And so we can draw some
resonance structures. So let's go ahead and show a
resonance structure for this. We could move these pi
electrons in over here. So let's go ahead and draw that. So we had a hydrogen up here. And you could just show
a nitro group as NO2. So I'm just going to go ahead
and do that to save some time. These pi electrons over
here are still there. And I'm saying that those pi
electrons moved over to here. So let me go ahead
and highlight those. So these pi electrons in
blue move over to here, took a bond away
from that carbon. So now we can put a
plus 1 formal charge at that carbon like that. We can draw yet another
resonance structure. So I could show these electrons
over here moving to here. So let me go ahead
and draw that. So we have our ring. We have our nitro group
already on our ring. We have some pi
electrons right here. And we have some more pi
electrons moving from here to here, which, of
course, takes a bond away from this top carbon. So that's where our positive
1 formal charge is now. So now we have our three
resonance structures. And remember, once again,
that the sigma complex is a hybrid of these three. And we're now ready
for our last step, so deprotonation of
our sigma complex. So if we go back up
to here, we think, what could function as a base? Well, the water molecule here
could function as a base. So a lone pair of electrons
on our water molecule are going to take
that proton, which would cause these
electrons to move in here to reform
your aromatic ring. So let's go ahead and show that. So we're going to reform
our benzene ring here. And we took off the proton. So deprotonation of
the sigma complex yields our product with a
nitro group substituted in. So let me go ahead and
highlight those electrons again. So this time I'll use green. So these electrons
right in here, when that sigma complex
is deprotonated, those electrons are
going to move in here to restore the
aromatic ring, and we have created our product. We have added in
our nitro group. So that's the mechanism
for nitration. Now, once you form a benzene
ring with a nitro group on it, sometimes when you're
doing synthesis, it's helpful to turn the
nitro group into an amine. So let's go ahead
and real quickly look at another
useful reaction here. So once you form your
nitro group like that, you can turn that nitro
group into an amine a couple of different ways. So one of the classic ways
to reduce a nitro group would be, in the first
step, to use either iron or tin and a source of
protons, so something like HCL will work well. And since you're doing this
in an acidic environment, in the second step,
you would need to neutralize it with something
like sodium hydroxide. And those steps will reduce
the nitro group to an amine. So let me go ahead and
show the product here. So instead of an NO2 on the
ring, now you have an NH2. So of course, this would
be the aniline molecule. And there are other
ways to reduce a nitro group to an amine. This is just one of the
classic ways to do it. So it could be useful for
synthesis problems, which we will study later
in this tutorial.