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Current time:0:00Total duration:7:41

In the last video, we
looked at the nitration of the nitrobenzene
molecule, and we saw that the nitro group
is a meta director because of this plus 1 formal
charge on the nitrogen right next to your ring. We also said that the nitro
group is a deactivator, which meant that the nitrobenzene
molecule reacted more slowly than benzene
itself in a nitration reaction. And so let's see if
we can figure out why the nitro group is
a deactivator in terms of both the inductive
and a resonance effect. And we'll start with
induction first, which we know is due to differences
in electronegativity, and so we could think
about this nitrogen here compared to this
carbon on the ring. The nitrogen being much
more electronegative, and so it can pull
electrons closer to itself, and so we can show the movement
of electrons towards nitrogen. A positively-charged nitrogen is
extremely electron withdrawing, and it could even
withdraw electron density from your ring itself. That makes the
ring more positive, which would destabilize
the sigma complex that results in the mechanism
for electrophilic aromatic substitution. Since the sigma complex
is less likely to form, that would deactivate the ring
towards electrophilic aromatic substitution. Also, you could think about
the electron withdrawing effect of this nitro group making
the ring less nucleophilic. And so both of those factors
mean that the nitro group is a deactivator towards
electrophilic aromatic substitution. So induction says that the
nitro group is a deactivator. Let's look at resonance next. So we could draw a
resonance structure, right, where we take these pi
electrons and move them in here, which would push these
electrons off onto your oxygen. So let's go ahead and draw that. We have our ring, we have
pi electrons in our ring, now, the nitrogen is
double bonded to our ring, and we have an oxygen
on the right, which has a negative 1 formal charge. The nitrogen is bonded to
an oxygen on the left, which now has three lone
pairs of electrons and therefore, has a
negative 1 formal charge. Let me go ahead and highlight
those electrons here. So if these electrons in blue
move off onto the oxygen, that gives that oxygen a
negative 1 formal charge. This nitrogen has a
plus 1 formal charge. And let's go ahead and follow
these pi electrons too. So in red, these pi
electrons move out here to form this pi
bond taking a bond away from that carbon, therefore,
giving that carbon a plus 1 formal charge. And you could keep going and
draw more resonance structures, and they would have a
positive charge in the ring, but I'm going to stop
there because the point I'm trying to make is that the nitro
group is electron withdrawing, and it makes the
ring more positive, which would, of
course, deactivate the ring towards electrophilic
aromatic substitution. So both the inductive effect
and the resonance effect say that the nitro
group is a deactivator, and so we can say it's a
strong deactivator because of both of those effects. Let's look at another
molecule that's considered to be a
strong deactivator. So this molecule down here. And we can see that
there's a carbon bonded to our benzene
ring, and that carbon is bonded to three fluorines. So you could call this
trifluoromethyl group, you could call it CF3 group,
whatever you want to call it. In terms of
electronegativity difference, obviously, fluorine is much
more electronegative than carbon so you could think about these
fluorine atoms withdrawing a lot of electron density
away from that carbon, which would give that carbon a
partial positive charge. And so that's an
explanation for why the CF3 group is
a meta director. So once again, look
at the last video for more details about
why a positive charge next to your ring
favors a meta attack. In terms of the inductive
effect on the ring, these are very
electronegative atoms and so they can withdraw
some electron density away from your ring so you
could show the electrons of the rain moving
in that direction towards your
electronegative fluorines. And so that withdraws
electron density from the ring making the
ring more positive, which would, of course,
deactivate the ring towards electrophilic
aromatic substitution. And so this trifluoromethyl
group is both a meta director and deactivating, and
this inductive effect is so powerful, this
CF3 group turns out to be a strong deactivator. We can't have any kind
of a resonance effect here because these
lone pairs of electrons on one of the fluorine's I'll
just choose this one right here-- those are too
far away from the ring to participate in resonance. So only in inductive
effect, but it turns out to be a very powerful one. Let's get one more example
of a meta director, and this one is a
moderate deactivator. So first, let's
try to figure out why this is a meta director, and
so if we look at the atom that is directly bonded
to the ring that is double bonded to an oxygen. And we know that oxygen is
much more electronegative than this carbon so we could
think about this oxygen as being partially negative,
pulling some electron density towards it, withdrawing
some electron density away from this carbon,
giving this carbon a partial positive charge. And so once again, you have
a partial positive next to a ring, which makes
this group a meta director. In terms of being
deactivating, we need to think about
resonance structures. So let's think about moving
these electrons in this pi bond off on to your oxygen so let's
go ahead and draw that one. So we have our benzene
ring, and we have, now, our carbon with only a
single bond to this oxygen. This oxygen now has a negative
1 formal charge on it like that, and then this carbon is still
bonded to this R group here. We took a bond away
from that carbon so now this carbon has a full
plus 1 formal charge like that. So we can now show a
resonance structure for this. We have these pi electrons
here can move out into here to form a pi bond between
those two carbons. So let's go ahead and show
this resonance structure. So we have our ring,
these pi electrons. Now, there's a double
bond here, and this carbon is still bonded to this oxygen. So this oxygen has a
negative 1 formal charge. And there's an R group
on it to that carbon. Let's show those pi
electrons moving here so these pi electrons moved
into here taking a bond away from this carbon so we have
a plus 1 formal charge right here. And, of course, once
again, you could keep going and show more
resonance structures, but I'm only trying to point out
that the presence of this group increases the positive
charge on the ring, and so the ring is
more positive, which means it is deactivating
towards electrophilic aromatic substitution. And so this group is both a
meta director and a deactivator. Now, let's think
about a general way to recognize these
moderate deactivators here. So if I go back to this
first drawing on the left, we have an atom bonded
to a benzene ring, and that atom has a pi bond
to an electronegative atom. In this case, it's oxygen,
and so that's the pattern that you're looking for. So a pi bond next to
an electronegative atom will give you a
moderate deactivator.