Now that we understand ortho
para and meta directors, let's see what
happens when you have multiple substituents
on a benzene ring. So if I look at
this first reaction, I can tell that this is
a halogenation reaction. And it's going to put a
chlorine onto the ring. The question is where
will the chlorine go. To figure that out you have
to look at your substituents and see what kind of
directors they are. So we'll start right here
with this methyl group, which we know is an ortho
para director. So it's going to direct the
chlorine to the ortho and para positions. And so this would be
the ortho position. And this is also
an ortho position. But by symmetry these
are the same thing, if you think about
the molecule here. The para position is occupied
by this nitro group right here. And so we can't add anything
on para in this example. If we look at the nitro
group, right here, I think about what kind
of a director that is. I know that it's a meta
director from the reasons we talked about in previous videos. So the position that's
meta to this nitro group would, of course, end up
being the exact same carbons that we marked before,
which again by symmetry are the same ones here. So this turns out to be a case
where both of my substituents direct to the same spot. And so I can go ahead
and draw my product. So I have my have benzene ring. And I have my substituents
on there, my methyl group, and also my nitro group. And I could have picked either
one of these two carbons, so this one or this one. It doesn't matter which
one you choose again, because of symmetry, you will
get an identical product here. So we're going to put a chlorine
on in one of those positions, ortho to the methyl group,
and meta to the nitro group. And so that's the only
product that we will get. Let's do another one here. Let's look at this reaction. So I know that this is a
sulfonation reaction, right. So I'm going to put an SO3H
group onto my benzene ring. And the question is,
where will that group go. And so once again I
analyze my substituents. And first I look
at the OH group, which I know is an ortho
para director, right. So this is an ortho
para director here. And I'd look at where
that would be on my ring. Well, once again this would
be the ortho position. And again by symmetry, the
same ortho position over there. The para position is once again
taken up by this methyl group here. When I think about the
methyl group-- so let me just use a different
color for that one. The methyl group is also
an ortho para director. So what's ortho and para
to the methyl group. Well this would be ortho, right. These two spots
would be ortho, again identical because of symmetry. The para spot is taken
up by this OH here. And so now we have a case
where we have two ortho para directors directing
to different spots. And so the way to
figure out which one wins is to think
about the activating strength of these two
ortho para directors. So the strongest
activator is going to be the directing group. And the strongest activator
here is, of course, the OH group here. So this is a strong activator. And the methyl group we
saw is a weak activator from some of the
earlier videos here. So the spot that's
going to get the SO3H is this carbon, which
again by symmetry would be this one right here. So we can go ahead and draw
the product of this sulfonation reaction. So let me go ahead and
sketch in my benzene ring. Let me do another one here. That one wasn't very good. And we have our ring. We have our OH. We have our methyl group. And since the OH is the
strongest activator, we're going to go ahead and
put SO3H ortho to the OH. And that is our final product. Let's do one more example. So let's see what happens
with this reaction here. So once again I
can identify this as being a halogenation
reaction where I add a chlorine onto my ring. And let's go ahead and analyze
the substituents once again. So I have a methyl
group again, which I know is an ortho
para director. So I'll go ahead and
mark the spots that are ortho and para
to that methyl group. So this would be
the ortho position. This would be the
ortho position. And then this time the
para position is free, so we could possibly
put the chlorine there. If I look at what
else is on my ring, right-- so I have
another substituent here, which is a chlorine. We know that halogens are also
ortho para directors because of the lone pair of electrons
that are on the chlorine there. So an ortho para director
for the chlorine. So what's ortho to
this chlorine here. Well this spot is ortho. So is this spot and
so is this spot. And so we have a couple of
different possible products here. And let's go ahead and
start drawing them here. So if I think
about the product-- if we add on a chlorine
onto this position. Let me go ahead and draw that. So we would have our
benzene ring here. We would have our methyl group. We would have our chlorine. And it's possible
that a chlorine would add on to that position. Let's next look
at this position. So this could be another
possible product here. So let me go ahead and
draw that one as well. So we have our benzene ring. And we have our methyl group. And we had the chlorine that
was originally on our ring. And then we're adding on a
chlorine to this position. And finally, let's
go ahead and think about what would happen
if a chlorine tried to add to this final position
right here on the right. Well, it turns out that there's
just too much steric hindrance for a chlorine to add
on to this position. So if you think about this
methyl group being bulky and this chlorine
being bulky, that makes it difficult for another
chlorine to add to here. So this is not observed
in large amounts. So this product, we're going
to say, does not form here. And so we're left with two major
products for this reaction. And that's how to think
about synthesis-- directing effects in synthesis--
when you're talking about
multiple substituents. First think about where the
substituents will direct. And you have to think
about activating strength. And then finally, think
about steric hindrance for your products.