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Course: Organic chemistry > Unit 9
Lesson 5: Directing effectsOrtho-para directors I
Regiochemistry. Created by Jay.
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- Why is para considered the major product while ortho the minor product?(5 votes)
- Products should always be more stable than the reactants. We determine the stability here by counting the number of resonance structures. The more the resonance structures, the more easily we can delocalise the charge within a benzene ring.
Ortho and Para have 4 resonance structures while meta has only 3 resonance structures. This means we can delocalise charge easily in ortho and para which also means that these two are more stable comparing to meta positions.
So now the competition is between the ortho and para.
We know that 2 large groups will have to face steric hindrance while coming into proximity. Since the para position is opposite to the substituent of the benzene, this arrangement will be more stable and will be easily formed.
Ortho products will have to overcome steric hindrance to form products. This is the reason why we see ortho as the minor product.
While para ones which have only minor resistance from spatial arrangement form the product easily and the para products are the major products.(9 votes)
- What are activators and de activators in the ortho para directors .
What does it mean ?(4 votes)- Activators make a benzene more reactive. They work by stabilizing the ring through hyperconjugation or resonance. Deactivators do the opposite. They also mostly direct Meta, with the exception being Halogens.(7 votes)
- In the meta resonant structures, why can't the oxygen form a double bond like the ortho and para resonant structures?(4 votes)
- In all three resonance structures that Sal has drawn at8:56in the meta example, the benzene carbon that is attached to the oxygen already has four bonds (one single bond to the oxygen, one single bond to another carbon in the ring, and a double bond to a second carbon in the ring). That is, that carbon already has a full octet so can't accept any more electrons to form a double bond with oxygen. (Note that it is possible to draw a resonance structure with the meta example that has a double bond between the carbon and oxygen, but this would involve three atoms with a formal charge and thus is not a good resonance structure (ie it's very unstable) so would not make a significant contribution to the resonance hybrid and thus is not considered).(3 votes)
- How would you do this for seven-membered rings?(4 votes)
- As far as I know, 7 membered rings are highly unstable and hardly exist.(2 votes)
- At5:28, Jay says that the lone pairs for Oxygen can form a pi bond between Oxygen and Carbon. Why can't they form pi electrons inside the Benzene ring?(2 votes)
- If the electrons in the C=O pi bond (from the oxygen lone pair) were pushed into the ring, that would leave oxygen with a +2 charge and an incomplete octet. This would be very unstable so does not happen.(6 votes)
- For a Phenol molecule there are 4 different resonance structures possible, and due to the positive mesomeric effect 3 of these resonance structures have a negative charge on the ortho and para carbons but none on the meta carbon. Do these negative charges make it more likely for an electrophile to attack at the ortho and para position ?(3 votes)
- why ortho and para are activating group while meta is deactivating group(3 votes)
- .
When we see resonance structure for attacking, we will see the most negative charge on benzene of activator groups are on ortho and para position.
While for deactivator groups make the meta position the most negative charge.
Good Luck.(1 vote)
- Can ortho and para substitutions occur on a same benzene ring, or do they always occur separately in different molecules?(3 votes)
- If the added substituent is activating, then you could easily get multiple substitutions (in fact this is one problem with doing Friedel-Crafts alkylations). However, where the additional groups are added will depend on interactions among the activating groups and steric effects.(0 votes)
- What if rather than the methoxy group there was an isopropyl group? Would the major product still be ortho? or would it be para? Would you consider isopropyl a sterically hindered group?(1 vote)
- So alkyl groups (which includes isopropyl) are known as activating ortho, para-directing substituents. Activating meaning they cause substitution reactions much faster than similar substitution reactions with ordinary unsubstituted benzene. And ortho, para-directing because they direct the new substituent to the ortho or para positions relative to the substituent obviously.
So if you did an electrophilic substitution on isopropylbenzene (or cumene), then the ortho and the para products would be considered the major products, while the meta product (which is still produced) would be considered the minor product. Now as far as the percentages go for each product, this can vary depending on the type of substitution (and even then there is some inconsistency among whether ortho- or para- are produced in larger amounts). But in general I would say that para- products are produced in greater quantity over the ortho- products becuase of the steric hindrance of the isopropyl group like you mentioned. To get an idea of the percentages, I've seen one reaction using an alkyl substituent resulting in 38% ortho, <1% meta, and 62% para products.
Hope that helps.(2 votes)
- what are the most common types of activators and deactivators(1 vote)
- Do you mean the groups themselves?
Activators: alkyl, NH2, OH
Deactivators: NO2, SO3H, SO2R, halogens(2 votes)
Video transcript
So far we've looked at
electrophilic aromatic substitution reactions
involving only benzene. But what happens when you
start with a substituent already on your benzene ring? So we'll look at this
molecule over here on the left, which
is methoxybenzene. So now we have a
methoxy substituent on our benzene ring. And if we react methoxybenzene
with concentrated nitric and sulphuric acids, we
should recognize that as being a nitration reaction,
which will install a nitro group onto your aromatic ring. The presence of that
methoxy substituent is going to affect where that
nitro group goes on your ring. So, for example, one of
the observed products is to put the nitro group
on this carbon, which is the carbon right
next to the carbon that has our methoxy substituent. We say that these two groups
are ortho to each other. So this would be
the ortho product. So if I go back over
here on this diagram I can label this carbon as
being the ortho position on my benzene ring. The other product
that we observe is a nitro group installed
on this carbon, which is on the opposite side
from the carbon containing our methoxy substituent. We call this the para product. So let me go ahead and
label the para position on our benzene ring over here. So this would be
the para position. Between the ortho
and the para product, it turns out that
the para product is the observed major
product in the reaction of the nitration
of methoxybenzene. And the ortho product
is the minor product. And this has to do
with mostly thinking about steric hindrance. And this methoxy group is
having some steric hindrance, which obviously would prevent
this nitro group from adding onto the ortho position easily. Obviously the para
position will have much less steric hindrance. And that's the reason
that you usually see for the para product
being the major product for this reaction. Now this isn't always the case. Sometimes your ortho product
is more than your para. But for this reaction the para
product is the major product. And once again steric
hindrance is one factor to think about when you're doing
these kinds of reactions here. So there's of course another
position on your aromatic ring. So if we installed the nitro
group on this position, we would call this
the meta product. And the meta product
is not observed in high yield for this reaction. So we say that the methoxy
group is an ortho para director. And we could also
label this as being the ortho position on this side. And we could say this is
the meta position because of symmetry. But the meta product is
not seen in a large yield, in a high yield. And let's go ahead
and look at why by drawing a bunch of
resonance structures and thinking about the mechanism
for electrophilic aromatic substitution. And so let's go ahead and
start with an ortho attack. And so we know that when
you're doing a nitration the sulfuric acid
acts as a catalyst to generate the nitronium
ion from nitric acid. And that functions
as the electrophile in your mechanism here. So if we're going to
do an ortho attack, we need to show the nitro group
adding onto the ortho position. So we need to show the nitro
group adding onto this carbon. And so if the nitro group is
going to add onto this carbon, then these are the
pi electrons that can function as a
nucleophile in our mechanism. So we have a nucleophile
electrophile reaction for the first step
of our mechanism. So the nucleophile,
these pi electrons are going to attack that
positively charged nitrogen, which kicks these electrons
off onto the oxygen. So if we draw the result of
that nucleophilic attack, we still have our methoxy
substituent up here. I'm showing the nitro group
adding onto the ortho position. And remember there's
still a hydrogen attached to that carbon. So I have pi electrons over
here, pi electrons over here. And I'm saying that these pi
electrons are the ones that formed a bond with this
nitrogen like that. That takes away a
bond from this carbon. So that carbon gets
a +1 formal charge. We can show some
resonance structures. So we can show some resonance
stabilization of this cation here. So I could show these pi
electrons moving over to here. And we could draw another
resonance structure. So let's go ahead and show the
movement of those pi electrons over to this position. So let me go ahead and draw
in the rest of the ion here. So we have a hydrogen here. We have an NO2 here. And we took these pi
electrons right here, moved them over
to this position, took a bond away
from that carbon. So we get a +1 formal
charge on this carbon. And that's another
resonance structure. We can draw another one. We can show the movement of
these pi electrons into here. So let's go ahead and show that. We have our ring. We have our methoxy group. We have, once again, the nitro
group in the ortho position. And we have these
pi electrons here. And now we show the
movement of those pi electrons over to here. So let me go ahead
and highlight those. These electrons in
red move down to here. I took a bond away
from this carbon. So that carbon is the one that
gets a plus 1 formal charge now. Since the oxygen is right
next to this carbon-- the oxygen has a lone
pair of electrons. And so that lone
pair of electrons can give us yet another
resonance structure. So these electrons
could move into here to draw a fourth
resonance structure. So the presence of that
methoxy substituent with the lone pair of
electrons on that oxygen allows you to draw a
fourth resonance structure. So this will give this
oxygen a +1 formal charge. We have these pi
electrons over here. We have our nitro group, once
again, in the ortho position. And let me just
go ahead and show the movement of those electrons. So these electrons--
I'll make them green-- these electrons
right here are going to move in to form
our pi bond like that. And we have a total of four
possible resonance structures. Remember the ion is actually a
hybrid of these four resonance structures, which we
call our sigma complex. And so, again, the presence
of this methoxy substituent with the lone pair of electrons
right next to our aromatic ring gives us an extra
resonance structure. This one is our extra
resonance structure. And so we have a total of
four resonance structures for an ortho attack. And the more resonance
structures you can draw, the more that positive
charge is delocalized. And the more stable your sigma
complex is, the more likely it is to form in your mechanism
for electrophilic aromatic substitution. And so because we can draw
four resonance structures for an ortho attack, that
is a favored carbocation. That's a stable sigma complex. And that's going to
form much more easily. And, of course, the last
step in your mechanism is to deprotonate
your sigma complex to reform your aromatic ring. But I'm not going to
show that step here. I just wanted you to see
the four possible resonance structures for an ortho attack. That's different
from a meta attack. So let's go ahead
and look at what would happen if we added
on our nitro group meta. So the meta position would be,
of course, this one right here. So we would use these pi
electrons, so nucleophilic attack. And it kicks these electrons
off onto our oxygen. So let's go ahead
and show the result of our nucleophilic
attack and adding on our nitro group
in the meta position. So if we're going to show
the nitro group in the meta position like that--
let's highlight these electrons here-- so these
pi electrons come off onto and form a bond, I should
say, with that nitrogen, taking a bond away
from that top carbon. So that top carbon is going
to get a plus 1 formal charge. And, of course, we have
our other pi electrons in our ring like that. So resonance structure-- I
could draw a resonance structure for this ion here. I could take these pi electrons
and move them over to here. So we'll go ahead and show
another resonance structure. So I have those pi electrons
moving over to there. I have these pi electrons here. I still have my nitro
group in the meta position to my original methoxy
substituent like that. And now my positive 1 formal
charge is on this carbon right here. So to save time I'm not
going to color coordinate these resonance structures. I can draw one more, right. I could show these pi
electrons moving over to here. And let's go ahead and draw
that resonance structure. So I have those pi
electrons move over to here. I have these pi electrons. I have my methoxy substituent. I have, once again,
my nitro group. And, of course, now
my positive charge moves down to this
carbon down here. And that's it. I can only draw three
resonance structures. So, once again, the
actual sigma complex is a hybrid of these three
resonance structures. And since I can only draw three,
only a total of three resonance structures for this
situation, this sigma complex is not as stable as the one
that we saw for an ortho attack. So the sigma complex
for an ortho attack was more stable because we could
draw four resonance structures. And here we can only draw three. Let's go ahead and
show a para attack. So if I want to show my
nitro group adding on to the para position. So here is the para position. So I'm going to use these
pi electrons, right. So nucleophilic attack
kicks these electrons off. So this would be
our para attack, adding on our nitro group
in the para position. So, once again, I
have my methoxy group. So this time I'm going to show
my nitro group in the para position. So let's go ahead and
follow these pi electrons. So these pi electrons in here. Those pi electrons formed
a bond with that nitrogen. And I took a bond
away from this carbon. So that is where my plus 1
formal charge is going to be. I still have these pi
electrons in my ring like that. So a resonance
structure for this one, I could show these electrons
in here moving over to there. And let's see what we would get. We would still have, once
again, this substituent. Our pi electrons
moved over to here. These pi electrons were here. I had my nitro group para. And now I took a bond
away from this carbon. So that is the carbon that
gets the plus 1 formal charge. So I can draw another
resonance structure. I could show these electrons
in here moving to there. So let's get a little more
room here for more resonance structures. So I can show my
ring and I can show these pi electrons have
moved over to here. I still have my substituent. And, of course, my nitro group
is still in the para position. I took a bond away
from this carbon. So I get a plus 1 formal charge. And once again the
presence of that lone pair of electrons on that oxygen
right next to my benzene ring allows me to draw another
resonance structure. So I could think about
these electrons on my oxygen moving into here
to form a pi bond and pushing these
electrons down to here. And so I can draw a
fourth resonance structure showing the oxygen now
double bonded to my ring. So there is still a lone pair of
electrons on that oxygen, which would give it a plus
1 formal charge. I still have pi electrons here. I showed pi electrons
moving over to here. And, once again, my nitro group
is still in the para position. And so I have a total of
four resonance structures if I show the nitro group adding
on in a para fashion as well. So the para attack also has
four resonance structures. And the ortho had four
resonance structures. And so that helps to explain
why the methoxy substituent functions as an
ortho para director. If you show an ortho
or a para attack, you can draw a total of four
resonance structures, which stabilizes the sigma complex
more than a meta attack. And that's the reason
for the regiochemistry that we see in the nitration
of methoxy benzene. Now I drew these
resonance structures because I started
with my benzene ring. I started with my electrons
in this position right here. Now if you started with
a different benzene ring-- so a resonant structure
of that-- then your resonance structures might look a little
bit different from mine. And you'll see
different versions in different textbooks. So make sure to, once
again, always look at what your professor does
in class or your textbook. And think about that
on exams if they have you draw resonance
structures for an ortho, para, or a meta attack.