In this video, we'll look at Resonance for benzene and substituted benzene molecules.
If you're new to resonance make sure you start at the beginning.
You can find the entire series, practice quiz, and cheat sheet on my website leah4sci.com/resonance.
Benzene looks like a cyclohexane with three pi-bonds.
And because the pi-bonds are sitting one carbon away from another pi-bond, they can resonate
clockwise or counterclockwise.
We can show this quickly or step by step.
The quick resonance will show everything that happens but as a series of arrows on the same
molecule as follows.
Say we start with the green electrons and move them over.
The blue electrons get kicked in the direction of purple.
Purple electrons get kicked in the direction of green.
Giving you a structure where every single pi-bond move over one position.
While doesn't show us exactly what happened where, it still demonstrates what makes benzene
so stable because the electrons were able to round, and round, and round.
In fact, the resonance hybrid for benzene is just the hexagon with the circle in the
middle representing these electrons going round, and round, and round.Let's take a look
at what happens step by step.
To see the resonance intermediates, we move one pi-bond one at a time, evaluate what happens,
determine formal charges and go to the next step.
We'll randomly start at this position with a purple electrons and move them in the direction
of the green electrons.
These purple electrons will form a pi-bond right here, and what you absolutely want to
avoid at all costs, is showing the structure that looks like this, where you have two pi-bonds
coming out of the same carbon molecule on benzene.
The reason for that is two fold: One, this carbon, if it had two pi-bonds without the
hydrogen atom would be sp hybridized with a bond angle of a 180 degrees, that would
be linear, yet we're showing something bent here.
The carbons on benzene are sp2 hybridized with a 120 degree bond angle, which allows
us to draw it in that shape rather than linear.
The other thing to keep in mind, is that even though we don't show it, unless benzene is
substituted there's an invisible hydrogen atom.
And if you have a hydrogen atom here, a quick count gives me 1, 2, 3, 4, 5 bonds to carbon
which cannot happen.
So how do we justify moving those purple electrons without forming too many bonds to carbon?
Well, we have to kick that pi-bond out and simply push it to the other carbon.
The pi-bond starts out on two carbon atoms, the carbon atom at the top will now be occupied
with the purple pi-bond, the carbon atom at the left can still hold on to both electrons
and so we show the first resonance intermediate where the purple bond moved over.
The green bond is now sitting as a lone pair on carbon and the entire molecule looks completely
different.
Next thing we have to do is a quick formal charge remembering that the net charge of
your molecule will always be the same.
Benzene starts out with the charge of zero.
It's neutral.
And that means if we have a carbanion forming here that's because carbon has a lone pair
and a negative charge, somewhere on the molecule we must also have a positive charge to balance
that.
How do we know that we have a carbon ion and that we have them positive somewhere else?
You can take the time to draw in that hydrogen atom and then do a quick formal charge using
the shortcut taught in the tutorial link below.
Carbon should have 4 electrons directly attached we have one, two, three, four, five.
Four minus five is negative one, and that's how we get the carbon ion, that's how we get
the negative one charge.
These two carbons and these two carbons looks exactly the same.
Formal charge doesn't change but this carbon here lost a pi-bond.
It still has that invisible hydrogen atom but that is all it has.
A quick formal charge gives us should have four minus three for a net charge of positive
one.
You can figure it out faster with this trick.
When resonating on benzene, if the carbon starts out neutral and you give it a lone
pair, it goes negative, if the carbon starts out neutral and it loses a pi-bond, it goes
positive.
Quickly looking at the structure, negative for the lone pair, positive for the carbon
that lost a pi-bond, plus one minus one equals zero, net charge is the same we are good to
go.
But that negative charge is not very happy so those electrons will continue moving around
the ring towards the blue pi-bond, causing too many bonds to this carbon atom.
We take those electrons and kick them out towards the right away from the incoming electrons,
giving us our next intermediate.
The purple bond hasn't moved.
The green electrons now form a pi bond and the blue electrons collapse as a lone pair
at the bottom of this molecule.
Using the trick, had a pi-bond now has a lone pair gets a charge of negative one.
This carbon here still has a positive charge because nothing changed and once again we
have a negative pair of electrons ready to move on.
Where do they go?
Well negative is attracted to positive so it moves directly towards that carbocation
to give us our final structure as follows: The purple bond hasn't moved, green hasn't
moved, the blue lone pair forms a pi-bond at the bottom and it looks like we reformed
our benzene.
Except for one little detail: Notice that the pi-bonds are now one position over from
where they initially started.
The purple one moved here, the green one moved here, and the blue one moved as well.
Whenever you see benzene always remember that the internal resonance will keep going so
if you're given a molecule where the pi-bonds represented like this and then later in the
reaction you see the pi-bonds like this, understand that it's the exact same thing because these
atoms can continue resonating.
Now what happens if you're asked to show resonance for benzene with the substituent?
The first thing to ask yourself is can that substituent participate a resonance?
If your substituent cannot participate in resonance, for example the methyl group and
toluene, ignore it and do the resonance in the ring just like you saw in the previous
example.
But if the substituent can participate in resonance, ask yourself if it participates
as a donor or as a receiver.
In the example of aniline, nitrogen has a lone pair that can resonate down into the
ring and in nitrobenzene, nitrogen actually has a positive charge and can accept resonance
out of the ring.
Many other molecules can fall into this categories of acceptor or donor so the pattern will be
the same.
Let's start with aniline.
When you have a lone pair outside of the ring, start your resonance from that lone pair and
resonate it into the ring.
This could be a neutral lone pair such as nitrogen or a negative lone pair such as a
negative oxygen.
We'll show this electrons collapsing down into the ring, forming a pi-bond between nitrogen
and carbon.
Do not make the mistake of leaving that pi-bond there and giving nitrogen and carbon a charge
because once again you're faced with a situation of carbon having too many bonds which cannot
happen.
Instead, if you're attacking a carbon that already has four bonds make sure you kick
out another bond to compensate for the incoming electrons.
Our intermediate will look as follows: two pi-bonds haven't been touched, nitrogen now
has a second bond to carbon for a formal charge of plus one, and the green electrons are now
sitting as lone pair of carbon for a formal charge of minus one.
The green electrons can continue resonating in the ring down towards the next pi-bond
which we'll show in blue and kick those electrons out onto the lower carbon.
Keep in mind that aniline started with a formal charge of zero.
And adding up plus one for the nitrogen and minus one for the carbanion, gives us a net
charge of zero.
Nitrogen hasn't changed but the green electrons are now sitting as a pi-bond on the left and
the blue electrons are not sitting as a lone pair with a negative charge at the bottom.
Once again, plus one for the nitrogen, minus one for the carbanion for a net charge of
zero.
But those blue electrons will continue moving around the ring kicking out the remaining
pi-bond which we'll show in orange collapsing onto the right carbon atom.
Our intermediate still has a plus one on nitrogen, green pi bond hasn't moved, the blue electrons
formed a pi bond and the orange electrons now sit as a lone pair with a negative charge
on the carbon.
These electrons can move over one more time for our final structure, but what do we kicked
out?
We have nothing else in the ring, so we give those purple electrons back to the nitrogen
as a lone pair.
This final step reforms the aromaticity, it gives us back three pi bonds in benzene as
follows.
The green and blue haven't moved, the orange lone pair reforms the pi bond and now the
purple electrons sit as a lone pair on a nitrogen atom.
We have no formal charges in this molecule so it has a net charge of zero.
Which brings us full circle.
But, when you're counting the number of resonance structures, this is not considered a fifth
structure.
Instead you want to recognize that the benzene with a neutral NH2 at the top is the same
exact molecule that we started with.
Even though the electrons inside the ring have moved over one position, these are still
considered to be the same structure when you're counting the number or resonance structures.
This gives us a total of one for the two that are the same, two, three, and four.
Here's your trick, benzene has a total of three resonance structures.
If there's a substituent that can participate in resonance, see how many times you can resonate
on that substituent and simply add one.
In this case NH2 gives us one more structure, one bonus resonance, for a total of four.
Three for benzene, a fourth when nitrogen participates.
And brackets around the whole thing to show resonance.
Now let's see what happens when we have a resonance capable electron withdrawing group
sitting on benzene.
NO2 looks very innocent, until you draw it out.
That's when you realize the nitro group itself is capable of resonance.
The nitro group is net neutral but that's because we have a positive nitrogen and negative
oxygen sitting on the molecule.
The positive nitrogen is going to attract the electrons from within the benzene ring
and that's where we're going to start the resonance.
But don't forget you can see resonance here as follows: the oxygen electrons can attack
nitrogen.
These pi electrons will collapse onto the other oxygen and these can go back and forth.
We're not going to focus on this, instead, we're going to focus on how the nitrogen interact
with benzene resonance.
Since nitrogen is positive, the resonance begins when nitrogen pulls a pi-bond out of
the benzene ring as follows: These green electrons are the pi-bond from the carbon touching the
nitrogen so we'll start by showing these forming a pi-bond between carbon and nitrogen.
Nitrogen would now have too many bonds so we take each pi-bond between nitrogen and
oxygen and collapse that onto the oxygen atom.
Let's see what happened.
Oxygen on the left had two lone pairs and now has a third lone pair for a formal charge
of minus one.
Nitrogen, instead of being double bound to oxygen, is now double bound to nitrogen and
still has a formal charge of plus one.
We have two pi-bonds in the ring that haven't moved and the carbon that lost its pi-bond
got nothing in return and gets a positive charge.
Our starting molecule had plus one, minus one for a net charge of zero.
Our resonance intermediate has plus two, minus two, once again for a net charge of zero.
Carbocation in the ring will attract the electrons nearby and so the blue pi-bond will reach
out and attack towards that carbocation.
The purple bond on the right hasn't moved.
The blue pi-bond simply moved over one position and sits on the left, leaving us with a carbocation
on the bottom.
This molecule has plus two, minus two, for a net charge of zero.
The carbocation will attract both the blue and purple electrons.
Never forget that these resonance structures can go both ways.
But it we show the blue electrons collapsing down, we're simply resuming the structure
we have before.
So we want to keep moving the same direction around the ring.
And that means taking the purple electrons and using them to attack the carbocation moving
that pi-bond down.
The resulting structure still has a blue pi-bond on the left, the purple pi-bond moved, and
the carbocation now sits on the right.
Once again, plus two, minus two, for a net charge of zero and the carbocation will pull
on the nearby electrons.
This time we're not taking a pi-bond from inside the ring, we're taking the pi-bond
between carbon and nitrogen and using that to attack the carbocation.
You have two options at this point, on your exam you’d probably just draw a double headed
arrow between what we just show and your starting molecule since we're about to get the same
thing.
But I'm gonna draw it out to show you exactly what happened.
The green electrons are not sitting as a pi-bond on the ring, along with the purple and blue.
This reformed our benzene ring, which is why I see this like the starting molecule.
In order to do that, we would show these electrons on oxygen going back towards nitrogen because
you want to avoid an incomplete octet on nitrogen and so we give it pi-bond, once again, for
a formal charge of plus one.
Net charge on this molecule, plus one minus one, zero just like the starting structure.
Don't forget to put brackets around your intermediates when you're showing resonance.
Discounting the individual resonance on the nitro group, we have a total of one for the
starting and ending structure, two, three, and four contributing structures for nitrobenzene.
Now what happens when you have multiple groups on benzene that can contribute for example
NH2, an electron donating group, and NO2 the electron withdrawing group?
This should just one of the questions you'll find on the resonance practice quiz linked
below.
So make sure you start with the tutorial series, truly learn what's going on, and then try
the quiz to see how you do.
You can find all of this on my website Leah4Sci.com/Resonance.