Leah here from leah4sci.com and in this video we'll look at key arrow patterns to help you
figure out resonance structures. Say you're given a molecule like this and ask to show
all the contributing resonance structures, how do you even know where to begin? Before
you'll look at patterns to recognize, let's look at what you should never do when working
at resonance.
Rule number 1, only ever move electrons. I know this sounds self-explanatory but I
can't tell you how many times I see students trying to put arrows on positive charges or
even hydrogen atoms. Only add remove electrons and more specifically these electrons should
be lone pairs or pi electrons. Resonance is the movement of electrons between different
atoms, that means you do not change the structure, you don't move the atoms and you don't break
any sigma bonds. You can move the lone pairs and you can move the double or triple bond
pi electrons.
Here's the Enolene ion we looked at in video 1. And for this ion we can move the lone pair
of electron on Oxygen down towards the pi bond and we can also move the pi bond electrons
between the two carbon atoms to collapse them onto carbon. So we're moving a lone pair and
we're moving pi electrons. But when we draw the resulting resonance structure, the skeleton
stays the same. The atoms are in the same place, the sigma or single bonds haven't move,
the only thing that's different is the red lone pair of oxygen is now a pi bond between
an atom and Oxygen and the red pi bond between the two carbon atoms is now a lone pair on
the oxygen atom. And yes the charge move but not because we had an arrow starting on the
negative and pushing that elsewhere but rather because when the electrons shift, the atoms
give new formal charges, Oxygen has a new formal charge of zero, carbon has a new formal
charge of negative 1 and that's how the charge is moved.
Rule number 2 is another very very common mistake and that is to never resonate on toward
an sp3 carbon atom. Say I have a molecule that looks like this with a negative charge
on an oxygen. If I take those electrons on Oxygen and try to resonate them down to carbon,
carbon would get an overfull octet. Carbon starts out with a complete octet of 8 and
if I bring that pi bond down, I have 10 electrons. And the same thing happened here but as a
result of that new pi bond forming, we kick out the other pi bond so we still overfill
the octet but we simply shifted the pi bond. In this case because the carbon atom is sp3,
it's single bound to 4 different atoms. Don't forget there's an invisible hydrogen atom
on carbon. So if Oxygen forms a pi bond we don't have any other bond to break, that means
we can't resonate those electrons down so there is no resonance form for the starting
molecule. Which brings me to rule number 3.
Rule number 3 is just the continuation of rule number 2. But carbon needs its own special
reminder. Just keep an eye on your octets. If the atom in question already has a complete
octet, then ask yourself, by adding electrons to this atom am I violating the octet? Can
I kick anything out or is this allowed to have more than just 8 in the octet? For example
in the sulfate iron, sulfur already has a complete octet but I am allowed to bring the
electrons from the Oxygen and resonate them towards sulfur to form a pi bond. That's because
sulfur is in period 3, it's a larger atom and it can hold as many as 12 electrons so
we don't have the issue of violating the octet because it's an exception to the octet rule.
In fact, I can do this one more time to take sulfur from having 10 on its octet to a total
of 12. Giving me this 3 valid structures which can continue resonating even more if I'd start
out with one of the other negative oxygen atoms.
So if the atom's an exception to the octet rule and you can have more than 8 electrons,
that's okay. If it's not an exception and you bring too many, make sure that there's
something else you can kick out potentially in the form of breaking a pi bond. Also keep
an eye on having too few electrons in your octet. For example, with this ketone propanone
I can take this pi bond and collapse the electron onto oxygen giving me a resonance form as
follows. Oxygen had 2 blue lone pairs, still has that but the red pi bond now sits as a
lone pair on oxygen giving me a negative charge on oxygen for the formal charge of plus 1
on the carbon atom. Conservation of charge is present, we're starting with a neutral
molecule and here we have the plus and minus to cancel for a net neutral molecule. But
carbon does not have a complete octet making it less stable than the starting molecule.
Now we come across structures where the resonance will have less than a complete octet.So keep
in mind the electronegativity.The more electronegative the atom, the less stable it's going to be
with that incomplete octet. Carbon will be common with the positive charge but atoms
like Oxygen and Hydrogen should only have a positive charge for too many bonds with
a complete octet rather than having an incomplete octet. Now that you know what not to do, let's
take a look at the key arrow patterns.
Every book or professor will teach you differently and this is just my method. The first thing
I look for are negative electrons. If resonance is the movement of electrons and electrons
are negative, they'll be attracted to positive or partial positive so it makes sense to find
the most negative electrons and see where they want to move. For example if I'm given
this molecule and ask to show resonance, I have a choice between lone pair and Nitrogen
and the lone pair on carbon. But the lone pair on carbon is negative and Nitrogen is
neutral so I want to start with a negative electrons. I may want to move them down but
the problem is that this carbon here is an sp3 carbon has nothing to kick out so that
will violate the octet and is not possible. Instead I'm looking for an sp2 hybridize carbon,
one that has other carbocation meaning an incomplete octet or a pi bond so that I can
kick out those extra electrons and I wanna move the electrons to form the pi bond between
the negative carbon and the carbon double bound to Nitrogen.
In forming that pi bond, that carbon that's double bound to Nitrogen will now have too
many electrons but I can fix them by taking the pi bond and showing in arrow moving towards
the Nitrogen atom kicking out those electrons and collapsing them onto Nitrogen. Show the
double headed arrows and first draw your skeleton. Everything that hasn't change then look for
what changed. The red electrons are now sitting as a pi bond between the former negative and
the former double bound to Nitrogen carbon. The green electrons are now sitting on the
Nitrogen atom. Next we want to do a formal charge. It's hard to do this on carbon in
skeleton structure so remember the trick.
If carbon starts out with a lone pair, and then forms a pi bond, the lone pair has one
more electron directly attached to carbon compared to the carbon with the pi bond so
the charge has to go up by 1. If we start with the negative carbon we go up to zero
giving it no formal charge or formal charge of zero. And the carbon that has a pi bond
and still has a pi bond even if it's in a different direction will not change its charge
because the total number of electrons haven't changed. Nitrogen is much easier to see, we
should have 5 directly attached, I count 6, 5 minus 6 is negative 1 giving me formal charge
of minus 1 and don't forget your brackets. You quick conservation of charge starting
with negative 1, we end with negative 1, charge is conserved and we are good to go.
If your molecule doesn't have negative electrons, the next thing to look for are lone electron
pairs. But here's the key, you're not just looking for lone electron pairs, you're looking
for the electrons to be sitting on an atom that does not yet have a pi bond. Now this
thing is tricky because you'll see lone electron pairs on atoms that are already pi bound and
can't move. So make sure that if the atoms is neutral and has a lone electron pair, it
shouldn't already have a pi bond. What do I mean by that? These lone electrons can resonate,
it's on a negative Nitrogen but they can resonate back towards that carbon to reform the starting
structure. The lone electron here that remains a Nitrogen cannot resonate further. And the
reason for that is the carbon atom has nothing left to kick out and that means you'll be
violating the octet.
So make sure that if the atom has a lone electron pair, if it's already has a pi bond, ask yourself
if I form another pi bond are there any electrons to kick out? Let's take a look at this enol.
For this molecule we have two lone pairs on Oxygen so we can take one of those lone pairs
and collapse it down towards that double bound sp2 carbon atom. Carbon would have too many
electrons on its octet but we can take the pi bond and collapse it on a lone on a carbon
atom that will not have a pi bond to Oxygen. We draw the double headed arrows, re-draw
the skeleton and let's see what changed. We have a green lone pair that hasn't moved,
the purple lone pair that now formed the pi bond to carbon and the red lone pair sitting
as a lone pair on carbon.
A formal charge on Oxygen shows a plus one, and a formal charge on carbon shows negative
one. Notice that the electrons collapse onto carbon rather than moving to form a pi bond
to the end and that's because the terminal carbon is sp3 and you cannot resonate onto
sp3 carbon atoms. Don't forget your brackets and you will quick net charge analysis. We're
starting with the neutral molecule, our final structure has a plus one and minus one which
cancel out to give us a neutral molecule and we're good to go. Contrast this with a lone
pair of electrons on the ketone Oxygen. In this case, if you try to resonate the electrons
down, carbon is sp2 but the sp2 is, is sp2 double bound to Oxygen there's nothing to
kick out. And if you try to show a resonance here, carbon has too many electrons in its
octet. And that is not allowed.
If your molecule doesn't have negative lone electrons or partially negative lone electrons,
the next thing you want to look for are pi bonds especially in a conjugated system. A
conjugated system is when you have alternating pi bond such as 1,3-butadiene. In this case
we have no lone pairs, negative, or neutral. So the next best negative electrons to look
at are the pi electrons. I can take these pi electrons and resonate them towards the
third carbon atom so that will give me an overfull octet. So as a result I have to take
those electrons and collapse them as a lone pair on the end of the molecule. That gives
me a resonance structure like this and then for formal charge, if carbon used to have
pi bond, lost that electron and got nothing in return, it has a formal charge of +1. If
carbon used to have a pi bond and now has a lone pair, it gets a charge of -1. Don't
forget the brackets and the conservation of charge we start with neutral plus and minus
cancel out, once again it's neutral.
For this molecule you can break the resonance of even more to show it step by step. We can
start by showing the red electrons collapse onto carbon as a lone pair giving me an intermediate
resonance structure that still has a pi bond on the end but now has a lone pair on the
second carbon with a negative charge and a positive charge on the first carbon. We can
then show this lone electrons collapsing down towards the second pi bond causing the second
pi bond to collapse as a lone pair giving me the third structure that we've already
seen with positive and negative on other end. Both sets of resonance are correct, it's simply
a question of what your professor is looking for.
Rule number 4 isn't a rule, it's just something I want you to keep in mind. Till now it started
with negative or neutral atoms but you'll sometimes get a starting molecule that has
a carbocation or a positive charge. Since electrons are moving and electrons are negative,
they're going to be attracted to that positive charge. So for example, if we start with a
conjugated system but this time we have two pi bonds and a positive charge, the pi bond
closest to the positive charge has negative electrons that are attracted to that positive
charge so you show that as your first resonance where the electrons go towards the positive.
As a result we now have a pi bond on the end, the pi bond on the left hasn't moved and the
carbocation in the middle of the molecule. Once again, electrons are attracted to positive.
So we'll show the pi bond on the left now attracted to those positive electrons.
Show the double headed arrow and see what changed. The green pi bond hasn't move, the
blue pi bond now sets one carbon over, and carbon on the left has a positive charge.
Put brackets around it, check out the net charge is the same for every structure, we
have plus one, plus one, and plus one, we are good to go. Using all of this information,
I challenge you to try this question which came out of my resonance practice quiz. You
can find this entire video series, practice quiz, solutions and resonance, study guide
by visiting my website, leah4sci.com/Resonance.
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