>> Welcome to reactions of organic functional groups,
part 2, redox, a closer look.
We had a brief introduction to redox, which is oxidation
and reduction a few weeks ago
and now we're going to dig in deeper.
Let's get started.
Alright, so, we know from our previous lessons
that redox is a change in the oxidation number.
Okay. So that de oxidation number.
And oxidation number is really just a fancy way
to say the charge.
How many electrons are lost or gained.
So, back from exam 1 we learned about the oxidation
and reduction of inorganic, really compounds, to form ions.
And, so, we learned the expression, Leo Says GER.
And, right, so, LEO was loss of electrons, is oxidation.
Right? So, there's our LEO.
And the GER was Gain of electrons is reduction.
Okay. And, so, if this is feeling a little sketchy,
it wouldn't be a bad idea to go back
and rewatch that earlier video.
We're going to get a little more formal now.
Remember that there's always some elements that we're going
to know what charge they take.
Right? And these are our best friends.
Right? So, pure elements are neutral.
Fixed charge, right, we know our alkali, our alkaline earth
and aluminum, we always know those charges.
Another couple that we know, we always know silver is plus 1,
we always know zinc is plus 2.
Right? And then halogens, right, remember x is halogens.
That's the 7th group.
Those are always minus 1 for our course, because we're going
to only look at halogens when they are
in ionic compounds or bonded to carbon.
There are exceptions here beyond chem 300 -- excuse me, chem 309,
but we're not going to worry about those,
because we're chem 309.
Oxygen, for our class, will always be minus 2.
All right?
And whether it's bonded to, you know, in an ionic compound
or a covalent compound, I will never be asking you questions
with peroxide the exception.
And I will never ask you about hydrogen
as hydride, the minus 1.
Here is -- so we'll always see it as plus 1.
All right.
So, the first thing we need to do before we can go
to reactions is we have to --
we need to feel comfortable with compounds.
Now, you guys are already really good at figuring
out the oxidation state, right?
Lead is a variable, right, a variable charge metal.
So, we've been practicing this all along, right?
If we look at this compound, right,
this is a neutral compound.
So, we know that the cation
and the anion need to add up to zero.
So, using our table here, right,
so when we see the oxygen we know it's 0 2 minus
and the chemical formula tells us there's two of those.
So, overall we have a minus 4.
So, we've learned, right?
So, this lead must bring in the entire positive charge
so that we can equal zero.
So, the lead must be plus 4.
Right? So, the oxides are easy.
And then we had to determine the lead.
Now, in this next example sodium, right, our buddy,
the alkaline metal, we know sodium is plus 1.
And once again the oxide we know is minus 2
and then the chemical formula says there's three of them.
Right? So, the mystery here is the nitrogen.
We don't know what the charge is.
Once again, when we look this is a compound and it's neutral,
so everything needs to add to zero.
So, basically, we just want to set
up a very simple algebraic expression, right?
So, the sodium is plus 1
and then the nitrogen is what we don't know.
And then the oxide there were three of those
and they were each minus 2.
And when we are all done this is a neutral compound,
so it adds to zero.
So, now we would simply solve for the charge on the nitrogen.
So, we've got 1 and minus 6, so I'll just switch colors here,
so we have the question mark from minus 5 equals zero.
So, the nitrogen is going to be plus 5.
So, we've -- we already know the nitrogen or the sodium
and the oxide and now we know that nitrogen is plus 5.
There we go, running out of room.
So, that's not very -- right,
it needs to be a good superscript, it's much prettier.
Okay, now, let's do one more and this one's a little trickier,
because this one is an ion.
So, when we add up the oxidation states they're not going
to add to zero, right?
Now we have our minus 1.
This is our minus 1 ion.
All right.
So, what do we know?
We know hydrogen is plus one.
Carbon, once again, is the mystery.
And, then, our buddy oxygen is oxide.
And we have three as well.
So, we'll do our algebraic expression again.
So, keep our color coding, right.
So, the hydrogen was plus 1
and then the carbon is the question mark.
And there just happens to only in each
of these examples there's only one, right?
If we had multiple we'd have to put our coefficient.
And then we have the three oxides
and they are each minus -- oops, oxides blue -- minus 2.
All right.
And then what makes this one a little different, right,
is now because it's a minus 1 charge, they add, right,
all of these charges have to add to minus 1.
So, we do our algebra and --
so, we have the plus 1 and the minus 6.
So, the question mark minus 5 equals minus 1.
Add 5 to both sides.
And, so, the charge, the oxidation state
of the carbon is plus 4.
Right? So, I'm emphasizing the one we had to figure out.
But, in the question you would want to make sure
to list the oxidation state of all of the elements
in the compound or ion.
Alrighty. So, now that we're feeling really comfortable
with how to look at individual compounds, let's expand
and start to look at chemical reactions.
Alrighty. So, to help us recognize
which elements are being oxidized
and which elements are being reduced,
in reaction half reactions are going to be very helpful.
And the key to half reactions is half reactions only look
at the element that's being oxidized or reduced.
So, step one will be to assign the oxidation numbers
to each atom in the reaction,
just like we practiced on the previous page.
Then we'll look for changes in the oxidation number
and write the half reactions for those elements.
All right, let's practice this together.
Alrighty. So, step one, you need to write the oxidation states.
So, once again, our buddy oxygen, right,
that's going to be oxide minus 2.
And this one's pretty simple, since it's a 1 to 1 ratio.
We know the copper must be plus 2.
Now, the hydrogen, right, it's pure,
have no fear of ice cold beer.
One of our diatomic elements.
Right? So, this is one of the only --
the have no fear of ice cold beer gang,
of these diatomic elements, this is the only time we're going
to worry about subscripts.
Because this is the pure form.
Now, when we look at the copper notice that now it's a zero.
And here the hydrogen, because it's part of a compound, plus 1,
especially because it's with the more electronegative oxygen.
And then the oxygen still tried and true, minus 2.
Alrighty. So, let's write the half reaction for hydrogen.
Right? So, hydrogen started as a zero and it went to plus 1.
Now, because of the pure state being diatomic,
we'll put a coefficient here to keep our atoms balanced.
And now we're going to add the electrons to balance the charge.
So, a zero and a plus.
So, we're going to put the electrons here, right,
so the positive 2 and the negative 2 cancel.
So, we have zero on the left and zero on the right.
Now, let's look at the copper.
Copper is plus 2 and it goes to copper zero.
So, here, by looking at the charge,
we know to put the electrons on the left side, so plus 2
and minus 2 cancels to neural zero equals zero.
Alrighty. So, now oxidation and reduction always occur together,
because the electrons lost need to equal the electrons gained.
So, we can't have one process without the other.
There's got to be a handoff somewhere.
Okay. So, which element was oxidized?
Right. So, there's our LEO says GER
and maybe we should have reviewed that, right?
So, LEO says GER.
Right? Or OIL RIG if you like that one better.
Okay? So, here's the lost electrons.
So, hydrogen was oxidized.
Oops.
There we go.
Okay. And which element was reduced?
Copper. Okay.
Copper plus 2.
Notice here I'm asking you about the element.
Okay? So, be careful with the wording of the question.
Sometimes we want to put the actual compound,
other times we'll put the element.
We're going to build up.
So, I just want that to be on your radar.
Now, let's look at a reaction that's a little more complex.
All right?
So, iron and oxide, right,
so we've practiced this one quite a bit.
Oxide is minus 2 and now to properly name this we need
to know the charge on the iron.
So, there's three oxides, so overall there's a minus 6,
which means there's going to need to be a plus 6.
And there are two irons present, so this would be iron 3 plus.
So, iron 3 oxide reacts with carbon monoxide.
So, there's our buddy oxide again, minus 2.
And this one's pretty straightforward, right,
because it's a 1 to 1.
So, we know the carbon must be plus 2.
All right, and then let's put in our reactionary here
to make sure we've got our boundaries.
Now, the iron, right, it's in the pure state.
Woops, runaway pen.
All right.
So, iron when it's pure, zero.
And then, once again, right, using our table, always start
with the elements we know for sure.
We know, right, oxygen will be oxides minus 2.
The chemical formula shows that we have two present.
So, overall there's a minus 4, so that means
that the carbon has to bring in a plus 4.
All right?
So, we can see that the iron is changing oxidation state
and the carbon is changing oxidation states.
Let's go ahead and write the half reaction for carbon, right.
So, we just write the two forms of the element and then looking
at the charges, we're like, okay,
where do we put the electrons to balance the charge?
So, we're going to need to add two electrons on the right
to balance out, because we only have plus 2 on the left, right?
So, plus 2 equals 4 minus 2.
Let's go ahead and do iron.
Iron starts as a plus 3 and goes to the pure form of zero.
So, this time to balance the charge we're going to need
to put three electrons on the left.
Now, you might be a little concerned here
because I mentioned that electrons lost have
to equal electrons gained and it feels like --
it feels like I told you a false statement.
But, let's be patient, let's finish up with A and B
and then we'll look at the whole chemical reaction.
Remember, because half reactions only look
at the element being oxidized or reduced,
so these are the correct half reactions for the elements,
but we need to integrate this information
with the chemical formula.
And I'll show you that very quickly.
But let's wrap up which element is oxidized, right.
And that means lost electrons.
So, here they are, here's the lost electrons.
So, the carbon, the carbon plus 2 gets oxidized.
And who's reduced?
The iron. Who gained electrons?
The iron plus 3 or the iron.
Okay. So, now let's sort out, right?
Because these are the gained electrons.
All right, so now let's reconcile this.
What's going on here?
So, we look at the half reaction.
So, if we look at the half reaction for carbon,
so that's our oxidation reaction.
We see that we have --
now we look at the full chemical formula or chemical reaction.
From the chemical reaction we see there's the coefficient
of 3.
From the reaction.
All right?
And then from the half reaction we see
that each carbon lost 2 electrons.
Right? So, this came from the half reaction and the 3 came
from the full reaction.
So, when we put those together we see that we lost 6 electrons.
And, now, if we look at the reduction.
All right.
Now, we're going to focus on the iron.
So, if we look up here, here is the chemical formula
that the iron is present in.
so, from the chemical formula we get the 2 chemical formula.
Okay. And then from our half reaction we see there's
3 electrons.
Right? So, woops, make sure you guys can see this.
There's our half reaction.
See, so we bring together the information
from the full balanced reaction and the information
from our half reaction and we bring those together.
We see that 6 electrons were lost
and 6 electrons were gained, so, once again,
electrons lost equaled electrons gained.
Alrighty. So, that's pretty much everything we need to know
about inorganic oxidation and reduction.
So, now we are going to expand our understanding of redox
to organic compounds, right?
And organic molecules, right, are based on carbon, right?
That carbon backbone.
So, we're going to use a different set of rules
to recognize oxidation and reduction.
And since organic compounds are all about carbon,
carbon will be our focus.
We're going to look at the carbon to oxygen bonds
and the carbon to hydrogen bonds to identify when oxidation
and reduction are occurring.
All right?
So, for oxidation we'll use the symbol of O
in the square brackets.
And for reduction we will use hydrogen in the square brackets.
All right.
So, the emphasis here oxidization is a gain in bonds
to oxygen, right, and this is bonds of carbon to oxygen.
Or the loss of bonds between carbon and hydrogen.
All right.
So, let's look at an example.
Let's just look at a simple alcohol right here.
All right.
So, here is the carbon that's bonded to the oxygen.
So, notice that there's only 1 carbon to oxygen bond.
And now you get undergoes an oxidation.
We can go to the aldehyde.
All right.
So, here is the carbon that's still bonded.
But now notice there are 2 carbons to oxygen bonds.
So, notice that there's this same number of oxygen atoms,
but we have more bonds between the carbon and the oxygen.
So that's why this is an oxidation.
The reduction, we look at the number of carbon hydrogen bonds.
So, in this next example there won't even be any
oxygen present.
So, we'll go here.
All right?
So, we can take 2 butene and make butane
and this would be a reduction, because we have a gain
in the number of carbon hydrogen bonds.
All right?
Or we'll do some more examples and we'll see we have a loss
of carbon oxygen bonds.
So, you want to think of oxygen and hydrogen as opposites here,
that the redox compliments.
To help us see this a little more clearly,
where we have the change it can be a very good idea to write
in the hydrogen, right, because oxidation and reduction
of organic compounds could involve changes in the number
of hydrogen atoms and typically
in a skeletal line structure we don't show those.
So, here this would be showing the missing hydrogen
where we have the change and then notice in the reaction,
now, we are gaining 2 carbon hydrogens, right?
So, we have 2 new carbon to hydrogen bonds.
So, since we had a gain that's why it's a reduction.
Now, I do want to point out one thing here.
We recently studied acid based chemistry, right?
So, H plus is a hydrogen ion and that's acid based chemistry.
The H in the square brackets, that represents hydride.
And that's redox chemistry.
So, hydrogen is very tricky, right?
It's the only element that we're studying
that can be positive or negative.
And that's -- when it's positive it's taking a role of a proton
for acid based chemistry
and when it's negative it will be a reducing substance,
hydride.
All right.
So, let's practice applying this knowledge.
Let's build our wisdom.
So, here are three practice reactions.
So, this would be a great place to pause the video and try
to answer them on your own and then compare answers.
Okay, great.
Let's see how you did.
All right.
So, on this first one we see where the change's occurring,
so let's draw in the missing hydrogens
so we can look for change.
And then we can see this is actually the reverse
of our practice, right?
Here we lost carbon to hydrogen bonds.
And, so, that would be an oxidation.
All right?
So, here's the reduction and the oxidation,
they can go back and forth.
Here we are looking at the change.
So, we notice change here and here and so we notice
that here we just have one carbon oxygen and one
and now we've doubled, right?
So, we've increased the number of carbon to oxygen bonds.
So, this would also be an oxidation.
And then this last one here, we look for where the change is.
Any time you see the change it's a good idea to write in, right?
So, those are the original hydrogens and then we notice
that now we have two more.
So, we gained carbon to hydrogen bonds,
so this would be an example of a reduction.
Alrighty. Okay.
So, let's build on this a little bit and practice.
So, here this one is a little more advanced.
Now, this question's letting you know the process that's
occurring and we need to predict the products.
And then let's practice our functional group classification,
because we need to reconnect
with those organic functional groups.
All right?
So, here we have a aldehyde.
Right? And here's our carbon to oxygen.
So, if we are going to do an oxidation.
All right?
We have to increase the number of carbon to oxygen bonds.
So, the aldehyde will oxidize to the carboxylic acid.
All right?
Or we could undergo reduction, right, and take away one
of these carbon oxygen bonds.
Then, so, then, we would simply have an alcohol.
All right?
So, here we added a new carbon oxygen bond
and then here we took it away.
And, so, now this would be a primary alcohol.
Okay. So, aldehydes are one of the trickier groups,
because they can be oxidized or reduced.
Here we have a secondary alcohol.
And it helps to focus our attention
where the carbon oxygen bonds are occurring
or where we see double bonds where change could take place.
We want to oxidize, so we want to increase the number
of carbon oxygen bonds.
Well, here we have less options.
Right? This carbon is already pretty committed, right?
It's got a bond to a carbon on the left and a bond
to a carbon the right.
And, so, all we can do is put a double bond there
and there's no more options.
Carbon's limited to four bonds.
We're done.
And I'm sure that you recognize this as a ketone.
All right?
So, we will have a follow-up video that will really emphasize
and go through the oxidation of alcohol and their reduction
of these carbonyl compounds.
They're very important to our study of biochemistry
that we're working towards.
Here's an example now
that brings together what we've learned
about inorganic oxidation and reduction
and organic oxidation and reduction.
When this is the test that's done on breathalyzers,
so while I have no personal experience with this,
I do have some friends that bought some of the breathalyzers
for a party, just kind of for fun.
When we blow into them the breathalyzer starts red,
this compound is red.
And then if there's alcohol in our breath,
as we blow the alcohol into the red compound a redox reaction
occurs and the chromium compound will change.
I don't want to give away the answer
and then it becomes green.
So, the more green the breathalyzer tube becomes,
the more alcohol was in the person's breath.
So, there's a, you know, real life example.
Now, let's apply our chemical understanding.
Alrighty. So, if we look here we look at this carbon
and it has one carbon to oxygen bond and now it has three.
Right? So, using our organic rules this is definitely
the oxidation.
All right.
Now, so, we know that we can pretty much already know
that this is going to be the reduction,
but let's convince ourselves of that.
All right?
So, let's practice what we've learned.
Right? The hydrogens are always plus one.
The chromium we don't know.
Question mark.
And then our buddy oxygen will always be oxide.
Okay. So, remember, right now we make our algebraic expression,
so the chemical formula tells us there's 2 hydrogens
and they are each plus 1.
Here, right, the chemical formula tells us there's 2
chromiums and we're not sure what the charge is.
And there are 7 oxides and they are each minus 2.
Then when we add this all together we should be zero.
So, algebraically, right, we end up with 2
and then this would be -- I'll just make that into an X
and minus 14, right, and then we do the algebra
and we find out that X is plus 6.
So, chromium is plus 6.
Oops, sorry, I forgot to slide my page, right?
But you guys are good at algebra, so that's good.
Now, let's look over here, right?
There's 2 chromiums and we don't know their charge.
And then we're going to add that, right.
And then there's the 3 oxides, each minus 2.
And it's a neutral compound, adds to zero.
So, when we solve for X here, it's going to be plus 3.
So, chromium plus 2 -- oops, plus 3.
Let me write that together, make it pretty.
There we go.
All right, so, we can see looking here, right,
we did our half reaction, right?
Chromium would need to gain 3 electrons, chromium 6 would need
to gain 3 electrons to become chromium plus 3
and gaining electrons is reduction.
So, there's an example of how organic and inorganic oxidation
and reduction can come together.
Alrighty. So, we have one last page to go over.
There's one more important concept about oxidation
and reduction that we need to learn.
And this is about the idea
of oxidizing agents and reducing agents.
All right.
So, at first it feels a little clunky,
but we'll get used to it, right?
So, it's an oxidizing agent.
So, that means the reactant gets reduced, right?
It helps, it's an agent, it helps other compounds oxidize.
And, so, the way it does that, how does it do that, right?
By being reduced.
All right.
So, an oxidized compound wants to lose electrons.
So, somebody has to be able to gain them.
There has to be an electron handout somewhere.
So, that's where this idea comes from.
Okay? And then reducing agents will be the reverse.
This is the reactant that gets oxidized, right?
It helps other compounds reduce.
Right? And, so, it's just the reverse.
And how does it do that, right?
By being oxidized.
Right, so being the partner in the electron handout, right?
So, reducing is gaining electrons.
So, if a compound needs to gain electrons it has
to get them from somewhere.
So the reducing agent loses its electrons
so that they can be gained by another substance.
So, it's a partnership.
Alrighty. So, let's put this together
with one last practice problem.
If you are feeling confident go ahead
and pause the video and see how you do.
And then otherwise and compare answers,
otherwise we'll work together.
All right.
So, remember, just from the very beginning of this video,
step one is we need to determine the oxidation states
of every element.
Zinc, it's easy, it's pure.
Hydrogen, plus 1.
Halogens, minus 1.
Wow, this has been really easy so far.
Ah, zinc plus 2, so now we have a change and -- excuse me --
here's hydrogen, right, it's pure.
That's zero.
And then our buddy chloride, also renamed and changed.
So, we have another change here between the hydrogen, all right?
So, we have identified our two elements
with the change of oxidation states.
Now, I have -- I'm giving --
not giving you so much guidance here.
Now, we have to find the oxidation, right?
So, oxidation is loss of electrons.
So, who lost electrons?
Ah, the zinc, right?
The zinc is the LEO.
And the hydrogen had to gain electrons.
So, this is our GER, our reduction.
All right?
So, the zinc started at zero, oxidized to zinc plus 2
and there's our lost electrons.
Well, the hydrogen, right, it started as a plus 1
and then it goes to hydrogen that's pure,
have no fear of ice cold beer.
So, here we are going to have to put coefficients
in our half reactions and so the hydrogen gains 2 electrons.
So, it's nice and neat.
Electrons lost and there's electrons gained.
Okay. Now, so, who -- right now, here we go, reactant, right?
So, when we get to oxidizing agents
and reducing agents we're talking
about the full compound, the reactant.
So, which reactant is the oxidizing agent?
That would be the hydrochloric acid.
And which reactant is the reducing agent?
That would be the zinc metal.
Alrighty. So, oxidation and reduction, are big topics
and this would be a great time for you to go and work some
of your post video practice problems
to reinforce your understanding.