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Organic chemistry
Organic oxidation-reduction reactions
How to determine if a starting compound is oxidized, reduced, or neither.
Want to join the conversation?
- how did sal determine the presence of hydrogen in the product of 2nd example. it could have been a carbocation?(19 votes)
- If it would have been a carbocation then it would have a symbol (+ here)... When the symbol is absent then we can assume that its not a carbocation but hydrogen bond (which are neglected in condensed diagrams)(35 votes)
- In 3rd example the double bonded carbon is sp2 hybridized(33.3% s character) while carbon attached to it is sp3 hybridized(25% s character) ,then why both carbon has same electronegativity?(20 votes)
- Yes, an sp² carbon is slightly more electronegative than an sp³ carbon.\
But, the arbitrary rules for oxidation number say that electrons in bonds between identical atoms must be shared equally.(20 votes)
- Does oxidation always result in an increase in oxygen in the product of a reaction?(4 votes)
- No, oxidation can be generalised as an increase in oxidation number, Fe^2+ -> Fe^3+ is oxidation but no oxygen was added.(12 votes)
- Overall,I dont get why we always determine if a reaction is oxydation or reduction or neither by looking at the carbon? Is it because between the reagent and product,it was where the difference was seen?
For the last example,could we have chose to look at another carbon ?(7 votes)- We dont look at the other carbons in the first example because they didn't change. Therefore we know instantly it won't be oxidised or reduced.
However in the other example, you can see instantly that the two different carbons had changed in the product side which is why we look at them.(5 votes)
- Why was the hydrogen not accounted for in the last reaction? When I did this on my own, before you got to the solution, I computed an oxidation state of -1 on the C from the structure on the right because I had the hydrogen attached to the carbon in my drawing. Thanks in advance for your help!(4 votes)
- There is no hydrogen on the right hand carbon.
That carbon is attached to an O and three other carbon atoms.
There is no room for a hydrogen atom, unless you violate the octet rule.(8 votes)
- Atwhat is the reducing agent? Oxygen or hydrogen? 9:15(2 votes)
- Neither. Reduction/oxidation can only occur when we have a chemical reaction (known as a redox reaction). Jay is looking at one of the reactants and comparing it to one of the products to see whether or not carbon underwent oxidation or reduction. To identify the oxidizing agents and the reducing agents we have to examine the entire reaction and identify which reactant species have atoms which are oxidized and which atoms get reduced. The reactant that gets reduced is the oxidizing agent, and the one which is oxidized is the reducing agent; it is important to note that we don't consider the individual atoms within a molecule to be oxidizing or reducing agents, but instead the ENTIRE molecule. So to answer your question, we don't know what the reducing agent was, because we were not shown the reaction.(3 votes)
- In the first example - aldehyde oxidation to carboxylic acid, what could be the oxidizing agent?(2 votes)
- You could use potassium dichromate or potassium permanganate.(1 vote)
- In the last example, what happened to the hydrogen connected to oxygen on the right hand side?(2 votes)
- at,one carbon is oxidised but one carbon is reduced.This implies that there must have been a reducing as well as an oxidising agent simultaneously.How is that possible? 12:30(2 votes)
- To answer that, you really need to see the chemical reaction ("redox" reaction) which created the change in oxidation states. Remember, individual atoms within a molecule are not oxidizing or reducing agents, but the entire molecule ITSELF is the agent.(1 vote)
- For the last example, can someone please clarify it that equation can be considered as a disproportionation reaction in because a carbon undergoes both oxidation and reduction in that example when having one carbon atom going from 0 to -1 and 0 and +1 ? Thanks.(1 vote)
Video transcript
- Now that we know how to
assign oxidation states to Carbon let's look at some
organic redox reactions, and remember these definition
from general chemistry. Oxidation involves an increase
in the oxidation state and reduction involves a decrease or a reduction in the oxidation state. You might also remember that
loss of electrons is oxidation and gain of electrons is reduction. So, LEO the lion goes GER as
a good way to remember that. Let's look at this starting compound here and this reaction, and let's figure out whether the starting compound has been oxidized, reduced, or neither. If you look at these
three carbons on the left and these three carbons on the right. All right, there's no
change to those carbons so there should be no change in the oxidation states of those carbons but when you look at
this carbon right here, the one I just marked in yellow, right, on the left it's bonded to, let me go ahead and draw that out, Is double bonded to an oxygen, and on the right it's
bonded to a hydrogen, and on the left it's bonded to a carbon. But on the right here
that same carbon, right, is now bonded to an oxygen on the right. So, it still double bonded it to an oxygen and has a, has a bond
to a carbon on the left, but on the right now,
it has a bond to an OH. So, that carbon has likely
changed its oxidation state and we can go ahead and
find the oxidation state using what we learned in the last video. So, remember you need to put
in your bonding electrons. So, let's put in our
bonding electrons here. Each bond consists of two electrons. All right, so we know that when
we're doing oxidation states we need to think about
electronegativity differences and carbon is more
electronegative than hydrogen. So, we assign both of
those electrons to carbon. Oxygen is more electronegative than carbon so oxygen takes all
four of those electrons. And when we get to carbon versus carbon we can assume that those carbons have the same electronegativies and therefore for two electrons we're gonna split those two electrons up. We're gonna give one
electron to one carbon and the other electron
to the other carbon. So, this carbon is
surrounded by three electrons and we know carbon should have, we know carbon is supposed to have a number of four valence
electrons around it. So, carbon's supposed to have four. In our dot structure here
Carbon only has three. Let me highlight them.
One, two, and three. So, four minus three gives us plus one which is the oxidation
state for this carbon. So, this carbon right here
has an oxidation state of plus one on the left. What about on the right? Let's go ahead and put in
our bonding electrons, right. So, we know that each bond
consists of two electrons. Let me put those in here. And again, we think
about electronegativity. This time on the right, this carbon is bonded to an oxygen now so carbon loses those two electrons to oxygen. Right, oxygen's more electronegative. Oxygen takes, this top oxygen
takes those four electrons and again, we split these two electrons. So, this time carbon is
only surrounded by one. So, define carbon's oxidation state we know that carbon's supposed to have four valence electrons and from that we subtract the number of valence electrons around carbon once we've accounted
for electronegativity. And that's only one electron now. So, four minus one gives us an oxidation state of plus three. So, on the right this same carbon now has an oxidation state of plus three. So, what happened in this reaction? Carbon went from an
oxidation state of plus one to an oxidation state of plus three. An increase in the oxidation
state is oxidation. So, our starting compound
was oxidized here. So, our starting compound was oxidized. In order for this to be oxidized we would need some sort
of oxidizing agent. So, I'll write that down
here really quickly. We would need an oxidizing agent to accomplish this reaction. And it's the oxidizing agent that itself is being reduced because remember, whenever something is being oxidized something else has to be reduced. Let's think about the
other definitions for, the other definition I
should say for oxidation. We know that oxidation
involves the loss of electrons. So, let's look at that here. On the left, right, carbon
had three electrons around it, and when we added a bond to oxygen, right, carbon lost these
two electrons over here. It had those two electrons. On the left it had those two electrons. On the right it lost those two electrons and now it's only surrounded by one. And we see an increase
in the oxidation state. So, it's like carbon lost electrons. Also there's a shortcut that you can use. For this reaction, let's think about what that shortcut would be. Let me use, I'll use red still. On the left, this carbon with
an oxidation state of plus one has two bonds to oxygen. Let me go ahead and highlight them here. So, here's one bond and
here's the other bond. On the right, that same carbon now has three bonds to oxygen. So, here's one bond, here's two bonds, and then this bond over here. So, we've increased in the
number of bonds to oxygen and that can tell us we have
an oxidation really quickly without going through and doing
all these oxidation states. Notice we lost a bond to hydrogen. All right, so over here we
lost this bond to hydrogen and we replaced it with a bond to oxygen. So, an increase in the
number of bonds to oxygen or a decrease in number
of bonds to hydrogen can tell you that your
carbon is oxidized too. Let's do another one. So, our goal is to figure out whether this starting compound has been oxidized, reduced, or neither. So, again, these three
carbons on the left, right, are the same as these
three carbons on the right. And there's no change in those carbons. So, therefore we would expect no change in the oxidation state. The same with this carbon. So, we're gonna focus in on this carbon. So, on the left that
carbon has a double bond to oxygen so we draw that in there an then it's bonded to
a carbon on the left and a carbon on the right. And our product, that same carbon, is now bonded to only one oxygen. Let me go ahead and do that. Only one bond I should say to oxygen and it's still bonded
to a carbon on the left and still bonded to a carbon on the right, and it must have a bond to
hydrogen in here, right? It's not drawn in but it's
assumed that you know that. So, let's put in that bond to hydrogen. And let's find our oxidation states. So, we start on the left, and we draw in our bonding electrons, and next we think about
electronegativity differences. So, we know that oxygen's more
electronegative than carbon so oxygen takes those four electrons. Over here, right, we assume
that the electronegativities for these carbons are the same, and so we split up those two electrons. We give on electron to one carbon and one electron to the other carbon. And same over here. So, the oxidation state for carbon would be four valence electrons minus this, in this case we have two around carbon, so here's two. Four minus two gives us an
oxidation state of plus two. So, on the left this carbon has an oxidation state of plus two. What about on the right? Let's put in our bonding electrons and again, once we've
done that we think about electronegativity and we can assign an oxidation state to that carbon. So, we know that oxygen's
more elecrtonegative so oxygen takes those two electrons. Right, again, there's, we assume that these carbons have
the same electronegativity so we split up those electrons. We split up these electrons. But now carbon is bonded to hydrogen. So, carbon is a little
more electronegative than hydrogen so we just give both of those electrons to carbon. And now let's find our oxidation state. We know carbon is supposed to have four valence electrons around it and in our drawing here, once we've accounted for electronegativity carbon is surrounded by
one, two, three, and four. So, four minus four is equal to zero. And so this carbon, this carbon now has an oxidation state of zero. So, that carbon went
from an oxidation state of plus two to an oxidation state of zero. That's a decrease in the oxidation state. So, carbon was reduced. So, in this reaction our
starting compound was reduced. All right, so this was, our
starting compound was reduced which means we would need
some sort of reducing agent to accomplish this reaction. And it's the reducing agent
that would be oxidized because if something is reduced something else has to be oxidized. We can also think about our
other definition for reduction. Gain of electrons is reduction and let's look at what happened here. So, the electrons that were gained right, were these two electrons right here because on the left we had this carbon with a double bond to this oxygen. On the right, we have only
now one bond to oxygen. Right, so we gained these two electrons because we increased the
number of bonds to hydrogen. So, our other shortcut is to look at the number of bonds to oxygen. Like we talked about in the last example. On the left, carbon has
a double bond to oxygen. So, two bonds to oxygen. On the right we have
only one bond to oxygen. So, we decreased in the
number of bonds to oxygen so that's a reduction. Again, a fast way of figuring it out. Or we increased an number
of bonds to hydrogen. Let's do one more of these problems. So, our goal is to figure out if our starting compound on the left has been oxidized, reduced, or neither. And this time these are the two carbons that we are going to analyze. Those are the two carbons that have changed in this reaction. So, on the left let's draw what we have. We have a carbon with a
double bond to another carbon and both of those carbons are bonded to, directly bonded to two more carbons. On the right, those carbons now only have a single bond between them and this carbon on the left is bonded to a hydrogen. This carbon on the right is
directly bonded to an oxygen and let's put in those
other carbons, right? We have another carbon, carbon bonds. I mean, let's put in those
carbon, carbon bonds like that. Let's assign our oxidation states. So, we draw in our bonding electrons and we think about
electronegativity differences but we're going assume once again that our carbons have the same
electronegativity on the left and so when we're assigning electrons let's just pick one of those carbons. So, let's pick the carbon
on the left, right? We divide up those two
electrons in that bond. Give one electron to one carbon. One electron to the other. We do the same thing here. And for the double bond
with four electrons we divide up those four electrons. We give two electrons to each carbon. And so carbon is supposed to
have four valence electrons and around it, this carbon
has one, two, three, and four. So, four minus four is an
oxidation state of zero. So, this carbon has an
oxidation state of zero. Same thing for the carbon on the right side of the double bond, right? So, the same situation. So, it also has an
oxidation state of zero. What about for our product, right? Let's examine those two carbons. Let's put in our bonding electrons. So, we draw in our bonding electrons here. So, we put those in. A lot more bonding electrons to draw. And we think about electronegativity
differences, right? Things have changed. So, now, now let's focus in
on the carbon on the left. So, the carbon on the left
is now bonded to a hydrogen and carbon is a little
bit more electronegative. So, we give both of
those electrons to carbon and then we have carbon bonded to carbon, carbon bonded to carbon,
and carbon bonded to carbon. So, what's the oxidation state of that carbon on the left now? Carbon is supposed to have
four valence electrons and how many do have around it? Let's see, one, two, three, four, five. So, four minus five gives us an oxidation state of minus one. So, this carbon on the left now has an oxidation state of minus one. What about the carbon on the right? Well, we have a tie here, right? So, we divide up those
electrons, a tie here, but oxygen is more
electronegative than carbon. So, oxygen gets both of those electrons. So, the carbon on the right should have an oxidation state of for minus three. Here are the three electrons around carbon once we've accounted
for electronegativity. So, that's an oxidation state of plus one. So, this carbon has an
oxidation state of plus one. So, overall, what happened here? Well, let me use red. So, the carbon on the left went from an oxidation state of zero to an oxidation state of minus one. That is a decrease in the oxidation state. So, that carbon was reduced. What about the carbon on the right side of the double bond that
we started off with? It started off with an
oxidation state of zero and it went to an oxidation
state of plus one. That's an increase in the oxidation state. So, that carbon was oxidized. What can we say about
starting compounds overall? Well, the starting compound overall there's no net change in
the oxidation states, right? So, on the left for those two carbons we get a total of zero. On the right we had negative
one and positive one for a total of zero. There's no net change. So, this would be neither. The starting compound was, overall, it wasn't oxidized or
reduced. It's neither.