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- In the last video I promised you that I would show a
- concrete example of electrophilic aromatic
- substitution.
- So let's do that right here.
- So let's say we have some benzene.
- And it's a solution with some molecular bromine.
- So I'll draw the molecular bromine like this.
- So it's one bromine right there.
- It has one, two, three, four, five, six,
- seven, valence electrons.
- And it's bonded to another bromine that has one, two,
- three, four, five, six, seven electrons.
- These two guys in the middle are bonded to each other.
- They can kind of act as an electron pair or make bromines
- feel like they have eight electrons.
- This guy thinks he has a magenta one.
- That guy thinks he has the blue one.
- And let's say we have some iron bromide
- in our mix as well.
- And what we're going to see is that this is going to catalyze
- the reaction.
- So the iron bromide.
- We have some iron and it's bonded to three
- bromines just like that.
- So we haven't seen iron bromide much.
- But the way I think about it is, bromine molecules are much
- more electronegative than the iron.
- So even though these look like fair covalent bonds, if we had
- thought about oxidation, these guys are going to be hogging
- the electrons.
- This actually would have an oxidation number of three.
- And in reality they are hogging the electrons.
- They are more electronegative.
- So the way I think about is that this iron will have a
- slightly positive charge.
- Because the electrons are being hogged away from it.
- So it wouldn't mind to gain an electron.
- Or it might want to accept an electron.
- It might want to act as a Lewis acid.
- Remember a Lewis acid will accept an electron.
- So this might want to act as a Lewis acid.
- So who can it nab electrons from?
- Well maybe it can nab an electron from this bromine
- right here.
- And I'm not saying that this always going to occur, but
- under the right circumstances if they bump into each other
- with the right energies it can happen.
- So this electron-- let me do it in a different color that I
- haven't used yet, this green color-- let's say this
- electron right here gets nabbed by that iron.
- Then what do we have?
- Well then we have a situation, we have this bromine-- the
- blue one-- with one, two, three, four, five, six, seven
- valence electrons.
- We have the magenta bromine with one, two,
- three, four, five.
- Now it only has the sixth valence electron right here.
- The seventh got nabbed by the iron.
- So the iron has the seventh valence electron and then you
- have the rest of the molecule.
- So then you have your iron and it's attached, of course, to
- the three bromines.
- Just like that.
- And then our bonds, these guys were bonded.
- They still are bonded.
- And now these guys are bonded.
- These were in a pair.
- This electron jumped over to the iron.
- And now we have another bond.
- But because this bromine lost an electron-- it was neutral,
- it lost an electron-- it now has a positive charge.
- And the iron, now that it gained this electron, now has
- a negative charge.
- So let's think about what's going to happen now.
- Now we're going to bring the benzene into the mix.
- So let me redraw the benzene.
- And we have this double bond, that double bond, and then
- just to make things clear, let me draw this double bond with
- the two electrons on either end.
- So we have the orange electron, you have your green
- electron right over there, and I'll draw the double bond as
- being green.
- Now let's think about this molecule right here.
- We have a bromine with a positive charge.
- Bromines are really, really, really, electronegative.
- You might see them with a negative charge, but the
- positive charge, it really wants to grab an electron.
- And in the right circumstances, you can imagine
- where it really wants to grab that electron right there.
- So maybe, if there was just some way it
- could pull this electron.
- But the only way it could pull this electron is maybe if
- this-- because if it just took that electron than this
- bromine would have a positive charge, which isn't cool.
- So this bromine maybe would want to pull an electron.
- If this guy gets an electron, then this
- guy can get an electron.
- So you can imagine this thing as a whole really, really
- wants to grab an electron.
- It might be very good at doing it.
- So this is our strong electrophile.
- So what actually will happen in the bromination of this
- benzene ring-- let me draw some hydrogens here just to
- make things clear.
- We already have hydrogens on all of these carbons.
- Sometimes it's important to visualize this when we're
- doing electrophilic aromatic substitutions.
- So we already have a hydrogen on all of these molecules.
- So maybe this is so electrophilic it can actually
- break the aromatic ring, nab this electron right there.
- So maybe, this electron right there goes to the bromine.
- Maybe I should even do it this way.
- Just to make it clear.
- Kind of replaces that one.
- Although, obviously the electrons are a bit fungible.
- So maybe it goes over there.
- And then when it goes to this-- let me make it clear
- it's going to the blue bromine.
- So the electron right here goes to the blue bromine.
- If the blue bromine gets an electron then it can let go of
- this blue electron.
- So this blue electron can then go to this
- bromine right over here.
- And then what is our situation look like?
- Well if we have that, then let me draw our benzene ring
- first. We have our benzene ring first. So let me draw the
- benzene ring.
- This double bond, that double bond.
- Let me draw all of the hydrogens.
- I want to do them in purple.
- So we have one hydrogen, two hydrogens, three hydrogens,
- four hydrogens, five hydrogens, and six hydrogens.
- This orange electron is still with this carbon right here.
- But that electron got nabbed by this bromine.
- So that electron got nabbed by this bromine right over here.
- I've kind of flipped it around and now it has its other six
- valence electrons.
- One, two, three, four, five, six.
- The electron got taken away from this carbon.
- So now that carbon will have a positive charge.
- But we saw in the last video, it's actually resonance
- stabilized.
- That electron could jump there.
- That electron can jump there.
- So it's not as stable as a nice aromatic
- benzene ring like this.
- But it's not a ridiculous carbocation.
- It's stable enough for it to exist for some small amount of
- time while we kind of hit our transition state.
- And then this molecule over here, what's it
- going to look like?
- Well this bromine had a positive charge.
- Now it gained an electron.
- Let me draw it.
- So you have your bromine.
- It gained an electron.
- So now it is neutral again.
- So let's see, it had the one, two, three, four, five, six.
- Now it gained this blue electron.
- So now it's seven valence electrons.
- Back to being neutral.
- It's bonded to the iron bromide.
- Let me draw the bromines.
- One, two, three.
- And so we've given this blue bromine to the benzene ring.
- But it's not happy here.
- It doesn't want to break it's aromaticity.
- It wants an electron back.
- So how can it gain an electron back?
- Well this thing-- actually let me make it very clear.
- Let me make it clear.
- This thing had a negative charge.
- So you can imagine-- and we had this electron right here.
- So maybe this thing right over here can now act
- as an actual base.
- It can nab a proton off of the benzene ring.
- Just like we saw in the last video.
- This is now the base.
- This whole complex, to some degree, acted as a strong
- electrophile.
- Now that we got rid of one of bromines, this thing might
- want to grab a proton now, since it is positive and will
- act as our base.
- And will nab one of these protons.
- If it nabs the proton, than the leftover
- electron is still there.
- That electron is still there.
- Let me do that in a different color.
- I'll do it in red.
- That electron is still there.
- And then that electron can be given to that original
- carbocation and we'll have a nice aromatic ring again.
- So how would that look?
- So you could imagine a situation where this electron,
- this green electron right here, gets given to the
- hydrogen nucleus.
- If it's given to that hydrogen nucleus, then this red
- electron right here can then be given to the carbocation
- And then what are we left with?
- Well, we have our-- let me draw what we started with-- so
- we have our ring.
- We have this double bond and that double bond right there.
- Now let me draw all of our hydrogens.
- We have this hydrogen, that hydrogen, this one over here,
- this one over here, and we have this one over here.
- Now this hydrogen just now got nabbed.
- So this hydrogen over here got nabbed, it got given--
- actually the hydrogen nucleus got given this green electron.
- It got given this green electron, which is paired with
- this magenta one right over here.
- So it is now bonded to the bromine.
- We now have hydrogen bromide.
- And this had one, two, three, four, five, six
- other valence electrons.
- I'll keep the colors consistent.
- And this is now bonded.
- Now this electron went away from the iron.
- So the iron will now lose its negative charge.
- The iron bromide.
- So now it is back in its original form.
- So we have our iron bonded to three bromines.
- We are bonded to three bromines just like that.
- It has lost its negative charge.
- And this electron right here has now gone to the
- carbocation So that electron right there has now gone to
- the carbocation.
- It is right there.
- So this bond, you can imagine it now being this double bond.
- So it was magenta, so I'll draw it in magenta just there.
- And we can't forget the whole point of this whole video was
- the bromination.
- So we now have this bromo group there.
- So we have this orange electron is right over here.
- And it is bonded to that bromine.
- Just like that.
- And we have brominated the benzene ring using-- and just
- let's be clear-- we had a negative charge
- and a positive charge.
- Now it's canceled out.
- The thing with the negative charge gave an electron to the
- positive charge.
- It is now neutral.
- That is now neutral.
- Now let's be clear on what happened here.
- Just to map it to what we saw before.
- We had a benzene ring.
- Right here we have no strong electrophile,
- or strong base yet.
- It had to become a strong electrophile.
- When this thing gave an electron to the iron bromine,
- it became this larger molecule.
- Now this whole thing can act as a pretty strong
- electrophile.
- This grabbed an electron.
- It broke the aromaticity of the benzene ring.
- But just long enough for this bromine to form.
- But once this bromine, it gained an electron and then
- bonded to the benzene.
- And then gave up an electron to this other bromine that
- really wanted to get an electron because it had a
- positive charge.
- And then once it got it, now this whole thing
- acted as the base.
- So we kind of have the same molecule changing up a little
- bit, acting as an electrophile or acting as a base.
- And then once it acts as a base, this bromine, this
- magenta bromine, nabs the proton, allows this electron
- to go back to that carbocation and then we're left with the
- iron bromide again.
- So this thing really didn't change
- through the whole reaction.
- That's why we can call it a catalyst. It wasn't one of the
- reagents, one of the reactants in the reaction.
- Hopefully you found that vaguely interesting.