[narrator] Let’s recap alkene cleavage reactions, including ozonolysis and potassium permanganate
oxidation.
We’re going to talk about predicting the products and we’re going to talk about predicting
the alkene reactant given a pair of products, a pair of cleavage products.
We’re also going to talk about both the oxidative and reductive work up of ozonolysis.
The products you’d get from either one.
When you’re doing these reactions you just want to draw a line down the center of your
alkene and you’re just going to slide these apart.
I draw them exactly the same as they are oriented on the page, already, and I would just add
an oxygen in the middle to each side.
All we’ve done is slid the two apart and put the oxygen in.
For the reductive work up, which involves either zinc metal, an acetic acid, or it involves
dimethyl sulfide.
Either one but not both.
We’re going to leave all the hydrogens intact.
These hydrogens that are not drawn in on our skeletal structure they all stay intact.
We do this for each one, whether it’s disubstituted.
Of course, the R groups could be different from one another but for this example they’re
the same.
They just keep everything lined up the same.
We just slide them apart.
Everything matches, so if there’s a methylene CH2 at the end of the alkene we’re going
to get a formaldehyde as one of the products and a ketone as the other.
However, we could also get two aldehydes.
Again, I could also have just said let’s not move a single thing relative to the way
it was originally drawn.
Less chance for error, that way.
Either way.
Trisubstituted; we just keep adding the R groups.
Slide the alkene apart.
Add the oxygen’s in.
Should get the same number of carbons, add two oxygen, for the formation of the carbonyls.
Tetrasubstituted; we get two ketones.
This is all reductive work up.
Now, let’s look at the oxidative work up of ozonolysis which is exactly the same as
potassium permanganate oxidation.
We could just use potassium permanganate and not a two-step reaction.
In a two-step reaction where we have the one and the two that I just highlighted in green
we have to add ozone at -78°, wait a while, let the reaction occur, and then to form the
ozonide then add the second set of reagents at a later time.
So two steps matter.
The one and the two matter.
Oxidative work up; anywhere there’s a hydrogen we become a carboxylic acid, or we add a hydroxyl
to become a carboxylic acid.
If we have a formaldehyde it becomes a CO2 in the oxidative work up.
But, if there’s no hydrogen available we still get the ketone.
But, if we have an aldehyde the hydrogen of the aldehyde is replaced with an OH
to form the carboxylic acid.
Then with the tetrasubstituted there’s no difference there’s no difference in the
products.
What if we’re not given the starting alkene?
What if we’re asked to predict the products?
This is where we need to count our carbons.
Otherwise, we’re going to get confused.
If we go one, two, three, four, five, six--if I’m going to join these together I need
six carbons.
I’m just going to draw this, best I can, illustrating how we would have done it in
the previous reaction--previous example.
We line everything up and then merge the two.
This should end up being six carbons and we merge the carbonyls into an alkene.
Our product is cyclohexene.
Or, our starting material, alkene, working backwards again is cyclohexene.
Now, if we look at example where a ring is already in the compound, our product that
we have to predict the starting material for, we count from one to the other.
We’ll take the short way around, one, two, three, four, five, six, seven--and we can
make a seven membered ring.
We’re not used to drawing these but we can do an almost stop sign.
Okay, make a seven member ring.
Let’s just go ahead and say, “Okay, carbons one and seven, we’re just going to define
those here as forming the alkene.”
Then at carbon two--one, two--we see that we have a bridge between carbons two and four
that is one, two, and three carbons in size.
There’s three additional carbons associated with this.
What I’m going to do is indicate a little perspective here and then let’s go one,
two, three and connect.
There’s our one, two, three.
By keeping track of everything we see that we really have a one, two, three, four, five,
six, seven, eight, nine membered ring ridged by carbon three there.
Again, it’s a little tricky but if you tag everything with a number you can put it together.
All right, what about if we have two separate products?
Well, that means we had a diene.
With four to stitch together we have a diene so let’s make sure we tag everything between
the carbonyls.
Not necessarily the entire length, we’ll take care of those in a moment, so one, two,
three and let’s say four, five, six, seven, and eight.
We should have an eight membered ring with two double bonds so let’s draw our stop
sign down below, here.
Let’s put, between carbons one, two, three--so we have a double bond here between three and
four--five, six, seven, eight, and between one and eight.
Then at the one we also have a methyl group, here, and we have a methyl group at three,
as well.
After tagging everything we can keep track of where everything goes to get our product
or, excuse me, our starting alkene.