[narrator] Let’s look at the four types of carbocation rearrangements that we talked
about.
Several different features of the alkene are going to govern which one is relevant for
which reaction.
What I’ve shown is the first step, the acid addition, where we form the carbocation on
the more substituted carbon in each case and now we prep for our rearrangement.
We look next door and we see a tertiary carbon with three R groups and this is going to indicate
an H-shift, a hydride shift, for us.
If we look at our available hydrogen we can shift this.
I always use a loopy arrow but you can use a straight arrow, it’s fine.
I just use that to show there’s a rearrangement and we shift that next door to form a more
stable carbocation.
The driving force is the formation of a more stable cation in the tertiary position.
We then go from a secondary cation to a tertiary cation, with a hydride shift.
Next scenario is we have a quaternary carbon next door and a methyl available.
Let’s keep the same color there.
Quaternary carbon and a methyl is available.
We can get a methyl shift.
Either methyl works for this example.
I shift that methyl to form a tertiary cation, again, from a less
stable secondary carbocation, again there’s a driving force of stability there.
More stable cation.
Where our difference is quaternary carbon next door and a methyl available as opposed
to a tertiary carbon next door with a hydrogen available.
What about if there is a strained ring?
Three, four membered rings are most common.
Sometimes five membered rings with ring expand.
We look at our carbocation, we have a secondary.
What do we do?
We open the ring and relieve that strain so the idea is that we’re not forming a more
stable cation but we are forming a larger ring.
Here is where counting the carbons from your original cation are going to be important.
One, two, three, and four.
One, two, three, four and we have to make a four membered ring with a methyl group coming
off on the other side.
Now, where is the cation?
Well, if you look, carbon two loses its bond as part of the ring expansion so whenever
we remove that bond it’s not making enough bonds so carbon two, again one is where the
ring closes.
One, keep the same counter-clockwise count that we had there; one, two, three, four,
and so that cation is going to go on carbon two.
Still a secondary cation but it’s a four membered ring now, so less strain.
For our purposes we are going to say that we’re out of time and the nucleophile is
going to add, at this point.
We don’t have to worry about additional rearrangements but depending on who you ask,
if there is enough time, a ring expansion in this case may be followed by a hydride
shift to form this cation.
It’s not unprecedented to see that occur but with nucleophiles present the time it
takes to shift the cation, usually a nucleophile has already snipped out the positive charge
by then so maybe this happens.
Primarily the ring expansion is our main concern, one at a time.
Let’s show one carbocation rearrangement at a time.
But, reaction conditions can dictate a second shift from time to time, so what is the final
condition we can have?
We can have more than one pi bond and if that occurs and we form a carbocation in the presence
of acid, initially at one of the positions--any one, I just picked one of the two--at this
point we could have a ring closure that behaves just like step one, a markovnikov type cation
formation, but instead of an acid starting it it’s a carbocation that begins the process.
This is intramolecular reaction.
I said intramolecular, between two different functionalities of the same compound.
That’s a ring closure, so, again, let’s count everything.
Make sure we’re not missing anything.
One, two, three, four, and five.
We end up with a five membered ring where we closed the ring at carbon one and, again,
this isn’t naming this is just keeping track of our carbons.
I’m going to number left just like I did in the left hand side.
If you numbered right you would keep track of it that way.
But, I like to start where we close the ring.
One, two, three, four, five.
Markovnikov would put the cation right here in a normal acid addition, and that’s carbon
four as we’ve drawn it.
One, two, three, four.
The plus charge ends up here.
We still have a secondary cation.