When alcohols are protonated by an acid, they're converted into good leaving groups and we've
seen before that when the acid has a nucleophilic counterion like HCl, HI and HBr, we're going
to enhance substitution, however in this video, we're going to talk about when a non-nucleophilic
counterion pairs up with that acid, we're gonna actually enhance the elimination pathway
instead.
So let's have a look we can do this for primary alcohols, we have E2, or excuse me, an E2
type mechanism we can do this with secondary alcohols, sulfuric acid and heat, treat those
and we'll proceed via a probably E2, but arguably E1 mechanism is still a little bit debated
out there, we distrust the secondary cation though and then tertiary alcohols will undergo
E1 assuredly and so again the reagents are just sulfuric acid, perhaps in a solvent of
water in reflux conditions and we'll get the most stable product because we've got equilibrium
type conditions and so due to sequential reactions, we can actually have the pi bond that we get
in the first elimination move around until it finds the most stable position for that
alkene.
So we're gonna get a Zaitsev product even if the bond has to move little bit, so let's
look at an example that will illustrate this to us, a lot of energy in this system and
a lot of heat that can help us get over those barriers and the catalyst acids in there as
well.
So let's take a look at this first reaction that I've got here and so if we were going
to do this via as a line reaction this is our prep, we'd have a reactant alcohol reacting
with sulfuric acid and heat to give us the alkene that we see, but the double bond ends
up in a kind of an odd place, let's look and see how that happens by looking at the mechanism
of this reaction.
So first things first, sulfuric acid, a strong acid, there is a solvent of water and a lot
of times there is so you're not dealing with fuming sulfuric acid it's going to have already
transferred its proton to H3O plus so that's what I'm invoking here to accept or to donate
our protons, so it's a Bronsted acid so we're going to protonate our alcohol, now we've
got a good leaving group, well in the presence of more of that solvent, we can pull the beta
hydrogen anti-periplanar fashioned to do an E2 elimination we've got a good leaving group,
a very poor base, so this takes a while, it's very slow, a big delta plus has to form on
that carbon before it gets started, but it will go eventually given enough of a push.
Then we form an alkene that we see here, but if we remember back to our alkene reactions,
alkenes can actually react with sulfuric acid, H3O plus here, to form a tertiary Markovnikov
cation for this particular alkene and so then we undergo that reaction, a sequential reaction
after our E2, we initially had E2, but we have we have to look and see if we can get
to a better place with our alkene and if we can get more stable than us then a disubstituted
exocyclic double bond we have to and so we go to via a reaction here, the similar to
E1 we would see the formation of a more stable trisubstituted alkene where we have our thermodynamic
well it's the most stable form because it has a lower energy than the other alkene in
our potential energy diagram it's going to have a more favorable equilibrium constant
because of that lower energy so we're gonna settle on a major product that we have in
the square there.
If we have a secondary alcohol, we may have to check for that kind of bond migration,
not for this one though, we're just going to protonate and let's take a look at the
traditional view, we're gonna have an E1 elimination to form the hotly contested secondary carbocation
and if we do that without a good nucleophile present, our poor base, the solvent, can come
in, grab that proton and form a major product of the Zaitsev product the trans isomer here,
it's more stable due to the groups not interfering with each other due to torsional strain, even
if we had formed a Z isomer it would just react again in equilibrium conditions and
at a certain equilibrium constant settled on a major product for the trans isomer, you
get a little bit of this Z product, but it's going to be a minor product in these conditions.
Now the latest view on this secondary is that we're not going to go through that secondary
carbocation, we are going to still protonate to activate our leaving group, but we're going
to undergo E2 elimination instead, so yes that leaving group it's really going, before
it leaves, it's gonna be all the way out here okay, it's gonna be way way long there's a
big delta plus forming on that carbon if I can get that out there yeah, it's definitely
approaching a carbocation, but the carbocation so unstable most chemists don't believe it's
there and there's data to support it.
The data for E2 eliminations not as convincing as substitution data, but it is out there,
either way you decide to do it is fine on the exam, either one of these views that we
see here, the SN1 path or E1 path, or the E2 path will work.
They both give us the same answer they give us a Zaitsev product in all cases, even if
we had it stereochemically locked into the E2 elimination in a non-Zaitsev it would just
react again, form a planar carbocation and then proceed through E1, but that initial
step is likely E2 and then if it gets followed up by a carbocation reaction that's gonna
be part of the process eventually, but just not initially so we are kind of splitting
hairs there.
Alright, if we have a tertiary alcohol there's not much of a debate here, we're going through
an E1 type mechanism by most accounts of the data and so we see we get a proton transfer,
we get our good leaving group, the tertiary cation is stable enough to form in a reasonable
amount of time in these conditions with all the blasting of the heat and then we form
the more substituted double bond, again our weak base solvent, our ionizing solvent, if
we have a strong base we can't go through this path, we can't have a strong base in
the presence of a strong acid anyway, but the E1 path will complete to form our most
stable product, we'll draw it down here.
There we go, okay so we would end up with the more substituted alkene, again there's
no stereochemistry here because of the equal methyl groups, if there were the Zaitsev product
would form.
Now a few quick notes, the reaction rates are just like substitution tertiaries faster
than secondary it's faster than primary because of those positive charges building up, we
don't like those to build up so and then they take longer because of that.
Consequences of the equilibrium conditions are going to be that the pi bond can move
around a little bit because of sequential reactions like we saw in the primary alcohol
example.
Ring expansions and rearrangements are also possible, even with secondary systems if they
proceed via the old E1 path or the new newer consideration, the E2 path.
Let's look at one that does both of these things, a sequential reaction and a rearrangement
as quickly as we can, so we're going to protonate again via the via the same pathway, but look
at what product would get very different than we'd expect so if we look at this the old
SN1, or E1 path here would eliminate this leaving group giving us a secondary unstable
carbocation that the ring would then open up to and form if we count our carbons and
keep track of everything starting with the cation, one, two, three, four and five around
the ring, we're going to get five membered ring keeping track of everything here.
At position one, oops, at position one, two, three, four, five we have a methyl group,
there it is.
The cation forms at position three, one, two, three because that's where the bond left from
is that position even if you'd drawn it down here, it would still leave from the more substituted
carbon on that ring and then always trying to get back closer to our groups, we would
have a solvent molecule complete the E1 mechanism and form a an alkene here, but it's not done,
it's going to undergo an additional reaction because we're hot and we've got plenty of
time here and a catalyst at work and so now we can imagine getting closer to a more substituted
double bond by reacting at this position putting the hydrogen on the left and the plus charge
on the right and then we can get another E1 to give us our more stable alkene and the
final product.
The highest percentage product of this equilibrium system, the most stable alkene possible for
this arrangement of carbons.
Now if we were going by the old or the new way of thinking, we would avoid the secondary
carbocation, initially, this one and invoke instead neighboring group assistance, so we
would do everything in one step as this bond gets longer and starts to polarize, this cyclopropane
will nudge around, get close enough and then assist the leaving group leaving and what
it does is it gets us straight to this intermediate without going through the secondary carbocation,
initially.
There's enough energy pent up in this cyclopropane to generate the final secondary carbocation,
but it avoids that spontaneous production of a secondary cation, so either way you preferred,
both will work.