hey guys, Professor Dave here, let's talk a little bit about Zaitsev and Hoffman
elimination products.
sometimes in an elimination reaction there can be
more than one elimination product
depending on which beta proton is extracted by the base
so let's take a look at this example here
there are two different kinds of protons that can be extracted and will yield
two different products, so for example let's see if
a base takes this proton here and then the
electrons there go to form the double bond, kick up the bromine, we're going to get
this product here, that is sp2
trigonal planar, so this methyl group flattens out. or
we can extract the proton over on this side, form the double bond here
leaving this side of the molecule untouched so we would get that, this would remain
sp3, that would remain
a wedge bond for the methyl group. so these are different molecules, they're
different products and
and both are possible. most likely we're gonna get a mixture so we want to be
able to discuss these two products
and as it as it happens we have these terms
Zaitsev and Hoffman to
refer to which one is the more highly substituted
alkene product. so this alkene
is trisubstituted. now the way we want to understand that is
there are two carbons participating in that carbon-carbon
double bond and we want to be able to count the number of alkyl
groups projecting from the carbons participating in the double bond.
so this carbon has one alkyl group, is connected to one carbon atom
over here
and then there's the implied hydrogen. this carbon over here
has two alkyl groups, it has that methyl group and the rest of the ring over here
so that's three alkyl groups so this is trisubstituted
whereas over here this carbon
has this side of the ring and the implied hydrogen
this carbon has this side of the ring and the implied hydrogen
so that is two alkyl groups so it is disubstituted
so you can have unsubstituted, mono substituted, disubstituted
trisubstituted, the maximum would be tetrasubstituted
and we're going to label the more highly substituted
alkene as the Zaitsev product
the less substituted will be Hoffman, so it is not the case that
any trisubstituted alkene elimination product
would be labeled Zaitsev, it is simply that the more substituted of the two
is the Zaitsev product so it's certainly a relative term
because a trisubstituted alkene could be the Hoffman product
if the Zaitsev product were tetra substituted, or more
substituted. so we need to understand which product will dominate
and when, so in order to understand that we need to know
something about alkene stability and it is the case that the more
substituted an alkene is the more stable it is
now this is similar for reasons of carbocation stability
with hyperconjugation it's kind of like that so
a tetrasubstituted alkenes is going to be the most stable alkene
then trisubstituted then
E disubstituted, Z substituted
and then the monosubstituted and then the unsubstituted will be the least
stable alkene
now that means that the Zaitsev product
is more stable and therefore thermodynamically favored
it is the thermodynamic product, the thermodynamically favored product
and the kinetically favored product will be the Hoffman product so it is less stable
but it could happen more frequently
if the base is very sterically hindered. so let's take a look at two possible bases
if it was hydroxide that was acting as the base
we would probably expect more
of the Zaitsev product than the Hoffman product, this is because the base
is very sterically unhindered, it has relatively equal access to any proton on
the molecule
so it's not going to have a very hard time getting to that proton
and because it can approach any
portion of the molecule with equal ease then
it will have a tendency to form the lowest
energy product, the most thermodynamically stable product, so that
would be
the Zaitsev product. however let's say we're looking at something
a little bit more sterically hindered, say we're looking at
ethoxide, now that's not quite as sterically hindered as, say tert-butoxide
but that's still a fair amount of steric hindrance because we have an ethyl group
and there's
protons, you know, there's five different protons there
so this is not going to have as easy a time
approaching every proton in the molecule there is going to be
a lower activation energy associated
with approaching that hydrogen, so if we were looking at an energy diagram
we could say that there was a lower activation energy
approaching one of these protons as opposed to a higher activation energy
approaching that proton and so this is the kinetically favored product
because more collisions on this side of the molecule
will result in a reaction because more collisions
on this side will be able to surpass the smaller
activation energy so that's why the Hoffman product
is kinetically favored and will be favored
in the product mixture if the base is quite sterically hindered
whereas the Zaitsev product is thermodynamically favored
and will probably be favored in the product mixture if
the base is sterically unhindered. let's look at a couple examples to see what's
going on here
so let's look at this substrate, we're reacting with methoxide so
just to go back to the mechanism stuff we know that this is a tertiary substrate
so we cannot do SN2
this is a strong bace that is relatively unhindered
so we could do SN2 or E2, however we just said SN2 is impossible so
we know we're gonna do E2, we know we're gonna do an elimination here
however there are multiple protons that can be extracted that would
yield different elimination products so let's draw some of the
implied hydrogens in, let's say if we got this one
grab that, double bond there
lose the bromine, we're gonna get this product, the double bond will be here
leaving group leaves so that's what we would get, however we could also
extract the proton
from this methyl group up here so if we did that
we would get
this product here, but what is the degree of substitution of these products
well over here the left carbon has two alkyl groups
the one on the right has one and an implied hydrogen so that is a trisubstituted
alkene. over here
this carbon has two alkyl groups, the two sides to the ring
and this has none, those are two implied hydrogens so that is a disubstituted
alkene. trisubstituted versus disubstituted means this will be the Zaitsev product
and this will be the Hoffman product and
in this case Zaitsev is going to be favored because it is more
thermodynamically favorable
and it's a relatively unhindered base so it's gonna have access, it's gonna have
roughly equal access
to this secondary proton as it would to that primary proton
so in this case the Zaitsev product is going to dominate, we would still get
some Hoffman
but we'd be looking at the majority to be Zaitsev
now if the base was significantly sterically hindered such a tert-butoxide
then we might expect a Hoffman product to be able to
dominate but even though it is less thermodynamically favorable
because kinetics would then drive that reaction, a tert-butoxide would be
able to access
this proton with much more ease because there would be a lower activation energy
associated with that acid-base reaction
let's look at something like this here, let's say we're
using methoxide again, now the thing with the cyclic systems is we're gonna
have to draw
the chair so I went ahead and drew that
so here's the chair, there's that isopropyl
group extending upwards
there is the chloro group there, there is the methyl group going down there
so now the reason we had to draw the chair
is because we have to be able to see which protons are available for elimination
now we know that we're going to have to extract a proton from a carbon
adjacent to the carbon that bears the leaving group so that means a
proton from here or proton from here
but as we learned before we have to have the proton
that is eliminating anti to the leaving group
and on this carbon it's the isopropyl group that anti
to the leaving group so this proton is not available for elimination
that 's anti to the rest to of the ring, so this one is no good, we're not going to
get elimination there.
over here we do have two protons so
even though this one's not going to work out this one will do just fine
because if we do the Newman projection we would see that
because they're both axial those are anti to one another and
that's where the elimination can occur, so let's see that
in this case we're only gonna get one product and we don't have to analyze
Zaitsev vs. Hoffman
whereas you might think that you would have, if you just went based on the top-down
view, it would be very possible to draw in this
implied hydrogen there and assume
that that was available for beta elimination just as one of the ones on
this carbon
but when we draw the chair we see it is not the case, we do not get a mixture of
Zaitsev and Hoffman products, we get
only one product and that is the disubstituted alkene there
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