hey guys, Professor Dave here, today wanna talk about E2 on cyclic systems
let's revisit the E2 reaction and recall few things about this mechanism. so we
know the base is gonna extract the proton
the electrons in the carbon-hydrogen bond will form the pi bond
and the electrons between carbon and bromine will leave with bromine
and there's our alkene product. now one thing we didn't talk about before
it is the case that the proton that is being extracted
must be in an anti conformation with respect to the leaving group, so we take
a look at this Newman projection so let's say
that we're looking this way down the bond right there
so we're gonna see the hydrogen up, directly up, the methyl down to the
right, the hydrogen
down to the left and then on the back carbon that bromine is going to be down
hydrogen up and to the right, and the methyl is up and to the left
so this is the Newman projection looking down this bond
now it is the case that the proton being extracted
and the leaving group must be in anti conformation with respect to each other
so that's the a 180 degree
dihedral angle, now the reason for this is we know that there is a pi
bond forming
and if we recall what a pi bond is, it is a lateral overlap
of electron orbitals that are the unhybridized
p orbitals and so it is the case that
the only way that the overlap can happen
is if these bonds are anti to one another because that's
that's how the orbitals can overlap so as to form the pi bond
so in any other position it wouldn't work. let's say that the methyl group
instead let's say we rotate
120 so that the methyl
is up top here, we cannot have elimination in that configuration
now the thing is with a linear system we we know that
sigma bonds are rotating all the time so at any given
instant the methyl group were anti to
to the bromine that's okay because that is simply going to rotate
until a proton
is in the correct position to allow for elimination to occur but with cyclic systems
it's not so easy because we cannot have constant rotation
so let's take a look
here so we're gonna look down this bond on our
chair structure here and we're gonna see that on the front carbon we've
got a proton down
methyl up and to the right and the rest to the ring to the left. on the back
carbon we're gonna see we have the
bromine up and we have the rest of the ring down and to the left and then we have
the implied hydrogen, that's that one down and to the right
now so we just
figured out that in order for elimination to occur
the proton being extracted and the leaving group must be in anti
conformation with respect one another and on a ring, the only way for this to occur
is if the two groups are both axial
so the bromine has to be axial, the proton being extracted has to be axial
here because the equatorial positions
are always going to be anti to the rest of the ring so
here's the equatorial hydrogen back here
that is anti to the rest of the ring on the front part
here is the methyl group over here that's in the equatorial position on
this carbon
that is going to be anti to the rest of the ring in the back so
there is no way for a proton
in the equatorial position to participate in elimination
so we know that we're always in equilibrium between the two chairs
so there's a lot of things to think about in terms of
which chair is going to be the one that will promote
elimination so if the leaving group is equatorial in one chair
that's not going to work right, we have to have that axial and then
once we have it in the axial position it is only protons
that are on beta carbons and
also in the axial position so this hydrogen
this hydrogen and this hydrogen
are available for elimination but the other hydrogen on this carbon
that one's not going to work so it is no longer
the case that these two equivalent protons are both available for elimination
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