Let s visualize four E2 reaction mechanisms
with increasing difficulty to explain basic concepts like orbitals, conformations and
stereospecificity. After getting smart on different systems and
mechanisms, we will use what we ve learned to understand a more complicated, exotic reaction. The most important thing to know is that E2
reactions preferentially take place from anti-periplanar conformations. Why is that? We usually draw that deprotonation by base
kicks out the leaving group, so the pi-s
ystem forms via overlap of the sigma C-H and the
sigma star C-Br bond within the same plane. An anti-periplanar setup maximizes this overlap,
but syn-periplanar arrangements are theoretically possible. The orbitals are also in the same plane, but
you can see that overlap is weaker due to worse alignment. Such syn-conformations are also energetically
disfavoured due to the eclipsing bonds so we generally ignore them in E2 reactions. Based on this orbital alignment, we have a
bimolecular concerted
transition state just like in the SN2 substitution, and we have
no intermediates as opposed to E1 elimination. For every elimination, we always need to first
find the leaving group, second the electrofuge as we will see it s not always a proton and
third consider conformations. Our examples will show that energetically
preferred and reactive conformations are often different. Let s look at the simple elimination of 2-bromobutane
with base. The bromide will obviously eliminate, but
three types o
f protons are eligible. First up, the terminal CH3 group is most accessible. Because our base is not sterically hindered,
this kinetical product forms in 20%. Instead, thermodynamic preference is at play. We have more internal protons which give rise
to more substituted olefins. You will have heard that higher substituted
olefins are energetically more stable. The two isomers, (E) and (Z) arise from deprotonation
of two different protons. How can we rationalize the preferential formation
of the
(E) isomer? There are two reactive conformations that
position the hydrogen and bromide anti-periplanar. Newman projections show us what the conformation
and transition state look like. The (Z)-alkene is formed upon deprotonation
of H1, but this requires both methyls to be close to each other. If we rotate the single bond by 120 , we can
eliminate this unfavorable gauche interaction. Upon deprotonation of this H2, the methyl
groups are antiperiplanar and nicely avoid each other. The conformation
al equilibrium lies on the
right side, the (E) isomer is formed preferentially. How do we know if the concerted E2 mechanism
is correct? One piece of evidence are kinetic isotope
effects maybe you watched my previous video or heard of them already. Deuterium behaves like hydrogen but due to
its heavier mass, forms stronger bonds with carbon. Elimination of deuterated 2-bromopropane is
almost 7-times slower versus hydrogen, so we know that the rate-limiting step involves
breaking of this particul
ar bond. As a second argument for the concerted mechanism,
E2 reactions can be highly stereospecific. Let s look at these two diastereomeric educts
do you think they will eliminate to different products? We again know our leaving group and there
s actually just one hydrogen available let s figure out the conformation again by drawing
the Newman projection. If we move H and Br antiperiplanar, the big
phenyl groups also oppose each other. Because of the concerted mechanism, the reaction
maintains
the arrangement and stereospecifically yields the (E) alkene. What about the second one? In the only possible antiperiplanar conformation,
the phenyl groups are gauche to each other, encoding for a (Z) product. The visual shows us that they are big, so
never be fooled by abbreviations in chemistry! Like our very first example, the single bond
rotation equilibrium is clearly not favouring this setup. Most molecules are chilling in a more stable
conformation that does not allow for anti-eliminatio
n. This leads to a 10-times slower reaction. To put completely put this to bed: The first
example we saw exhibited stereoselectivity with formation of three different olefins
with different ratios. The second one had stereospecificity because
the diastereomers had no other choice but to yield the (E) or (Z) alkene, respectively. Before we continue, let me thank all my channel
members and also you for watching this video. I m experimenting with some basic but more
visual content so let me know an
y feedback and suggestions in the comments. Another favorite of many chemistry teachers
are cyclohexanes. You will know that substituents like to be
equatorial to avoid steric clashes via 1,3-diaxial interactions. However, because the leaving group would be
antiperiplanar to C-C bonds and not any C-H bonds, E2 is impossible. The ring needs to flip to position the leaving
group axially which could be energetically disadvantageous. To illustrate this point, let s consider the
elimination of these
di-substituted diastereomers, cis and trans. The huge tertbutyl group will dictate our
ground state conformation through its equatorial preference, characterized by a large A-value. Notice anything different about the isomers? If the bromide is cis, the most stable chair
conformation already has the C-Br axial and thus antiperiplanar to two hydrogens. We are ready to eliminate. The trans isomer is not ready we need a chair
flip from a double equatorial to a double axial conformation. This comes
with a hefty energy price, so this
elimination happens hundreds of times slower. How about this trimethylsilyl compound: Would
you expect different reactivity? The answer is straight-forward if you know
that TMS has a lower A-value but why is that? They might look similar on paper but again,
don t be fooled. The carbon-silicon bond is much longer than
carbon-carbon, so the system can avoid some of the unfavorable steric effects that arise
from axial positions. The di-axial setup is thus less dis
favoured
and would eliminate faster, albeit still pretty slow. To understand the final example, we need to
arm ourselves with some more advanced reactions. Some of you might know the Peterson olefination
which is a silicon-variant of Wittig-type reactions with two pathways. Basic conditions facilitate syn-elimination,
either concerted through a four-membered ring intermediate or a step-wise mechanism. The energetic driving force is the strong
silicon-oxygen bond from the relatively weak silicon-
carbon bond. As a second pathway, acidic conditions convert
the hydroxy group into a better leaving group which eliminates in E2 anti fashion similar
to what we ve seen before. Another reaction facilitated by silicon-oxygen
bond formation is the Brook rearrangement. Here, an oxy-anion intramolecularly attacks
a silyl group sitting at a carbon, leading to a silyl migration. This leaves behind a carbanion that is typically
simply protonated. Almost five decades ago, chemists reported
a surprising
result when studying beta-hydroxysilanes that also have an alpha-hydroxyl group. You might expect a Peterson syn elimination
to the enolate. Instead, the reaction gave the (E) silyl enol
ether selectively. It looks like we have a hybrid pathway at
hand! Our model can help visualize the stereochemistry. A Brook migration of the silyl group occurs,
but instead of protonation of the anion we can eliminate the beta-hydroxy group. If we assume that this conserves the configuration
of the educt, we ca
n nicely predict that the antiperiplanar arrangement encodes the (E)
olefin in the product. As always, we should check the other diastereomer
as well. This one leads to the opposite (Z) olefin
with a slightly lower selectivity of 2:1. This could suggest that the pathway is energetically
more challenging, potentially proceeding through a more stepwise mechanism which erodes the
selectivity. This concludes this video on basics and some
more advanced features of E2 reactions. If you learned somethi
ng, please like and
subscribe to my channel. Let me know what topics you re interested
most and as always, until next time.
Comments