Download Alkenes 3 - ChemWeb (UCC)

Cis- and Trans- Revisited: The Sequence Rules and the E- and Zconvention for describing isomerism about the double bond in
structurally complex alkenes:
CH3
H
H
H
C
C
C
C
CH3
trans-2-butene
CH3
CH3CH2 CH2
H
H
CH3
CH3
cis-2-butene
H
C
C
C
C
CH3
cis- or trans- ?
CH3CH2 CH2
CH3
CH3
cis- or trans- ?
In order to describe the structure of complex alkenes the Cahn-IngoldPrelog Rules or Sequence Rules (already used to assign R and S
configurations to chiral molecules with stereogenic centres) are used to
assign a priority to each of the substituents on the alkene carbons.
Be clear that use of the Sequence Rules for describing alkene structure
has nothing whatever to do with chirality and/or R or S configurations.
When the substituents of highest priority on each alkene carbon are
both on the same side of the double bond the alkene is designated Z(German: 'zusammen' = together) and when they are both on opposite
sides of the double bond the alkene is designated E- (German:
'entgegen' = opposite).
Cl
C
C
H
CH3
Cl
Br
C
Br
H
CH3
E-
H2N
C
C
H
Z-
CH3
OCH3
H 2N
C
C
C
H
CH2 H
C
H
Z-
Cl
C
CH2CH3
C
H
C
H
C
Z-
CH2CH2 CH3
Br
CH2CH2 Br
H
E-
CH2CH3
CH2 H
EBr
OCH3
CH3
E-
Cl
C
C
C
CH2CH2 Br
CH2CH2 CH3
Z-
H2C
CH3
CH3
CH
C
C
CH3
Z-
C
CH3
H2C
H
CH3
CH3
CH
C
C
H
C
CH3
CH3
E-
Reading assignment: Work through the practice problem on p. 200 of
McMurry (5th Ed.) and problems 6.9 to 6.11 on p. 201.
Industrial Preparation of Alkenes:
Thermal 'cracking' of high molecular weight alkanes:
CnH2n+2
ca. 900 °C
H2 + CH4 + H2C=CH2 + CH3CH=CH2
n = 2-8
+ CH3 CH2CH=CH2
Laboratory Preparation of Alkenes:
There are four standard laboratory methods for preparing alkenes from
non-alkene starting materials:
(1) From alcohols (dehydration, 1,2- or -elimination of H2O).
(2) From alkyl halides and related compounds of the type RX (1,2- or
-elimination of HX).
(3) From aldehydes or ketones via the Wittig Reaction (This reaction
will be covered in detail in module CM3001)
(4) From alkynes.
(1) Alkenes from alcohols (dehydration, 1,2- or elimination of H2O)
N.B. OH - is
a poor leaving
group (good
nucleophile).
OH
H3PO4
Heat
H
H
+ H2 O
H3PO4
+
OH2
H2O is a better
leaving group
than OH- (i.e. a
weaker
nucleophile).
H
H
+ [H2 PO4]-
- H2 O
O
+
H
H
-
O
P
OH
OH
Unsymmetrical alcohols may yield more than one product:
H
CH3
H
OH
H
H
CH3
+
H3 PO4
Heat
- H2 O
H
H
- H+
+
Zaitsev (or Saytzeff) rule: the most substituted (i.e. most stable) alkene
is favoured.
(2) Alkenes from alkyl halides and related compounds of the type RX
(1,2- or -elimination of HX).
H
C C
  X
1 2
Base
- HX
C
C
+ B:H+ + X
-
X = Cl, Br, I, OTs (tosylate) but X ° F
B = OH-, OR- (alkoxide) etc.
In the reaction illustrated above an alkene is formed by removal of H
and X are from an alkyl halide, i.e. a small molecule, HX, is eliminated
and the reaction is referred to as an elimination:
Some important points:
This reaction illustrated above is called a 1,2- or -elimination to
indicate that the groups being eliminated are located on adjacent atoms
in the starting material as compared to a 1,1- or -elimination where
both are located on the same carbon atom This, in itself, tells you
nothing about the actual mechanism of the elimination process.
Notice that the requirements for a -elimination reaction are:
(i) A base
(ii) A leaving group in the substrate such as halide, tosylate etc. and ...
(iii) A C-H bond - to the leaving group.
We have already seen that most strong bases are also good nucleophiles
- and vice versa.
For this reason nucleophilic substitution and -elimination reactions can
often compete with one another if the substrate organic halide has a C-
H group - to the leaving group.
Alkene formation via elimination of HX from alkyl halides can follow
two different mechanistic pathways identified as E2 and E1.
Here is the E2 mechanism:
B:

H
B
R
C
H
R
C
R
R
C
R
C

R
R
X
R
X
Transition State
R
R
C
R
+ B-H + X:
C
R
E because it is an 'elimination' - a small molecule (here HX) is lost or
'eliminated' from one of the reactants in order to form the product.
E2 because the reaction is bimolecular, i.e. two species - the base and
the organic halide ot tosylate - come together to form the transition
state.
Therefore the reaction follows a second-order rate law:
Rate = k x [RX] x [Base]
The E2 reaction proceeds - like SN2 - in a single step with no
intermediates. The transition state is illustrated in greater detail below:
Notice that the transition state prefers a geometry in which:
(i) The base, the proton being removed, the two carbon atoms and the
leaving group all lie in the same plane and ...
(ii) The proton being removed and the departing leaving group lie on
opposite sides of the connecting C-C bond.
This preferred geometry of the reacting H-C-C-X system is called anti
coplanar or anti periplanar geometry.
As the proton and leaving group depart, the hybridisation at carbon
changes from sp3 to sp2.
The sp3 orbitals of the C-H and C-X bonds must rehybridise to p and be
in the same plane to overlap in a sideways-on fashion and form the component of the double bond.
This developing -overlap is most efficient if the C-H and C-X
bonds - and hence the developing p-orbitals - are already in the
same plane in the transition state.
A syn periplanar conformation also produces the desired coplanar
orientation of the developing p-orbitals but requires an eclipsed
relationship between the substituents on the adjacent carbon atoms and
is therefore much less favourable than the anti periplanar
conformation as shown by the following Newman projections:
H
HX
•
•
X
anti periplanar
staggered - lower energy
syn periplanar
eclipsed - higher energy
The syn and anti periplanar geometries produce different
stereochemical outcomes in the E2 reaction and this can be used to
show that the anti periplanar conformation is preferred:
H
Br
C
CH3
Ph
C
H
H
CH3
CH3
Ph
H
C
C
Br
anti
staggered
syn
eclipsed
Base
Base
CH3
C
CH3
C
Ph
CH3
CH3
H
Ph
H
C
E-
C
CH3
Z-
E2 reactions are favoured by the same factors that favour SN2
substitution - 1° or 2° C-X, strong base (= strong nucleophile) - so that
the two different process often compete under appropriate conditions:
Br
+ Na+ OEt
EtOH
53 °C
OEt
+
E2
SN 2
12%
88%
As we have seen, strong bases are normally good nucleophiles.
However a strong base which is sterically demanding is likely to be a