1
Stereoselectivity in E1 (E or Z)
• Potential for two products, E or Z alkene
• E normally favoured as substituents far apart
OH
H
Ph
• First cation formation
• Second bond rotation
HO
H
Me
Ph
H
H
H
with Me
..& Ph on opposite faces
..is lower in energy
• Favoured
• Conformation with Me
..& Ph on same face is
..higher in energy as they
..are closer together
• Disfavoured
Ph
+
Ph
E-alkene
95 %
Z-alkene
5%
to align empty p orbital & C–H σ bond
H2O
H
H
Me
Ph
H
• Conformation
H
H
H
H
Me
Ph
H
H
H2O
H2O
H
Ph
Ph
Me
H
H
Me
H
H
Me
Ph
H
H
Ph
Me
2
Regioselectivity in E1
OH
HBr, H2O
major product
•
•
•
•
minor product
Two possible products depending on position of double bond
Most stable alkene is formed
Reason for stability is controversial...
Overlap with π* creates stronger sp2-sp3 bond - more stable
H
H
H
H
H
H
CH3
H
H3C
CH3
H
π*
H
H
H
π*
H
H
H
H
H
no C–H bonds
parallel with π*
most stable
R
R1
C C
R2
R3
tetrasubstituted
H
σ
>
R
C C
H
R3
trisubstituted
>
H
H
R
H
C C
H
H
σ
H
R1
H π*
H
R3
disubstituted
H
>
σ
H
H
least stable
R
H
C C
H
H
monosubstituted
3
Stereochemistry in E2
•
•
•
•
New π bond formed by overlap of C–H σ orbital and C–X σ* orbital
Optimum overlap if orbitals are parallel
Allows selectivity
2 possibilities...
Syn-periplanar
BASE
orbital
overlap
σ
σ∗
orbital
orbital
HX
R
R
Base
R
R
H
X
R
R
R
R
R
R
R
R
4
Stereochemistry in E2 II
Anti-periplanar
BASE
σ
orbital
σ∗
orbital
overlap
H
R
R
R
R
•
•
•
•
•
X
orbital
Base
H
R
R
R
R
R
R
R
R
X
E2 occurs via anti-periplanar transition state when possible because...
Orbitals are truly parallel
Staggered conformation more stable
Base and leaving group are on opposite faces, out of each others way
Electron flow occurs via all-backside displacements like SN2
5
Stereoselectivity in E2
• Stereoselectivity - mechanistically there is a choice of two
•
products but one is favoured - there is a choice
If two protons can be eliminated the reaction will proceed via the
anti-periplanar transition state that suffers least steric hindrance
H
H H
Me
Me
NaOEt
Me
Me
Me
Br H
Me
H
H
minor
Me
Me
H
H
major
H
Me
or
Me
H
minor
H
H
H
Br
two methyl
groups gauche
more hindered
Me
H
H
H
Me
Br
two methyl antiperiplanar
less hindered
Me
Me
H
major
6
Stereospecificity in E2
•
Stereospecific - mechanistically only one outcome; different
stereoisomers of starting material give different stereoisomer of
alkene - there is no choice
If only one proton can be eliminated then geometry of alkene
depends on stereochemistry of starting material
•
Me
H
Ph
Ph
Br
NaOH
FAST
H
H
Ph
Me
Ph
Me
Ph
rotation
NaOH
Ph
Br
SLOW
H
Me
Ph
rotation
Br
Me
Ph
H
HO
Ph
Br
Ph
H
Br
Me
Ph
Ph
H
H
H & Br
anti-periplanar
(1S,2R) gives E alkene
Ph
Me
H
HO
Br
Ph
H
Ph
Me
Ph
H
H
H & Br
anti-periplanar
(1S,2S) gives Z alkene
7
E2 Eliminations from cyclohexanes
H
H
HO
H
H
H
X
H
H
H
C–X antiperiplanar to C–C
E2 impossible
H
X
C–X antiperiplanar to C–H
E2 possible
H
• Necessity for anti-periplanar conformation has a huge effect in cyclohexanes
• For E2 we must have C–H & C–X both axial (trans-diaxial)
• Below shows the effect of this requirement...
Me
Me
NaOEt
Me
+
Cl
1:3
Me
Me
NaOEt
Cl
250 times
SLOWER
8
E2 Eliminations from cyclohexanes II
favoured: i-Pr equatorial
E2 possible
2 anti-periplanar C–H
disfavoured: i-Pr axial
E2 not possible
no anti-periplanar C–H
CH3
Me
≡
H
HO
ring
inversion
OH
H
H
H
Cl
CH3
H
Cl
Cl
Me
Me
H
≡
Cl
Me
ring
inversion
H
Cl
CH3
H
favoured: i-Pr equatorial
E2 not possible
no anti-periplanar C–H
CH3
Cl
H
H
H
OH
disfavoured: i-Pr axial
but E2 possible
Slow as rarely in this
conformation
9
Regiochemistry in E2
OH
Cl
H3PO4
120˚C
KOCEt3
E2
E1
• E1 gives thermodynamically more stable - more substituted alkene
• E2 can give both - more hindered the system, the more of the less
substituted alkene
Cl
Cl
CH3
H
H
X
H
H
H
OR
OR
methyl hydrogen
easily accessible
ring hydrogens hindered
by 1,3-diaxial interaction
Hindered base can change selectivity
base
Br
NaOEt
69%
31%
t-BuOK
28%
72%
10
Summary
Poor nucleophile
(e.g. H2O, ROH)
acid conditions
Weakly basic
nucleophile
(e.g. I–, RS–)
Strongly basic,
unhindered
nucleophile (e.g.
RO–)
Stgrongly basic,
hindered nucleophile
(e.g. DBU, DBN,
t-BuO–)
X
methyl
no reaction
SN2
SN2
SN2
X
primary
(unhindered)
no reaction
SN2
SN2
E2
no reaction
SN2
E2
E2
SN2
E2
E2
SN1, E1
E2
E2
H3C
X
primary
(hindered)
SN1, E1 (slow)
X
secondary
X
tertiary
E1 or SN1
11
Summary II
•
•
•
•
•
•
•
Methyl halides will never eliminate (no protons in correct place)
Increasing branching (more substituents) on substrate will favour
elimination
Strongly basic hindered nucleophiles will always eliminate unless
no option
Good nucleophiles will go via SN2 unless substrate tertiary then
E1 and SN1 compete
Weaker bases that are good nucleophiles give substitution
Normally observe E1 products when SN1 occurs!
Best LG give weakest bases as products
These mechanisms are just extremes
Real life can be somewhere inbetween!
There are other substitution and elimination mechanisms!