CHEM 109A
CLAS
Elimination Reactions - KEY
1. Predict the products of the following reactions
(a-c E2, d-f E1 – KEY focuses only on elimination products, in most cases there
will also be substitution products.)
Br
O
-O
a.
Br
O
LG must be axial
O
HO
Br-
-O
H
are the same
H
-O
b.
Br
H
HO
H
-O
Br-
major P (E > Z
and more
substituted over
less substituted alkene)
Br
-O
c.
Cl
H
HO
-O
Cl-
major P
(more substituted
alkene)
Cl
LG must be axial
H
O
d.
I
H
H
H
I
O
H
IH 3O+
major P
Page 1 of 7
CHEM 109A
CLAS
Elimination Reactions - KEY
Br
H
O
e.
H
Br
H
Me shift
H
O
H
2o carbocation
3o carbocation
H 3O +
Br-
major P
HO
Br
f.
H
Hydride shift
Br
H
O
H
H 2O +CH 3
Br-
major P
2. Which reactant in each of the following pairs will undergo an elimination reaction
more rapidly? Explain your choice and show the mechanism.
OH-
OH Cl
Br
H 2O
H 2O
a.
or
E2 conditions and Br better LG so more rapid reaction.
Page 2 of 7
CHEM 109A
CLAS
Elimination Reactions - KEY
CH3O -
CH3O -
CH 3OH
Br
CH 3OH
Br
b.
or
E2 conditions and LG must be axial for E2 reaction.
3. For each of the following, give the major elimination product; if the product can
exist as stereoisomers, indicate which stereoisomer is obtained in greater yield.
a. (R)-2-bromohexane + high concentration HO- →
H
H
-OH
Br
H
Br-
H 2O
H
trans is major P (cis and monosub alkene are minor)
E2
b. (R)-2-bromohexane + H2O →
H
Br
O
H
H
H
H
H 3O +
BrH
trans is major P (cis and monosub alkene are minor)
E1
c. trans-1-chloro-2-methylcyclohexane + high concentration CH3O- →
Cl
H
Will remove H on C6 b/c it is axial
-OCH 3
E2
d. trans-1-chloro-2-methylcyclohexane + methanol →
Page 3 of 7
CHEM 109A
CLAS
Elimination Reactions - KEY
H
O
Cl
H
Hydride shift
H2O +CH3
Cl -
E1
e. (R)-2-bromohexane + high concentration sodium tert-butoxide →
Na+
H
Br
H
H
H
H
H
H
-O
H
H
H
H
O
(R)-2-bromohexane
Na+
H
Br-
E2
f. (2-Bromo-3-methyl-butyl)-benzene/2-bromo-3-methyl-1-phenylbutane +
high concentration methoxide →
H
Br
-O
CH3
H
H
H
(2-Bromo-3-methyl-butyl)-benzene
H
O
Br-
CH 3
H
H
H
major P (dbl bond is less substituted,
conjugated with benzene ring)
E2
Additional Information:
Definitions
Elimination rxn: in general, removal of H-X molecule and formation of double bond
(alkene). A.K.A. β-elimination, 1,2-elimination
Zaitsev’s (Saitseff’s) Rule: The more substituted alkene is obtained when a proton is
removed from the β-C w/fewest Hs.
EXCEPTIONS:
If R has double bond or benzene ring, the conjugated alkene (NOT more sub) will
be more stable/major P.
If bulky B is used, alkene w/ bulky groups opposite each other (NOT more sub)
will be the more stable/major P.
Page 4 of 7
CHEM 109A
CLAS
Elimination Reactions - KEY
E2
Mech: Bimolecular, 1 step/concerted reaction: B removes proton from β-C, e- go to form
double bond and α-C – X bond breaks heterolytically so LG leaves as X-.
Br
H 2O
Br-
H
H
H
-OH
Rel Reactivity:
Based on LG: RI > RBr > RCl > RF
Based on R: 3o > 2o > 1o
Regioselectivity:
Usu. follows Zaitsev’s rule
Br
H
-OH
major P, more substituted alkene
unless opportunity for formation
of conjugated double bonds
I
major P, conjugated diene
minor P, more substituted,
but isolated diene
Page 5 of 7
CHEM 109A
CLAS
Elimination Reactions - KEY
major P, dbl
bond conjugated
w/ benzene ring
Cl
minor P, more
substituted, but
dbl bond isolated
from benzene
ring
OR steric hinderence for B reacting w/ H on β-C (i.e. bulky base)
Br
major P, b/c less
steric hinderence
-O
Tbl 9.1
If alkyl halide (R) is NOT sterically hindered and Base is…
Little/Moderately hindered
Very hindered
More sub P
Less sub P
W/ alkyl fluoride major P is the less substituted alkene b/c F is such a poor LG that a neg
charge develops on the α-C in the T.S. ~ carbanion (1o > 2o > 3o, carbanion stability b/c
destabilized by e- donating alkyl groups).
Stereoselectivity:
Since elimination of H and X occurs in 1 step, the H-C and X-C bonds have to be in the
same plane (a.k.a. periplanar)
Syn-elimination (from eclipsed conformer) – “front attack”
Anti-elimination (from staggered conformer) – “back attack”, favored b/c it is the
more stable conformer & allows for back-side attack so base doesn’t feel the LG as much
→ get more E than Z.
E1
Mech: 2-step rxn 1) LG leaves to form carbocation (r.d.s.) 2) B removes H from β-C
(delocalization of e- b/c of hyperconjugation dec B.D.E. of H attached to sp3 C).
H 3O+
Br
Br-
H
O
H
H
*WATCH for 1,2-hydride shifts, 1,2-methyl shifts (carbocation rearrangements) to form
more stable carbocation before deprotonation.
Page 6 of 7
CHEM 109A
CLAS
Elimination Reactions - KEY
Rel Reactivity:
Based on LG: RI > RBr > RCl > RF
Based on R = rel stability of carbocations: 3o benzylic ~ 3o allylic > 2o benzylic ~ 2o
allylic ~ 3o > 1o benzylic ~ 1o allylic ~ 2o > 1o > vinyl
Regioselectivity:
Same as E2
Stereoselectivity:
W/ carbocation intermediate, both anti and syn elim possible → major P has bulkiest
substituents opposite each other rel to the double bond.
Cyclohexanes
E2 – H and X must be axial (more stable conformer will NOT go E2)
E1 – carbocation intermediate w/ possibility of rearrangement, usually follows Zaitsev’s
Rule.
Substitution and Elimination Compete
So look at rxn conditions
Tbl 9.4, 9.5and 9.6
Rxn Conditions
Alkyl Halide 1o
2o
3o
Rel Reactivity
Stereochemistry
SN1 vs.
E1
SN2 vs.
Poor Nuc/wk B
N/A
Favored
Both
Both
Both
N/A
o
o
o
o
o
o
o
3 > 2 > 1 1 > 2o > 3o
3 >2 >1
Both E & Z
(more w/
Inverted
bulky
Both R & S
only
groups
opposite)
E2
Strong Nuc/strong B
W/ sterically hindered R or Nuc/B
W/stronger B, bulkier B, higher T
Only
o
3 > 2o > 1o
Both E & Z (more w/ bulky groups
opposite, unless β-C bound to 1 H and
then only single stereoisomer formed)
Page 7 of 7