E1 and E2 Reactions
E1 and E2 Reactions
E1 and E2 Reactions
Definitions
SN2
S – substitution
N – nucleophilic
2 - bimolecular reaction
H
HH H
Nu- C X Nu C + X-
H H
Backside attack
Inversion of Configuration: As the nucleophile attacks, the three groups attached
to the carbon undergo inversion, that is, they flip to the opposite side of the carbon.
Kinetics
CH3X methyl 3 x 10 6
CH3CH2X 10 1 x 105
CH3CH2CH2X 10 4 x 104
(CH3)2CH-X 20 2.5 x 103
(CH3)3CCH2-X 10 1
(CH3)3C-X 30 ~0
Reaction Energy Diagram for SN2 Mechanism
TS
Ea
E
R-X + Nu R-Nu + X
Reaction Coordinate
Why this order of reactivity? What controls the relative rate of reaction of the
various substrates?
H3C
H3C
Nu- No Reaction
C X
H3C
The bulky methyl groups prevent backside attack by sterically hindering the
nucleophile from attacking the electrophilic carbons. Contrast this situation to
that of the CH3X group.
Nucleophiles
Problems:25-39, 43,47,54
Goals:
1. Know the detailed mechanisms of SN1, SN2, E1 and E2
2. Know what is a good nucleophile and what is a poor nucleophile
3. Understand the concept of inversion of configuration (SN2)
4. Know the kinetics associated with each reaction
5. Know Zaitzev’s Rule and how it applies to the elimination reactions.
6. Know what is a good leaving group in a reaction.
Reaction:
Conclusion: The strongest base is not the best nucleophile. In other words,
basicity does not control nucleophilicity.
HH
HH
I- C X I- C X
H
H
Electrons loosely held by
the nucleus and easily Transition state – very good
distorted bonding between carbon
and iodine atom. Lower energy
2. Solvation
H-bond H
H
H
..O
Generally, the larger nucleophile is the better one in a given group.
Halogen Nucleophiles:
What happens when a single enantiomer with a reactive chiral carbon undergoes
SN2 reaction???
CH3 CH3
H3C H
H H
HO- C Br HO C X + X
HO C
CH3CH2
CH2CH3
CH2CH3
(S)-2-bromobutane (R)-2-butanol
CH3 CH3
H3C C Br + H2O H3C C OH + H Br
CH3 CH3
CH3 CH3
H3C C+ + H2O H3C C OH2+
CH3 CH3
CH3
CH3
H3C C OH2+ Br
CH3
H3C C OH + H Br
CH3
Reaction Energy Diagram for SN1
TS#1 TS#2
R+ + X-
E
NuH
TS#3
R-X R-NuH+ + X-
R-Nu + H-X
Reaction Coordinate
Reaction Coordinate
Allylic and Benzylic Substrates are very reactive in SN1 reactions.
Why so??
Allyl bromide
H2O
OH OH2+
Br
H Br
?
+ H2O
Leaving Groups: The same factors that favor SN2 leaving groups also
Favor SN1 leaving groups, i.e. if a LG is good for SN2, it is good for SN1.
CH3OH
H3C C
CH2CH3
B (CH3)2CH
Br
+ S isomer
H3C C H3C C CH2CH3
(CH3)2CH ~50%
CH2CH3
(CH3)2CH
A
S isomer
carbocation OCH3
The carbocation is attacked both from the top and the bottom by R isomer
the nucleophilic methanol, resulting in a near racemic mixture of ~50%
enantiomers as products
Reactions that proceed by the SN1 reaction often undergo rearrangements.
Why?
Carbocations are intermediates and may undergo 1,2 ~H or 1,2 ~CH3 shifts.
Br OCH2CH3 OCH2CH3
CH3CH2OH
CH3CHCHCH3 CH3CHCHCH3 + CH3CHCHCH3
CH3 CH3 CH3
+ H Br
Practice: Write a mechanism for this reaction to account for both products.
Solvents in SN1 and SN2 Reactions
Rearrangements No Yes
E2 Elimination
Hß
Cß C
X
The ß-hydrogen is bonded to the ß-carbon, which is bonded to the -carbon
which is bonded to the leaving group X!
Weakest Strongest
Mechanism: The E2 mechanism is a one-step mechanism with bond-breaking and
Bond-making taking place at the same time; termed a concerted mechansim. In
Addition, the Hß and the leaving group X must be anti-coplanar for rapid reaction.
OH-
X
+ H2O + X
Notice that H and X are anti to each other and the four atoms, H-C-C-X,
are coplanar in the reactant. This situation stabilizes the transition state
leading to the alkene.
Overlap develops in the T.S. if the
anti-coplanar relationship is maintained.
Kinetics
30 > 20 > 10
R R H R H R H R H H
R R R R R H H H H H
OH-
H H
H3C H H3C H
H X H
+ H2O + X
H
HO-
a
b a
H H H H3C H H H
H3C C C C H
C C + C C
H CH3 H3C CH3
H Br H
b 81%
+ H2O + Br
CH3CH2 H
C C
H H
19%
The more stable alkene predominates in the product mixture.
Ph
s Ph Ph H
Br H OH- H
H3C s H
or
Ph H3C Ph Ph CH3
Ph
H3C Br Ph Br
rotate
Br H H H H3C
H3C H Ph
Ph Ph H H
Ph
Note: H and Br are
anti-coplanar in a staggered
conformation.
Ph Br
H3C Ph Ph
Ph H3C H
H H
OH-
Homework Problem: Show that the (1R,2R) yields the cis and the (1R,2S)
yields only the trans.
E2 in Cyclohexane Systems
Br
OH-
The equatorial conformation if favored, but it does not provide the necessary
anti, coplanar relationship of the H and Br.
+ H2O + Br-
H
Br
Br
trans
CH3 CH3
base, E2
Br
cis
E1 Mechanism
CH3 CH3
CH3OH
H3C C Br H2C C + HBr
CH3 CH3
CH3
slow CH3
H3C C Br H3C C + Br
CH3 CH3
CH3OH
H
CH3 CH3
H2C C + H2C C + CH3OH2+
CH3 CH3
CH3 CH3
H3C C Br H3C C + Br
CH3 CH3
SN1 E1
Mixture of Products
Summary of E1 and E2
Topic E2 E1
Rearrangements No Yes