Chapter 8 – 5th ed
CHEM 321 Organic Chemistry I - Professor Kathleen Kilway
"Organic Chemistry" by Maitland Jones, 5th edition
Chapter 8 – 1, 2, 3, 5, 7, 8, 15, 20, 21, 22, 24, 26, 27, 28.
CHAPTER 8: Elimination Reactions: The E1 and E2 Reactions
Section 8.1
I. Preview
A- Elimination versus Substitution
Section 8.2
II. The Unimolecular Elimination Reaction: E1
A- Rate Law 1- Rate determining step is ionization of substrate.
a- Same as for SN1, Rate (v) = k [R-L].
b- First order reaction.
c- First step yields carbocation intermediate (same as in SN1 reaction).
2- Instead of adding a nucleophile as is done during the second step of a SN1 reaction, the
substrate is deprotonated. An alkene is formed.
a- The nucleophile is acting as Brønsted base.
Figure 1
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Chapter 8 – 5th ed
3- Mixture of E1 and SN1 products depends on nucleophile’s basicity.
a- Strong Brønsted bases - E1.
b- Good nucleophiles favor SN1.
B- Selectivity in the E1 Reaction 1- More stable alkene is favored, thus more substituted.
2- Saytzeff elimination - produces the most substituted alkene possible [versus Hofmann
elimination - formation of the less stable alkene].
3- Competes with the SN1 reaction.
4- See Figure 8.3 on page 334 for an example.
C- Dehydration
Section 8.3
III. The Bimolecular Elimination Reaction: E2
A- Rate Law
1- Second order process, dependent on substrate and nucleophile.
2- Rate (v) = k [R-L] [Nu-]
Figure 2
3- Strong base attacks proton attached to β-carbon (carbon adjacent to the carbon bearing
the leaving group).
4- E2 competes with SN2.
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Chapter 8 – 5th ed
B- Effect of Substrate Branching on the E2-SN2 Mix 1- The more branching of substrate, the more likely reaction will proceed E2.
2- Tertiary substrates only undergo E2 reactions under E2-SN2 conditions.
C- Stereochemistry of the E2 Reaction 12345-
Only syn (0 °) and anti (180 °) arrangements of substrate can lead to π-bond formation.
Syn elimination - elimination where the dihedral angle between C-H and C-L is 0 °.
Anti-elimination - elimination where the dihedral angle between C-H and C-L is 180 °.
Anti elimination is favored over syn elimination (See Figure 8.15 on page 340).
Formation of the more substituted alkene is also favored.
Figure 3
D- Selectivity in the E2 Reaction
1- The larger the size of the base the more likely 1:1 (Hofmann/ Saytzeff) product ratio.
2- Regiochemistry - refers to the outcome of the reaction in which more than one orientation
of substituents is possible in the product.
3- Hofmann elimination - formation of the less stable alkene and is favored when:
a- The leaving group is fluoride, ammonium (R3N+), and sulfonium (R2S+).
b- The base is a strong one like an alkoxide (RO-).
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Chapter 8 – 5th ed
c- For R3N and R2S as leaving groups, the Hofmann product formed is due to a steric
effect.
d- See Figure 8.19 on page 343 for an example.
+
+
Figure 4 Figure 4
(Also see figure 8.23 on page 348).
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Chapter 8 – 5th ed
Section 8.4
IV. Transition States: Thermodynamics versus Kinetics
Energy Barriers in Chemical Reactions: The Transition State and Activation Energy
1-It is the difference in free energy between the starting material and the
transition state in a reaction.
2-This energy must be supplied in order for a reaction to occur.
3-Note that we now use ∆G‡to denote activation energy and not Ea
4-SN2-For the forward reaction:
a- also see figures 8.16 and 8.17 on page 344.
Figure
2-For the reverse reaction:
Figure
A- Thermodynamics
1-Thermodynamics is the study of energy relationships.
B- Kinetics
1-This is the determination of the rates of reaction.
C- Thermodynamic Control
1-A thermodynamically controlled reaction is a reaction in which the product is
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Chapter 8 – 5th ed
determined by the relative energies of the products.
D- Kinetic Control
1-A kinetically controlled reaction is a reaction in which the product distribution is
determined by the heights of the different transition states leading to products.
E- Hammond Postulate
1-The transition state for an endothermic reaction will resemble the product.
2-It can be equivalently stated as, "The transition state for an exothermic
reaction will resemble the starting material."
Section 8.5
V. Rearrangements of Carbocations
A- Hydride and Methyl Shifts
1-Only happen when going through carbocation.
2-Therefore, cannot happen in SN2 and E2 reactions.
3- It is in the case when there is a secondary carbocation, which can undergo
one of these shifts to make it tertiary.
4- Need to know that it only happens once.
5- Shifts hydride first (if available), then a methyl (second choice) but notehr
larger than a methyl group.
Section 8.6
VI. Special Topic: Other Eliminations
A- Thermal syn Eliminations – when no other option is available
B- Eliminations through loss of small molecules
Section 8.7
VII. Special Topic: Enzymes and Reaction Rates
A- Enzymes
1-Enzymes act as catalysts to lower the activation energy, thereby increasing reaction
rates in biological systems, at a much lower heat than would be required in the
laboratory.
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Chapter 8 – 5th ed
Section 8.8
VIII. Special Topic: Why Are Rearrangements of Carbocations Fast?
Summary of Substitution versus Elimination
A- SN1 Reaction competes with E1 Reaction (3o and some 2o Carbons) but Nu is neutral (e.g.,
H2O or HOR)
B- SN2 Reaction (THF – good solvent- nonpolar) competes with E2 Reaction (2o, 1o, and methyl
Carbons) – Nu must be charged (e.g., NaOH/H2O or NaOR/HOR).
C- Exceptions are:
Substitution only for azide and cyanide
E2 only when tertiary R-L and charged Nu ()
Tert-butoxide (CH3)3CO- only favors E2 for secondary and tertiary R-L
Combination of charged Nu and LG (sulfonium, ammonium, or fluoride) – more stable alkene
is minor; less stable alkene is major.
SN1/E1 – can have hydride and methyl shifts for a secondary carbocation
D- In order to decide which reaction is favored, follow the following flow chart:
1- Is the R of the alkyl halide methyl (a), primary (b), secondary (c), or tertiary (d)?
2- Is the nucleophile a good base?
Yes or no?
3- Is the solvent polar?
Yes or no?
E-­‐ Flow Chart (a) methyl carbon, no - not a good base, no - not a polar solvent = SN2 only SN2 possible
(b), yes, no = E2
(b), no, no = SN2
(c), yes, no = E2
(c), no, no = SN2
(c), no, yes = SN1
(d), yes, yes = E1
(d), no, yes = SN1
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Chapter 8 – 5th ed
Section 8.9
IX. Summary - Vocabulary
Equilibrium constant
Gibbs free energy
Thermodynamic vs. Kinetic Control
Activation Energy
Entropy
Enthalpy
Dependence of rate on temperature, pressure, and solvent
Transition States
Microscopic reversibility
Hammond Postulate
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SN1, SN2, E1, or E2
R-L: where R contains C such as CH3; CH3-L is methyl; CH3CH2-L is 1o; (CH3)2CH-L is 2o; (CH3)3C-L is 3o
L is a leaving group such as a halide or F-, Cl-, Br-, or I- (best LG is I-)
Nu: can be neutral such as H2O or ROH or charged such as NaOH or NaOR this can add to C (SN1 or SN2) or act as a base
to deprotonate (E1 or E2)
R= methyl or 1o
SN2 only
R = 2o
Nu = -:CN: or -:N3 (e.g., NaCN or NaN3)
SN2 and SN1
R = 3o
SN1 only
R= methyl or 1o
no reaction
Nu = neutral H2O or ROH
R = 2o and 3o
SN1 and E1; for 2o hydride/methyl shifts may occur
R= methyl
Nu = charged such as NaOH or NaOR
R=
2o
R=
3o
or
SN2 only
1o
SN2 and E2
E2 only
For Elimination Reactions, normally more/most substituted alkene - major product(s) (Saytzeff versus least stable Hofmann products)
H
H
H
>
H >
H>
H
H
4 sp2-sp3
or 4C - 0H = 4
3 sp2-sp3
or 4C - 1H = 3
>
H
H
H
2 sp2-sp3
3 sp2-sp3
or 4C - 2H = 2 or 4C - 3H = 1
0 sp2-sp3
or 4C - 4H = 0
Hoff - major product when Nu: charged such as NaOH or NaOR (E2) and L = sulfonium (SR2), ammonium (NR3), or F
S(CH3)2 Lookin down C2 and C3, must have
Hb anti to have E2 so Hb has to be anti
S(CH3)2
but gauche interaction is not good
to have Saytzeff product
H
Hb
Ha
Hb
Ha
S(CH3)2
Ha
H
Ha
Looking down C1 and C2, must have
anti to have E2 so Ha has to be anti which is
possible with 3 Ha