Nucleophilic Substitution & Elimination Chemistry
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This current chapter continues the theme of the previous acid/base chapter. As much as possible,
try to focus on the similarities: electron pair donation and acceptance. The approach taken in this chapter
is somewhat simplistic. There is a slight sacrifice in accuracy, but it reduces much of the complicated
subtlety and ambiguity that makes this topic so difficult for students. Our approach still leads to mostly
correct predictions for the outcome of the reactions under study. Even in trying to keep it simple, there
are a lot of details to learn and keep track of.
All of the reactions to be presented in this chapter involve two electron transfers from an electron
pair donor to an electron pair acceptor. The electron pair donors look very similar to the bases presented
previously. In fact, electron pair donors will still attack protons (where we still call them bases), but there
will be a competitor site to attack: carbon atoms. When attack occurs at carbon (or any nonhydrogen
atom, in general), the electron pair donors are called nucleophiles.
There are four reactions to be studied in this chapter: two types of substitution reactions (SN2 and
SN1) and two types of elimination reactions (E2 and E1). The main competition in these four reactions, as
we view them, is “SN2 versus E2” and “SN1 versus E1”. “S” stands for substitution, “E” stands for
elimination and “N” stands for nucleophilic. The “1” and “2” are from the kinetics of the reactions, and
represent the order of each reaction. A “2” means that there are two participants in the slow step of a
bimolecular reaction (an RX compound and an electron pair donor). A “1” means that the slow step of a
reaction depends on only one reactant (just the RX compound). The reaction conditions for the “2”
reactions will be presented as sharply different from the reaction conditions for the “1” reactions, even
though this is not completely accurate. Your choices will be more straight forward when presented this
way, and you can fine tune the details if you continue your studies beyond this first year course.
The “2” reaction conditions will have strong electron pair donors present. We will view all anions
as strong electron pair donors. This allows a clear demarcation from weak electron pair donors, which we
will view as neutral molecules. There are two neutral exceptions that we will have to learn. Neutral
nitrogen compounds (ammonia, primary, secondary and tertiary amines) and neutral sulfur compounds
(thiols and sulfides) are considered strong electron pair donors in this chapter. Polar aprotic solvents will
enhance the reactivity of strong nucleophiles in SN2 reactions. Typical “1” conditions generally involve a
neutral polar protic solvent, such as H2O, ROH and RCO2H (or mixtures of these).
Common to all four reactions (SN2/E2 and SN1/E1) is a carbon atom with a leaving group, which
we will identify as the “R-X” compound. We will present just five types of leaving groups, “X”, in this
chapter. They are chloride, bromide, iodide, tosylate and water. There are other possible leaving groups
that will appear during the rest of the book, but by the time you see them, they will just be a part of your
general approach to mechanisms. Our reason for calling the leaving group “X” is to think of them all just
alike, and only learn one strategy for the entire group. There are actually small differences among the
leaving groups, but at this point of your organic career it is more important for you to see the similarities.
The feature that makes something a good leaving group, is that it is stable on its own. We know that
chloride, bromide, iodide, sulfonates and water are stable conjugate bases from discussions in the
previous acid/base chapter. All of them have very strong conjugate acids (pKa’s < 0). The carbon
portion, “R”, will favor or prevent certain of the reaction choices. The carbon atom attached to the
leaving group (called the C-alpha carbon, Cα) will be classified by its substitution pattern as methyl,
primary, secondary, tertiary, allylic or benzylic. It will also be necessary to examine the features of any
carbon atoms attached to Cα. These carbon atoms are called C-beta carbons (Cβ,) and there can be
anywhere from zero to three of them.
These are most of the details you will need to sort through in order to make reasonable predictions
about how these reactions work. Many of the reactions you will study later in this book are variations of
the mechanisms you learn hear. Spend the time to learn this chapter well and it will help make
subsequent chapters easier to learn.
Nucleophilic Substitution & Elimination Chemistry
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Four new mechanisms to learn: SN2 vs E2 and SN1 vs E1
S = substitution = a leaving group (X) is lost from a carbon atom (R) and replaced by nucleophile (Nu:)
N = nucleophilic = nucleophiles {Nu:) donate two electrons in a manner similar to bases (B:)
E = elimination = two vicinal groups (adjacent) disappear from the skeleton and are replaced by a pi bond
1 = unimolecular kinetics = only one concentration term appears in the rate law expression, Rate = k[RX]
2 = bimolecular kinetics = two concentration terms appear in the rate law expression, Rate = k[RX] [Nu: or B:]
SN2 competes with E2
SN1 competes with E1
These electrons
always leave with X.
SN 2
Competing
Reactions
Nu
B
R
X
SN1
H
H
B
(weak)
(strong)
Leaving
Group
Carbon
Group
E2
Nu
Competing
Reactions
E1
Nu: / B: = is an electron pair donor to carbon (= nucleophile) or
to hydrogen (= base). It can be strong (SN2/E2) or weak (SN1/E1).
R = methyl, primary, secondary,
tertiary, allylic, benzylic
X = -Cl, -Br, -I, -OSO2R (possible leaving groups in neutral,
basic or acidic solutions)
X = -OH2
(only possible in acidic solutions)
Problem 1 - Draw all of the "R" patterns listed above connected to “X”.
SN2 and E2 reactions are one step reactions. The key bonds are broken and formed
simultaneously, without any intermediate structures. These are referred to as concerted reactions. The
SN2 and E2 mechanisms compete with one another in consuming the R-X compound. Approach of the
nucleophile/base is always from the backside in SN2 reactions and mainly from the backside in E2
reactions (always from the backside in this chapter).
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SN2 and E2 Competition – One Step Reactions
E2 approach (usually
from the backside in
an "anti" conformation,
"syn" is possible, but
not common)
B
H
Cβ
Cα
Nu
One step,
concerted
reactions,
from the
backside.
H
≈ Nu
strong base ≈ strong nucleophile
S = substitution
Terms
X
X
Nu
SN2 approach (always
from the backside,
resulting in inversion
of configuration)
Cβ
Cβ
B
E = elimination
Cα
SN2 product
(a nucleophile
substitutes for
a leaving group)
Nu: = strong nucleophile
stable
leaving
group
Cα
E2 product
(a pi bond forms)
B: = strong base
X = leaving group
2 = bimolecular kinetics (second order, rate in slow step depends on RX and Nu: /B: )
R-X = R-Cl, R-Br, R-I, R-OTs, ROH2+
SN1 and E1 reactions are multistep reactions and also compete with one another. Both of these
reactions begin with the same rate-limiting step of carbocation formation from an R-X compound.
Carbocations (R+) are very reactive electron deficient carbon intermediates that typically follow one of
three possible pathways leading to two ultimate outcomes: 1. add a nucleophile or 2. lose a beta hydrogen
atom. The additional competing pathway for carbocation intermediates is rearrangement, in which atoms
in a carbocation change positions to form a similar or more stable carbocation. Once formed, a new
carbocation is analyzed in a similar manner to the previous one it came from. It may possibly rearrange
again, but ultimately, the final step will be to either add a nucleophile at the carbocation carbon (SN1) or
to react with a base at a beta hydrogen atom (E1).
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SN1 and E1 Competition – Multistep Reactions
SN1 and E1 reactions
begin exactly the same
way. The leaving group,
X, leaves on its own,
forming a carbocation.
X
Cβ
Cα
H
SN1 and E1 reactions
are multistep reactions.
B
Nu
=
weak
base
Nu
H
B
addition of a weak
nucleophile (SN1
type reaction)
X
Cβ
Cα
Loss of a beta hydrogen
atom, (Cβ-H in most E1
reactions that we study).
H
The R-X bond
ionizes with help
from the polar, protic
solvent, which is also
usually the weak
nucleophile/base.
H
H
solvated
carbocation
H
Nu
Nu
Cβ
solvated
carbocation
SN1 approach is from
either the top face or
the bottom face.
S N1
Cα
Nu
H
Nu
Cβ
Cα
Nu
H
Nu
Cβ
E1 approach comes
from parallel Cβ-H with
either lobe of 2p orbital.
S = substitution
Terms
E1
E = elimination
Cα
Cβ
Cα
X
H-B: = weak base
Nu
SN1 product
(a nucleophile
substitutes for
a leaving group)
H
E1 products
(pi bond forms)
H-Nu: = weak nucleophile
Cα
H
H
B
weak
base
Cβ
H
Often a final
proton transfer
is necessary.
weak
nucleophile
H
H
H
Cα
H
H
lose beta H (E1)
rearrange
weak
nucleophile
Cβ
add Nu: (SN1)
rearrangement to a new
carbocation of similar or
greater stability
R
stable
leaving
group
X = leaving group
1 = unimolecular kinetics (first order reaction, the rate in the slow step depends only on RX)
R-X = R-Cl, R-Br, R-I, R-OTs, ROH2+
Nu
H = H
B
= usually a polar, protic solvent (or mixture) of H2O, ROH or RCO2H
The above pairs of reactions (SN2/E2 and SN1/E1) look very similar overall, but there are some
key differences. The nucleophile/base is a strong electron pair donor in SN2/E2 reactions (that’s why they
participate in the slow step of the reaction) and a weak electron pair donor in SN1/E1 reactions (that’s why
they don’t participate in the slow step of the reaction). This leads to differences in reaction mechanisms,
which show up in the kinetics of the rate law expression (bimolecular = 2 and unimolecular = 1). It is
recommended that you look at the reaction conditions first to decide what mechanisms are possible. You
cut you choices in half when you decide that the electron pair donor is strong (SN2/E2) or weak (SN1/E1).
Problem 2 - Draw each of the “X” groups, as they would appear as products. Indicate why they are good
leaving groups.
Nucleophilic Substitution & Elimination Chemistry
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Problem 3 - Fill in an example of each of the following patterns below.
(tosylate)
-Cl
-Br
-I
(protonated alcohol)
-OH2
-OTs
Me = methyl = CH31o = primary = RCH22o = secondary = R2CH3o = tertiary = R3Callyl =
CH2=CHCH2-
benzyl = C6H5CH2-
Important details to be determined in deciding the correct mechanisms of a reaction.
1. Is the nucleophile/base considered to be strong or weak? We simplistically view strong electron pair
donation as coming from anions of all types and neutral nitrogen and sulfur atoms. Weak electron
pair donors will typically be neutral solvent molecules, usually water (H2O), alcohols (ROH),
mixtures of the two, or simple, liquid carboxylic acids (RCO2H).
2. What is the substitution pattern of the R-X substrate at the Cα carbon attached to the leaving group,
X? Is it a methyl, primary, secondary, tertiary, allylic, or benzylic carbon? What about any Cβ carbon
atoms? How many additional carbon atoms are attached at a Cβ position (none, one, two or three)?
Answers to these questions will determine SN2, E2, SN1 and E1 reactivities and alkene substitution
patterns and relative stabilities in E2 and E1 reactions.
R
H3C
X
methyl (Me)
R
H2
C
X
primary (1o)
R
H
C
X
R
R
C
X
H2C
C
H
H2
C
H2
C
X
R
secondary (2o)
tertiary (3o)
C-beta carbon
(0-3 of these)
allylic
C-alpha carbon
(only 1 of these)
Cβ
Cα
X
benzylic
X
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3. Are the necessary “ anti ” Cβ-H/Cα-X bond orientations possible to allow E2 reactions to occur? Also
possible, but much less common, are “ syn ” E2 reactions, which are not emphasized in this book.
syn Cβ-H/Cα-X bonds
anti Cβ-H/Cα-X bonds
Two possible conformations
for an E2 reaction to occur.
X
Cβ
Cα
H
B
strong
base
Cβ
concerted reaction
(anti elimination is
usual and common)
In E2 reactions the base approach occurs
most commonly from an anti direction at
Cβ-H because the conformation is
staggered. The electrons from the Cβ-H bond
force off the leaving group, X, as the base, B:,
forces them away from the Cβ hydrogen atom.
Cα
Different E2
products
are possible.
B
strong
base
syn
elimination
is rare
H
X
Cβ
Cα
In E2 reactions the base approach occurs
rarely from a syn direction at Cβ-H
because the conformation is eclipsed.
The Cβ-H bond and the Cα-X bonds need to be parallel so the pi bond of the
product can form as the reaction proceeds. Only two conformations allow
this, anti and syn. Anti is highly preferred (> 99%) due to its staggered
conformation. We will emphasize anti eliminations in E2 reactions in this
book.
SN2 / E2 mechanisms
The rate laws for both SN2 and E2 reactions are bimolecular (that’s what the “2” means). In fact
they look very similar, except that the values of the rate constants; kSN2 and kE2, are different. That means
the activation energies (Ea) are different [k = exp –(Ea/RT)]. The reaction with the larger rate constant
(lower Ea) will occur faster and will form more product. That’s because the reactions are usually
occurring under kinetic control (what forms fastest in a single step), instead of thermodynamic control
(what is most stable in repeated steps).
SN2 rate law
Rate = kSN2 [RX][Nu: ]
E2 rate law
Rate = kE2 [RX][B: ]
kSN2 ≠ kE2
and
(Ea)SN2 ≠ (Ea)E2
B: is the same as Nu: , only
the position of attack is different.
...except by coincidence
Simplistic pictures of an SN2 mechanism and an E2 mechanism are shown below. Initially, both
the strong nucleophile/base and the R-X electrophile must shed portions of their solvation shells to
encounter one another. In our approach, solvation factors are usually ignored (though we will briefly
discuss them later in this chapter). The effects on the reactions from many solvent molecules are very
complicated and since the solvent does not itself undergo change in the overall reaction, it is often easier
to leave it out at first. However, it is often one of the most important details of a reaction mechanism.
Nucleophilic Substitution & Elimination Chemistry
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Simplistic Picture of a General SN2 mechanism (with the solvent)
"Transition State"
S
S
S
S
S
S
S
S
S
S
S
S
H
H
S
S
S
S
Br
S
S
S
S
S
S
S
S
S
S
S
S
C
S
S
H
H
S
S
S
S
= solvent molecule
Br
H H
S
The reactants must encounter one another
with the proper orientation (backside attack
by nucleophile in SN2 reactions).
C
S
H
S
HO
HO
S
S
S
S
S
S
S
S
H
S
S
C
S
S
S
S
S
S
H
H O
S
S
S
S
S
S
S
Br
S
S
S
The SN2 transition state is highly conjested
as the sp2 carbon inverts in configuration.
The carbon atom has high PE because it
has 10 electrons around the carbon, Cα. It
is very sensitve to steric effects.
The products diffuse away from one another
after the nucleophile is added and the leaving
group is gone. The Cα carbon has inverted its
configuration at the reaction site. The reaction
succeeds because the products are more stable
than the reactants and there is sufficient energy
to crossthe transition state energy barrier (Ea)
Same picture without the solvent
"Transition State"
H
H
H O
C
H
H
Br
C
HO
H
Br
HO
Br
C
H
H
H H
Potential Energy vs. Progress of Reaction Diagram
"Transition State"
H
HO
C
Br
H H
sp2 carbon transition state as carbon
inverts configuration forms a high PE
carbon with 10 electrons at carbon.
This is a concerted, one-step reaction.
Ea - this energy difference determines
how fast the reaction occurs: "kinetics".
Rate = kSN2 [RX][Nu] = bimolecular reaction
PE
-Ea
Ea
H
H O
C
H
H
reactants
kSN2 = 10
2.3RT
Ea = -2.3RT log(kSN2)
Br
H
∆G - this energy difference determines the
∆G
extent of the reaction; the ratio of products
versus reactants at equilibrium (when
kinetics allows the reaction to proceed.
∆Go = -2.3RT log(Keq)
Thermodynamics is determined mostly by the
-∆Go
strengths of the bonds and solvation energies of
2.3RT
the reactant and product speces: "thermodynamics". Keq = 10
POR = progress of reaction
HO
C
H
H
products
Br
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Important Details of SN2 Reactions
1. SN2 reactions occur in one step. They are concerted reactions that go directly from reactants to
products.
2. Both reactants (RX and Nu:-/B:-) are involved in the slow (…and only) step. The transition state,
(T.S.) is the high potential energy point of the reaction and determines the activation energy (Ea).
This is called the rate determining step (R.D.S.).
3. Bond formation and bond breakage are occurring at the same time, which keeps the Ea lower.
4. The high potential energy T.S. carbon is approximately sp2, trigonal planar, where carbon is
surrounded by ten electrons in violation of the octet rule.
5. The reacting carbon (Cα) inverts its configuration. Generally, if this is a stereogenic center, R Æ S or
S Æ R, or if in a disubstituted ring, cis Æ trans or trans Æ cis. Inversion is also assumed to occur at
achiral RX centers, even if it is not observable (such as a methyl or primary RX center).
6. Generally, strong base/nucleophiles are used in SN2 reactions. We simplistically indicate this with
negatively charged nucleophile to represent anions of all types, (and neutral nitrogen and sulfur
atoms).
7. The HOMO/LUMO overlap is important in molecular orbital theory (but not discussed here).
Relative rates of SN2 reactions with different substitution patterns at Cα of the R-X structure
Substitution at Cα and Cβ (relative to the carbon with X) slows the rate of SN2 reactions because
the substituent groups block the backside approach of the nucleophile to the Cα-X carbon (…and increase
the activation energy by increasing the congestion in the pentavalent transition state). Both Cα and Cβ
substitution patterns show their most dramatic decreases in rate of reaction when all possible carbon
positions are switched from hydrogen to carbon (or other large group). As long as there is at least one
hydrogen atom at the Cα and Cβ positions, there remains a possible path of approach by the nucleophile to
the backside of the carbon.
Relative Rates of SN2 Reactions - Steric hindrance at the Cα carbon slows down the rate of SN2 reaction.
H
H
Cα
CH3
H
X
H
k 30
methyl (unique)
CH3
Cα
X
H
k 1
ethyl (primary)
Reference compound
CH3
Cα
CH3
X
H
k 0.025
isopropyl (secondary)
1
40
CH3
Cα
X
CH3
k 0
t-butyl (tertiary)
(very low)
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primary RX
methyl RX
secondary RX
R
H
H
Nu
Nu
X
Cα
H
Cα
R
X
Cα
Nu
H
H
Methyl has three easy paths of
approach by the nucleophile. It is
the least sterically hindered carbon
in SN2 reactions, but it is unique.
X
R
H
Secondary substitution allows one
easy path of approach by the
nucleophile. It reacts slowest of the
possible SN2 substitution patterns.
Primary substitution allows two easy paths
of approach by the nucleophile. It is the
least sterically hindered "general" substitution
pattern for SN2 reactions.
tertiary RX
R
H
H
Cα
Nu
X
=
R
Nu
H
R
Tertiary substitution has no easy path of approach
by the nucleophile from the backside. We do not
propose any SN2 reaction at tertiary RX centers.
Cβ
H
Cβ Cα
H H
Cβ
H
H H
X
When the Cα carbon is completely substituted the
nucleophile cannot get close enough to make a bond
with the Cα carbon. Even Cβ-H sigma bonds block the
nucleophile's approach
Relative rates of SN2 reactions with different substitution patterns at Cβ of the R-X structure
Steric hindrance at the Cβ carbon also slows down the rate in SN2 reactions.
H
CH3
H
H
CH3
C β CH2 X
C β CH2 X
CH3
k 0.4
propyl
C β CH2 X
CH3
H
H
H
k 1
ethyl
Reference compound
CH3
C β CH2 X
CH3
k 0.00001
2,2-dimethylpropyl
(neopentyl)
k 0.03
2-methylpropyl
All of these structures are primary R-X compounds at Cα, but substituted differently at Cβ.
CH3
CH3
H H
C
Nu
H
Cβ
Cα
H
H
A completely substituted Cβ
carbon atom also blocks the
Nu: approach to the backside
X of the C-X bond. A large group
is always in the way at backside
of the C-X bond.
If even one bond at Cβ has
a hydrogen then approach
by Nu: to backside of the
C-X bond is possible and an
X SN2 reaction is possible.
CH3
CH3
H
Nu
Cβ
Cα
H
H
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Especially reactive SN2 patterns
The excess electron density in the sp2 transition state of an SN2 reaction (ten electrons at carbon)
can be delocalized into an adjacent pi system, when present. This helps lower the activation energy, Ea.
This is an especially powerful effect when an oxygen can take on some of the excess negative charge.
Especially reactive SN2 patterns include allyl RX, benzyl RX and X at a Cα position of a carbonyl group.
O
H 2C
CH3 CH2 X
C CH2
X
CH2
CΗ CH2 X
X
R
k 1
ethyl
Reference compound
k 40
allyl RX
k 120
benzyl RX
k
100,000
α substituted carbonyl
All of these structures are primary R-X compounds, but the
pi patterns are much more reactive than a simple ethyl RX.
The adjacent pi bond can take on some of the excess negative charge density in the transition
state, because of its parallel orientation with the transient 2p orbital at Cα in the transition state. This will
lower the activation energy, Ea, and increase the rate of the reaction.
See data above.
Nu
C
X
transition
state
C
O
O
R
Relative
Rates
Delocalization of excess negative
charge in an SN2 transition state,
lowers Ea and makes the rate
constant larger (faster rate).
C
CH3
CH2
CH2
X
=
X
100,000
1
Inversion of Configuration
Inversion of configuration at the Cα position (R Æ S or S Æ R) in SN2 reactions is so important, it
gets its own name, “Walden Inversion”.
4
4
S
Nu
Cα
3
2
X
Nu
Cα
3
X
R
2
If Cα is a chiral center and X is #1 in priority (the other groups are #2, #3 and #4 above), then S
absolute configuration typically inverts to R (and vice versa). This observation assumes there is no
relative change in the order of priorities of the groups at the chiral center (X = #1 priority and Nu = #1
priority).
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SN2 Inversion of Configuration in Rings
A cyclic RX compound with no other substituents does not allow us to detect inversion of
configuration (versus retention of configuration or racemization of configuration).
Nu
inversion of
configuration
X
C
?
H
Nu
retention of
configuration
same
product
X
C
?
H
We cannot tell the top from the bottom without some other marker in the ring.
However, if the RX ring were substituted with another group that did not interfere with the SN2
reaction, it would allow the observation of backside attack by the nucleophile. A very simple methyl
substituent would allow us to see which side the nucleophilic attack occurs from. We can observe if the
opposite side or the same side…or both sides are attacked by comparing the initial relative position of the
leaving group to the final relative position of the nucleophile.
H3C
X
H 3C
H
C
C
H
H
Nu
H
Nu
cis = initial stereochemistry
(of methyl and X)
X
trans = final stereochemistry
(of methyl and Nu) from
backside attack by nucleophile
...leads to...
Problem 4 - How can you tell whether the SN2 reaction occurs with front side attack, backside attack or
front and backside attack? Use the two molecules to explain you answer. Follow the curved arrow
formalism to show electron movement.
a
b
H3C
H
O
I
H3C
H
Br
C
O
H
CH3CH2 C Br
H3C
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Ring Size Effects on the Rate of SN2 Reactions
Each type of ring (C3, C4, C5, C6,…and substituted variations) will have its own peculiar effect on
how the SN2 reaction behaves. As previously shown, if there is some conformation that blocks approach
of the nucleophile from the backside of the C-X bond, this will slow the SN2 reaction.
If the sp2 hybridization of the reaction carbon in the transition state (TS) is restricted, this will also
slow down the rate of SN2 reaction. Since rings only have a limited number of possible conformations,
each ring system will have to be examined individually to understand experimental data on its relative
rate of SN2 reaction. Bond angle changes in small rings can be very inhibiting to the SN2 reaction,
because of the angle strain [sp3 (109o) Æ sp2 (120o) = T.S.]. Some typical SN2 rate data are shown below.
Br
Br
krelative
(SN2)
1.0
0.00001
Br
Br
Br
Br
0.008
1.6
0.01
0.97
Br
0.22
Problem 5 - Why might C3 and C4 rings react so slowly in SN2 reactions? (Hint-think about bond angles
in the transition state versus bond angles in the starting ring structure.)
Problem 6 - Why might C6 rings react slower in SN2 reactions? What are the possible conformations from
which a reaction is expected? Trace the path of approach for backside attack across the cyclohexane ring
to see what positions block this approach. Which conformation would have the leaving group in a more
reactive position (axial or equatorial)? Is this part of the difficulty (which conformation is preferred)?
These two chair
conformations
interconvert in a
very fast equilibrium.
Problem 7 - In each of the following pairs of nucleophiles one is a much better nucleophile than its
closely related partner. Propose a possible explanation.
a. relative rates
b.
250/1
c.
H
H C H
H
H
N
N
O
O
H C
O
H
methoxide
H C
C O
H C
H
H
H
t-butoxide
Nucleophilic Substitution & Elimination Chemistry
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Problem 8 - The following two reaction sequences have not been specifically studied, but are analogous
to reactions we are studying. Part of the fun of organic chemistry is to discover the mechanistic logic in a
newly seen reaction. See if you can explain the observations below.
a. The first reaction occurs in two steps, forming initially two ionic species which react further with one
another. Propose an arrow pushing mechanism and list the related analogy mechanism. Look at the new
bonds formed in the products and the old bonds broken in the reactants to decide what atoms joined
together and split apart. You also have to decide which atoms donated and accepted the electron pairs.
R' CH2 S CH2 R'
step
1
R1
P
R1 R
1
R' CH2 S
step
2
S CH2 R'
What is the
mechanism?
What is the
mechanism?
R1
P
S
resonance
R1 R
1
Hint: The first mechanism type is the
same as the second mechanism type
and the same as the topic we are
currently discussing.
R1
P
S
R1 R
1
Add in any formal charge.
b. A single dipolar species (two charges) is formed in the first step of the following reaction. What is the
leaving group in this first step? Normally such a leaving group would be poor, but in this reaction it is
OK. Why? (How small is the ring?) This intermediate undergoes a rotation about the center bond and
then reacts in a second step. The two ionic sites bond together and make a four-membered ring, which
makes the phosphorous exceed the octet rule. This is acceptable because of phosphorous’ available 3d
orbitals. The ring ruptures, forming two pi bonds (P=O and C=C). Propose an arrow pushing
mechanism, which rationalizes the overall transformation. Be specific in your third step to indicate
whether the elimination occurs in a syn or anti manner.
O
H3C C
H
R1
P
R1 R
1
step
1
C H
CH3
rotate center
bond by 180o
"SN2 like
mechanism"
still two ionic
charges, P+ and O-
two ionic charges in a
single intermediate
structure, (P+ and O-)
R1
P
R1 R
1
O
resonance
R1
P
R1 R
1
add in any formal charge
O
H
H
C
H3C
C
CH3
O- attacks P+ to form
a four atom ring
step
3
internal
elimination
mechanism
step
2
Nucleophilic Substitution & Elimination Chemistry
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Simplistic Picture of a General E2 mechanism – with the solvent
S
S
S
"Transition State"
S
S
S
H O
S
H
S
S
C
S
S
S
S
S
S
S
S
S
Br
S
CH3
S
The reactants must encounter one another with the
proper orientation (anti conformation of Cβ-H and
Cα-X) for the E2 reaction to occur. There may be
more than one type of anti Cβ-H that can react
leading to different E2 products. The high potential
energy of the strong base drives the reaction to
completion.
S
C
S
C
S
S
H 3C
S
S
S
Br
S
CH3
S
S
S
S
S
The transition state developing the pi bond
requires parallel overlap for developing 2p
orbitals. Anti and syn allow this, but anti is
much preferred due to its staggered
conformation.
S
H
S
S
S
S
H
S
H
H3C
S
S
O H
S
H
C
S
Br
S
S
S
C
CH3
CH3
S
S
S
S
C
H
H O
S
S
H
S
S
S
S
S
S
H
H
S
S
S
S
S
The products diffuse away from one another
after reacting. The carbon hybridizations
have changed from sp3 to sp2. Less common,
but possible is a change from sp2 to sp. The
most stable alkene (most substituted) has the
lower Ea and froms at a faster rate (usually).
Same picture without the solvent
The transition State - requires parallel overlap
of the two 2p orbitals forming the pi bond.
This is easiest when Cβ-H is anti to Cα-X.
H O
H
C
H
H
H
H O
H
H
H O
H
H
C
H
C
C
H 3C C
CH3
C
H 3C
Br
Br
CH3
Br
H 3C
CH3
When present E or Z configurations are
fixed by the anti Cβ-H/Cα-X
conformation(s).
Important Details
1. E2 reactions occur in one step. They are concerted reactions that go directly from reactants to products.
2. Both reactants (RX and Nu:-/B:-) are involved in the slow (…and only) step. The transition state, (T.S.)
is the high potential energy (PE) point of the reaction and determines the activation energy (Ea). This is
called the rate determining step (R.D.S.).
3. Bond formation (O-H and C=C) and bond breakage (Cβ-H and Cα-Br) are occurring at the same time,
which keeps the activation energy (Ea) lower. This requires parallel overlap of these two bonds.
4. Two sp3 carbon atoms are both becoming the sp2 carbon atoms of the pi bond.
5. Stereochemistry (E/Z) at C=C is fixed by anti orientation of Cβ-H and Cα-Br bonds
6. Generally, strong base/nucleophiles are used. (This is simplistically shown as negative charge, although
sterically hindered neutral nitrogen is also a good base. Neutral sulfur is not a good base.) To
emphasize E2 > SN2, bulky sterically hindered, strong bases are used (such as potassium
t-butoxide).
Nucleophilic Substitution & Elimination Chemistry
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The competition in SN2 versus E2 reactions is between electron donation to a carbon atom (SN2) or
electron donation to hydrogen atom (E2). We will see competition between these two sites many times in
organic chemistry.
Problem 9 - a. What dihedral angles are possible for Cβ-H versus Cα-X in typical conformations. Use
Newman projections to illustrate these.
θ
0o
θ = dihedral angle
60o
120o
180o
240o
300o
b. Considering the fact that the new pi bond has to form simultaneously with (Base-H) bond formation
and Cβ-H and Cα-X bond breakage, which of the above conformations look best to allow all of this to
occur at once? If more than one conformation permits the two 2p orbitals forming the pi bond to overlap,
is one conformation better than another? Why?
Problem 10 - Draw a Newman projection of the two possible conformations leading to E2 reaction. Show
how the orientation of the substituents about the newly formed alkene compares. Are they the same?
Assume priorities are R2 > R1 and R4 > R3. Specify the absolute configuration of each alkene as E or Z.
H
anti E2
C
reaction R
1
R2
R3
R4
C
X
reactant conformation 1
rotate about
center bond
alkene stereochemistry?
Newman
projections
syn E2
reaction
alkene stereochemistry?
reactant conformation 2
Nucleophilic Substitution & Elimination Chemistry
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Keeping Track of the Cβ Hydrogens
At least one Cβ position, with a hydrogen atom, is necessary for an E2 reaction to occur. In more
complicated systems there may be several different types of hydrogen atoms on different Cβ positions.
After all, in E2 reactions there can be anywhere from one to three Cβ carbons, and each Cβ carbon can
have up to three hydrogen atoms.
H
H
H
Cα
Cβ
H
Cα
H
Cβ
X
Cα
one Cβ position
(1 RX, usually SN2 > E2)
(2o RX, SN2/E2 both competitive)
o
Cβ
Cα
Cβ
H
X
two Cβ positions
X
zero Cβ position
(unique methyl, only SN2)
Cβ
Cβ
X
three Cβ positions
(3o RX, only E2)
With proper rotations, each Cβ-H may potentially be able to assume an anti conformation
necessary for an E2 reaction to occur. There are a lot of details to keep track of and you must be
systematic in your approach to consider all possibilities. Let’s consider a moderately complicated
example.
CH3
CH3
R
CH2
H3C
CH
H3C
Cα
CH2
O
R
OH
H
H3C
Na
H
CH3
Redraw structure to explicitely show all
of the Cβ hydrogen atoms. Notice the
base/nucleophile is strong, so SN2 or E2
reaction are possible. The RX pattern is
tertiary, so no SN2 is possible. There are
three Cβ carbon atoms and all of them
have hydrogen atoms on them. That leaves
E2 reaction as the only possible choice.
X
H3C
Cβ1
CH2
Cβ2
Cα
X
H
H
Cβ3
H
H
There are two chiral centers, Cα and
Cβ1, that make this structure more
complicated than we are treating it.
Possible stereoisomers are RR, RS,
SR and SS.
Problem 11 - How many total hydrogen atoms are on Cβ carbons? How many different types of hydrogen
atoms are on Cβ carbons? How many different products are possible? Hint - Be careful of the simple
CH2 above. The two hydrogen atoms appear equivalent, but E/Z (cis/trans) possibilities are often present.
(See below for relative expected amounts.)
B
H
Cβ
Cα
X
This sketch will work for every
possibility above, IF you fill in
the blank positions correctly.
Nucleophilic Substitution & Elimination Chemistry
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Stability of Pi Bonds
In elimination reactions, more substituted alkenes are normally formed in greater amounts.
Greater substitution of carbon groups in place of hydrogen atoms at alkene carbons (and alkyne carbon
atoms too) translates into greater stability (lower potential energy). Alkene substitution patterns are
shown below. There are three types of disubstituted alkenes and their relative stabilities are usually as
follows: geminal ≈ cis < trans.
Relative stabilities of substitued alkenes.
H
H
C
H
C
1
<
H
H
R
C
C
H
R
R
C
<
H
2
R
≈
C
H
H
R
H
C
R
3
<
C
H
C
<
C
H
4
R
H
R
R
R
C
R
6
5
<
C
H
R
C
C
R
R
7
The more substituted alkenes are usually more stable than less substitued alkenes. Substitution
here, means an R group for a hydrogen atom at one of the four bonding positions of the alkene.
1 = unsubstituted alkene (ethene is unique)
2 = monosubstituted alkene
3 = cis disubstituted alkene
4 = geminal disubstituted alkene
5 = trans disubstituted alkene
6 = trisubstituted alkene
7 = tetrasubstituted alkene
Saytzeff's rule: More stable alkenes
tend to form faster (because of lower
Ea) in dehydrohalogenation reactions.
They tend to be the major alkene product,
though typically a little of every alkene
product possible is obtained.
Possible explanations for greater stability with greater substitution of the pi bond
1. A fairly simple-minded explanation for the relative alkene stabilities is provided by considering the
greater electronegativity of an sp2 orbital over an sp3 orbital. An alkyl group (RÆ) inductively
donates electron density better than a simple hydrogen. From the point of view of a more
electronegative sp2 alkene carbon, it is better to be connected to an electron donating alkyl carbon
group than a simple hydrogen atom. The more R groups at the four sp2 positions of a double bond,
the better. However, be aware that other features such as steric effects or resonance effects can
reverse expected orders of stability.
R
C
C
H
...inductively donating "R" substituent
is more stable than unsubstituted "H"...
C
C
Nucleophilic Substitution & Elimination Chemistry
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2. Hyperconjugation can be viewed as a form of “sigma” resonance from a C-H sigma bonding orbital
into the adjacent pi-bond. (HOMO/LUMO orbitals can be used to explain this donation too.)
H
H
2D
H
H
H
C
C
C
C
C
H
hyperconjugation
(sigma resonance?)
3D
H
H
H
C
C
C
C
C
H
C
H
C
H
H
H
Problem 12 – Order the stabilities of the alkynes below (1 = most stable).
H
C
C
H
R
C
C
H
R
C
C
R
"R" = a simple alkyl group
Problem 13 - Match the heats of formation, ∆Hfo’s, with the appropriate isomeric alkenes. Which alkene
is most stable in each group?
a
∆Hof's = -0.2 , -1.9 , -3.0
∆Hof's = -8.6 , -10.1
b
Problem 14 - Calculate the following heats of hydrogenation, ∆Horxn’s, (hydrogenation = addition of H2 to
a pi bond). Do these values make sense, given the above patterns of alkene stabilities? Explain your
answers.
a
H2
catalyst
∆Hof = -8.6
∆Hof = -7.9
∆Hof = -35.1
∆Hof = -5.3
∆Hof = -36.9
H2
catalyst
c
H2
catalyst
b
H2
catalyst
d
∆Hof = -35.1
∆Hof = -6.9
∆Hof = -35.1
Problem 15 -Provide an explanation for any unexpected deviations from our general rule for alkene
stabilities above.
∆Hof = -18.7
∆Hof = -22.7
∆Hof = -19.9
Nucleophilic Substitution & Elimination Chemistry
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Problem 16 - Reconsider the elimination products expected in the problem on page 16 and identify the
ones that you now expect to be the major and minor products.
CH3
H
H3C
H
Cβ1
CH2
H3C
Cβ2
Cα
H
H
Cβ3
R
H
O
R
OH
Na
H
X
Problem 17 - What are all of the possible products resulting from E2 elimination of 2-bromobutane?
Assume B:-- is the base/nucleophile. Consider the total number of beta C-H’s and the different types of
beta C-H. Which products are expected to be more stable? What is the most likely competing reaction?
Possible sawhorse and Newman projections are provided below. Fill in the appropriate atoms for each
possibility, according to our understanding of the E2 mechanism.
H
H
H
H
H
H
C C
C C
C C
Problem 18 - (2R,3S)-2-bromo-3-deuteriobutane when reacted with potassium ethoxide produces
cis-2-butene having deuterium and trans-2-butene not having deuterium. The diastereomer
(2R,3R)-2-bromo-3-deuteriobutane under the same conditions produces cis-2-butene not having
deuterium and trans-2-butene having deuterium present. Explain these observations by drawing the
correct 3D structures, rotating to the proper conformation for elimination and showing an arrow pushing
mechanism leading to the observed products. (Protium = H and deuterium = D; H and D are isotopes.)
O
K
O
cis-2-butene
(D present)
K
and
and
(2R,3S)
trans-2-butene
(no D present)
cis-2-butene
(no D present)
trans-2-butene
(D present)
(2R,3R)
Nucleophilic Substitution & Elimination Chemistry
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Substituted Cyclohexanes and E2 reactions
Studies of substituted cyclohexane structures have reinforced the idea that an anti Cβ-H/Cα-X
conformation is necessary to allow an E2 reaction to succeed. Chair conformations of cyclohexane
structures have well defined positions for all of their atoms. When a leaving group is in an axial position,
the adjacent axial positions are anti as required for E2 elimination. A hydrogen atom at either of these
positions will allow for easy E2 elimination. The other cyclohexane conformation will put the leaving
group into the equatorial position. In this conformation the anti positions are the carbons of the
cyclohexane ring. No elimination is possible from this conformation. The combined effect of all
substituents on the ring, as well as with each other, that will determine the preferred position for X at
equilibrium (axial or equatorial). One cyclohexane conformation is potentially reactive (X when axial)
and one conformation is nonreactive (X when equatorial).
If is a hydrogen atom,
then E2 is possible because
is in an anti position
from X. SN2 also works H
better in this conformation
as long as Cα is not tertiary
and Cβ positions are not
quaternary.
H
H
4
3
5
3
4
H
H
5
6
1
2
H
X
H
H
H
2 1
H
6
H
H
X
H
represents anti positions
H
E2 is not possible with
in an equatorial position
because the ring carbons
C3 and C5 are anti to X
(bolded bonds). SN2 is
inhibited because of the
axial positions at C3 and C5.
Problem 19 - What are the cyclohexane conformational relationships in adjacent axial/axial positions,
axial/equatorial positions and equatorial/equatorial positions (syn, gauche or anti)? Which
combination(s) allow for easy E2 reaction to occur? Draw these representations clearly as bond-line
formulas and as Newman projections.
Problem 20 - The structures of neomenthyl chloride and menthyl chloride are given below. Neomenthyl
chloride reacts very quickly and produces two cyclohexane E2 products in the indicated amounts, while
menthyl chloride reacts very slowly and only produces one cyclohexane E2 product as indicated. Explain
these observations. What would be the observed product if any SN2 product were observed? Redraw the
structures in a more realistic 3D representation (two chair conformations for each isomer) and point out
the problems leading to these observations.
H
CH3
H
CH
H
Cl
H
CH3
H
CH3
R O
R O H
CH3
CH
CH3
CH3
CH3
Cl
CH
H
H
H
H
menthyl chloride
very
slow
R O
R O H
CH3
H
H
H
3-menthene
75%
2-menthene
25%
neomenthyl chloride
CH3
H
CH3
H3C
CH
H3C
fast
H
H
CH3
H
CH3
CH
H
CH3
2-menthene
100%
Nucleophilic Substitution & Elimination Chemistry
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Problem 21 - A stronger base (as measured by a higher pKa of its conjugate acid) tends to produce more
relative amounts of E2 compared to SN2, relative to a second (weaker) base/nucleophile. Greater
substitution at Cα and Cβ also increases the proportion of E2 product, because the greater steric hindrance
slows down the competing SN2 reaction. Use this information to make predictions about which set of
conditions in each part would produce relatively more elimination product. Briefly, explain your
reasoning. Write out all expected elimination products. Are there any examples below where one
reaction (SN2 or E2) would completely dominate?
a.
Cl
Cl
I
I
...versus...
b.
O
Cl
Cl
O
...versus...
O
pKa = 5
pKa = 16
c.
Cl
Cl
...versus...
d.
Cl
Cl
N
N
C
C
...versus...
e.
Br
R
Br
O
R
O
...versus...
f.
Cl
N
...versus...
pKa = 25
C
Cl
pKa = 9
Problem 22 – One of the following reactions produces over 90% SN2 product and one of them produces
about 85% E2 product. Match these results with the correct reaction and explain why.
O
pKa = 16
Cl
O
...versus...
pKa = 19
Cl
Nucleophilic Substitution & Elimination Chemistry
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Problem 23 - Propose an explanation for the following table of data. Write out the expected products and
state by which mechanism they formed. Nu: -- /B: -- = CH3CO2 -- (a weak base, but good nucleophile).
O
percent
substitution
percent
elimination
H3C
H2
C
Br
100 %
0%
H3C
H
C
Br
100 %
0%
Br
11 %
89 %
Br
0%
100 %
O
O
O
CH3
H3C
O
H
C
CH
O
H3C
CH3
CH3
O
H3C
C
O
CH3
Problem 24 - Explain the following observations. Models would probably help.
a
H Br
CH3
O
CH3
CH3O
CH3OH
CH3
Br
O
O
O
O
H
b
H
H
H
CH3O
CH3OH
O
O
O
H
H
Problem 25 - Provide a step-by-step mechanism (curved arrows, formal charge, electron pairs, etc.) for
each of the following observations and explain the observed differences in product formation. Hint - What
cyclohexane conformation is present in the bridged ether product below? How could it get that way?
O
a. trans-4-chlorocyclohexanol + HO
OH
b. cis-4-chlorocyclohexanol + HO
OH
HO
OH