Organic Chemistry
Nucleophilic Substitution
H. D. Roth
Chemistry 307
Chapter 7
When we discussed the SN2 reaction, we had considered
two possible mechanisms, involving either a single step
(nucleophilic substitution with second order kinetics, SN2) or
two steps, i.e.,
+
1.
R–LG
→
R + LG
2.
R + Nu–
→
R–Nu
+
–
The reactions you learned about so far followed the onestep mechanism; you learned about the kinetic order and about
the role of nucleophile, leaving group, the structure of the
substrate, and of the solvent. These reactions proceeds via
transition states.
The second mechanism we considered was a two-step
reaction: in the first step, the haloalkane undergoes ionic
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
dissociation forming a carbocation; in the second step reaction
with a nucleophile completes the substitution. This mechanism
would have first-order kinetics: the relative slow formation of
+
the carbocation, R (step 1), would determine the rate (k α[R–
LG]); once the cation is generated, it would capture the
nucleophile rapidly, forming the product. Does this alternative
type of nucleophilic substitution, involving two steps, actually
occur?
We can answer this question by comparing the reactions
of two nucleophiles with two different substrates:
+
–
a) the reaction of iodide ion (Na I ) with bromoalkanes in
acetone (Table 6.8); bromomethane reacts much faster
+ –
with Na I than 2-bromo-2-methylpropane; this result is
exactly what you learned to expect from an SN2 reaction;
b) the reaction of water, H2O, with the same bromoalkanes
(Table 7.1); this reaction proceeds much faster for 2bromo-2-methylpropane than for bromomethane; the
reversed order of reactivity is incompatible with an SN2
reaction and requires a different mechanism.
A more detailed study shows that the reaction with water
follows a first order rate, as expected for the two-step reaction.
The reaction also has different stereochemical consequences:
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
the stereochemistry at the reacting carbon is randomized, i.e.,
iodide
ion
converts
enantiomerically
pure
3-bromo-3-
methylhexane to racemic 3-iodo-3-methylhexane. This result
requires an achiral intermediate, identified as the planar
carbocation (cf., Figure 7.6).
The two-step mechanism of these reactions and the
nature of the intermediate causes a range of important features:
1. the reaction proceeds fastest for tertiary substrates;
2. the reaction requires a good leaving group, LG;
3. the solvent influences the rate significantly;
4. the nucleophile does not play a significant role
[because it is not involved in the rate-determining
step].
1. The fast reaction of tertiary substrates is related to
carbocation stability; carbocations with alkyl substituents are
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
stabilized by hyperconjugation (remember?), that is, donation
of electron density from a C–H σ-bond to the empty p orbital
(Figure 7.5). The relative stability of carbocations follows the
order:
+
(CH3)3C > (CH3)2CH
+
>
CH3CH2
+
>
CH3
+
The formation of the most stable (lowest-energy)
carbocation intermediate has the lowest-lying transition state
and, thus, the fastest rate. This is opposite to the SN2 reaction,
where the least hindered substrate has the lowest transition
state. Salient features of SN2 and SN1 reactions are compared
below.
2. The SN1 reaction works best with good leaving
groups, e.g.,
–OSO2
–
>
–
I
>
4
–
Br
>
–
Cl
Organic Chemistry
Nucleophilic Substitution
H. D. Roth
3. Polar solvents accelerate the reaction; polar protic
solvents are optimal. For example, the reaction of water with
2-bromo-2-methylpropane is 400,000 times as fast in pure
(100%) water than in acetone containing 10% water.
The role of the solvent can be understood by comparing
the rate determining transition states for the SN2 and SN1
reactions (Figure 7.4): in the SN2 reaction the negative charge
is distributed between Nu and LG; in the SN1 reaction a polar
bond is broken with separation of two opposite charges, + for
the carbocation, – for the LG.
4. The nucleophile does not play a significant role in the
SN1 reaction. Even weak nucleophiles, such as water or
alcohols, react with carbocations. Reactions in which water or
alcohols serve as solvents and nucleophiles are called
hydrolysis and solvolysis, respectively. These reactions
proceed in three steps (Figure 7.2):
a. dissociation (rate determining);
b. addition (capture) of an oxygen lone pair unto the
empty pi orbital, forming an oxonium ion;
c. deprotonation of the oxonium ion forming an alcohol
(when water is the nucleophile) or ether (when an
alcohol serves as nucleophile).
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
While the nature of the nucleophile does not affect the
rate of an SN1 reaction, it determines the nature of the
products; we call this “product determining”. In solutions
containing more than one nucleophile, substitution products
are formed with all nucleophiles present in competition with
one another. Good nucleophiles react faster than poor
nucleophiles (hardly surprising).
In summary, we now have two reactions of haloalkanes
with nucleophiles, depending on the nature of the substrate, a
bimolecular, one-step (SN2) and a unimolecular two-step
substitution (SN1).
Please NOTE that for this general scheme R1 and/or R2 may
be alkyl or H (especially for the SN2 reaction)
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
The simple features discussed above cannot be applied,
without caution, to all reactions; some reactions may be more
complicated.
–
–
Complication 1: a large leaving group (I , Br ) may
interfere with the approach of the nucleophile, causing the rate
of substitution with “retention” (from the side left open by the
leaving group, LG) to be slightly lower than the rate of
substitution with “inversion” (from the backside). As a result
the product may be slightly optically active, indicating some
degree of “inversion” (NOT due to an SN2 reaction).
Sufficient separation of ions - unhindered access - racemization
H
R3
R3
O
Br–
+
Br
R2
R1
H
O
H
R2
H
R1
inversion
retention
Insufficient separation of ions - access for retention impeded
incomplete racemization
H
H
+
Br
R2
R1
O
R3
R3
H
O
H
R2
Br–
R1
Complication 2: We have seen that the reaction of 3bromo-3-methylhexane with iodide ion leads to racemic 37
Organic Chemistry
Nucleophilic Substitution
H. D. Roth
iodo-3-methylhexane. What happens to substrates with two
chiral centers, such as 2-bromo-3-methylpentane? The
configuration at C-2 is randomized, but C-3 is not affected; it
remains unchanged. As a result we obtain two diastereomers
of 2-iodo-3-methylpentane.
C2H5
H3C
C2H5
H3C
C
+
C
H
CH3
C
Nu
C
H
CH3
H
H
CH3
C2H5
H
H3C
H
C
C
Nu
Diastereomers
Complication 3: We know that the strength of a
nucleophile correlates with its base strength and that typical
nucleophiles are also bases, that is, H+ acceptors. For that
reason, nucleophiles may abstract H+ from the beta-carbon
(Cβ) of a carbocation, competing with "nucleophilic capture",
that is, the second step of the SN1 reaction. This is a new
reaction, an “elimination, which we call the E1 reaction
(because it is unimolecular; Table 7.3, Figure 7.7).
1.
R–LG
3.
R + Nu–
+
→
+
R + LG
→
8
–
alkene + H–Nu
Organic Chemistry
Nucleophilic Substitution
H. D. Roth
The E1 and the SN1 reaction proceed via the same
carbocation intermediate (formed in the rate determining step,
the dissociation, 1). The distribution of the corresponding
products is determined by the nature of the base/nucleophile(s)
in solution. The nature of the leaving group affects the rate, but
NOT the distribution of products.
RateE1 = k [R–LG]
Adding a base to the reaction increases the fraction of
E1 over SN1, but the overall rate of product formation {Rate α
[R-Hal]} remains unchanged [because the different products
are formed via the same intermediate.
Complication 4: Addition of a weak base does not affect
the rate law but changes the ratio of E1 to SN1. In contrast,
addition of a strong base may change the rate law as well as
the ratio of substitution to elimination. This again is a new
reaction, a bimolecular “concerted” reaction with second-order
kinetics, which we call the E2 reaction.
RateE2 = k [R–LG] [B–]
Both the loss of the leaving group and the deprotonation
occur in the rate-determining step; the mechanism is shown in
Figure 7.8. Because of the bimolecular nature of this reaction,
the base strength and the leaving group ability are important,
9
Organic Chemistry
R–I
Nucleophilic Substitution
>
R–Br
>
H. D. Roth
R-Cl
The reaction works best when the Cα–LG bond and the
Cβ–H bond are anti-parallel to each other; we call this an antiperiplanar arrangement and the reaction an anti elimination.
The E2 elimination from cis- and trans-1-bromo-4-(1,1dimethylethyl)cyclohexane is a good example: it occurs
readily for the trans- but not for cis-isomer.
The existence of two elimination reactions with different
rate laws requires that you consider several reaction features
before you can assign a reaction product with any degree of
confidence. For example, sterically hindered nucleophile/bases
favor elimination (E2) over substitution (SN2).
potassium tert-butoxide
lithium diisopropylamide
Considering the alkoxide ions, methoxide is unhindered
and, therefore, a good nucleophile (as well as a base). Ethoxide
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
and isopropoxide have increasing bulk and increasing steric
hindrance. tert-Butoxide is no longer a nucleophile.
Bulk of Alkoxide Anions
H
H
H
C
H
H
C
O–
C
H
H
H
H
C
O–
O–
C
H
H
H
H
H
H
H
O–
C
H
C
C
H
H
C
H
H
C
H
H
H
In summary we have learned that haloalkane can undergo
four reactions of with nucleophiles (bases): SN2, SN1, E1, E2.
Please note that for this general scheme R1 and/or R2 may be
alkyl or H (especially for the SN2 reaction)
Hokusai,
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Organic Chemistry
Nucleophilic Substitution
Hokusai,
Miscellaneous SN2, SN1, E2, E1
12
H. D. Roth
Organic Chemistry
Nucleophilic Substitution
H. D. Roth
Factors governing the regiochemistry of the E2 reaction
Regiochemistry of the E2 reaction with a small base:
K OH
Br
70%
30%
Small bases preferentially abstract the more acidic H+
from the more highly substituted carbon forming the more
highly substituted (more stable) alkene. This was recognized
by Saytzev (19th century) who formulated a rule that “the H is
abstracted from the carbon with fewer Hs” (Saytzev’s rule).
Saytzev based his rule on a difference between the substrates
that he noted. You understand today that the course of the
reaction is determined by the energetics.
Regiochemistry of the E2 reaction with a bulky base
CH3
H3C
CH3
CH3
H3C
–OC(CH )
3 3
Br
CH2
72 %
CH3
CH3
H3C
28 %
For bulky bases (t-butoxide) the approach to the more
highly substituted carbon is sterically hindered; the less highly
substituted alkene is favored, even though it is less stable.
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
Some cyclohexane derivatives show particularly
interesting features (limited rotation around the key C–C bond).
For example two stereo-isomers, cis- and trans-1-bromo2-methylcyclohexane, generate different products by E2
elimination.
–OCH
3
Br
–OCH
3
Br
Can you think of a good reason? Drawing the chair forms
of the isomers with the Br in the axial position (because of the
stereochemistry required for the E2 reaction), you will notice
that the trans-isomer (right) has only one antiperiplanar H
eligible for abstraction; therefore, it can form only the less
highly substituted, less stable alkene (anti-Saytzev). The
stereochemistry required for an E2 reaction overrides the
energetic considerations. The cis-isomer (left) has two
antiperiplanar H atoms and follows the favorable energetics.
H
H
H
CH3
CH3
Br
Br
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
The E2 reactions of cis- and trans-1-bromo-4-tertbutylcyclohexane take place with vastly different rates.
(H3C)3C
–OH
(H3C)3C
Br
(H3C)3C
–OH
(H3C)3C
Br
Why would one isomer react significantly faster than the
other, and which one? Can you think of a good reason?
The trans-isomer has antiperiplanar Hs only in the
conformer with an axial t-butyl group; any coformer with two
bulky axial substituents is very unstable. Moreover, the
approach of the base is further disfavored by the bulky t-butyl
group.
Br
stable conformer
2 antiperiplanar Hs
H
H
H
H
H
H
Br
Br
highly unstable conformer
2 antiperiplanar Hs
stable conformer
0 antiperiplanar Hs
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
Stereochemistry of the E2 reaction – the general case –
free rotation around the C–C bond which of the “eligible” Hs is
chosen and which alkene is formed?
B–
R2
R1
H
R
H'
C
C
R
H
C
C
H
R
C
C
R2
R1
Br
rotate around C–C by 120°
B–
R2
R1
H'
C
C
H
R2
R1
Br
R
If the substrate contains bulky groups, the different
conformers may have significantly different stabilities, e.g.,
anti-Elimination
B
H
B
t-Bu
H
H
t-Bu
C
C
CH3
t-Bu
CH3
t-Bu
H
Br
Br
H
t-Bu
C
C
t-Bu
CH3
B
B
H
H
C
t-Bu
H
t-Bu
t-Bu
CH3
CH3
C
Br
H
t-Bu
Br
t-Bu
H
C
16
C
t-Bu
CH3
Organic Chemistry
Nucleophilic Substitution
H. D. Roth
Both conformers have H and Br in an antiperiplanar
arrangement; the top conformer is much less hindered (more
stable) than the bottom one and it forms the more stable alkene.
SN2 at two differently substituted centers
Na+ –CN
H Cl
Br H
H CN
NC H
2R,4R
2S,4S
Na+ –CN
Cl H
Br H
H CN
H CN
meso
2S,4R
Carbocation rearrangements
Some reactions involving carbocations yield unexpected
products. The “normal” (expected) product is obtained in low
yield and an unexpected product in higher yield.
H
H 3C
I
H
CH3OH
H
H3C
OCH3
H
H3CO
H3C
CH3
CH3
CH3
CH3
expected product
minor
H
H
CH3
CH3
unexpected product
major
These reactions occur, because 2° carbocations formed
–
by dissociation of I from an iodoalkane can form more stable
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
–
carbocations by rearrangement. Migration (1,2-shifts) of H or
–
R yield the "unexpected” substitution products.
I
H
H
H 3C
CH3OH
CH3
H
CH3
CH3
[the symbol “
H
+
H3C
H
CH3
H 3C
+
H 3C
H
CH3
“ signifies rearrangement] Both
carbocations react with methanol but the intramolecular
rearrangement is faster than the intermolecular “capture” by
the nucleophile.
In order to understand the mechanism by which these
rearrangements occur, we recall the concept of hyperconjugation
stabilizing a free radical or a carbocation (cf., Figure 9.2).
Hyperconjugation involves distorting or shifting the electrons of
a C–H bond in the direction of the electron-poor carbon. For the
rearrangement the hydrogen atom migrates along with the
electron pair; this can be described as a hydride shift. The
positive charge ends up at the carbon from where the hydride left.
We will discuss three different types of rearrangement:
(i)
1,2-hydride shift
Rearrangement by 1,2-hydride shift is exemplified by the
first reaction in this section, involving conversion of a 2° to a
3° carbocation.
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
Another example, also involving conversion of a 2° to a
3° carbocation, shows in addition that the 1,2-hydride shift can
be followed either by capture (SN1) or by deprotonation (E1);
however, the cyclohexene derivative is a minor product.
Br
H
CH3
CH3
+
CH3OH
CH3
+
o
–Br–
CH3
CH3
CH3
CH3OH
OCH3
H
– CH3+OH2
CH3
CH3
CH3
OCH3
CH3
CH3
CH3
The curly arrow indicates random stereochemistry, as
you had expected for a reaction involving a carbocation.
(ii) 1,2-alkyl shift
In some cases, secondary carbocations undergo
rearrangement by 1,2-methyl shift, followed by either
nucleophilic capture (SN1) or by deprotonation (E1).
H3 C
CH3
CH3OH
CH3
H 3C
–Br–
H3 C
CH3
CH3OH
CH3
H3 C
H3 C
H 3C
+
CH3
OCH3
Br
CH3
H 3C
CH3
CH3
Br–
CH3
CH3
–HBr
+
CH3OH
CH3
H 3C
CH3
19
H 3C
H 3C
OCH3
CH3
CH3
Organic Chemistry
Nucleophilic Substitution
H. D. Roth
(iii) ring expansion releasing angle strain
Bromoethylcyclobutane generates a 2° carbocation with three
neighboring groups that can migrate. The 1,2-hydride shift forms a
3° carbocation with increased ring strain; shift of one of the two
equivalent alkyl groups forms a 2° carbocation with greatly
reduced ring strain. Stay tuned for more in future chapters.
+
Br
–Br–
H
H
H
+
H
+
+
H
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
Summary
(parting shot?)
SN2, SN1, E2, E1
Concerning the competition between SN2, SN1, E2, and E1:
Nature of the substrate:
1° substrates clearly favor SN2 and E2
3° substrates clearly favor SN1 and E1
Solvent:
Polar aprotic solvents favor SN2 and E2
Polar protic solvents favor SN1 and E1
Leaving Group:
Only good leaving groups undergo SN1 and E1
Poor leaving groups disfavor SN1 and E1
The leaving group does not affect the ratio of
SN1 vs. E1
Nucleophile/Base:
Strong nucleophiles favor SN2
Strong bases favor E2
Strong unhindered alkoxides favor SN2
Strong hindered alkoxides favor E2
SN1 and E1 with rearrangement
2° carbocations formed may be stabilized by
hydride shift, alkyl shift, ring enlargement;
SN1 is typically preferred, E1 becomes more
pronounced if the rearranged carbocation has
an adjacent 3° carbon.
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Organic Chemistry
Nucleophilic Substitution
H. D. Roth
Some further examples:
Secondary carbocations whose rearrangements generate
tertiary carbocations with an adjacent tertiary carbon produce
greater yields of elimination products.
H
H
H
CH3OH
+
Br
OCH3
o
OCH3
+
H
H
22
H
Organic Chemistry
Nucleophilic Substitution
H. D. Roth
Overall we are dealing with the following palette of reactions
Here is what your book says:
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Organic Chemistry
Nucleophilic Substitution
Kin-kaku-ji (Kyoto)
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H. D. Roth