About The Authors

Chapter 6
These PowerPoint Lecture Slides were created and prepared by Professor
William Tam and his wife, Dr. Phillis Chang.

Nucleophilic Substitution Professor William Tam received his B.Sc. at the University of Hong Kong in
1990 and his Ph.D. at the University of Toronto (Canada) in 1995. He was an
NSERC postdoctoral fellow at the Imperial College (UK) and at Harvard
and Elimination Reactions University (USA). He joined the Department of Chemistry at the University of
Guelph (Ontario, Canada) in 1998 and is currently a Full Professor and
Associate Chair in the department. Professor Tam has received several awards
of Alkyl Halides in research and teaching, and according to Essential Science Indicators, he is
currently ranked as the Top 1% most cited Chemists worldwide. He has
published four books and over 80 scientific papers in top international journals
such as J. Am. Chem. Soc., Angew. Chem., Org. Lett., and J. Org. Chem.

Dr. Phillis Chang received her B.Sc. at New York University (USA) in 1994, her
Created by M.Sc. and Ph.D. in 1997 and 2001 at the University of Guelph (Canada). She
lives in Guelph with her husband, William, and their son, Matthew.
Professor William Tam & Dr. Phillis Chang
Ch. 6 - 1 Ch. 6 - 2

1. Organic Halides Carbon-Halogen Bond Lengths

Carbon-
and Bond Strength
H H H H
δ+ δ− H C F H C Cl H C Br H C I
C X H H H H
C–X Bond
X = Cl, Br, I Length (Å)
1.39 1.78 1.93 2.14

increase

Halogens are more electronegative
than carbon C–X Bond
Strength 472 350 293 239
(kJ/mol)

Ch. 6 - 3
decrease Ch. 6 - 4

1A. Physical Properties of Organic Halides: Physical Properties of Organic Halides:

Boiling Point (bp/oC) Density (ρ
(ρ)
Group Fluoride Chloride Bromide Iodide Group Fluoride Chloride Bromide Iodide

Me -78.4 -23.8 3.6 42.5 Me 0.84(-60) 0.92 1.73(0) 2.28

Et -37.7 13.1 38.4 72 Et 0.72 0.91(15) 1.46 1.95

Bu 32 78.4 101 130 Bu 0.78 0.89 1.27 1.61

s s
Bu - 68 91.2 120 Bu - 0.87 1.26 1.60
i i
Bu - 69 91 119 Bu - 0.87 1.26 1.60
t t
Bu 12 51 73.3 100(dec) Bu 0.75(12) 0.84 1.22 1.57(0)

Ch. 6 - 5 Ch. 6 - 6
Different Types of Organic Halides
Vinyl halides (Alkenyl halides)
sp2

Alkyl halides (haloalkanes)
X
C X sp3

Aryl halides
sp2
Attached to Attached to Attached to X
1 carbon atom 2 carbon atoms 3 carbon atoms
C C C
benzene or aromatic ring

Acetylenic halides (Alkynyl halides)

C Cl C Br C I sp
a 1o chloride a 2o bromide a 3o iodide X
Ch. 6 - 7 Ch. 6 - 8

sp3

Prone to undergo 2. Nucleophilic Substitution Reactions
δ− Nucleophilic Substitutions
δ+ C X (SN) and Elimination δ+ δ−
Nu + C X Nu C + X
Reactions (E) (the focus (nucleophile) (substrate) (product) (leaving
Alkyl halides of this Chapter) group)

The Nu⊖ The Nu⊖ uses

The bond The LG
donates its e⊖ pair to
between gains the
sp2 an e⊖ pair form a new
sp2 sp C and LG pair of e⊖
to the covalent bond
X X X breaks,
with the originally
substrate giving both
substrate C bonded

Different reactivity than alkyl halides, e⊖ from the in the
and do not undergo SN or E reactions bond to LG substrate
Ch. 6 - 9 Ch. 6 - 10

Timing of The Bond Breaking & Bond ● 2nd type: SN1 (stepwise mechanism)
Making Process Step (1):
R (k 1 ) R

Two types of mechanisms R C Br R C + Br
slow
● 1st type: SN2 (concerted mechanism) R
r.d.s.
R
Step (2) k1 << k2 and k3
R (k 2 ) R H
R
R R R C + H 2O R C O
δ+ δ− δ −
δ−
R fast R H
HO C Br HO C Br HO C
R R R
R Step (3)
R
transition state (T.S.)
R R H (k3) R
+ Br- R C O + H 2O R C OH + H3O+
R H fast R
Ch. 6 - 11 Ch. 6 - 12
3. Nucleophiles
Examples:
H H H H

A reagent that seeks a positive center
HO + C C + Cl

Has an unshared pair of e⊖ CH3 Cl CH3 OH
e.g.: HO , CH3O , H2N (negative charge) (Nu ) (substrate) (product) (L.G.)

H H H H
H2O, NH3 (neutral)
H
O
H
+ C C H + Cl
CH3 Cl CH3 O
δ+ δ− (Nu ) (substrate) H (L.G.)
This is the positive
center that the C X H H
Nu⊖ seeks (product)
C + H3O
Ch. 6 - 13 CH3 OH Ch. 6 - 14

4. Leaving Groups 5. Kinetics of a Nucleophilic Substitution

Reaction: An SN2 Reaction

To be a good leaving group, the substituent
must be able to leave as a relatively stable, HO + CH3 Br HO CH3 + Br
weakly basic molecule or ion -
e.g.: I⊖, Br⊖, Cl⊖, TsO⊖, MsO⊖, H2O, NH3 Rate = k[CH3Br][OH ]

O
The rate of the substitution reaction is

OTs = O S CH3 (Tosylate) linearly dependent on the
O
concentration of OH⊖ and CH3Br
O
Overall, a second-order reaction ⇒
OMs = O S CH3 (Mesylate) bimolecular
O
Ch. 6 - 15 Ch. 6 - 16

5A. How Do We Measure the Rate of Graphically…

Graphically…
This Reaction?
Concentration, M

e.g.: [CH3Cl] ↓
H H [CH3OH] ↑
HO + C Cl HO C + Cl
H H
(Nu ) H H (leaving
(substrate) (product) group)

The rate of reaction can be measured by
● The consumption of the reactants Time, t
(HO⊖ or CH3Cl) or ∆[CH3Cl] [CH3Cl]t=t − [CH3Cl]t=0
● The appearance of the products Rate = =−
(CH3OH or Cl⊖) over time ∆t Time in seconds
Ch. 6 - 17 Ch. 6 - 18
Initial Rate
Example:
60oC
[CH3Cl]t=0 HO + Cl CH3 HO CH3 + Cl
Concentration, M
H2O

[CH3Cl]t=t [OH⊖]t=0 [CH3Cl]t=0

Initial rate
Result
mole L-1, s-1
1.0 M 0.0010 M 4.9 × 10-7
[CH3Cl]
1.0 M 0.0020 M 9.8 × 10-7 Doubled
Time, t
2.0 M 0.0010 M 9.8 × 10-7 Doubled

Initial Rate [CH3Cl]t=t − [CH3Cl]t=0 2.0 M 0.0020 M 19.6 × 10-7 Quadrupled

=−
(from slope) ∆t
Ch. 6 - 19 Ch. 6 - 20

Conclusion: The Kinetic Rate Expression

60oC 60oC
HO + Cl CH3 H2O
HO CH3 + Cl HO + Cl CH3 H2O
HO CH3 + Cl

● The rate of reaction is directly Rate α [OH⊖][CH3Cl]

proportional to the concentration of
either reactant. Rate = k[OH⊖][CH3Cl]

Initial Rate
● When the concentration of either k=
[OH⊖][CH3Cl]
reactant is doubled, the rate of
reaction doubles. = 4.9 × 10-7 L mol-1 s-1
Ch. 6 - 21 Ch. 6 - 22

6. A Mechanism for the SN2 Reaction 7. Transition State Theory:

H
Free Energy Diagrams
H H
+ − δ− δ−
HO
δ
C
δ
Br HO C Br HO C
A reaction that proceeds with a
H H H
H negative free-energy change (releases
H H
transition state (T.S.) energy to its surroundings) is said to
negative OH⊖ + Br-
O–C bond be exergonic
brings an e⊖ O–C bond
partially formed;
A reaction that proceeds with a positive
pair to δ+ C; δ– C–Br bond formed; Br⊖
Br begins to departed. free-energy change (absorbs energy
partially broken.
move away with Configuration of Configuration from its surroundings) is said to be
an e⊖ pair C begins to invert of C inverted endergonic
Ch. 6 - 23 Ch. 6 - 24

At 60oC (333 K) CH3 Br + OH CH3 OH + Cl
CH3 Br + OH CH3 OH + Cl ● Its equilibrium constant (Keq) is
∆Go = -100 kJ/mol ∆Go = –RT ln Keq
–∆Go
● This reaction is highly exergonic ln Keq =
RT
–(–100 kJ/mol)
∆Ho = -75 kJ/mol =
(0.00831 kJ K-1 mol-1)(333 K)
● This reaction is exothermic = 36.1

Keq = 5.0 ╳ 1015

Ch. 6 - 25 Ch. 6 - 26

A Free Energy Diagram for a Hypothetical SN2
The reaction coordinate indicates the
Reaction That Takes Place with a Negative ∆Go progress of the reaction, in terms of the
conversion of reactants to products

The top of the energy curve corresponds to
the transition state for the reaction

The free energy of activation (∆G‡) for
the reaction is the difference in energy
between the reactants and the transition
state

The free energy change for the
reaction (∆Go) is the difference in energy
between the reactants and the products
Ch. 6 - 27 Ch. 6 - 28

A Free Energy Diagram for a Hypothetical 7A. Temperature, Reaction Rate,

Reaction with a Positive Free-
Free-Energy Change and the Equilibrium Constant

A 10°C increase
in temperature
will cause the
reaction rate to
double for many
Distribution of energies at two
reactions taking
different temperatures. The number place near room
of collisions with energies greater
than the free energy of activation is temperature
indicated by the corresponding
shaded area under each curve.
Ch. 6 - 29 Ch. 6 - 30

The relationship
A reaction with a
between the rate lower free energy of
constant (k) and ∆G‡ activation (∆G‡) will
is exponential : occur exponentially
‡
k = k0 e −∆G /RT faster than a
reaction with a
e = 2.718, the base of
natural logarithms higher ∆G‡, as
dictated by
k0 = absolute rate
‡
Distribution of energies at two
constant, which equals
Distribution of energies at two k = k0 e −∆G /RT
different temperatures. The number the rate at which all different temperatures. The number of
of collisions with energies greater transition states proceed collisions with energies greater than
than the free energy of activation is the free energy of activation is
indicated by the corresponding to products (At 25oC, indicated by the corresponding
shaded area under each curve. k0 = 6.2 ╳ 1012 s−1 ) shaded area under each curve.
Ch. 6 - 31 Ch. 6 - 32

Free Energy Diagram of SN2 Reactions 8.The Stereochemistry of SN2 Reactions

T.S.
Inversion of configuration

∆G = free energy of
ee Energy

∆G activation CH3
HO- + CH3Br ∆Go = free energy
change HO + C Br
∆Go H (inversion)
Fre

CH3OH + Br- CH2CH3

Reaction Coordinate (R) CH3
HO C + Br

Exothermic (∆Go
is negative) H

Thermodynamically favorable process (S) CH 2 CH 3
Ch. 6 - 33 Ch. 6 - 34

Example:
Example:
Nu⊖ attacks from the TOP face.
Nu⊖ attacks from the BACK face.

CH3 I
+ OCH3 + CN + Br
Br CN
(inversion of configuration) (inversion of
configuration)

CH3 OCH3 +
I
Ch. 6 - 35 Ch. 6 - 36
 9. The Reaction of tert-Butyl Chloride 9A. Multistep Reactions & the Rate-
Rate-
with Hydroxide Ion: An SN1 Reaction Determining Step
CH3 CH3
 In a multistep reaction, the rate of the
CH3 C Br + H2O CH3 C OH + HBr
overall reaction is the same as the rate
CH3 CH3
of the SLOWEST step, known as the
 The rate of SN1 reactions depends only on rate-determining step (r.d.s)
concentration of the alkyl halide and is
independent on concentration of the Nu⊖  For example:
Rate = k[RX] Reactant
k1 k2 k3
Intermediate Intermediate Product
In other words, it is a first-order reaction (slow)
1
(fast)
2
(fast)

⇒ unimolecular nucleophilic substitution

Ch. 6 - 37
k1 << k2 or k3 Ch. 6 - 38

The opening A is 10. A Mechanism for the SN1 Reaction

much smaller than
openings B and C
A multistep process

The overall rate at
which sand reaches Step (1):
A
to the bottom of
the hourglass is CH3 (k 1 ) CH3
B limited by the rate
CH3 C Br CH3 C + Br
at which sand falls
CH3 slow CH3
C through opening A
Opening A is r.d. step


(ionization
analogous to the
rate-determining of alkyl
step of a multistep halide)
reaction Ch. 6 - 39 Ch. 6 - 40

Free Energy Diagram of SN1 Reactions Step (2)

T.S. (1) CH3 (k 2 ) CH3 H

T.S. (2)
CH3 C + H2O CH3 C O
fast
Energy

T.S. (3)
(CH3)3C
CH3 CH3 H
-
+ Br
Free E

∆G1 (CH3)3C -OH2

+ Br-
(CH3)3CBr intermediate (CH3)3C-OH
+ H2O
+ Br-
Reaction Coordinate
Ch. 6 - 41 Ch. 6 - 42
 Free Energy Diagram of SN1 Reactions Step (2)

T.S. (1) CH3 (k 2 ) CH3 H

T.S. (2)
CH3 C + H2O CH3 C O
fast
Energy

T.S. (3)
(CH3)3C
CH3 CH3 H
-
+ Br
Free E

∆G1 (CH3)3C -OH2 Step (3)

+ Br-
(CH3)3CBr intermediate CH3 H (k 3 ) CH3
(CH3)3C-OH
+ H2O
+ Br- CH3 C O + H2O CH3 C OH
fast
Reaction Coordinate CH3 H CH3
Ch. 6 - 43 + H3O+
Ch. 6 - 44

Free Energy Diagram of SN1 Reactions Step (2)

T.S. (1) CH3 (k 2 ) CH3 H

T.S. (2)
CH3 C + H2O CH3 C O
fast
Energy

T.S. (3)
(CH3)3C
CH3 CH3 H
-
+ Br
k1 << k2 and k3
Free E

∆G1 (CH3)3C -OH2 Step (3)

-
+ Br
(CH3)3CBr intermediate CH3 H (k 3 ) CH3
(CH3)3C-OH
+ H2O
+ Br- CH3 C O + H2O CH3 C OH
fast
Reaction Coordinate CH3 H CH3
Ch. 6 - 45 + H3O+
Ch. 6 - 46

11. Carbocations

2 intermediates and 3 transition states
(T.S.) 11
11A.
A. The Structure of Carbocations

Carbocations are

The most important T.S. for SN1 trigonal planar

The central carbon
reactions is T.S. (1) of the rate- atom in a carbocation
determining step (r.d.s.) H3C is electron deficient; it
C CH3 has only six e⊖ in its
valence shell
CH3 H3C
The p orbital of a
δ +
δ− carbocation contains
CH3 C Br no electrons, but it can
accept an electron pair
sp2-sp3 π bond when the carbocation
CH3 undergoes further
reaction
Ch. 6 - 47 Ch. 6 - 48
 11
11B.
B. The Relative Stabilities of
Stability of cations
Carbocations most stable (positive inductive effect)

General order of reactivity (towards SN1 R R R H

> > >
reaction) C C C C
● 3o > 2o >> 1o > methyl R R R H H H H H

Resonance stabilization of allylic and

The more stable the carbocation benzylic cations
formed, the faster the SN1 reaction
CH2 CH2
etc.
Ch. 6 - 49 Ch. 6 - 50

12. The Stereochemistry of SN1 Reactions
Example: racemic mixture

( 1 : 1 )
Ph Ph Ph H2O
(R) (R) (S)
CH3OH CH3OH +
CH3
C CH3 C OCH3 (SN1)
Br Br OH OH
CH2CH3 CH3 CH2CH3 CH2CH3
(S) (trigonal planar) (R) and (S) (one enantiomer)
racemic mixture attack from H2O
slow TOP face H2O
CH3OH CH3OH r.d.s.
attack from left 50:50 attack from right
chance H2O O
Ph Ph H H
CH3 CH3
CH3O OCH3 H2O attack from O
CH2CH3 (1 : 1) CH2CH3 H H
(R) (S ) (carbocation) BOTTOM face
Ch. 6 - 51 Ch. 6 - 52

Example: 13. Factors Affecting the Rates of

I OMe Me SN1 and SN2 Reactions
t
Bu Me MeOH tBu Me + Bu t
OMe

The structure of the substrate

MeOH
slow Me H MeOH
The concentration and reactivity of the
O
r.d.s. nucleophile (for SN2 reactions only)
t
Bu Me
MeOH
⊕ CH3
Me
The effect of the solvent
t t H
Bu ⊕ Bu O
MeOH Me
trigonal planar
The nature of the leaving group
Ch. 6 - 53 Ch. 6 - 54
13
13A.
A. The Effect of the Structure of the Substrate
For example:
R Br + HO R OH + Br

General order of reactivity (towards SN2 Relative Rate (towards SN2)
reaction) CH3 CH3
CH3 Br CH3CH2 Br CH3CH Br CH3 C CH2Br CH3 C Br
CH3 CH3 CH3
● Methyl > 1o > 2o >> 3o > vinyl or aryl
methyl 1o 2o neopentyl 3 o

DO NOT 2 × 106 4 × 104 500 1 <1

undergo Most Least
SN2 reactions reactive reactive
Ch. 6 - 55 Ch. 6 - 56

Compare H H
δ+ δ−
H δ+ δ−
H HO C Br HO C + Br
HO C Br + Br t very t
HO C Bu Bu
faster H slow
H H CH3
H H

H δ+ δ−
H CH3 δ+ δ−
CH3
HO C Br HO C + Br HO C Br HO C + Br
slower extremely
CH3 CH3 CH3 CH3
slow
CH3 CH3 CH3 CH3
Ch. 6 - 57 Ch. 6 - 58

Note NO SN2 reaction on sp2 or sp Reactivity of the Substrate in SN1

carbons Reactions
e.g. sp2
H I
General order of reactivity (towards SN1
+ Nu No reaction reaction)
H H ● 3o > 2o >> 1o > methyl
sp2
I

The more stable the carbocation
+ Nu No reaction
formed, the faster the SN1 reaction
sp

I + Nu No reaction
Ch. 6 - 59 Ch. 6 - 60

Stability of cations
Resonance stabilization for allylic and
most stable (positive inductive effect) benzylic cations
R R R H
> > >
C C C C
R R R H H H H H

Allylic halides and benzylic halides also

CH2 CH2
undergo SN1 reactions at reasonable
etc.
rates I
Br
an allylic bromide a benzylic iodide
Ch. 6 - 61 Ch. 6 - 62

13
13B.
B. The Effect of the Concentration
& Strength of the Nucleophile

For SN1 reaction
For SN2 reaction

Recall: Rate = k[RX] Recall: Rate = k[RX][RX]
● The Nu⊖ does NOT participate in ● The rate of SN2 reactions depends
the r.d.s. on both the concentration and
● Rate of SN1 reactions are NOT the identity of the attacking Nu⊖
affected by either the
concentration or the identity of
the Nu⊖
Ch. 6 - 63 Ch. 6 - 64

Identity of the Nu⊖
The relative strength of a Nu⊖ can be

● The relative strength of a Nu⊖ (its correlated with 3 structural features
nucleophilicity) is measured in ● A negatively charged Nu⊖ is always a
terms of the relative rate of its SN2 more reactive Nu⊖ than its conjugated
reaction with a given substrate acid
 e.g. HO⊖ is a better Nu⊖ than H2O
rapid
CH3O + CH3I CH3OCH3 + I and RO⊖ is better than ROH
⊖ ● In a group of Nu⊖s in which the
Good Nu nucleophilic atom is the same,
Very nucleophilicities parallel basicities
CH3OH + CH3I slow CH3OCH3 + I  e.g. for O compounds,
Poor Nu⊖ RO⊖ > HO⊖ >> RCO2⊖ > ROH > H2O
Ch. 6 - 65 Ch. 6 - 66
 ● When the nucleophilic atoms are 13
13C.
C. Solvent Effects on SN2 Reactions:
different, then nucleophilicities may Polar Protic & Aprotic Solvents
not parallel basicities

Classification of solvents
 e.g. in protic solvents HS⊖, CN⊖,
and I⊖ are all weaker bases than Non-polar solvents
(e.g. hexane, benzene)
HO⊖, yet they are stronger Nu⊖s
than HO⊖ Solvents Polar protic solvents
HS⊖ > CN⊖ > I⊖ > HO⊖ (e.g. H2O, MeOH)
Polar
solvents
Polar aprotic solvents
(e.g. DMSO, HMPA)
Ch. 6 - 67 Ch. 6 - 68

SN2 Reactions in Polar Aprotic Solvents  Polar aprotic solvents tend to

● The best solvents for SN2 reactions solvate metal cations rather than
are nucleophilic anions, and this
 Polar aprotic solvents, which results in “naked” anions of the
have strong dipoles but do not Nu⊖ and makes the e⊖ pair of
have OH or NH groups the Nu⊖ more available
 Examples
O
DMSO
CH3 O CH3O Na CH3O + DMSO Na
O
H N P NMe "naked anion"
S Me2N NMe2 2
CH3 CH3 CH3 CH3CN
(DMSO) (DMF) (HMPA) (Acetonitrile)
Ch. 6 - 69 Ch. 6 - 70

Tremendous acceleration in SN2


SN2 Reactions in Polar Protic Solvents
reactions with polar aprotic ● In polar protic solvents, the Nu⊖
solvent anion is solvated by the surrounding
CH3Br + NaI CH3I + NaBr protic solvent which makes the e⊖
pair of the Nu⊖ less available and
Solvent Relative Rate thus less reactive in SN2 reactions

MeOH 1 OR
H
DMF 106 RO H Nu H OR
H
OR
Ch. 6 - 71 Ch. 6 - 72

Halide Nucleophilicity in Protic Solvents
Halide Nucleophilicity in Polar Aprotic
● I⊖ > Br⊖ > Cl⊖ > F⊖ Solvents (e.g. in DMSO)
OR
RO δ+ δ+ OR H ● F⊖ > Cl⊖ > Br⊖ > I⊖
H H  Polar aprotic solvents do not solvate
δ+ δ+
RO H F -
H OR RO H I- anions but solvate the cations
+ +
δ δ
H H The “naked” anions act as the Nu⊖
RO OR H 
OR
(strongly solvated) (weakly solvated)
 Since F⊖ is smaller in size and the
 Thus, I⊖
is a stronger in proticNu⊖ charge per surface area is larger
solvents, as its e⊖ pair is more available than I⊖, the nucleophilicity of F⊖ in
to attack the substrate in the SN2 reaction. this environment is greater than I⊖
Ch. 6 - 73 Ch. 6 - 74

13
13D.
D. Solvent Effects on SN1 Reactions:
Polar protic solvents stabilize the
The Ionizing Ability of the Solvent development of the polar transition
state and thus accelerate this rate-

Solvent plays an important role in SN1 determining step (r.d.s.):
reactions but the reasons are different
δ+ OR
from those in SN2 reactions H
CH3 H3C +
slow δ δ−
CH3 C Cl CH3 C Cl
Solvent effects in SN1 reactions are due r.d.s.

CH3 δ− CH3 δ+
R O H
largely to stabilization or destabilization H OR
of the transition state CH2
-
Cl + CH3 C
CH3
Ch. 6 - 75 Ch. 6 - 76

13
13E.
E. The Nature of the Leaving Group
Examples of the reactivity of some X⊖:

The better a species can stabilize a CH3O
+ CH3–X → CH3–OCH3 + X

negative charge, the better the LG in Relative Rate:
an SN2 reaction OH
, Worst X⊖ Best X⊖
SN1 Reaction: NH2
, <<F
< Cl
< Br
< I
< TsO

slow δ+ δ− RO

C X C X C + X
r.d.s.
~0 1 200 10,000 30,000 60,000
SN2 Reaction:  Note: Normally R–F, R–OH, R–NH2,
slow δ− δ−
C X Nu C X Nu C +X R–OR’ do not undergo SN2
r.d.s.
reactions.
Nu: Ch. 6 - 77 Ch. 6 - 78
 14. Organic Synthesis: Functional Group
Nu Transformation Using SN2 Reactions
R OH  R Nu + OH
a poor a strong OH
leaving group basic anion
H
CN CN HO Me

MeO
H Nu Br

✔
R O R Nu + H2O
H MeS
HS
a good weak
leaving group base SMe SH
Ch. 6 - 79 Ch. 6 - 80

Me
Examples:

O NaOEt,??DMSO
I Me C C O Me Br O
I
MeCOO
Br
I SMe
N3 NaSMe,?? DMSO
Me3N

N3 NMe3 Br
Ch. 6 - 81 Ch. 6 - 82

Examples:
?? ??

I CN I CN

(optically active, chiral) (optically active, chiral) (optically active, chiral)

(optically active, chiral)

● Need SN2 reactions to control NaBr NaCN

stereochemistry DMSO DMSO
● But SN2 reactions give the inversion of
configurations, so how do you get the (SN2 with Br (SN2 with
“retention” of configuration here?? inversion) inversion)
● Solution: (Note: Br⊖ is a stronger Nu than
“double inversion” ⇒ “retention” I⊖ in polar aprotic solvent.)
Ch. 6 - 83 Ch. 6 - 84
14
14A.
A. The Unreactivity of Vinylic and
Examples
Phenyl Halides
NaCN
X Br No Reaction
C C DMSO

X
vinylic halide phenyl halide I NaSMe
No Reaction

Vinylic and phenyl halides are generally HMPA
unreactive in SN1 or SN2 reactions

Ch. 6 - 85 Ch. 6 - 86

15. Elimination Reactions of Alkyl
Substitution reaction (SN) and

Halides elimination reaction (E) are processes
in competition with each other

Substitution
H OCH3 H OCH3 e.g.
C C C C + Br- t
(acts as a BuOK t
Br I O Bu +
Nu ) t
BuOH

Elimination SN2: 15% E2: 85%
H OCH3
+ + -
H
C C (acts as a C C CH3OH Br
Br base)
Ch. 6 - 87 Ch. 6 - 88

15
15A.
A. Dehydrohalogenation 15
15B.
B. Bases Used in Dehydrohalogenation

β hydrogen
Conjugate base of alcohols is often used

H α carbon as the base in dehydrohalogenations
C C Na
β carbon R−O⊖ + Na⊕ + H2
X halide as LG
R−O−H
NaH
Br R−O⊖ + Na⊕ + H2
LG t
β
BuOK
+ KBr + tBuOH
e.g.
α t o
BuOH, 60 C
H t
β hydrogen EtO Na BuO K
⊖OtBu sodium ethoxide potassium tert-butoxide
Ch. 6 - 89 Ch. 6 - 90
16. The E2 Reaction Mechanism for an E2
E2 Reaction
Br Et O Et O
EtO + + EtOH + Br H CH3 H CH3
δ− H CH3
C H
α
H Cβ C H C C C
H H δ− H H
H Br H Br +

Rate = k[CH3CHBrCH3][EtO⊖]
EtO⊖ removes Et OH + Br
Partial bonds in
a β proton; the transition C=C is fully

Rate determining step involves both C−H breaks; state: C−H and formed and
the alkyl halide and the alkoxide anion new π bond C−Br bonds the other
forms and Br break, new π products are
begins to C−C bond forms EtOH and Br⊖

A bimolecular reaction depart
Ch. 6 - 91 Ch. 6 - 92

Free Energy Diagram of E2

E2 Reaction 17. The E1 Reaction
T.S.
E1: Unimolecular elimination
E2 reaction has ONE
∆G‡ CH3 CH3 CH3
Frree Energy

transition state H2 O
CH3CHBrCH3 CH3 C Cl CH3 C OH + CH2 C
+ EtO- CH3 CH3 CH3
CH2=CHCH3
+ EtOH + Br-
slow (major (SN1)) (minor (E1))
r.d.s
Reaction Coordinate
CH3 H2O as H2O as
Rate = k[CH3CHBrCH3][EtO⊖] nucleophile base
CH C 3

Second-order overall ⇒ bimolecular CH3
Ch. 6 - 93 Ch. 6 - 94

Mechanism of an E1
E1 Reaction Free Energy Diagram of E1
E1 Reaction
α carbon T.S. (1)
β hydrogen T.S. (2)
H
H 2O
Energy

H2O
Cl + H3O (CH3)3C
slow fast
r.d.s. (E1 product) + Cl-
Free E

∆G1
fast H2O (CH3)2C=CH2
+ H3O + Cl-
(CH3)3CCl
H H O + H2O
2
O OH + H3O
H Reaction Coordinate
(SN1 product)
Ch. 6 - 95 Ch. 6 - 96
Step (1): Free Energy Diagram of E1
E1 Reaction
T.S. (1)
CH3 H 2O CH3 T.S. (2)
CH3 C Cl CH3 C + Cl

Energy
(k 1 )
CH3 CH3 (CH3)3C
slow + Cl-

Free E
Aided by the r.d. step Produces relatively ∆G1
polar solvent, a stable 3o carbocation (CH3)2C=CH2
chlorine departs and a Cl⊖. The ions + H3O + Cl-
(CH3)3CCl
with the e⊖ pair are solvated (and
that bonded it to stabilized) by + H2O
the carbon surrounding H2O
Reaction Coordinate
molecules
Ch. 6 - 97 Ch. 6 - 98

Step (2) 18. How To Determine Whether

Substitution or Elimination Is Favoured
H 3C H H 3C
( k 2)
All nucleophiles are potential bases and
C C H + H 2O CH2
fast all bases are potential nucleophiles
H 3C H H 3C
H
H2O molecule removes one of Substitution reactions are always in
the β hydrogens which are + H O

acidic due to the adjacent H competition with elimination reactions

positive charge. An e⊖ pair
moves in to form a double Produces alkene and
Different factors can affect which type
bond between the β and α hydronium ion
carbon atoms
of reaction is favoured
Ch. 6 - 99 Ch. 6 - 100

18
18A.
A. SN2 vs. E2
E2 Primary Substrate

With a strong base, e.g. EtO⊖
● Favor SN2
(a) H C
+X
(b) SN2 Nu C OEt
H C SN2: 90%
Nu
C X NaOEt
(a) (b) C Br +
+ Nu H + X EtOH
E2 C
E2: (10%)

Ch. 6 - 101 Ch. 6 - 102

Secondary Substrate Tertiary Substrate

With a strong base, e.g. EtO⊖
With a strong base, e.g. EtO⊖
● Favor E2 + ● E2 is highly favored

NaOEt
E2: 80% +
NaOEt EtOH
+ Br OEt
Br EtOH E2: 91% SN1: 9%

OEt
SN2: 20%
Ch. 6 - 103 Ch. 6 - 104

Base/Nu⊖: Small vs. Bulky Basicity vs. Polarizability

Unhindered “small” base/Nu⊖ O
O
NaOMe O CH3
Br MeOH + CH3 C O
OMe +
SN2: 99% E2: 1% (weak base)
Br SN2: 100% E2: 0%

Hindered “bulky” base/Nu⊖
OEt
t EtO
KO Bu + +
Br t OtBu (strong base)
BuOH SN2: 15% E2: 85% SN2: 20% E2: 80%
Ch. 6 - 105 Ch. 6 - 106

Tertiary Halides: SN1 vs. E1

E1 & E2
E2 19. Overall Summary

EtO OEt SN1 SN2 E1 E2

+
(strong CH3X ─ Very fast ─ ─

base) E2: 100% SN1: 0% Hindered bases give

Br RCH2X ─ Mostly ─ mostly alkenes;

e.g. with tBuO⊖

R' Very little;

Solvolysis possible;
Mostly SN2 with
weak bases; Very little
Strong bases
promote E2;
RCHX
e.g. with H2O;
e.g. with CH3COO⊖ e.g. with RO⊖, HO⊖
EtOH OEt MeOH
+
heat R' Very favorable
Strong bases
E1 + E2: 20% SN1: 80% RCX with weak bases;
e.g. with H2O;
─
Always competes
with SN1
promote E2;
e.g. with RO⊖, HO⊖
MeOH
R"
Ch. 6 - 107 Ch. 6 - 108
Review Problems
(3) CH3 CH3 Cl
OH HCl Cl CH3
(1) Br +
Na CN CN t t t
o Bu Bu Bu
t DMF, 25 C ( 50 : 50)
Bu t
SN2 with inversion Bu
Cl⊖ attacks Cl⊖ attacks
(2)
H NaH
O from top face from bottom
I O Et2O CH3 face
H CH3
O
H
H⊖ t
Bu t
Bu sp2 hybridized
I O carbocation
SN1 with racemization
Intramolecular SN2 Ch. 6 - 109 Ch. 6 - 110

 END OF CHAPTER 6 

Ch. 6 - 111