SUBSTITUTION AND ELIMINATION
REACTIONS OF ALKYL HALIDES
SN1, SN2, E1 & E2 REACTIONS
1
Reactions of Alkyl Halides (R-X): [SN1, SN2, E1, & E2 reactions]
d
d
H3C
F
d
H3C
d
d
Cl
d
EN (F-C) =
(4.0 – 2.5) = 1.5
EN (Cl-C) =
(3.0 – 2.5) = 0.5
H3C
Br
EN (Br-C) =
(2.8 – 2.5) = 0.3
H3C
I
EN (I-C) =
(2.5 – 2.5) = 0.0
The -carbon in an alkyl halide is electrophilic (electron accepting) for either
or both of two reasons…
a) the C to X (F, Cl, Br) bond is polar making carbon d+
b) X (Cl, Br, I) is a leaving group
decreasing basicity, increasing stability
The best
leaving
groups are
the weakest
bases.
pKb = 23
pKb = 22
pKb = 21
pKb = 11
pKb = -1.7
I30,000
Br 10,000
Cl 200
F1
HO 0
increasing leaving ability
The poorest
leaving
groups are
the strongest
bases.
2
Reactions of Alkyl Halides (R-X): [SN1, SN2, E1, & E2 reactions]
When a nucleophile (electron donor, e.g., OH-) reacts with an alkyl halide, the
halogen leaves as a halide
Nu:
..
Br
.. :
R
..
: Br
.. :
There are two competing reactions of alkyl halides with nucleophiles….
1) substitution
Nu:
H
H
-
C
+
C

C
The Nu:- replaces the halogen on the -carbon.
2) elimination
H
Nu:
X-
Nu
X
-
+
C
+
b
C

C
X
C
C
+
-
X
+
Nu
The Nu:- removes an H+ from a b-carbon &
the halogen leaves forming an alkene.
H
3
2nd Order Nucleophilic Substitution Reactions, i.e., SN2 reactions
There are two kinds of substitution reactions, called SN1 and SN2.
As well as two kinds of elimination reactions, called E1 and E2.
Let’s study SN2 reactions first. SN2 stands for Substitution,
Nucleophilic, bimolecular. Another word for bimolecular is ‘2nd
order’.
 Bimolecular (or 2nd order) means that the rate of an SN2 reaction is
directly proportional to the molar concentration of two reacting
molecules, the alkyl halide ‘substrate’ and the nucleophile:
Rate = k [RX] [Nu:-] (This is a rate equation and k is a constant).
 The mechanism of an SN2 reaction is the one shown on slide #2:
Nu:
-
+
H
H
C
C

C
X
C
+
X-
Nu
 Note that the nucleophile must hit the back side of the -carbon.
The nucleophile to C bond forms as the C to X bond breaks.
No C+ intermediate forms. An example is shown on the next slide.
4
2nd Order Nucleophilic Substitution Reactions, i.e., SN2 reactions
..
..
:O
H
H +
C
H
H
..
..
Br :
H
..
..O
C
H
..
Br :
..
H
..
..O
C
HH
H
H
H
..
+ : Br :
..
transition state
The rate of an SN2 reaction depends upon 4 factors:
1.
The nature of the substrate (the alkyl halide)
2.
The power of the nucleophile
3.
The ability of the leaving group to leave
4.
The nature of the solvent
1. Consider the nature of the substrate:
 Unhindered alkyl halides, those in which the back side of the -carbon is
not blocked, will react fastest in SN2 reactions, that is:
Me°
>>
1°
>>
2°
>>
3°
 While a methyl halides reacts quickly in SN2 reactions, a 3° does not react.
The back side of an -carbon in a 3° alkyl halide is completely blocked.
5
Effect of nature of substrate on rate of SN2 reactions:
H3C
H3C
H3C
Br
methyl bromide
H3C
CH2
CH
Br
H3C
Br
Br
H3C
H3C
ethyl bromide
C
isopropyl bromide
t-butyl bromide
SPACE FILLING MODELS SHOW ACTUAL SHAPES AND RELATIVE SIZES
Back side of -C
of a methyl halide
is unhindered.
Me°>>
Back side of -C of a
1° alkyl halide is
slightly hindered.
1°
>>
Back side of -C of a
2° alkyl halide is
mostly hindered.
2°
>>
decreasing rate of SN2 reactions
Back side of -C of a
3° alkyl halide is
completely blocked.
3°
6
Effect of the nucleophile on rate of SN2 reactions:
The -carbon in vinyl and aryl halides, as in 3° carbocations, is completely
hindered and these alkyl halides do not undergo SN2 reactions.
H2C
CH
Br
Br
vinyl bromide
bromobenzene
Nu:H2C
CH
Nu:Br
Br
The overlapping p-orbitals that form the p-bonds in vinyl and aryl halides
completely block the access of a nucleophile to the back side of the
-carbon.
7
Effect of nature substrate on rate of SN2 reactions:
2. Consider the power of the nucleophile:
Reactivity
Nu:-
Relative Reactivity
very weak
HSO4-, H2PO4-, RCOOH
< 0.01
weak
ROH
1
HOH, NO3-
100
F-
500
Cl-, RCOO-
20  103
NH3, CH3SCH3
300  103
N3-, Br-
600  103
OH-, CH3O-
2  106
CN-, HS-, RS-, (CH3)3P:, NH2- ,RMgX, I-, H-
> 100  106
fair
good
very good
increasing
 The better the nucleophile, the faster the rate of SN2 reactions.
 The table below show the relative power or various nucleophiles.
 The best nucleophiles are the best electron donors.
8
Effect of nature of the leaving group on rate of SN2 reactions:
3. Consider the nature of the leaving group:
The leaving group usually has a negative charge
 Groups which best stabilize a negative charge are the best leaving groups,
i.e., the weakest bases are stable as anions and are the best leaving groups.
 Weak bases are readily identified. They have high pKb values.
pKb = 23
pKb = 22
pKb = 21
pKb = 11
pKb = -1.7
pKb = -2
pKb = -21
I-
Br -
Cl-
F-
HO-
RO-
H2N-
30,000
10,000
200
1
0
0
0
Increasing leaving ability
 Iodine (-I) is a good leaving group because iodide (I-) is non basic.
 The hydroxyl group (-OH) is a poor leaving group because hydroxide (OH-)
is a strong base.
9
Effect of the solvent on rate of SN2 reactions:
4. Consider the nature of the solvent:
There are 3 classes of organic solvents:

Protic solvents, which contain –OH or –NH2 groups. Protic solvents
slow down SN2 reactions.

Polar aprotic solvents like acetone, which contain strong dipoles but no
–OH or –NH2 groups. Polar aprotic solvents speed up SN2 reactions.

Non polar solvents, e.g., hydrocarbons. SN2 reactions are relatively
slow in non polar solvents.
Protic solvents (e.g., H2O, MeOH, EtOH, CH3COOH, etc.) cluster around the
Nu:- (solvate it) and lower its energy (stabilize it) and reduce its reactivity via
+ H-bonding.
d
d
OR
H
RO H
+
d
d
-
X:
+ dd
H
H OR
+ dd
OR
A solvated anion (Nu:-) has reduced nucleophilicity,
reduced reactivity and increased stability
A solvated nucleophile has
difficulty hitting the -carbon.
10
CH3
C
: O:
N:
acetonitrile
Effect of the solvent on rate of SN2 reactions:
C
H3C
CH3
acetone
 Polar Aprotic Solvents solvate the cation counterion of the nucleophile but
not the nucleophile.
 Examples include acetonitrile (CH3CN), acetone (CH3COCH3),
dimethylformamide (DMF) [(CH3)2NC=OH], dimethyl sulfoxide, DMSO
[(CH3)2SO], hexamethylphosphoramide, HMPA {[(CH3)2N]3PO} and
dimethylacetamide (DMA).
:O :
H
..
C
: O:
CH3
N
H3C
CH3
S
..
.. ..
[(CH3)2N]3P O
..
CH3
HMPA
d
Polar aprotic solvents solvate metal cations
leaving the anion counterion (Nu: -) bare and
thus more reactive
:O:
.. _
+
:
CH3C O
.. Na
-
C
CH3
N
+
d
d
C N:
DMA
+
d
C CH3
N
..
H3C
CH3CN :
H3C
..
CH3
DMSO
DMF
:O:
:O:
Na
+
: N C CH3
- +
d
d
+
.. _
:
CH3C O
..
..
N C CH3
- +
d
d
11
Effect of the solvent on rate of SN2 reactions:
Non polar solvents (benzene, carbon tetrachloride, hexane, etc.) do not
solvate or stabilize nucleophiles.
 SN2 reactions are relatively slow in non polar solvents similar to that
in protic solvents.
Cl
Cl
Cl
CH3CH2CH2CH2CH2CH3
C
Cl
benzene
n-hexane
carbon
tetrachloride
12
1st Order Nucleophilic Substitution Reactions, i.e., SN1 reactions
CH3
H3C
3°
C
CH3
rapid
Br
+
Na+ I-
H3C
C
I
+
Na+ Br-
CH3
CH3
 3 alkyl halides are essentially inert to substitution by the SN2 mechanism
because of steric hindrance at the back side of the -carbon.
 Despite this, 3 alkyl halides do undergo nucleophilic substitution reactions
quite rapidly , but by a different mechanism, i.e., the SN1 mechanism.
 SN1 = Substitution, Nucleophilic, 1st order (unimolecular).
SN1 reactions obey 1st order kinetics, i.e., Rate = k[RX].
 The rate depends upon the concentration of only 1 reactant, the
alkyl halide-not the nucleophile
 The order of reactivity of substrates for SN1 reactions is the reverse of SN2
3
R3C-Br
>
2
R2HC-Br
>
1
RH2C-Br
>
vinyl
CH2=CH-Br
increasing rate of SN1 reactions
>
phenyl
-Br
>
Me°
H3C-Br
13
Mechanism of SN1 reactions
The mechanism of an SN1 reaction occurs in 2 steps:
CH3
3°
H3C C
CH3
..
Br
.. :
CH3
1.
- Br -
H3C
C+
CH3
CH3
rapid
2.
..
Na+ :I :..
H3C
C
..
..I :
+
Na+ Br-
CH3
3° C+
Reaction Steps …
1.
the slower, rate-limiting dissociation of the alkyl halide forming a C+
intermediate
2.
a rapid nucleophilic attack on the C+
Note that the nucleophile is not involved in the slower, rate-limiting step.
14
The Rate of SN1 reactions
The rate of an SN1 reaction depends upon 3 factors:
1. The nature of the substrate (the alkyl halide)
2. The ability of the leaving group to leave
3. The nature of the solvent
The rate is independent of the power of the nucleophile.
1. Consider the nature of the substrate:
Highly substituted alkyl halides (substrates) form a more stable C+.
increasing rate of SN1 reactions
more
stable
less
stable
CH3
CH3
H3C
C+
H3C
C+
H
H
CH3
tertiary
3º
CH3
>
secondary
2º
H
C+
H
H
>
primary
1º
C+
H
>
methyl
15
Stability of Carbocations
Alkyl groups are weak electron donors.
 They stabilize carbocations by donating electron density by induction
(through s bonds)
CH3
H3C
Inductive effects:
Alkyl groups donate (shift) electron
density through sigma bonds to
electron deficient atoms.
This stabilizes the carbocation.
C+
CH3
 They stabilize carbocations by hyperconjugation (by partial overlap of the
alkyl C-to-H bonds with the empty p-orbital of the carbocation).
vacant p orbital
of a carbocation
overlap (hyperconjugation)
sp 2
hybridized
carbocation
..
C
+
C
H
Csp 3-Hs
sigma bond
orbital
H
H
HYPERCONJUGATION
16
Stability of Carbocations
 Allyl and benzyl halides also react quickly by SN1 reactions because
their carbocations are unusually stable due to their resonance forms
which delocalize charge over an extended p system
H2C
CH
+
CH2
H2C+
HC CH2
H2C
CH
1º allyl carbocation
+
+
H2C
CHR
HC CHR
2º allyl carbocation
H
1º benzylic
H
+
C
C
H
+
+
C
H
2º benzylic
H
H
H
H
C
H
+
C
R
+
17
Relative Stability of All Types of Carbocations
Increasing C+ stability and rate of SN1 reaction
+
C H2
+
C HR
+
C R2
1º benzylic
2º benzylic
3º C +
3º benzylic
CH3
C+
CH3
3º allylic
1º C +
CH3
H
CH3
>
+
CH2 CH CR2
2º C +
+
CH2 CH CHR
2º allylic
>
CH3
C+
H
>
CH3
C+
H
H
+
>
+
CH2 CH
vinyl C
>
>
+
phenyl
H C+
H
+
m ethyl C
+
CH2 CH CH2
1º allylic
Note that 1° allylic and 1° benzylic C+’s are about as stable as 2°alkyl C+’s.
Note that 2° allylic and 2° benzylic C+’s are about as stable as 3° alkyl C+’s.
Note that 3° allylic and 3° benzlic C+’s are more stable than 3° alkyl C+’s
Note that phenyl and vinyl C+’s are unstable. Phenyl and vinyl halides do not
usually react by SN1 or SN2 reactions
18
Effect of nature of the leaving group on rate of SN1 reactions:
2. Consider the nature of the leaving group:
The nature of the leaving group has the same effect on both SN1 and SN2
reactions.
The better the leaving group, the faster a C+ can form and hence the faster will
be the SN1 reaction.
The leaving group usually has a negative charge
 Groups which best stabilize a negative charge are the best leaving groups,
i.e., the weakest bases are stable as anions and are the best leaving groups.
 Weak bases are readily identified. They have high pKb values.
pKb = 23
pKb = 22
pKb = 21
pKb = 11
pKb = -1.7
pKb = -2
pKb = -21
I-
Br -
Cl-
F-
HO-
RO-
H2N-
30,000
10,000
200
1
0
0
0
Increasing leaving ability
 Iodine (-I) is a good leaving group because iodide (I-) is non basic.
 The hydroxyl group (-OH) is a poor leaving group because hydroxide (OH-) is
a strong base.
19
Effect of the solvent on rate of SN1 reactions:
3. Consider the nature of the solvent:
 For SN1 reactions, the solvent affects the rate only if it influences the
stability of the charged transition state, i.e., the C+. The Nu:- is not
involved in the rate determining step so solvent effects on the Nu:- do not
affect the rate of SN1 reactions.
 Polar solvents, both protic and aprotic, will solvate and stabilize the
charged transition state (C+ intermediate), lowering the activation energy
and accelerating SN1 reactions.
 Nonpolar solvents do not lower the activation energy and thus make SN1
reactions relatively slower
The relative rates of an SN1 reaction due to solvent effects are given
(CH3)3C-Cl + ROH  (CH3)3C-OR + HCl
H2O
100,000
20% EtOH (aq)
14,000
40% EtOH (aq)
100
EtOH
1
reaction rate increases with polarity of solvent
20
Effect of the solvent on rate of SN1 reactions:
 Solvent polarity is usually expressed by the “dielectric constant”, , which
is a measure of the ability of a solvent to act as an electric insulator.
 Polar solvents are good electric insulators because their dipoles surround
and associate with charged species.
 Dielectric constants of some common solvents are given in the following
table
name
dielectric constant
name
aprotic solvents
dielectric constant
protic solvents
hexane
1.9
acetic acid
6.2
benzene
2.3
acetone
20.7
diethyl ether
4.3
ethanol
24.3
chloroform
4.8
methanol
33.6
HMPA
30
formic acid
58.0
DMF
38
water
80.4
DMSO
48
21
Effect of the nucleophile on rate of SN1 reactions:
Consider the nature of the Nucleophile:
 Recall again that the nature of the nucleophile has no effect on the rate of
SN1 reactions because the slowest (rate-determining) step of an SN1
reaction is the dissociation of the leaving group and formation of the
carbocation.
 All carbocations are very good electrophiles (electron acceptors) and even
weak nucleophiles, like H2O and methanol, will react quickly with them.
 The two SN1 reactions will proceed at essentially the same rate since the
only difference is the nucleophile.
CH3
CH3
H3C
H3C
3°
C
Br
+
Na+ I-
H3C
C
I
CH3
CH3
CH3
CH3
3°
C
CH3
Br
+
K+ F-
H3C
C
F
+
Na+ Br-
+
K+ Br-
CH3
22
Elimination Reactions, E1 and E2:
We have seen that alkyl halides may react with basic nucleophiles such as NaOH
via substitution reactions.
..
..
:O
H
H +
C
H
H
..
..
Br :
H
..
..O
C
H
..
Br :
..
H
..
..O
HH
H
C
H
H
..
+ : Br :
..
transition state
Also recall our study of the preparation of alkenes. When a 2° or 3° alkyl halide
is treated with a strong base such as NaOH, dehydrohalogenation occurs
producing an alkene – an elimination (E2) reaction.
KOH in ethanol
+
Br
KBr
+
H2O
-HBr
bromocyclohexane
+
KOH

cyclohexene (80 % yield)
Substitution and elimination reactions are often in competition. We shall
consider the determining factors after studying the mechanisms of elimination.
23
E2 Reaction Mechanism
There are 2 kinds of elimination reactions, E1 and E2.
E2 = Elimination, Bimolecular (2nd order). Rate = k [RX] [Nu:-]
 E2 reactions occur when a 2° or 3° alkyl halide is treated with a strong
base such as OH-, OR-, NH2-, H-, etc.
H
OH-
+
b
C
C

Br
C
C
+
Br- +
HO
H
The Nu:- removes an H+ from a b-carbon &
the halogen leaves forming an alkene.
 All strong bases, like OH-, are good nucleophiles. In 2° and 3° alkyl halides
the -carbon in the alkyl halide is hindered. In such cases, a strong base
will ‘abstract’ (remove) a hydrogen ion (H+) from a b-carbon, before it hits
the -carbon. Thus strong bases cause elimination (E2) in 2° and 3° alkyl
halides and cause substitution (SN2) in unhindered methyl° and 1° alkyl
halides.
24
E2 Reaction Mechanism
In E2 reactions, the Base to H s bond formation, the C to H s bond breaking,
the C to C p bond formation, and the C to Br s bond breaking all occur
simultaneously. No carbocation intermediate forms.
B:H
R
C
R
C
R
R
d
B
+
R
H
R
C
R
R
C
C
R
X
R
B H +
X-
R
R
Xd
+
C
Reactions in which several steps occur simultaneously are called ‘concerted’
reactions.
Zaitsev’s Rule:
Recall that in elimination of HX from alkenes, the more highly substituted
(more stable) alkene product predominates.
Br
CH3CH2O-Na+
CH3CH
CH3CH2CHCH3
EtOH
CHCH3
2-butene
major product
( > 80%)
+
CH3CH2CH
CH2
1-butene
minor product
( < 20%)
25
E2 Reactions are ‘antiperiplanar’
 E2 reactions, do not always follow Zaitsev’s rule.
 E2 eliminations occur with anti-periplanar geometry, i.e., periplanar
means that all 4 reacting atoms - H, C, C, & X - all lie in the same
plane. Anti means that H and X (the eliminated atoms) are on opposite
sides of the molecules.
 Look at the mechanism again and note the opposite side & same plane
orientation of the mechanism:
B:H
R
C
R
C
R
R
X
d
B
+
R
H
R
C
C
C
R
R
Xd
R
R
R
+
C
B H +
X-
R
26
Antiperiplanar E2 Reactions in Cyclic Alkyl Halides
 When E2 reactions occur in open chain alkyl halides, the Zaitsev
product is usually the major product. Single bonds can rotate to the
proper alignment to allow the antiperiplanar elimination.
 In cyclic structures, however, single bonds cannot rotate. We need to
be mindful of the stereochemistry in cyclic alkyl halides undergoing E2
reactions.
See the following example.
 Trans –1-chloro-2-methylcyclopentane undergoes E2 elimination with
NaOH. Draw and name the major product.
H3C
H
H
Na+ OH-
H3C
H
H
H
H
Cl
E2
H
H
Non Zaitsev product
is major product.
3-methylcyclopentene
+
NaCl
+
HOH
H3C
H
H
Little or no Zaitsev (more stable)
product is formed.
1-methylcyclopentene
27
E1 Reactions
 Just as SN2 reactions are analogous to E2 reactions, so SN1 reactions have
an analog, E1 reaction.
 E1 = Elimination, unimolecular (1st order); Rate = k  [RX]
CH3
CH3
C Br
CH3
CH3
slow
- Br-
CH3
H
rapid
C+
C
H
H
CH3
C
H
CH3
+
C
B H
+
Br
H
B:-
 E1 eliminations, like SN1 substitutions, begin with unimolecular
dissociation, but the dissociation is followed by loss of a proton from the
b-carbon (attached to the C+) rather than by substitution.
 E1 & SN1 normally occur in competition, whenever an alkyl halide is
treated in a protic solvent with a nonbasic, poor nucleophile.
 Note: The best E1 substrates are also the best SN1 substrates, and
mixtures of products are usually obtained.
28
-
E1 Reactions
 As with E2 reactions, E1 reactions also produce the more highly
substituted alkene (Zaitsev’s rule). However, unlike E2 reactions where no
C+ is produced, C+ rearrangements can occur in E1 reactions.
 e.g., t-butyl chloride + H2O (in EtOH) at 65 C  t-butanol + 2-methylpropene
CH3
CH3
H2O, EtOH
C Cl
CH3
65ºC
CH3
CH3
C OH
CH3
64%
S N1
product
H
CH3
C
+
C
H
CH3
36%
E1
product
 In most unimolecular reactions, SN1 is favored over E1, especially at low
temperature. Such reactions with mixed products are not often used in
synthetic chemistry.
 If the E1 product is desired, it is better to use a strong base and force the
E2 reaction.
 Note that increasing the strength of the nucleophile favors SN1 over E1.
Can you postulate an explanation?
29
Predicting Reaction Mechanisms
1. Non basic, good nucleophiles, like Br- and I- will cause substitution not
elimination. In 3° substrates, only SN1 is possible. In Me° and 1°
substrates, SN2 is faster. For 2° substrates, the mechanism of
substitution depends upon the solvent.
2. Strong bases, like OH- and OR-, are also good nucleophiles.
Substitution and elimination compete. In 3° and 2° alkyl halides, E2 is
faster. In 1° and Me° alkyl halides, SN2 occurs.
3. Weakly basic, weak nucleophiles, like H2O, EtOH, CH3COOH, etc.,
cannot react unless a C+ forms. This only occurs with 2° or 3°
substrates. Once the C+ forms, both SN1 and E1 occur in competition.
The substitution product is usually predominant.
4. High temperatures increase the yield of elimination product over
substitution product. (G = H –TS) Elimination produces more
products than substitution, hence creates greater entropy (disorder).
5. Polar solvents, both protic and aprotic, like H2O and CH3CN,
respectively, favor unimolecular reactions (SN1 and E1) by stabilizing
the C+ intermediate. Polar aprotic solvents enhance bimolecular
reactions (SN2 and E2) by activating the nucleophile.
30
Predicting Reaction Mechanisms
-
-
-
alkyl
halide
(substrate)
good Nu
strong base
e.g., bromide
e.g., ethoxide
strong bulky base
e.g., t-butoxide
Br
C2H5O
(CH3)3CO
Me
SN2
SN2
SN2
no reaction
1°
SN2
SN2
E2 (SN2)
no reaction
2
SN2
E2
E2
SN1, E1
3
SN1
E2
E2
SN1, E1
-
-
good Nu
-
good Nu
nonbasic
-
very poor Nu
nonbasic
e.g., acetic acid
CH3COOH
Strong bulky bases like t-butoxide are hindered. They have difficulty hitting
the -carbon in a 1° alkyl halide. As a result, they favor E2 over SN2
products.
31
Predicting Reaction Mechanisms
The nucleophiles in the table on slide 30 are extremes. Some
nucleophiles have basicity and nucleophilicity in between these extremes.
The reaction mechanisms that they will predominate can be interpolated
with good success.
Predict the predominant reaction mechanisms the following table.
alkyl
halide
(substrate)
-
v. gd. Nu………..
moderate
……….. base
fair Nu
………..
weak base
………..
e.g., cyanide
e.g., alkyl sulfide
e.g., carboxylate
CN
RS , also HS
RCOO
9
pkb = ………
v. gd. Nu………..
moderate
………. .base
-
4.7
pkb = ………
-
-
6.0 / 7.0
pkb = ………
-
alcohol
(substrate)
HI
HBr
HCl
Me
SN2
SN2
SN2
Me
SN2
1
SN2
SN2
SN2
1
SN2
2
SN2
SN2
E2
2
SN1
3
E2
E2
E2
3
SN1
HCl, HBr and HI are assumed to be in aqueous solution, a protic solvent.
32
Alkylation of Alkynides
Recall the preparation of long alkynes.
1. A terminal alkyne (pKa = 25) is deprotonated with a very strong base…
R-C  C-H
+
NaNH2 
R-C  C:- Na+ + NH3
2. An alkynide anion is a good Nu:- which can substitute (replace) halogen
atoms in methyl or 1° alkyl halides producing longer terminal alkynes 
R-C  C: - Na+ + CH3CH2-X

R-C  C-CH2CH3
+
NaX
 The reaction is straightforward with Me and 1 alkyl halides and proceeds
via an SN2 mechanism
 Alkynide anions are also strong bases (pKb = -11) as well as good Nu:-’s,
so E2 competes with SN2 for 2 and 3 alkyl halides
CH3(CH2)3CC:- Na+ + CH3-CH(Br)-CH3  CH3(CH2)3CCCH(CH3)2 (7% SN2)
+ CH3(CH2)3CCH + CH3CH=CH2 (93% E2)
33
Preparation of Alkyl Halides from Alcohols
Alkyl halides can be prepared from alcohols by reaction with HX, i.e., the
substitution of a halide on a protonated alcohol.
3º
..
(Lucas Test)
(CH
)
C
OH
33
SN1
..
+
H
+
(CH3)3C OH
.. 2
Cl
- H2O
(CH3)3C +
..
(CH3)3C Cl :
..
+
H2O
.. : Cl :
..
Rapid. 3° C+ is stabilized by protic sovent (H2O)
 OH- is a poor leaving group, i.e., is not displaced directly by nucleophiles.
Reaction in acid media protonates the OH group producing a better leaving
group (H2O). 2 and 3 alcohols react by SN1 but Me° and 1 alcohols react
by SN2.
1º
SN2
CH3CH2
..
OH
..
+
H
Cl

CH3CH2
+
OH2
.. : Cl :
..
CH3CH2Cl
+
H2O
Very slow. Protic solvent inhibits the nucleophile.
 Draw the mechanism of the reaction of isopropyl alcohol with HBr.
 What products form if concentrated H2SO4 is used in place of aq. HCl?
34
Preparation of Alkyl Halides from Alcohols
Alternative to using hydrohalic acids (HCl, HBr, HI), alcohols can be
converted to alkyl halides by reaction with PBr3 which transforms OH- into
a better leaving group allowing substitution (SN2) to occur without
rearrangement.
Br
:P
1º
SN2
CH3(CH2)4CH2
..
OH
..
Br
Br
CH3(CH2)4CH2
ether
+ PBr2
O
..
H
CH3(CH2)4CH2Br
Br
-
35
Preparation of Alkenes from Alkyl Halides
On Slide 22 we noted that 2° and 3° alkyl halides can be dehydrohalogenated
with a strong base such as OH- producing an alkene.
KOH in ethanol
+
Br
KBr
+
H2O
-HBr
bromocyclohexane
+
KOH

cyclohexene (80 % yield)
Clearly, this is an E2 reaction.
 Predict the mechanism that occurs with a Me° or 1° alkyl halide.
 Predict the products and mechanism that occur with isopentyl chloride and
KOH
36
Summary of SN /Elimination Reactions
Alkyl Halide Substrate Reactivity:
H
methyl
1º
2º
3º
H
H
CH3
CH3
C Br
H
CH3
C Br
H
unhindered substrates favor S N2
do not form a stable C +
do not react by S N1 or E1
CH3
C Br
H
CH3
C Br
CH3
hindered substrates. S N2 increasingly unfavorable, E2 is OK
form increasingly stable C +
favors SN1 and E1. E2 is OK
E2 reactions possible with strong bases
E2 reactions possible with strong bulky bases (t-butoxide)
37
Summary of SN /Elimination Reactions
Reactivity of Nucleophiles:
HS-
CN-
I-
CH3O-
HO-
NH3
Cl-
H2O
125,000
125,000
100,000
25,000
16,000
1000
700
1
good nucleophiles which are
weak bases favor SN reactions
good nucleophiles which are
also strong bases favor elimination
Note that poor nucleophiles that are also weak bases (H2O, ROH, CH3COOH,
etc.) do not undergo any reaction unless a C+ is formed first. If a C+ can form
(as with a 2º, 3º, any benzylic, or any allylic halides), then E1 and SN1
generally occur together.
Leaving Group Activity:
pKb = 23
pKb = 22
pKb = 21
pKb = 11
pKb = -1.7
pKb = -2
pKb = -21
I-
Br -
Cl-
F-
HO-
RO-
H2N-
30,000
10,000
200
1
0
0
0
good leaving groups favor both
substitution and elimination reactions
poor leaving groups make both
substitution and elimination
reactions unfavorable
38