CEB 1013 – ORGANIC CHEMISTRY
ALKYL HALIDE AND
ELIMINATION REACTION
LECTURE 5
DR. MOHD DZUL HAKIM WIRZAL
mdzulhakim.wirzal@utp.edu.my
1
COURSE OUTCOME
At the end of this course, students should be able to:
1 Draw mechanism for some fundamental reactions of organic chemistry
Identify chemical reactivity from knowledge of acid/base and
2 nucleophile/electrophile properties.
Interpret the important relationship between structure and
5 reactivity of organic molecules.
2
LEARNING OUTCOME
At the end of this lecture, students should be able to:
To correlate the type of alkyl halide (primary, secondary, or tertiary) and
1 correlate it with their reactivity toward E1 and E2 reaction.
To establish the relationship between chemical structure
2 of base and E1-E2 reaction.
To construct elimination reaction from the given reactant or target
3 molecules.
3
ELIMINATION REACTIONS
▪ All elimination reactions involve loss of elements from the starting material to
form a new π bond in the product.
▪ Alkyl halides undergo elimination reactions with Brønsted–Lowry bases. The
elements of HX are lost and an alkene is formed.
4
ELIMINATION REACTIONS (CONT.)
▪ Equations [1] and [2] illustrate examples of elimination reactions. In both
reactions a base removes the elements of an acid, HBr or HCl, from the organic
starting material.
The by-products usually
omitted from organic
reaction’ equation
5
DEHYDROHALOGENATION
▪ Removal of the elements of HX, called dehydrohalogenation, is one of the most
common methods to introduce a π bond and prepare an alkene.
▪ Dehydrohalogenation is an example of a elimination, because it involves loss of
elements from two adjacent atoms: the ` carbon bonded to the leaving group X,
and the a carbon adjacent to it.
▪ Three curved arrows illustrate how four bonds are broken or formed in the
process.
6
DEHYDROHALOGENATION (CONT.)
▪ The base (B:) removes a proton on the β carbon, thus forming H –B+.
▪ The electron pair in the β C – H bond forms the new π bond between the α and
β carbons.
▪ The electron pair in the C – X bond ends up on halogen, forming the leaving
group :X–.
7
DEHYDROHALOGENATION (CONT.)
▪ The most common bases used in elimination reactions are negatively charged
oxygen compounds such as –OH and its alkyl derivatives, –OR, called alkoxides.
The same bases used
for SN2. There will be
competition between
substitution and
elimination.
These two “big and bulky”
bases are exclusively used
for elimination
8
DEHYDROHALOGENATION (CONT.)
To draw any product of dehydrohalogenation:
▪ Find the ` carbon—the sp3 hybridized carbon bonded to the leaving group.
▪ Identify all a carbons with H atoms.
▪ Remove the elements of H and X from the α and β carbons and form a π bond.
9
CLASS ACTIVITY 5.1
1. Label the α and β carbons in each alkyl halide. Draw all possible elimination
products formed when each alkyl halide is treated with K+ –OC(CH3)3.
2. Draw all possible constitutional isomers formed by dehydrohalogenation of
each alkyl halide.
10
CLASS ACTIVITY 5.1 (CONT.)
3. Identify the alkyl halide forms each of the following alkenes as the only product
in an elimination reaction?
11
THE MECHANISM OF ELIMINATION
▪ There are two mechanisms for elimination—E2 and E1—just as there are two
mechanisms for nucleophilic substitution—SN2 and SN1.
I. The E2 mechanism (bimolecular elimination).
II. The E1 mechanism (unimolecular elimination).
▪ The E2 and E1 mechanisms differ in the timing of bond cleavage and bond
formation, analogous to the SN2 and SN1 mechanisms.
▪ In fact, E2 and SN2 reactions have some features in common, as do E1 and SN1
reactions.
12
THE E2 MECHANISM
▪ The most common mechanism for dehydrohalogenation is the E2 mechanism.
▪ For example, (CH3)3CBr reacts with –OH to form (CH3)2C–CH2 via an E2
mechanism.
RECALL
By this time, you should be
able to identify the type of
alkyl halide - 1°, 2°, or 3°.
13
THE E2 MECHANISM (CONT.)
Kinetics
1 ▪ An E2 reaction exhibits second-order kinetics; that is, the reaction is
bimolecular and both the alkyl halide and the base appear in the
rate equation.
14
THE E2 MECHANISM (CONT.)
One-Step Mechanism
2 ▪ The most straightforward explanation for the second-order kinetics is
a concerted reaction: all bonds are broken and formed in a single
step.
▪ The base –OH removes a proton from the β carbon, forming H2O (a
by-product).
▪ The electron pair in the β C – H bond forms the new π bond.
▪ The leaving group Br– comes off with the electron pair in the C– Br
bond.
15
THE E2 MECHANISM (CONT.)
The Base
3 ▪ The base appears in the rate equation, so the rate of the E2 reaction
increases as the strength of the base increases.
▪ E2 reactions are generally run with strong, negatively charged bases
like –OH and –OR.
▪ Two strong, sterically hindered nitrogen bases, called DBN and DBU,
are also sometimes used.
16
THE E2 MECHANISM (CONT.)
The Leaving Group
4 ▪ Because the bond to the leaving group is partially broken in the
transition state, the better the leaving group the faster the E2
reaction.
The Solvent
5 ▪ Polar aprotic solvents increase the rate of E2 reactions.
17
THE E2 MECHANISM IN SUMMARY
Characteristics of the E2 Mechanism.
18
E2 VERSUS SN2
The SN2 and E2 mechanisms differ in how the R group affects the reaction rate.
▪ As the number of R groups on the carbon with the leaving group increases, the
rate of the E2 reaction increases.
▪ This trend is exactly opposite to the reactivity of alkyl halides in SN2 reactions,
where increasing alkyl substitution decreases the rate of reaction.
19
E2 VERSUS SN2 (CONT.)
▪ Why does increasing alkyl substitution increase the rate of an E2 reaction?
▪ In the transition state, the double bond is partially formed, so increasing the
stability of the double bond with alkyl substituents stabilizes the transition state
(i.e., it lowers Ea), which increases the rate of the reaction.
▪ Increasing the number of R groups on the carbon with the leaving group forms
more highly substituted, more stable alkenes in E2 reactions.
20
E2 VERSUS SN2 (CONT.)
▪ For example, the E2 reaction of a 1° alkyl halide (1-bromobutane) forms a
monosubstituted alkene, whereas the E2 reaction of a 3° alkyl halide (2-bromo-
2-methylpropane) forms a disubstituted alkene.
▪ The disubstituted alkene is more stable, so the 3° alkyl halide reacts faster than
the 1° alkyl halide.
21
THE ZAITSEV RULE
▪ The Zaitsev rule - The major product in a elimination has the more substituted
double bond.
▪ For example, elimination of the elements of H and I from 1-iodo-1-
methylcyclohexane yields two constitutional isomers: the trisubstituted alkene
A (the major product) and the disubstituted alkene B (the minor product).
22
THE ZAITSEV RULE (CONT.)
▪ More example, E2 elimination of HBr from 2-bromo-2-methylbutane yields
alkenes C and D. D, having the more substituted double bond, is the major
product
23
THE ZAITSEV RULE (CONT.)
▪ When a mixture of stereoisomers is possible from dehydrohalogenation, the
major product is the more stable stereoisomer.
▪ For example, dehydrohalogenation of alkyl halide X forms a mixture of trans
and cis alkenes, Y and Z. The trans alkene Y is the major product because it is
most stable.
24
THE ZAITSEV RULE (CONT.)
Example of test question
▪ Predict the major product in the following E2 reaction.
25
THE ZAITSEV RULE (CONT.)
Solution
▪ The alkyl halide has two different β C atoms (labeled β1 and β2), so two
different alkenes are possible: one formed by removal of HCl across the α and
β1 carbons, and one formed by removal of HCl across the α and β2 carbons.
▪ Using the Zaitsev rule, the major product should be A, because it has the more
substituted double bond.
26
CLASS ACTIVITY 5.2
1. Rank the alkyl halides in each group in order of increasing reactivity in an E2
reaction.
2. Draw all constitutional isomers formed in each E2 reaction and predict the
major product using the Zaitsev rule
27
THE E1 MECHANISM
▪ The dehydrohalogenation of (CH3)3CI with H2O to form (CH3)2C–CH2 can be used
to illustrate the second general mechanism of elimination, the E1 mechanism.
RECALL
By this time, you should be
able to identify the type of
alkyl halide - 1°, 2°, or 3°.
28
THE E1 MECHANISM (CONT.)
Kinetics
1 ▪ An E1 reaction exhibits first-order kinetics.
▪ Like the SN1 mechanism, the kinetics suggest that the reaction
mechanism involves more than one step, and that the slow step is
unimolecular, involving only the alkyl halide.
29
THE E1 MECHANISM (CONT.)
A Two-Steps Mechanism
2 ▪ The most straightforward explanation for the observed first-order
kinetics is a two-step reaction: the bond to the leaving group breaks
first before the π bond is formed.
▪ Heterolysis of the C – I bond forms an intermediate carbocation.
This is the same fi rst step as the SN1 mechanism. It is responsible for
the first-order kinetics because it is rate-determining.
30
THE E1 MECHANISM (CONT.)
A Two-Steps Mechanism
2 ▪ The most straightforward explanation for the observed first-order
kinetics is a two-step reaction: the bond to the leaving group breaks
first before the π bond is formed.
▪ A base (such as H2O) removes a proton from a carbon adjacent to
the carbocation (a β carbon). The electron pair in the C – H bond is
used to form the new π bond.
31
THE E1 MECHANISM (CONT.)
A Two-Steps Mechanism
2 ▪ The most straightforward explanation for the observed first-order
kinetics is a two-step reaction: the bond to the leaving group breaks
first before the π bond is formed.
▪ A base (such as H2O) removes a proton from a carbon adjacent to
the carbocation (a β carbon). The electron pair in the C – H bond is
used to form the new π bond.
32
THE E1 MECHANISM (CONT.)
The Carbocation
3 ▪ The rate of an E1 reaction increases as the number of R groups on
the carbon with the leaving group increases.
33
THE E1 MECHANISM (CONT.)
The Base
4 ▪ Because the base does not appear in the rate equation, weak bases
favour E1 reactions.
▪ The strength of the base usually determines whether a reaction
follows the E1 or E2 mechanism.
▪ Strong bases like –OH and –OR favour E2 reactions, whereas weaker
bases like H2O and ROH favour E1 reactions.
34
THE E1 MECHANISM (CONT.)
The Zaitsev Rule
5 ▪ The Zaitsev rule applies to E1 reactions, too.
▪ For example, E1 elimination of HBr from 1-bromo-1-
methylcyclopentane yields alkenes A and B. A, having the more
substituted double bond, is the major product.
35
THE E1 MECHANISM IN SUMMARY
Characteristics of the E1 Mechanism
36
E1 VERSUS SN1
▪ SN1 and E1 reactions have exactly the same first step—formation of a
carbocation. They differ in what happens to the carbocation.
▪ The same conditions that favor substitution by an SN1 mechanism also favor
elimination by an E1 mechanism.
▪ As a result, both reactions usually occur in the same reaction mixture to afford
a mixture of products.
▪ Because E1 reactions often occur with a competing SN1 reaction, E1 reactions
of alkyl halides are much less useful than E2 reactions.
37
E1 VERSUS SN1 (CONT.)
▪ For example, draw the SN1 and E1 products formed in the reaction of (CH3)3CBr
with H2O.
38
E1 VERSUS SN1 (CONT.)
The second step - reaction of the carbocation with H2O as a nucleophile affords
the substitution product (Reaction [1]). Alternatively, H2O acts as a base to
remove a proton, affording the elimination product (Reaction [2]). Two products
are formed.
39
E1 VERSUS E2
Given a particular starting material and base, how do we know whether a reaction
occurs by the E1 or E2 mechanism?
▪ The strength of the base is the most important factor in determining the
mechanism for elimination.
▪ Strong bases favour the E2 mechanism. Weak bases favour the E1 mechanism.
40
CLASS ACTIVITY 5.3
1. Draw the major product formed from each alkyl halide in an E1 reaction?
2. Draw both the SN1 and E1 products of each reaction.
41
CLASS ACTIVITY 5.3
3. Draw all constitutional isomers formed in each elimination reaction. Label the
mechanism as E2 or E1.
42
43
WHEN IS THE REACTION SN1, SN2, E1, AND E2
Unfortunately, there is no easy answer, and often mixtures of products result.
▪ Two generalizations help to determine whether substitution or elimination
occurs.
1. Classify the alkyl halide as 1°, 2°, or 3°.
2. Classify the base or nucleophile as strong, weak, or bulky.
44
WHEN IS THE REACTION SN1, SN2, E1, AND E2
1 3° Alkyl halides (R3CX react by all mechanisms except SN2.)
▪ With strong base
45
WHEN IS THE REACTION SN1, SN2, E1, AND E2
1 3° Alkyl halides (R3CX react by all mechanisms except SN2.)
▪ With weak base
46
WHEN IS THE REACTION SN1, SN2, E1, AND E2
2 1° Alkyl halides (RCH2X react by SN2 and E2 mechanisms.)
▪ With strong and small base
47
WHEN IS THE REACTION SN1, SN2, E1, AND E2
2 1° Alkyl halides (RCH2X react by SN2 and E2 mechanisms.)
▪ With strong but bulky base
48
WHEN IS THE REACTION SN1, SN2, E1, AND E2
3 2° Alkyl halides (R2CHX react by all mechanisms.).
▪ With strong base and small base.
49
WHEN IS THE REACTION SN1, SN2, E1, AND E2
3 2° Alkyl halides (R2CHX react by all mechanisms.).
▪ With strong but bulky base.
50
WHEN IS THE REACTION SN1, SN2, E1, AND E2
3 2° Alkyl halides (R2CHX react by all mechanisms.).
▪ With weak base.
51
WHEN IS THE REACTION SN1, SN2, E1, AND E2
Summary Chart
52
WHEN IS THE REACTION SN1, SN2, E1, AND E2
▪ Example of test question - draw the products of the following reaction, and
include the mechanism showing how each product is formed.
53
WHEN IS THE REACTION SN1, SN2, E1, AND E2
Solution
1. Classify the halide as 1°, 2°, or 3° and the reagent as a strong or weak base (and
nucleophile) to determine the mechanism. In this case, the alkyl halide is 3°
and the reagent (CH3OH) is a weak base and nucleophile, so products of both
SN1 and E1 mechanisms are formed.
2. Draw the steps of the mechanisms to give the products. Both mechanisms
begin with the same first step: loss of the leaving group to form a carbocation.
54
WHEN IS THE REACTION SN1, SN2, E1, AND E2
a. For SN1: The carbocation reacts with a nucleophile. Nucleophilic attack of
CH3OH on the carbocation generates a positively charged intermediate that
loses a proton to afford the neutral SN1 product.
b. For E1: The carbocation reacts with a base (CH3OH or Br–). Two different
products of elimination can form because the carbocation has two different β
carbons.
55
WHEN IS THE REACTION SN1, SN2, E1, AND E2
b. For E1: The carbocation reacts with a base (CH3OH or Br–). Two different
products of elimination can form because the carbocation has two different β
carbons.
Three products are formed: one from an SN1 reaction and two from E1 reactions.
56
CLASS ACTIVITY 5.4
1. Draw the products in each reaction.
2. Draw a stepwise mechanism for the following reaction.
57
SELF ASSESSMENT
SCORE
Class Activity LO1 LO2 LO3
1 2 3 4 5
5.1 1 x
2 x
3 x
5.2 1 x
2 x x
5.3 1 x
2 x x
3 x x
5.4 1 x
2 x
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