Transcript 32.6

Chapter 32
Halogeno-compounds
32.1
32.2
32.3
32.4
32.5
32.6
32.7
32.8
1
Introduction
Nomenclature of Halogeno-compounds
Physical Properties of Halogeno-compounds
Preparation of Halogeno-compounds
Reactions of Halogeno-compounds
Nucleophilic Substitution Reactions
Elimination Reactions
Uses of Halogeno-compounds
New Way Chemistry for Hong Kong A-Level Book 3A
32.1 Introduction (SB p.169)
• Haloalkanes are organic compounds having one or
more halogen atoms replacing hydrogen atoms in
alkanes
• Haloalkanes are classified into primary, secondary and
tertiary, based on the number of alkyl groups attached to
the carbon atom which is bonded to the halogen atom
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32.1 Introduction (SB p.169)
Halobenzenes are organic compounds in which the halogen
atom is directly attached to a benzene ring
e.g.
 not a halobenzene, because the
chlorine atom is not directly attached to
the benzene ring
3
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32.2 Nomenclature of Halogeno-compounds (SB p.170)
• Naming haloalkanes are similar to those for naming
alkanes
• The halogens are written as prefixes: fluoro- (F), chloro(Cl), bromo- (Br) and iodo- (I)
e.g.
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32.2 Nomenclature of Halogeno-compounds (SB p.170)
When the parent chain has both a halogen and an alkyl
substituent, the chain is numbered from the end nearer the first
substituent regardless of what substituents are
e.g.
5
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32.2 Nomenclature of Halogeno-compounds (SB p.171)
In case of halobenzenes, the benzene ring is numbered so as
to give the lowest possible numbers to the substituents
e.g.
6
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32.2 Nomenclature of Halogeno-compounds (SB p.171)
Example 32-1
Draw the structural formulae and give the IUPAC names
of all isomers with the following molecular formula.
(a) C4H9Br
Solution:
Answer
(a)
7
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32.2 Nomenclature of Halogeno-compounds (SB p.171)
Example 32-1
Solution:
Draw the structural formulae and give the IUPAC names
(b)
of all isomers with the following molecular formula.
(b) C4H8Br2
Answer
8
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32.2 Nomenclature of Halogeno-compounds (SB p.172)
Check Point 32-1
Draw the structural formulae and give the IUPAC names for
all the structural isomers of C5H11Br.
Answer
9
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32.2 Nomenclature of Halogeno-compounds (SB p.172)
10
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32.3 Physical Properties of Halogeno-compounds (SB p.173)
Name
Formula
Chloro-derivatives:
Chloromethane
Chloroethane
1-Chloropropane
1-Chlorobutane
1-Chloropentane
1-Chlorohexane
(Chloromethyl)benzene
Chlorobenzene
CH3Cl
CH3CH2Cl
CH3(CH2)2Cl
CH3(CH2)3Cl
CH3(CH2)4Cl
CH3(CH2)5Cl
C6H5CH2Cl
C6H5Cl
11
Melting
Boiling
Density at
point (°C) point (°C) 20°C (g cm–3)
–97.7
–136
–123
–123
–99
–83
–39
–45.2
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–23.8
12.5
46.6
78.5
108
133
179
132
—
—
0.889
0.886
0.883
0.878
1.100
1.106
32.3 Physical Properties of Halogeno-compounds (SB p.173)
Name
Formula
Bromo-derivatives:
Bromomethane
Bromoethane
1-Bromopropane
1-Bromobutane
1-Bromopentane
1-Bromohexane
(Bromomethyl)benzene
Bromobenzene
CH3Br
CH3CH2Br
CH3(CH2)2Br
CH3(CH2)3Br
CH3(CH2)4Br
CH3(CH2)5Br
C6H5CH2Br
C6H5Br
12
Melting Boiling
Density at
point
point
20°C (g cm–3)
(°C)
(°C)
–93.7
–119
–109
–113
–95
–85
–3.9
–30.6
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3.6
38.4
70.8
101
129
156
201
156
—
1.460
1.354
1.279
1.218
1.176
1.438
1.494
32.3 Physical Properties of Halogeno-compounds (SB p.173)
Name
Formula
Iodo-derivatives:
Iodomethane
Iodoethane
1-Iodopropane
1-Iodobutane
1-Iodopentane
1-Iodohexane
(Iodomethyl)benzene
CH3I
CH3CH2I
CH3(CH2)2I
CH3(CH2)3I
CH3(CH2)4I
CH3(CH2)5I
C6H5CH2I
13
Melting
Boiling
Density at
point (°C) point (°C) 20°C (g cm–3)
–66.5
–108
–101
–103
–85.6
—
24.5
42.5
72.4
102
130
155
181
decompose
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2.279
1.940
1.745
1.617
1.517
1.437
1.734
32.3 Physical Properties of Halogeno-compounds (SB p.174)
Boiling Point and Melting Point
14
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32.3 Physical Properties of Halogeno-compounds (SB p.174)
Haloalkanes have higher b.p. and m.p. than alkanes
∵ dipole-dipole interactions are present between
haloalkane molecules
m.p. and b.p. increase in the order:
RCH2F < RCH2Cl < RCH2Br < RCH2I
∵ larger, more polarizable halogen atoms increase the
dipole-dipole interactions between the molecules
No. of carbon   m.p. and b.p. 
15
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32.3 Physical Properties of Halogeno-compounds (SB p.174)
Density
•
Relative molecular mass 
 density 
∵
closer packing of the smaller molecules in the
liquid phase
•
Bromo and iodoalkanes are all denser than water at
20°C
16
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32.3 Physical Properties of Halogeno-compounds (SB p.174)
Solubility
Although C — X bond is polar, it is not polar enough to
have a significant effect on the solubility of haloalkanes
and halobenzenes
 Immiscible with water
 Soluble in organic solvents
17
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32.4 Preparation of Halogeno-compounds (SB p.175)
Preparation of Haloalkanes
Substitution of Alcohols
• Prepared by substituting –OH group of alcohols
with halogen atoms
• Common reagents used: HCl, HBr, HI, PCl3 or PBr3
• The ease of substitution of alcohols:
3° alcohol > 2° alcohol > 1° alcohol > CH3OH
• This is related to the stability of the reaction
intermediate (i.e. stability of carbocations)
18
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32.4 Preparation of Halogeno-compounds (SB p.175)
Reaction with Hydrogen Halides
• Dry HCl is bubbled through alcohols in the presence
of ZnCl2 catalyst
• For the preparation of bromo- and iodoalkanes, no
catalyst is required
19
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32.4 Preparation of Halogeno-compounds (SB p.176)
• The reactivity of hydrogen halides: HI > HBr > HCl
• e.g.
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32.4 Preparation of Halogeno-compounds (SB p.176)
Reaction with Phosphorus Halides
Haloalkanes can be prepared from the vigorous reaction
between cold alcohols and phosphorus(III) halides
21
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32.4 Preparation of Halogeno-compounds (SB p.177)
Addition of Alkenes and Alkynes
Addition of halogens or hydrogen halides to an alkene or
alkyne can form a haloalkane
e.g.
22
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32.4 Preparation of Halogeno-compounds (SB p.177)
Preparation of Halobenzenes
Halogenation of Benzene
Benzene reacts readily with chlorine and bromine in the
presence of catalysts (e.g. FeCl3, FeBr3, AlCl3)
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32.4 Preparation of Halogeno-compounds (SB p.177)
From Benzenediazonium Salts
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32.4 Preparation of Halogeno-compounds (SB p.178)
Check Point 32-2
State the major products of the following reactions:
(a) CH3CHOHCH2CH3 + PBr3 
(b) CH3CH = CH2 + HBr 
(c) CH3C  CH + 2HBr 
(d)
(a) CH3CHBrCH2CH3
(b) CH3CHBrCH3
(c) CH3CBr2CH3
(d)
25
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Answer
32.5 Reactions of Halogeno-compounds (SB p.178)
• Carbon-halogen bond is polar
• Carbon atom bears a partial positive charge
• Halogen atom bears a partial negative charge
26
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32.5 Reactions of Halogeno-compounds (SB p.178)
• Characteristic reaction:
Nucleophilic substitution reaction
• Alcohols, ethers, esters, nitriles and amines can be
formed by substituting – OH, – OR, RCOO –, – CN
and – NH2 groups respectively
27
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32.5 Reactions of Halogeno-compounds (SB p.179)
• Another characteristic reaction:
Elimination reaction
Haloalkane
Base
Alkene
• Bases and nucleophiles are the same kind of reagents
• Nucleophilic substitution and elimination reactions
always occur together and compete each other
28
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32.6 Nucleophilic Substitution Reactions (SB p.179)
Reaction with Sodium Hydroxide
The reactions proceed in 2 different reaction mechanisms:
bimolecular nucleophilic substitution (SN2)
unimolecular nucleophilic substitution (SN1)
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32.6 Nucleophilic Substitution Reactions (SB p.180)
Bimolecular Nucleophilic Substitution (SN2)
Example: CH3 – Cl + OH–  CH3OH + Cl–
Experiment
number
1
2
3
4
Initial
[CH3Cl]
(mol dm–
3)
0.001
0.002
0.001
0.002
Initial
Initial rate
[OH–]
(mol dm–3 s–1)
–3
(mol dm )
1.0
1.0
2.0
2.0
4.9  10–7
9.8  10–7
9.8  10–7
19.6  10–7
Results of kinetic study of reaction of CH3Cl with OH–
Rate = k[CH3Cl][OH–]
Order of reaction = 2
 both species are involved in rate determining step
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32.6 Nucleophilic Substitution Reactions (SB p.181)
Reaction mechanism of the SN2 reaction:
• The nucleophile attacks from the backside of the
electropositive carbon centre
• In the transition state, the bond between C and O is
partially formed, while the bond between C and Cl
is partially broken
31
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32.6 Nucleophilic Substitution Reactions (SB p.181)
Transition state involve both
the nucleophile and substrate
 second order kinetics of the
reaction
Energy profile of the reaction of CH3Cl and OHby SN2 mechanism
32
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32.6 Nucleophilic Substitution Reactions (SB p.182)
Stereochemistry of SN2 Reactions
• The nucleophile attacks from the backside of the
electropositive carbon centre
• The configuration of the carbon atom under attack
inverts
33
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32.6 Nucleophilic Substitution Reactions (SB p.182)
Unimolecular Nucleophilic Substitution (SN1)
Example:
• Kinetic study shows that:
Rate = k[(CH3)3CCl]
• The rate is independent of [OH–]
• Order of reaction = 1
 only 1 species is involved in the rate
determining step
34
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32.6 Nucleophilic Substitution Reactions (SB p.183)
Reaction mechanism of SN1 reaction involves 2 steps
and 1 intermediate formed
Step 1:
• Slowest step (i.e. rate determining step)
• Formation of carbocation and halide ion
35
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32.6 Nucleophilic Substitution Reactions (SB p.183)
Step 2:
• Fast step
• Attacked by a nucleophile to form the product
36
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32.6 Nucleophilic Substitution Reactions (SB p.183)
•
Rate determining step involves the
breaking of the C – Cl bond to
form carbocation
•
Only 1 molecule is involved in the
rate determining step
 first order kinetics of the
reaction
Energy profile of the
reaction of (CH3)3CCl
and OH- by SN1
mechanism
37
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32.6 Nucleophilic Substitution Reactions (SB p.184)
Stereochemistry of SN1 Reactions
• The carbocation formed has a trigonal planar structure
• The nucleophile may either attack from the frontside
or the backside
38
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32.6 Nucleophilic Substitution Reactions (SB p.184)
For some cations, different products may be formed by
either mode of attack
e.g.
The reaction is called racemization
39
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32.6 Nucleophilic Substitution Reactions (SB p.184)
The above SN1 reaction leads to racemization
∵ formation of trigonal planar carbocation intermediate
40
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32.6 Nucleophilic Substitution Reactions (SB p.185)
The attack of the nucleophile from either side of the planar
carbocation occurs at equal rates and results in the formation
of the enantiomers of butan-2-ol in equal amounts
41
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32.6 Nucleophilic Substitution Reactions (SB p.185)
Factors Affecting the Rates of SN1 and SN2 Reactions
Most important factors affecting the relative rates of SN1
and SN2 reactions:
42
1.
The structure of the substrate
2.
The concentration and strength of the nucleophile
(for SN2 reactions only)
3.
The nature of the leaving group
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32.6 Nucleophilic Substitution Reactions (SB p.186)
The Structure of the Substrate
1. SN2 reactions
• The reactivity of haloalkanes in SN2 reactions:
CH3X > 1° haloalkane > 2° haloalkane > 3° haloalkane
• Steric hindrance affects the reactivity
∵
bulky alkyl groups will inhibit the approach of
nucleophile to the electropositive carbon centre
 energy of transition state  activation energy 
43
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32.6 Nucleophilic Substitution Reactions (SB p.186)
Steric effects in the SN2 reaction
44
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32.6 Nucleophilic Substitution Reactions (SB p.187)
2. SN1 reactions
• Critical factor: the relative stability of the carbocation
formed
• Tertiary carbocations are the most stable
∵ 3 electron-releasing alkyl groups stabilize the
carbocation by releasing electrons
• Methyl, 1°, 2° carbocation have much higher energy
 activation energies for SN1 reactions are very large and
rate of reaction become very small
45
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32.6 Nucleophilic Substitution Reactions (SB p.187)
The Concentration and Strength of the Nucleophile
• Only affect SN2 reactions
• Concentration of nucleophile   rate 
46
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32.6 Nucleophilic Substitution Reactions (SB p.187)
•
Relative strength of nucleophiles can be correlated
with two structural features:
(I) A negatively charged nucleophile (e.g. OH–) is
always a stronger nucleophile than a neutral
nucleophile (e.g. H2O)
(II)In a group of nucleophiles in which the
nucleophilic atom is the same, the order of
nucleophilicity roughly follows the order of
basicity:
e.g. RO– > OH– >> ROH > H2O
•
47
Strength   rate 
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32.6 Nucleophilic Substitution Reactions (SB p.188)
The Nature of Leaving Group
• Halide ion departs as a leaving group
• For the halide ion, the ease of leaving:
I– > Br– > Cl– > F–
• This is in agreement with the order of bond enthalpies
of carbon-halogen bonds
48
Bond
Bond enthalpy (kJ mol–1)
C–F
+484
C – Cl
+338
C – Br
+276
C–I
+238
C – I bond is
weakest  I– is
the best leaving
group
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32.6 Nucleophilic Substitution Reactions (SB p.188)
• Uncharged or neutral compounds are better
leaving groups
e.g.
The ease of leaving of oxygen compounds:
H2O >> OH– > RO–
• Strongly basic ions rarely act as leaving group
e.g.
49
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32.6 Nucleophilic Substitution Reactions (SB p.188)
When an alcohol is dissolved in a strong acid, it can
react with a halide ion
∵
the acid protonates the –OH group, and the
leaving group becomes a neutral water molecule
e.g.
50
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32.6 Nucleophilic Substitution Reactions (SB p.188)
Comparision of Rates of Hydrolysis of
Haloalkanes and Halobenzene
1. Experiment 1 : Comparison of the rates of hydrolysis
of 1-chlorobutane, 1-bromobutane and 1-iodobutane
(a) Objective
To study the effect of the nature of the halogen leaving
group on the rate of hydrolysis of haloalkanes
51
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32.6 Nucleophilic Substitution Reactions (SB p.189)
(b) Procedure
52
•
Put 2 cm3 of ethanol and 1 cm3 of 0.1 M aqueous silver nitrate into
each of three test tubes
•
Place them in a water bath at 60°C
•
After 5 mins, add 5 drops of 1-chlorobutane the test tube A, 5 drops of
1-bromobutane to B and 5 drops of 1-iodobutane to C
•
Shake each test tube and observe for 10 mins
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32.6 Nucleophilic Substitution Reactions (SB p.189)
(c) Result and Observation
A precipitate of silver halide is formed in each of the
three test tubes
Test tube A
AgCl(s)
53
Test tube B
AgBr(s)
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Test tube C
AgI(s)
32.6 Nucleophilic Substitution Reactions (SB p.190)
(d) Discussion
• Water molecule is the nucleophile of the reaction
• Haloalkanes react with water by nucleophilic substitutions
• The halide ion departs as the leaving group
• The ease of leaving of halide ions decreases: I– > Br– > Cl–
• The order of precipitates appeared tends to follow the order
of ease of leaving of the halide ions, which subsequently
form precipitates with Ag+ ions from AgNO3
Ag+(aq) + X–(aq)  AgX(s)
54
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32.6 Nucleophilic Substitution Reactions (SB p.190)
2. Experiment 2: Comparison of the rates of hydrolysis of
primary, secondary and tertiary haloalkanes and
halobenzene
(a) Objective
To study the effect of the structure of haloalkanes on
the rate of hydrolysis of them and to compare the
rates of hydrolysis of haloalkanes and halobenzene
55
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32.6 Nucleophilic Substitution Reactions (SB p.190)
(b) Procedure
56
•
Put 2 cm3 of ethanol and 1 cm3 of 0.1 M aqueous silver nitrate into
each of four test tubes
•
Add 5 drops of 1-chlorobutane the test tube D, 5 drops of
2-chlorobutane to E, 5 drops of 2-chloro-2-methylpropane to F and
5 drops of chlorobenzene to G
•
Shake each test tube well and observe for 10 mins
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32.6 Nucleophilic Substitution Reactions (SB p.190)
(c) Result and Observation
Except test tube G, a white precipitate of silver chloride
was formed in each of test tubes D, E and F.
Test tube D
57
Test tube E
Test tube F
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Test tube G
32.6 Nucleophilic Substitution Reactions (SB p.191)
(d) Discussion
• The halogen-compounds used in the experiment are of
different classes:
• The rate of formation of the white precipitate of silver
chloride decreases in the order:
2-chloro-2-methylpropane > 2-chlorobutane 
chlorobutane >> chlorobenzene
58
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1-
32.6 Nucleophilic Substitution Reactions (SB p.191)
• The rate of hydrolysis of halogeno-compounds is related to
the structure of the substrate around the carbon which is
being attacked
•
The experimental condition favours SN1 reactions
∴ tertiary haloalkane reacts at the fastest rate while
primary haloalkane proceeds at a slower rate
•
59
Chlorobenzene can be hydrolyzed to phenol under severe
conditions (cannot be carried out in school laboratory)
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32.6 Nucleophilic Substitution Reactions (SB p.192)
Unreactivity of Halobenzene
• Halobenzenes are comparatively unreactive to
nucleophilic substitution reactions
∵ the p orbital on the carbon atom of the benzene ring
and that on the halogen atom overlap side-by-side to
form a delocalized  bonding system
60
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32.6 Nucleophilic Substitution Reactions (SB p.192)
∵ Delocalization of  electrons throughout the ring and
halogen atom
•
∴ The C – X bond has partial double bond character
 stronger than that of haloalkane
 larger amount of energy is required to break the
bond
 substitution reactions become more difficult to
occur
•
61
∵ Delocalization of  electrons makes the polarity of
C – X bond 
 electropositive carbon center is less
susceptible to nucleophilic attack
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32.6 Nucleophilic Substitution Reactions (SB p.192)
•
Delocalized electrons repel any approaching
nucleophiles
 unreactive towards SN2 reactions
•
Benzene cations are highly unstable because of loss
of aromaticity
 unreactive towards SN1 reactions
62
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32.6 Nucleophilic Substitution Reactions (SB p.192)
Example 32-2
The reactions between three bromine-containing
compounds and aqueous silver nitrate at room conditions
are summarized in the following table:
Compound
Reaction with aqueous silver nitrate
Sodium bromide
A pale yellow precipitate appears
immediately
1-Bromobutane
No reaction at first; a pale yellow
precipitate appears after several
minutes
Bromobenzene
No reaction even after several hours
Solution:
(a) What is
the pale yellow precipitate produced in the
Answer
reaction
silver nitrate and sodium bromide?
(a)between
Silver bromide
63
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32.6 Nucleophilic Substitution Reactions (SB p.192)
Example 32-2
(b) Write an ionic equation for the reaction.
Answer
Solution:
(b) Ag+(aq) + Br–(aq)  AgBr(s)
64
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32.6 Nucleophilic Substitution Reactions (SB p.192)
Example 32-2
(c) Why does silver nitrate produce no immediate
precipitate with 1-bromobutane, even though it
contains bromine? Why is there the formation of the
pale yellow precipitate after several minutes?
Answer
Solution:
(c) The hydrolysis of 1-bromobutane takes time.
Precipitation of AgBr occurs only after OH– from
water has replaced Br– from 1-bromobutane.
65
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32.6 Nucleophilic Substitution Reactions (SB p.192)
Example 32-2
(d) Briefly explain why bromobenzene does not give any
precipitate with aqueous silver nitrate.
Answer
Solution:
(d) The C – Br bond of bromobenzene is strengthened
due to the delocalization of  electrons throughout the
benzene ring and the halogen atom. As the breaking of
the C – Br bond of bromobenzene requires a larger
amount of energy than 1-bromobutane, the substitution
reaction becomes more difficult to occur. Thus,
bromobenzene does not give any precipitate with
aqueous silver nitrate.
66
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32.6 Nucleophilic Substitution Reactions (SB p.193)
Example 32-3
Which is the stronger nucleophile in each of the following
pairs? Explain your choice briefly.
(a) OH– and H2O
(b) OH– and CH3CH2O–
Answer
Solution:
(a) OH– is a stronger nucleophile than H2O because it
carries a negative charge while H2O is electrically neutral.
(b) CH3CH2O– is a stronger nucleophile than OH–. It is
because the ethyl group (CH3CH2–) is an electronreleasing group, this increases the electron density on
the oxygen atom. This makes CH3CH2O– to be a stronger
nucleophile than OH–.
67
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32.6 Nucleophilic Substitution Reactions (SB p.194)
Check Point 32-3
Predict whether the following substitution reaction follows
mainly SN1 or SN2 pathway. Briefly explain your answer.
(a) CH3I + OH–  CH3OH + I–
Answer
(a) The reaction follows mainly the SN2 mechanism
because
of the following reasons. The haloalkane (CH3I) is a methyl
halide. There is little steric hindrance for the nucleophile to
attack the carbon atom of the molecule. On the other hand, if
the reaction follows the SN1 mechanism, the carbocation
(CH3+) formed is not stabilized by the inductive effects of
alkyl groups. Thus the SN1 mechanism for this reaction is
unfavourable.
68
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32.6 Nucleophilic Substitution Reactions (SB p.194)
Check Point 32-3
Predict whether the following substitution reaction follow
mainly SN1 or SN2 pathway. Briefly explain your answer.
(b)
69
(b) The reaction follows mainly the SN1 mechanism. It is
because the haloalkane is a secondary haloalkane with a
bulky phenyl group attached directly to the carbon atom
bearing the halogen atom. The bulky phenyl group exerts a
dramatic steric hindrance to the approaching nucleophile.
Therefore, the SN2 mechanism for this reaction is not
favoured. On the other hand, the carbocation formed inAnswer
the
SN1 reaction is stabilized by both the inductive effect of the
electron-releasing ethyl group and the resonance
effect of the phenyl group.
New Way Chemistry for Hong Kong A-Level Book 3A
32.6 Nucleophilic Substitution Reactions (SB p.194)
Reaction with Potassium Cyanide
A nitrile is formed when a haloalkane is heated under reflux
with an aqueous alcoholic solution of potassium cyanide
e.g.
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New Way Chemistry for Hong Kong A-Level Book 3A
32.6 Nucleophilic Substitution Reactions (SB p.194)
• Cyanide ion (CN–) acts as a nucleophile
• Halobenzenes do not react with potassium cyanide
• The reaction is very useful because the nitrile can be
hydrolyzed to carboxylic acids which can be reduced
to alcohols
•
71
A useful way of introducing a carbon atom into an
organic molecule, so that the length of the carbon
chain can be increased
New Way Chemistry for Hong Kong A-Level Book 3A
32.6 Nucleophilic Substitution Reactions (SB p.195)
Reaction with Ammonia
When a haloalkane is heated with an aqueous alcoholic solution
of ammonia under a high pressure, an amine is formed
e.g.
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New Way Chemistry for Hong Kong A-Level Book 3A
32.6 Nucleophilic Substitution Reactions (SB p.195)
• Ammonia is a nucleophile because the presence of a
lone pair of electrons on the nitrogen atom
• As the lone pair electrons on nitrogen atom in ethylamine
are still available, the ethylamine will compete with
ammonia as the nucleophile.
• A series of further substitutions take place
• A mixture of products is formed
73
New Way Chemistry for Hong Kong A-Level Book 3A
32.6 Nucleophilic Substitution Reactions (SB p.195)
• The reaction stops at the formation of a quaternary
ammonium salt
• The competing reactions can be minimized by using an
excess of ammonia
74
New Way Chemistry for Hong Kong A-Level Book 3A
32.6 Nucleophilic Substitution Reactions (SB p.195)
Example 32-4
Give the reagents and reaction conditions needed for each
of the following conversions:
(a) (CH3)3CBr  (CH3)3COH
(b) CH3I  CH3OC2H5
(c) CH3I  (CH3)4N+I–
Solution:
(a) Dilute NaOH
(b) C2H5O–Na+ or Na in C2H5OH
(c) NH3 in excess CH3I
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New Way Chemistry for Hong Kong A-Level Book 3A
Answer
32.6 Nucleophilic Substitution Reactions (SB p.196)
Check Point 32-4
Give the name(s) and structural formula(e) of the major
organic product(s) formed in each of the following reactions.
(a)
(a)
(b)
(b) CH3NH2
(c)
Methylamine
(c)
Answer
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.196)
Formation of Alkenes
The elimination of HX from adjacent atoms of a haloalkane
is widely used for synthesizing alkenes
e.g.
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.196)
The elements of a hydrogen halide are eliminated from a
haloalkane in this way, the reaction is called
dehydrohalogenation
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.196)
Dehydrohalogenation of most haloalkanes yields more than
one product
e.g.
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.197)
• The major product will be the more stable alkene
• The more stable alkene has the more highly substituted
double bond
• Elimination follows the Saytzeff’s rule when the elimination
occurs to give the more highly substituted alkene as the
major product
• The stabilities of alkenes:
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.197)
Elimination Versus Substitution
• Nucleophiles are potential bases
• Bases are potential nucleophiles
• In SN2 pathway, elimination and nucleophilic
substitution compete each other
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.198)
• Substitution is favoured when the substrate is primary
alcohol and the base is hydroxide ion
• Elimination is favoured when the substrate is secondary
alcohol
82
New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.198)
• With tertiary haloalkanes, SN2 reactions cannot take place
 Elimination is highly favoured especially at high
temperatures
 Substitution occurs through SN1 mechanism only
83
New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.198)
Eliminations will be favoured when using:
1. higher temperatures
2. strong sterically hindered bases (e.g. (CH3)3CO–)
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.199)
CH3X
Methyl
RCH2X
1°
R2CHX
2°
R3CX
3°
Gives SN2
reactions only
Gives mainly SN2
and gives mainly
E with a strong
sterically hindered
base (e.g.
(CH3)3CO–)
Gives mainly
SN2 with a weak
base (e.g. I–, CN–
, RCO2–) and
gives mainly E
with a strong
base (e.g. RO–)
No SN2 reaction.
In hydrolysis,
gives SN1 or E.
At low
temperatures,
SN1 is favoured.
When a strong
base (e.g. RO–)
is used or at
high
temperatures, E
predominates.
Summary of the reaction pathways for the substitution and
elimination reactions of simple haloalkanes
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.199)
Formation of Alkynes
• Alkynes can be produced by dehydrohalogenation of
dihaloalkanes
• Two molecules of hydrogen halides are eliminated
e.g.
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.199)
Example 32-5
(a) Hot and concentrated alcoholic potassium hydroxide
can eliminate hydrogen iodide from the compound
CH3CH2CHICH3. Suggest and name two possible
products.
Answer
Solution:
(a)
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.199)
Example 32-5
(b) Draw
the structural formulae and give the names of all
Solution:
possible products formed by elimination of hydrogen
(b)
bromide from the dibromoalkane, CH3CHBrCHBrCH3.
Answer
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.200)
Check Point 32-5
(a) Notice how the hydrogen and halogen atoms come off
from adjacent carbon atoms in an elimination reaction.
Could (iodomethyl)benzene undergo an elimination to
give a HI molecule? Why?
(a) No, because there is no hydrogen available
on the carbon atom adjacent to the carbon
atom that is directly bonded to the iodine
atom.
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New Way Chemistry for Hong Kong A-Level Book 3A
Answer
32.7 Elimination Reactions (SB p.200)
Check Point 32-5
(b) 2-Iodo-2-methylbutane gives two elimination products:
one is 2-methylbut-2-ene, what is the other one?
Answer
(b) 2-Methylbut-1-ene
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New Way Chemistry for Hong Kong A-Level Book 3A
32.7 Elimination Reactions (SB p.200)
Check(c)Point 32-5
(c) Arrange the following compounds in order of increasing
tendency towards elimination reactions:
2-bromo-2-methylbutane, 1-bromopentane and
2-bromopentane
Answer
The rate of elimination depends on the stability of the
alkene formed. A more highly substituted alkene is more
stable and is formed more readily.
91
New Way Chemistry for Hong Kong A-Level Book 3A
32.8 Uses of Halogeno-compounds (SB p.200)
As Solvents in Dry-cleaning
• Chlorinated hydrocarbons are good solvents for oil and
greases
 widely used in the dry-cleaning industry
e.g. trichloroethene, CCl2 = CHCl
tetrachloroethene, CCl2 = CCl2
• Properties that favour the use:
1. Relatively non-flammable
92
2.
Volatile
3.
Little or no structural effect on fabrics
New Way Chemistry for Hong Kong A-Level Book 3A
32.8 Uses of Halogeno-compounds (SB p.201)
As Raw Materials for Making Addition Polymers
Poly(chloroethene) (also known as PVC):
• Produced by means of the addition polymerization of
the chloroethene monomers in the presence of a
peroxide catalyst
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New Way Chemistry for Hong Kong A-Level Book 3A
32.8 Uses of Halogeno-compounds (SB p.201)
• Polar C – Cl bond results in dipole-dipole interactions
between polymer chains, making PVC hard and brittle
and used to make pipes and bottles
Products made of PVC without
plasticizers
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New Way Chemistry for Hong Kong A-Level Book 3A
32.8 Uses of Halogeno-compounds (SB p.201)
• PVC becomes flexible when plasticizer is added
• Used to make shower curtains, raincoats, artificial
leather, insulating coating of electrical wires
Products made of PVC with
plasticizers
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New Way Chemistry for Hong Kong A-Level Book 3A
32.8 Uses of Halogeno-compounds (SB p.201)
Poly(tetrafluoroethene) (PTFE, ‘Teflon’):
•
Produced through addition polymerization of the
tetrafluoroethene monomers under high pressure and
in the presence of catalyst
96
New Way Chemistry for Hong Kong A-Level Book 3A
32.8 Uses of Halogeno-compounds (SB p.201)
• Teflon has a high melting point and is chemically inert
• Used to make non-stick frying pans
97
New Way Chemistry for Hong Kong A-Level Book 3A
The END
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New Way Chemistry for Hong Kong A-Level Book 3A