New Way Chemistry for Hong Kong A-Level Book 3A1 Halogeno-compounds 32.1Introduction...

New Way Chemistry for Hong Kong A-Level Book 3A

1

Halogeno-compounds32.132.1 IntroductionIntroduction

32.232.2 Nomenclature of Halogeno-compoundsNomenclature of Halogeno-compounds

32.332.3 Physical Properties of Halogeno-compoundsPhysical Properties of Halogeno-compounds

32.432.4 Preparation of Halogeno-compoundsPreparation of Halogeno-compounds

32.532.5 Reactions of Halogeno-compoundsReactions of Halogeno-compounds

32.632.6 Nucleophilic Substitution ReactionsNucleophilic Substitution Reactions

32.732.7 Elimination ReactionsElimination Reactions

32.832.8 Uses of Halogeno-compoundsUses of Halogeno-compounds

Chapter 32Chapter 32


2

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


3

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


4

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.


5


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.


6


In case of halobenzenes, the benzene ring is numbered so as t

o give the lowest possible numbers to the substituents

e.g.


7

Example 32-1Example 32-1Draw the structural formulae and give the IUPAC names of all isomers with the following molecular formula.

(a) C4H9Br

Answer


Solution:

(a)


8

Example 32-1Example 32-1Draw the structural formulae and give the IUPAC names of all isomers with the following molecular formula.

(b) C4H8Br2

Answer


Solution:

(b)


9

Check Point 32-1 Check Point 32-1

Draw the structural formulae and give the IUPAC names for all the structural isomers of C5H11Br.

Answer



10



11

32.3 Physical Properties of Halogeno-compounds (SB p.173)

Name Formula Melting

point (°C)

Boiling point (°C)

Density at 20°C (g cm–3)

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

–97.7

–136

–123

–123

–99

–83

–39

–45.2

–23.8

12.5

46.6

78.5

108

133

179

132

—

—

0.889

0.886

0.883

0.878

1.100

1.106


12



point (°C)

Boiling point (°C)

Density at 20°C (g cm–

3)

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

–93.7

–119

–109

–113

–95

–85

–3.9

–30.6

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


13



point (°C)

Boiling point (°C)

Density at 20°C (g cm–3)

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

–66.5

–108

–101

–103

–85.6

—

24.5

42.5

72.4

102

130

155

181

decompose

2.279

1.940

1.745

1.617

1.517

1.437

1.734


14


Boiling Point and Melting PointBoiling Point and Melting Point


15


Haloalkanes have higher b.p. and m.p. than alkanes

∵ dipole-dipole interactions are present between halo

alkane molecules

m.p. and b.p. increase in the order:

RCH2F < RCH2Cl < RCH2Br < RCH2I

larger, more polarizable halogen atoms increase th∵e dipole-dipole interactions between the molecules

No. of carbon m.p. and b.p.


16


DensityDensity

• 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


17


SolubilitySolubility

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


18

32.4 Preparation of Halogeno-compounds (SB p.175)

Preparation of HaloalkanesPreparation of Haloalkanes

• 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)

Substitution of Alcohols


19


• Dry HCl is bubbled through alcohols in the presence of ZnCl2 catalyst

Reaction with Hydrogen Halides

• For the preparation of bromo- and iodoalkanes, no catalyst is required


20


• The reactivity of hydrogen halides: HI > HBr > HCl

• e.g.


21


Haloalkanes can be prepared from the vigorous reaction

between cold alcohols and phosphorus(III) halides

Reaction with Phosphorus Halides


22


Addition of halogens or hydrogen halides to an alkene or alkyne can form a haloalkane

e.g.

Addition of Alkenes and Alkynes


23


Preparation of HalobenzenesPreparation of Halobenzenes

Benzene reacts readily with chlorine and bromine in the

presence of catalysts (e.g. FeCl3, FeBr3, AlCl3)

Halogenation of Benzene


24


From Benzenediazonium Salts


25


State the major products of the following reactions:

(a) CH3CHOHCH2CH3 + PBr3

(b) CH3CH = CH2 + HBr

(c) CH3C CH + 2HBr

(d)Answer


(a) CH3CHBrCH2CH3

(b) CH3CHBrCH3

(c) CH3CBr2CH3

(d)


26

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


27


• Characteristic reaction:

Nucleophilic substitution reaction

• Alcohols, ethers, esters, nitriles and amines can be for

med by substituting – OH, – OR, RCOO –, – CN and

– NH2 groups respectively


28


• Another characteristic reaction:

Elimination reaction

• Bases and nucleophiles are the same kind of reagents

• Nucleophilic substitution and elimination reactions

always occur together and compete each other

Haloalkane Base Alkene


29

32.6 Nucleophilic Substitution Reactions (SB p.179)

The reactions proceed in 2 different reaction mechanisms:

bimolecular nucleophilic substitution (SN2)

unimolecular nucleophilic substitution (SN1)

Reaction with Sodium HydroxideReaction with Sodium Hydroxide


30


Example: CH3 – Cl + OH– CH3OH + Cl–

Bimolecular Nucleophilic Substitution (SN2)

4.9 10–7

9.8 10–7

9.8 10–7

19.6 10–7

1.0

1.0

2.0

2.0

0.001

0.002

0.001

0.002

1

2

3

4

Initial rate (mol dm–3 s–1)

Initial [OH–]

(mol dm–3)

Initial [CH3Cl]

(mol dm–3)

Experiment number

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


31


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


32


Energy profile of the reaction of CH3Cl and OH- by SN2 mechanism

Transition state involve both the nucleophile and substrate second order kinetics of the reaction


33


• The nucleophile attacks from the backside of the electropositive carbon centre

• The configuration of the carbon atom under attack inverts

Stereochemistry of SN2 Reactions


34


Example:

Unimolecular Nucleophilic Substitution (SN

1)

• 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


35


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


36


Step 2:

• Fast step

• Attacked by a nucleophile to form the product


37


Energy profile of the reaction of (CH3)3CCl and OH- by SN1 mechanism

• 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


38


• The carbocation formed has a trigonal planar structure

• The nucleophile may either attack from the frontside or the backside

Stereochemistry of SN1 Reactions


39


For some cations, different products may be formed by either mode of attack

e.g.

The reaction is called racemization


40


The above SN1 reaction leads to racemization

∵ formation of trigonal planar carbocation intermediate


41


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


42


Most important factors affecting the relative rates of SN1

and SN2 reactions:

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

Factors Affecting the Rates of SN1 and SN2 Reactions


43


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

The Structure of the Substrate


44


Steric effects in the SN2 reaction


45


2. SN1 reactions

• Critical factor: the relative stability of the carbocation f

ormed

• Tertiary carbocations are the most stable

∵ 3 electron-releasing alkyl groups stabilize the carbocat

ion 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


46


• Only affect SN2 reactions

• Concentration of nucleophile rate

The Concentration and Strength of the Nucleophile


47

• 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

• Strength rate



48


• 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

The Nature of Leaving Group

BondBond 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


49


• 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.


50

When an alcohol is dissolved in a strong acid, it can re

act with a halide ion

∵ the acid protonates the –OH group, and the l

eaving group becomes a neutral water molecule

e.g.



51


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 leav

ing group on the rate of hydrolysis of haloalkanes

Comparision of Rates of Hydrolysis of Haloalkanes and Halobenzene


52


(b) Procedure

• 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


53


(c) Result and Observation

A precipitate of silver halide is formed in each of the three test tubes

Test tube A

AgCl(s)

Test tube B

AgBr(s)

Test tube C

AgI(s)


54


(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)


55


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


56


(b) Procedure

• 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


57


(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 GTest tube D Test tube E Test tube F


58


(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 1-chlorobutane >> chlorobenzene


59


• 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

• Chlorobenzene can be hydrolyzed to phenol under severe conditions (cannot be carried out in school laboratory)


60


• 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

Unreactivity of Halobenzene


61


• ∵ 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

• ∵ Delocalization of electrons makes the polarity of C – X bond

electropositive carbon center is less susceptible to nucleophilic attack


62


• Delocalized electrons repel any approaching nucleo

philes

unreactive towards SN2 reactions

• Benzene cations are highly unstable because of loss

of aromaticity

unreactive towards SN1 reactions


63

Example 32-2Example 32-2The reactions between three bromine-containing compounds and aqueous silver nitrate at room conditions are summarized in the following table:

(a) What is the pale yellow precipitate produced in the reaction between silver nitrate and sodium bromide? Answer


Compound Reaction with aqueous silver nitrate

Sodium bromideA pale yellow precipitate appears immediately

1-BromobutaneNo reaction at first; a pale yellow precipitate appears after several minutes

Bromobenzene No reaction even after several hours

Solution:

(a) Silver bromide


64

Example 32-2Example 32-2(b) Write an ionic equation for the reaction.

Answer


Solution:

(b) Ag+(aq) + Br–(aq) AgBr(s)


65

Example 32-2Example 32-2(c) Why does silver nitrate produce no immediate precipit

ate 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.


66

Example 32-2Example 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.


67

Example 32-3Example 32-3Which 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 electron-releasing group, this increases the electron density on the oxygen atom. This makes CH3CH2O– to be a stronger nucleophile than OH–.


68


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 h

alide. There is little steric hindrance for the nucleophile to att

ack the carbon atom of the molecule. On the other hand, if th

e reaction follows the SN1 mechanism, the carbocation (CH3

+) formed is not stabilized by the inductive effects of alkyl gr

oups. Thus the SN1 mechanism for this reaction is unfavoura

ble.



69


Predict whether the following substitution reaction follow mainly SN1 or SN2 pathway. Briefly explain your answer.

(b)

Answer


(b) The reaction follows mainly the SN1 mechanism. It is

because the haloalkane is a secondary haloalkane with a bu

lky phenyl group attached directly to the carbon atom beari

ng the halogen atom. The bulky phenyl group exerts a dra

matic steric hindrance to the approaching nucleophile. The

refore, the SN2 mechanism for this reaction is not favoured.

On the other hand, the carbocation formed in the SN1 reacti

on is stabilized by both the inductive effect of the electron-

releasing ethyl group and the resonance

effect of the phenyl group.


70

Reaction with Potassium CyanideReaction with Potassium Cyanide

e.g.


A nitrile is formed when a haloalkane is heated under reflux wit

h an aqueous alcoholic solution of potassium cyanide


71

• 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

• A useful way of introducing a carbon atom into an organic molecule, so that the length of the carbon chain can be increased


72

Reaction with AmmoniaReaction with Ammonia


When a haloalkane is heated with an aqueous alcoholic solution of ammonia under a high pressure, an amine is formed

e.g.


73

• 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



74


• The reaction stops at the formation of a quaternary ammonium salt

• The competing reactions can be minimized by using an excess of ammonia


75

Example 32-4Example 32-4Give 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– Answer


Solution:

(a) Dilute NaOH

(b) C2H5O–Na+ or Na in C2H5OH

(c) NH3 in excess CH3I


76


Give the name(s) and structural formula(e) of the major organic product(s) formed in each of the following reactions.

(a)

(b)

(c)

Answer


(a)

(b) CH3NH2 Methylamine

(c)


77

Formation of AlkenesFormation of Alkenes

32.7 Elimination Reactions (SB p.196)

The elimination of HX from adjacent atoms of a haloalkane

is widely used for synthesizing alkenes

e.g.


78


The elements of a hydrogen halide are eliminated from a haloa

lkane in this way, the reaction is called dehydrohalogenation


79


Dehydrohalogenation of most haloalkanes yields more than o

ne product

e.g.


80


• 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:


81

• Nucleophiles are potential bases

• Bases are potential nucleophiles

• In SN2 pathway, elimination and nucleophilic substituti

on compete each other

Elimination Versus Substitution



82

• Substitution is favoured when the substrate is primary

alcohol and the base is hydroxide ion


• Elimination is favoured when the substrate is secondary

alcohol


83

• With tertiary haloalkanes, SN2 reactions cannot take place

Elimination is highly favoured especially at high

temperatures

Substitution occurs through SN1 mechanism only



84

Eliminations will be favoured when using:

1. higher temperatures

2. strong sterically hindered bases (e.g. (CH3)3CO–)



85


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


86

Formation of AlkynesFormation of Alkynes


• Alkynes can be produced by dehydrohalogenation of dihaloalkanes

• Two molecules of hydrogen halides are eliminated

e.g.


87

Example 32-5Example 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)


88

Example 32-5Example 32-5(b) Draw the structural formulae and give the names of all

possible products formed by elimination of hydrogen bromide from the dibromoalkane, CH3CHBrCHBrCH3.

Answer


Solution:

(b)


89


(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?

Answer


(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.


90


(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



91


(c) Arrange the following compounds in order of increasing tendency towards elimination reactions:

2-bromo-2-methylbutane, 1-bromopentane and 2-bromopentane

Answer


(c)

The rate of elimination depends on the stability of the alk

ene formed. A more highly substituted alkene is more sta

ble and is formed more readily.


92

As Solvents in Dry-cleaningAs Solvents in Dry-cleaning

32.8 Uses of Halogeno-compounds (SB p.200)

• 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

2. Volatile

3. Little or no structural effect on fabrics


93

As Raw Materials for Making Addition PolymersAs Raw Materials for Making Addition Polymers


Poly(chloroethene) (also known as PVC):

• Produced by means of the addition polymerization of t

he chloroethene monomers in the presence of a peroxi

de catalyst


94

• 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


95


• 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


96


Poly(tetrafluoroethene) (PTFE, ‘Teflon’):

• Produced through addition polymerization of the tetra

fluoroethene monomers under high pressure and in the

presence of catalyst


97

• Teflon has a high melting point and is chemically inert

• Used to make non-stick frying pans



98

The END

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New Way Chemistry for Hong Kong A-Level Book 3A1 Halogeno-compounds 32.1Introduction...

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