2. New Way Chemistry for Hong Kong A-Level Book 3A1 Halogeno-compounds 32.1Introduction 32.2Nomenclature of Halogeno-compounds 32.3Physical Properties of Halogeno-compounds 32.4Preparation of Halogeno-compounds 32.5Reactions of Halogeno-compounds 32.6Nucleophilic Substitution Reactions 32.7Elimination Reactions 32.8Uses of Halogeno-compounds Chapter 32

3. New Way Chemistry for Hong Kong A-Level Book 3A2 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

4. New Way Chemistry for Hong Kong A-Level Book 3A3 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

5. New Way Chemistry for Hong Kong A-Level Book 3A4 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.

6. New Way Chemistry for Hong Kong A-Level Book 3A5 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.

7. New Way Chemistry for Hong Kong A-Level Book 3A6 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.

8. New Way Chemistry for Hong Kong A-Level Book 3A7 Example 32-1 Draw the structural formulae and give the IUPAC names of all isomers with the following molecular formula. (a)C 4 H 9 Br Answer 32.2 Nomenclature of Halogeno-compounds (SB p.171) Solution: (a)

9. New Way Chemistry for Hong Kong A-Level Book 3A8 Example 32-1 Draw the structural formulae and give the IUPAC names of all isomers with the following molecular formula. (b)C 4 H 8 Br 2 Answer 32.2 Nomenclature of Halogeno-compounds (SB p.171) Solution: (b)

10. New Way Chemistry for Hong Kong A-Level Book 3A9 Check Point 32-1 Draw the structural formulae and give the IUPAC names for all the structural isomers of C 5 H 11 Br. Answer 32.2 Nomenclature of Halogeno-compounds (SB p.172)

11. New Way Chemistry for Hong Kong A-Level Book 3A10 32.2 Nomenclature of Halogeno-compounds (SB p.172)

12. New Way Chemistry for Hong Kong A-Level Book 3A11 32.3 Physical Properties of Halogeno-compounds (SB p.173) NameFormula 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 CH 3 Cl CH 3 CH 2 Cl CH 3 (CH 2 ) 2 Cl CH 3 (CH 2 ) 3 Cl CH 3 (CH 2 ) 4 Cl CH 3 (CH 2 ) 5 Cl C 6 H 5 CH 2 Cl C 6 H 5 Cl –97.7 –136 –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

13. New Way Chemistry for Hong Kong A-Level Book 3A12 32.3 Physical Properties of Halogeno-compounds (SB p.173) NameFormula Melting 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 CH 3 Br CH 3 CH 2 Br CH 3 (CH 2 ) 2 Br CH 3 (CH 2 ) 3 Br CH 3 (CH 2 ) 4 Br CH 3 (CH 2 ) 5 Br C 6 H 5 CH 2 Br C 6 H 5 Br –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

14. New Way Chemistry for Hong Kong A-Level Book 3A13 32.3 Physical Properties of Halogeno-compounds (SB p.173) NameFormula Melting 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 CH 3 I CH 3 CH 2 I CH 3 (CH 2 ) 2 I CH 3 (CH 2 ) 3 I CH 3 (CH 2 ) 4 I CH 3 (CH 2 ) 5 I C 6 H 5 CH 2 I –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

15. New Way Chemistry for Hong Kong A-Level Book 3A14 32.3 Physical Properties of Halogeno-compounds (SB p.174) Boiling Point and Melting Point

16. New Way Chemistry for Hong Kong A-Level Book 3A15 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: RCH 2 F < RCH 2 Cl < RCH 2 Br < RCH 2 I ∵ larger, more polarizable halogen atoms increase the dipole-dipole interactions between the molecules No. of carbon   m.p. and b.p. 

17. New Way Chemistry for Hong Kong A-Level Book 3A16 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

18. New Way Chemistry for Hong Kong A-Level Book 3A17 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

19. New Way Chemistry for Hong Kong A-Level Book 3A18 32.4 Preparation of Halogeno-compounds (SB p.175) Preparation of Haloalkanes Prepared by substituting –OH group of alcohols with halogen atoms Common reagents used: HCl, HBr, HI, PCl 3 or PBr 3 The ease of substitution of alcohols: 3° alcohol > 2° alcohol > 1° alcohol > CH 3 OH This is related to the stability of the reaction intermediate (i.e. stability of carbocations) Substitution of Alcohols

20. New Way Chemistry for Hong Kong A-Level Book 3A19 32.4 Preparation of Halogeno-compounds (SB p.175) Dry HCl is bubbled through alcohols in the presence of ZnCl 2 catalyst Reaction with Hydrogen Halides For the preparation of bromo- and iodoalkanes, no catalyst is required

21. New Way Chemistry for Hong Kong A-Level Book 3A20 32.4 Preparation of Halogeno-compounds (SB p.176) The reactivity of hydrogen halides: HI > HBr > HCl e.g.

22. New Way Chemistry for Hong Kong A-Level Book 3A21 32.4 Preparation of Halogeno-compounds (SB p.176) Haloalkanes can be prepared from the vigorous reaction between cold alcohols and phosphorus(III) halides Reaction with Phosphorus Halides

23. New Way Chemistry for Hong Kong A-Level Book 3A22 32.4 Preparation of Halogeno-compounds (SB p.177) Addition of halogens or hydrogen halides to an alkene or alkyne can form a haloalkane e.g. Addition of Alkenes and Alkynes

24. New Way Chemistry for Hong Kong A-Level Book 3A23 32.4 Preparation of Halogeno-compounds (SB p.177) Preparation of Halobenzenes Benzene reacts readily with chlorine and bromine in the presence of catalysts (e.g. FeCl 3, FeBr 3, AlCl 3 ) Halogenation of Benzene

25. New Way Chemistry for Hong Kong A-Level Book 3A24 32.4 Preparation of Halogeno-compounds (SB p.177) From Benzenediazonium Salts

26. New Way Chemistry for Hong Kong A-Level Book 3A25 Check Point 32-2 State the major products of the following reactions: (a)CH 3 CHOHCH 2 CH 3 + PBr 3  (b)CH 3 CH = CH 2 + HBr  (c)CH 3 C  CH + 2HBr  (d) Answer 32.4 Preparation of Halogeno-compounds (SB p.178) (a)CH 3 CHBrCH 2 CH 3 (b)CH 3 CHBrCH 3 (c)CH 3 CBr 2 CH 3 (d)

27. New Way Chemistry for Hong Kong A-Level Book 3A26 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

28. New Way Chemistry for Hong Kong A-Level Book 3A27 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 – NH 2 groups respectively

29. New Way Chemistry for Hong Kong A-Level Book 3A28 32.5 Reactions of Halogeno-compounds (SB p.179) 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 HaloalkaneBaseAlkene

30. New Way Chemistry for Hong Kong A-Level Book 3A29 32.6 Nucleophilic Substitution Reactions (SB p.179) The reactions proceed in 2 different reaction mechanisms: bimolecular nucleophilic substitution (S N 2) unimolecular nucleophilic substitution (S N 1) Reaction with Sodium Hydroxide

31. New Way Chemistry for Hong Kong A-Level Book 3A30 32.6 Nucleophilic Substitution Reactions (SB p.180) Example: CH 3 – Cl + OH –  CH 3 OH + Cl – Bimolecular Nucleophilic Substitution (S N 2) 4.9  10 –7 9.8  10 –7 19.6  10 –7 1.0 2.0 0.001 0.002 0.001 0.002 12341234 Initial rate (mol dm –3 s –1 ) Initial [OH – ] (mol dm –3 ) Initial [CH 3 Cl] (mol dm –3 ) Experiment number Results of kinetic study of reaction of CH 3 Cl with OH – Rate = k[CH 3 Cl][OH – ] Order of reaction = 2  both species are involved in rate determining step

32. New Way Chemistry for Hong Kong A-Level Book 3A31 32.6 Nucleophilic Substitution Reactions (SB p.181) Reaction mechanism of the S N 2 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

33. New Way Chemistry for Hong Kong A-Level Book 3A32 32.6 Nucleophilic Substitution Reactions (SB p.181) Energy profile of the reaction of CH 3 Cl and OH - by S N 2 mechanism Transition state involve both the nucleophile and substrate  second order kinetics of the reaction

34. New Way Chemistry for Hong Kong A-Level Book 3A33 32.6 Nucleophilic Substitution Reactions (SB p.182) The nucleophile attacks from the backside of the electropositive carbon centre The configuration of the carbon atom under attack inverts Stereochemistry of S N 2 Reactions

35. New Way Chemistry for Hong Kong A-Level Book 3A34 32.6 Nucleophilic Substitution Reactions (SB p.182) Example: Unimolecular Nucleophilic Substitution (S N 1) Kinetic study shows that: Rate = k[(CH 3 ) 3 CCl] The rate is independent of [OH – ] Order of reaction = 1  only 1 species is involved in the rate determining step

36. New Way Chemistry for Hong Kong A-Level Book 3A35 32.6 Nucleophilic Substitution Reactions (SB p.183) Reaction mechanism of S N 1 reaction involves 2 steps and 1 intermediate formed Step 1: Slowest step (i.e. rate determining step) Formation of carbocation and halide ion

37. New Way Chemistry for Hong Kong A-Level Book 3A36 32.6 Nucleophilic Substitution Reactions (SB p.183) Step 2: Fast step Attacked by a nucleophile to form the product

38. New Way Chemistry for Hong Kong A-Level Book 3A37 32.6 Nucleophilic Substitution Reactions (SB p.183) Energy profile of the reaction of (CH 3 ) 3 CCl and OH - by S N 1 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

39. New Way Chemistry for Hong Kong A-Level Book 3A38 32.6 Nucleophilic Substitution Reactions (SB p.184) The carbocation formed has a trigonal planar structure The nucleophile may either attack from the frontside or the backside Stereochemistry of S N 1 Reactions

40. New Way Chemistry for Hong Kong A-Level Book 3A39 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

41. New Way Chemistry for Hong Kong A-Level Book 3A40 32.6 Nucleophilic Substitution Reactions (SB p.184) The above S N 1 reaction leads to racemization ∵ formation of trigonal planar carbocation intermediate

42. New Way Chemistry for Hong Kong A-Level Book 3A41 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

43. New Way Chemistry for Hong Kong A-Level Book 3A42 32.6 Nucleophilic Substitution Reactions (SB p.185) Most important factors affecting the relative rates of S N 1 and S N 2 reactions: 1.The structure of the substrate 2.The concentration and strength of the nucleophile (for S N 2 reactions only) 3.The nature of the leaving group Factors Affecting the Rates of S N 1 and S N 2 Reactions

44. New Way Chemistry for Hong Kong A-Level Book 3A43 32.6 Nucleophilic Substitution Reactions (SB p.186) 1.S N 2 reactions The reactivity of haloalkanes in S N 2 reactions: CH 3 X > 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

45. New Way Chemistry for Hong Kong A-Level Book 3A44 32.6 Nucleophilic Substitution Reactions (SB p.186) Steric effects in the S N 2 reaction

46. New Way Chemistry for Hong Kong A-Level Book 3A45 32.6 Nucleophilic Substitution Reactions (SB p.187) 2.S N 1 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 S N 1 reactions are very large and rate of reaction become very small

47. New Way Chemistry for Hong Kong A-Level Book 3A46 32.6 Nucleophilic Substitution Reactions (SB p.187) Only affect S N 2 reactions Concentration of nucleophile   rate  The Concentration and Strength of the Nucleophile

48. New Way Chemistry for Hong Kong A-Level Book 3A47 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. H 2 O) (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 > H 2 O Strength   rate  32.6 Nucleophilic Substitution Reactions (SB p.187)

49. New Way Chemistry for Hong Kong A-Level Book 3A48 32.6 Nucleophilic Substitution Reactions (SB p.188) 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

50. New Way Chemistry for Hong Kong A-Level Book 3A49 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: H 2 O >> OH – > RO – Strongly basic ions rarely act as leaving group e.g.

51. New Way Chemistry for Hong Kong A-Level Book 3A50 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. 32.6 Nucleophilic Substitution Reactions (SB p.188)

52. New Way Chemistry for Hong Kong A-Level Book 3A51 32.6 Nucleophilic Substitution Reactions (SB p.188) 1.Experiment 1 : Comparison of the rates of hydrolysis of 1-chlorobutane, 1-bromobutane and 1-iodobutane (a)Objective To study the effect ofthe nature of the halogen leaving group on the rate of hydrolysis of haloalkanes Comparision of Rates of Hydrolysis of Haloalkanes and Halobenzene

53. New Way Chemistry for Hong Kong A-Level Book 3A52 32.6 Nucleophilic Substitution Reactions (SB p.189) (b)Procedure Put 2 cm 3 of ethanol and 1 cm 3 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

54. New Way Chemistry for Hong Kong A-Level Book 3A53 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) Test tube B AgBr(s) Test tube C AgI(s)

55. New Way Chemistry for Hong Kong A-Level Book 3A54 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 AgNO 3 Ag + (aq) + X – (aq)  AgX(s)

56. New Way Chemistry for Hong Kong A-Level Book 3A55 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

57. New Way Chemistry for Hong Kong A-Level Book 3A56 32.6 Nucleophilic Substitution Reactions (SB p.190) (b)Procedure Put 2 cm 3 of ethanol and 1 cm 3 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

58. New Way Chemistry for Hong Kong A-Level Book 3A57 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 GTest tube DTest tube ETest tube F

59. New Way Chemistry for Hong Kong A-Level Book 3A58 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  1- chlorobutane >> chlorobenzene

60. New Way Chemistry for Hong Kong A-Level Book 3A59 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 S N 1 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)

61. New Way Chemistry for Hong Kong A-Level Book 3A60 32.6 Nucleophilic Substitution Reactions (SB p.192) 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

62. New Way Chemistry for Hong Kong A-Level Book 3A61 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 ∵ Delocalization of  electrons makes the polarity of C – X bond   electropositive carbon center is less susceptible to nucleophilic attack

63. New Way Chemistry for Hong Kong A-Level Book 3A62 32.6 Nucleophilic Substitution Reactions (SB p.192) Delocalized electrons repel any approaching nucleophiles  unreactive towards S N 2 reactions Benzene cations are highly unstable because of loss of aromaticity  unreactive towards S N 1 reactions

64. New Way Chemistry for Hong Kong A-Level Book 3A63 Example 32-2 The 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 32.6 Nucleophilic Substitution Reactions (SB p.192) CompoundReaction 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 BromobenzeneNo reaction even after several hours Solution: (a)Silver bromide

65. New Way Chemistry for Hong Kong A-Level Book 3A64 Example 32-2 (b)Write an ionic equation for the reaction. Answer 32.6 Nucleophilic Substitution Reactions (SB p.192) Solution: (b)Ag + (aq) + Br – (aq)  AgBr(s)

66. New Way Chemistry for Hong Kong A-Level Book 3A65 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 32.6 Nucleophilic Substitution Reactions (SB p.192) Solution: (c)The hydrolysis of 1-bromobutane takes time. Precipitation of AgBr occurs only after OH – from water has replaced Br – from 1-bromobutane.

67. New Way Chemistry for Hong Kong A-Level Book 3A66 Example 32-2 (d)Briefly explain why bromobenzene does not give any precipitate with aqueous silver nitrate. Answer 32.6 Nucleophilic Substitution Reactions (SB p.192) 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.

68. New Way Chemistry for Hong Kong A-Level Book 3A67 Example 32-3 Which is the stronger nucleophile in each of the following pairs? Explain your choice briefly. (a)OH – and H 2 O (b)OH – and CH 3 CH 2 O – Answer 32.6 Nucleophilic Substitution Reactions (SB p.193) Solution: (a)OH – is a stronger nucleophile than H 2 O because it carries a negative charge while H 2 O is electrically neutral. (b)CH 3 CH 2 O – is a stronger nucleophile than OH –. It is because the ethyl group (CH 3 CH 2 –) is an electron- releasing group, this increases the electron density on the oxygen atom. This makes CH 3 CH 2 O – to be a stronger nucleophile than OH –.

69. New Way Chemistry for Hong Kong A-Level Book 3A68 Check Point 32-3 Predict whether the following substitution reaction follows mainly S N 1 or S N 2 pathway. Briefly explain your answer. (a)CH 3 I + OH –  CH 3 OH + I – Answer (a)The reaction follows mainly the S N 2 mechanism because of the following reasons. The haloalkane (CH 3 I) 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 S N 1 mechanism, the carbocation (CH 3 + ) formed is not stabilized by the inductive effects of alkyl groups. Thus the S N 1 mechanism for this reaction is unfavourable. 32.6 Nucleophilic Substitution Reactions (SB p.194)

70. New Way Chemistry for Hong Kong A-Level Book 3A69 Check Point 32-3 Predict whether the following substitution reaction follow mainly S N 1 or S N 2 pathway. Briefly explain your answer. (b) Answer 32.6 Nucleophilic Substitution Reactions (SB p.194) (b)The reaction follows mainly the S N 1 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 S N 2 mechanism for this reaction is not favoured. On the other hand, the carbocation formed in the S N 1 reaction is stabilized by both the inductive effect of the electron-releasing ethyl group and the resonance effect of the phenyl group.

71. New Way Chemistry for Hong Kong A-Level Book 3A70 Reaction with Potassium Cyanide e.g. 32.6 Nucleophilic Substitution Reactions (SB p.194) A nitrile is formed when a haloalkane is heated under reflux with an aqueous alcoholic solution of potassium cyanide

72. New Way Chemistry for Hong Kong A-Level Book 3A71 Cyanide ion (CN – ) acts as a nucleophile 32.6 Nucleophilic Substitution Reactions (SB p.194) 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

73. New Way Chemistry for Hong Kong A-Level Book 3A72 Reaction with Ammonia 32.6 Nucleophilic Substitution Reactions (SB p.195) When a haloalkane is heated with an aqueous alcoholic solution of ammonia under a high pressure, an amine is formed e.g.

74. New Way Chemistry for Hong Kong A-Level Book 3A73 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 32.6 Nucleophilic Substitution Reactions (SB p.195)

75. New Way Chemistry for Hong Kong A-Level Book 3A74 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

76. New Way Chemistry for Hong Kong A-Level Book 3A75 Example 32-4 Give the reagents and reaction conditions needed for each of the following conversions: (a)(CH 3 ) 3 CBr  (CH 3 ) 3 COH (b)CH 3 I  CH 3 OC 2 H 5 (c)CH 3 I  (CH 3 ) 4 N + I – Answer 32.6 Nucleophilic Substitution Reactions (SB p.195) Solution: (a)Dilute NaOH (b)C 2 H 5 O – Na + or Na in C 2 H 5 OH (c)NH 3 in excess CH 3 I

77. New Way Chemistry for Hong Kong A-Level Book 3A76 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) (b) (c) Answer 32.6 Nucleophilic Substitution Reactions (SB p.196) (a) (b)CH 3 NH 2 Methylamine (c)

78. New Way Chemistry for Hong Kong A-Level Book 3A77 Formation 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.

79. New Way Chemistry for Hong Kong A-Level Book 3A78 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

80. New Way Chemistry for Hong Kong A-Level Book 3A79 32.7 Elimination Reactions (SB p.196) Dehydrohalogenation of most haloalkanes yields more than one product e.g.

81. New Way Chemistry for Hong Kong A-Level Book 3A80 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:

82. New Way Chemistry for Hong Kong A-Level Book 3A81 Nucleophiles are potential bases Bases are potential nucleophiles In S N 2 pathway, elimination and nucleophilic substitution compete each other Elimination Versus Substitution 32.7 Elimination Reactions (SB p.197)

83. New Way Chemistry for Hong Kong A-Level Book 3A82 Substitution is favoured when the substrate is primary alcohol and the base is hydroxide ion 32.7 Elimination Reactions (SB p.198) Elimination is favoured when the substrate is secondary alcohol

84. New Way Chemistry for Hong Kong A-Level Book 3A83 With tertiary haloalkanes, S N 2 reactions cannot take place  Elimination is highly favoured especially at high temperatures  Substitution occurs through S N 1 mechanism only 32.7 Elimination Reactions (SB p.198)

85. New Way Chemistry for Hong Kong A-Level Book 3A84 Eliminations will be favoured when using: 1.higher temperatures 2.strong sterically hindered bases (e.g. (CH 3 ) 3 CO – ) 32.7 Elimination Reactions (SB p.198)

86. New Way Chemistry for Hong Kong A-Level Book 3A85 32.7 Elimination Reactions (SB p.199) CH 3 X Methyl RCH 2 X 1° R 2 CHX 2° R 3 CX 3° Gives S N 2 reactions only Gives mainly S N 2 and gives mainly E with a strong sterically hindered base (e.g. (CH 3 ) 3 CO – ) Gives mainly S N 2 with a weak base (e.g. I –, CN –, RCO 2 – ) and gives mainly E with a strong base (e.g. RO – ) No S N 2 reaction. In hydrolysis, gives S N 1 or E. At low temperatures, S N 1 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

87. New Way Chemistry for Hong Kong A-Level Book 3A86 Formation of Alkynes 32.7 Elimination Reactions (SB p.199) Alkynes can be produced by dehydrohalogenation of dihaloalkanes Two molecules of hydrogen halides are eliminated e.g.

88. New Way Chemistry for Hong Kong A-Level Book 3A87 Example 32-5 (a)Hot and concentrated alcoholic potassium hydroxide can eliminate hydrogen iodide from the compound CH 3 CH 2 CHICH 3. Suggest and name two possible products. Answer 32.7 Elimination Reactions (SB p.199) Solution: (a)

89. New Way Chemistry for Hong Kong A-Level Book 3A88 Example 32-5 (b)Draw the structural formulae and give the names of all possible products formed by elimination of hydrogen bromide from the dibromoalkane, CH 3 CHBrCHBrCH 3. Answer 32.7 Elimination Reactions (SB p.199) Solution: (b)

90. New Way Chemistry for Hong Kong A-Level Book 3A89 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? Answer 32.7 Elimination Reactions (SB p.200) (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.

91. New Way Chemistry for Hong Kong A-Level Book 3A90 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 32.7 Elimination Reactions (SB p.200)

92. New Way Chemistry for Hong Kong A-Level Book 3A91 Check 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 32.7 Elimination Reactions (SB p.200) (c) 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.

93. New Way Chemistry for Hong Kong A-Level Book 3A92 As 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, CCl 2 = CHCl tetrachloroethene, CCl 2 = CCl 2 Properties that favour the use: 1.Relatively non-flammable 2.Volatile 3.Little or no structural effect on fabrics

94. New Way Chemistry for Hong Kong A-Level Book 3A93 As Raw Materials for Making Addition Polymers 32.8 Uses of Halogeno-compounds (SB p.201) Poly(chloroethene) (also known as PVC): Produced by means of the addition polymerization of the chloroethene monomers in the presence of a peroxide catalyst

95. New Way Chemistry for Hong Kong A-Level Book 3A94 Polar C – Cl bond results in dipole-dipole interactions between polymer chains, making PVC hard and brittle and used to make pipes and bottles 32.8 Uses of Halogeno-compounds (SB p.201) Products made of PVC without plasticizers

96. New Way Chemistry for Hong Kong A-Level Book 3A95 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

97. New Way Chemistry for Hong Kong A-Level Book 3A96 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

98. New Way Chemistry for Hong Kong A-Level Book 3A97 Teflon has a high melting point and is chemically inert Used to make non-stick frying pans 32.8 Uses of Halogeno-compounds (SB p.201)

99. New Way Chemistry for Hong Kong A-Level Book 3A98 The END