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按一下以編輯母片標題樣式 Halogeno-compounds and Nucleophilic Substitution 1 New Way Chemistry for Hong Kong A-Level Book 3A 1 Classification 按一下以編輯母片標題樣式 1. Halogenoalkanes (alkyl halides) 2. Unsaturated halides 2 New Way Chemistry for Hong Kong A-Level Book 3A Halogenoalkanes 按一下以編輯母片標題樣式 • Alkanes with one or more hydrogen atoms replaced by halogen atoms. R, R’, R’’ : alkyl groups X : halogen 3 New Way Chemistry for Hong Kong A-Level Book 3A 3 Unsaturated halides 按一下以編輯母片標題樣式 H H H H R R X vinylic halide R X X allylic halide 4 New Way Chemistry for Hong Kong A-Level Book 3A 4 Unsaturated halides -carbon 按一下以編輯母片標題樣式 X X aryl halide halobenzene benzylic halide 5 New Way Chemistry for Hong Kong A-Level Book 3A 5 按一下以編輯母片標題樣式 Physical Properties of Halogenocompounds 6 New Way Chemistry for Hong Kong A-Level Book 3A 6 Boiling Point and Melting Point 按一下以編輯母片標題樣式 • CX bond is polar in nature • Molecules are held by dispersion forces and dipole-dipole attractions. • Haloalkanes > alkanes of comparable relative molecular masses • All RX(X=Cl, Br, I) are liquids or solids at room conditions except CH3Cl, CH3Br, C2H5Cl 7 New Way Chemistry for Hong Kong A-Level Book 3A 7 Boiling Point and Melting Point 按一下以編輯母片標題樣式 CH X < C H X < C H X < C H X < … 3 2 5 3 7 4 9 RCH2F < RCH2Cl < RCH2Br < RCH2I ∵ Dispersion forces as molecular size 8 New Way Chemistry for Hong Kong A-Level Book 3A 8 Variation of boiling points with the number of carbon atoms of straight-chain haloalkanes 按一下以編輯母片標題樣式 9 New Way Chemistry for Hong Kong A-Level Book 3A 9 Density 按一下以編輯母片標題樣式 CH X > C H X > C H X > C H X > … 3 2 5 3 7 4 9 ∵ packing efficiency as molecular size packing efficency is the dominant factor 10 New Way Chemistry for Hong Kong A-Level Book 3A 10 Density 按一下以編輯母片標題樣式 RCH F & RCH Cl less dense than water 2 2 RCH2Br, RCH2I and ArX denser than water 11 New Way Chemistry for Hong Kong A-Level Book 3A 11 Solubility • 按一下以編輯母片標題樣式 All are immiscible with water but miscible with organic solvents 12 New Way Chemistry for Hong Kong A-Level Book 3A 12 按一下以編輯母片標題樣式 Preparation of Halogenocompounds 13 New Way Chemistry for Hong Kong A-Level Book 3A 13 A. Preparation of Halobenzenes X (refer to pp.50-51) 按一下以編輯母片標題樣式 X2 FeX3 B. Preparation of Haloalkanes 14 New Way Chemistry for Hong Kong A-Level Book 3A Preparation of Haloalkanes 按一下以編輯母片標題樣式 1. Free radical halogenation of alkanes R – H + X2 light R – X + HX Not suitable for preparing mono-substituted haloalkanes as further substitution may occur. CH4 + Cl2 light CH3Cl + CH2Cl2 + … 15 New Way Chemistry for Hong Kong A-Level Book 3A 15 Q.28(b) 按一下以編輯母片標題樣式 Cl2 Cl Light, 25C Cl 64% 36% 16 New Way Chemistry for Hong Kong A-Level Book 3A Q.28(b) 按一下以編輯母片標題樣式 Br2 Light, 127C Br Br trace 17 >99% Lower reactivity but higher selectivity New Way Chemistry for Hong Kong A-Level Book 3A 2. SN reactions of ROH with HX 按一下以編輯母片標題樣式 Reactivity : HI > HBr > HCl 3o > 2o > 1o 18 New Way Chemistry for Hong Kong A-Level Book 3A 18 Reaction conditions : (i) Chlorination 按一下以編輯母片標題樣式 - heating under reflux - HCl(g) or HCl(aq) - ∵ least reactive ZnCl2 is always used as catalyst 19 New Way Chemistry for Hong Kong A-Level Book 3A 19 Lucas test 按一下以編輯母片標題樣式 - To distinguish different types of R-OH with not more than 6 carbon atoms R-OH conc. HCl ZnCl2 R-Cl 1 R-OH no apparent reaction in 5 minutes 2 R-OH reaction mixture turns cloudy in 5 minutes 3 R-OH reaction mixture turns cloudy almost immediately. 20 New Way Chemistry for Hong Kong A-Level Book 3A Lucas test 按一下以編輯母片標題樣式 - To distinguish different types of R-OH with not more than 6 carbon atoms R-OH conc. HCl R-OH conc. HCl 1 ZnCl2 ZnCl2 R-Cl + R-OH2 soluble in H2O if R has no more than 6 C atoms 21 New Way Chemistry for Hong Kong A-Level Book 3A Lucas test 按一下以編輯母片標題樣式 - To distinguish different types of R-OH with not more than 6 carbon atoms R-OH conc. HCl R-OH conc. HCl 2/3 ZnCl2 ZnCl2 R-Cl R-Cl insoluble in H2O 22 New Way Chemistry for Hong Kong A-Level Book 3A turns cloudy Reaction conditions : (ii) Bromination 按一下以編輯母片標題樣式 - heating under reflux - HBr(aq) is generated in situ R-OH NaBr(s) conc. H2SO4 or H3PO4 R-Br NaBr + H2SO4 NaHSO4 + HBr NaBr + H3PO4 NaH2PO4 + HBr 23 R-OH + HBr R-Br + H2O New Way Chemistry for Hong Kong A-Level Book 3A 23 (ii) Bromination NaBr(s) 按一下以編輯母片標題樣式 R-OH R-Br + H2O conc. H2SO4 The functions of conc. H2SO4 include : 1. generating HBr for the reaction 2. increasing the yield of the reaction by shifting the equilibrium position to the right (by removing H2O) 3. increasing the rate of the reaction by protonating the –OH group. 24 New Way Chemistry for Hong Kong A-Level Book 3A 24 - Br protonation 按一下以編輯母片標題樣式 H H O + H O H Protonating the –OH group weakens the C – O bond. It causes the H2O to leave more easily than the HO 25 New Way Chemistry for Hong Kong A-Level Book 3A Br + H2O Reaction conditions : (iii)Iodination 按一下以編輯母片標題樣式 - heating under reflux - HI(aq) is generated in situ R-OH NaI(s) conc. H3PO4 R-I NaI + H3PO4 NaH2PO4 + HI R-OH + HI R-I + H2O 26 New Way Chemistry for Hong Kong A-Level Book 3A 26 Reaction conditions : (iii)Iodination 按一下以編輯母片標題樣式 R-OH NaI(s) conc. H3PO4 R-I Conc. H3PO4 is preferred to H2SO4 as the latter may oxidize HI to give I2 NaI + H2SO4 NaHSO4 + HI HI + H2SO4 I2 + …… 27 New Way Chemistry for Hong Kong A-Level Book 3A 27 3. Reactions with PX3, PCl5 or SOCl2 Red 按一下以編輯母片標題樣式 P + Br2 / I2 PBr3 / PI3 (in situ preparation) 3R – OH + PBr3 / PI3 reflux 3RBr/3RI + H3PO3 R – OH + PCl5 RCl + HCl + POCl3 R – OH + SOCl2 RCl + HCl + SO2 faster At 25C Substitution reactions but NOT SN reactions Less side products(arise from rearrangements) than SN reactions of ROH with HX 28 New Way Chemistry for Hong Kong A-Level Book 3A 28 4. Electrophilic Addition of Alkenes and Alkynes (pp.41-46) 按一下以編輯母片標題樣式 CH2 = CH2 + HBr CH3CH2Br CH2 = CH2 + Br2 CH2BrCH2Br CH CH + HBr CH2 = CHBr CH2 = CHBr + HBr CH3CHBr2 CH CH + 2Br2 CHBr2CHBr2 29 New Way Chemistry for Hong Kong A-Level Book 3A 29 按一下以編輯母片標題樣式 Reactions of Halogenocompounds 30 New Way Chemistry for Hong Kong A-Level Book 3A 30 Reactions of Halogeno-compounds Two按一下以編輯母片標題樣式 types : 1. Nucleophilic substitution reaction (SN reaction) Nucleophiles : OH, CN, RO (anions) NH3, H2O (molecules with lone pairs) 31 New Way Chemistry for Hong Kong A-Level Book 3A 31 CX bond is polar partial +ve charge(+) on carbon atom 按一下以編輯母片標題樣式 susceptible to nucleophilic attack Nu Nu substrate 32 New Way Chemistry for Hong Kong A-Level Book 3A C + X leaving group 32 (alcohol) (ether) 按一下以編輯母片標題樣式 H2O (nitrile) ROH -H+ C2H5OH, reflux ROH2 (amine) 33 New Way Chemistry for Hong Kong A-Level Book 3A (alcohol) (ether) 按一下以編輯母片標題樣式 (nitrile) C2H5OH as a common solvent for RX and the nucleophiles(OH, CN…) (amine) 34 New Way Chemistry for Hong Kong A-Level Book 3A Reactions of Halogeno-compounds Two按一下以編輯母片標題樣式 types : 2. Elimination reaction heat HX is eliminated Bases : OH, NH3 35 New Way Chemistry for Hong Kong A-Level Book 3A 35 Reactions of Halogeno-compounds 按一下以編輯母片標題樣式 The attacking species (OH, NH …) can act as 3 nucleophiles or bases depending on the reaction conditions. Elimination reactions always competes with SN reactions 36 New Way Chemistry for Hong Kong A-Level Book 3A 36 As a base 按一下以編輯母片標題樣式 H C C X OHH C As a nucleophile 37 New Way Chemistry for Hong Kong A-Level Book 3A C OH + X- Mechanisms of SN reactions A. 按一下以編輯母片標題樣式 S 1 reactions N First Order Unimolecular Nucleophilic Substitution Reactions B. SN2 reactions Second Order Bimolecular Nucleophilic Substitution Reactions 38 New Way Chemistry for Hong Kong A-Level Book 3A 38 SN1 reactions 按一下以編輯母片標題樣式 C C Br slow + C fast HO C + 50% 39 OH Br New Way Chemistry for Hong Kong A-Level Book 3A C OH 50% 39 Rate = k[ C Br ] 2-step, 1st order, unimolecular 按一下以編輯母片標題樣式 C C Br slow + C fast HO C + 50% 40 OH Br New Way Chemistry for Hong Kong A-Level Book 3A C OH 50% 40 * C OHBr HO *C + * C C H OH, reflux 按一下以編輯母片標題樣式 2 5 Optically active Optically inactive Racemic mixture 41 New Way Chemistry for Hong Kong A-Level Book 3A OH SN2 reactions 按一下以編輯母片標題樣式 C Br HO C OH Rate = k[OH][ C Br ] 1-step, 2nd order, bimolecular 42 New Way Chemistry for Hong Kong A-Level Book 3A + Br SN2 reactions 按一下以編輯母片標題樣式 5-coordinated (Trigonal bipyramidal) 43 New Way Chemistry for Hong Kong A-Level Book 3A SN2 reactions 按一下以編輯母片標題樣式 C Br HO C + Br OH To minimize the repulsion between OH and the leaving group, Br backside attack is preferred inversion of configuration 44 New Way Chemistry for Hong Kong A-Level Book 3A SN2 reactions 按一下以編輯母片標題樣式 inversion * C Br HO *C OH (+) 45 New Way Chemistry for Hong Kong A-Level Book 3A (-) + Br 按一下以編輯母片標題樣式 []=+9.9 []=34.6 OH []=9.9 46 New Way Chemistry for Hong Kong A-Level Book 3A Q.44(a) 按一下以編輯母片標題樣式 H < 0 ∵ E(C – O) > E(C – Cl) 47 New Way Chemistry for Hong Kong A-Level Book 3A 47 Q.44(b) H < 0 按一下以編輯母片標題樣式 ∵ E(C – O) > E(C – Cl) 48 New Way Chemistry for Hong Kong A-Level Book 3A 48 Factors Affecting the Rates of SN Reactions at a saturated (sp3) carbon 按一下以編輯母片標題樣式 A. The structure of the substrate B. The concentration of the nucleophile C. The reactivity of the nuecleophile D. The nature of the leaving group E. The nature of solvent 49 New Way Chemistry for Hong Kong A-Level Book 3A 49 The Structure of the Substrate 按一下以編輯母片標題樣式 1. SN 1 Reactivity depends on the stability of the carbocaion formed in the r.d.s. Stability of carbocations : Benzylic > allylic > 3 > 2 > 1 > CH3+ 50 New Way Chemistry for Hong Kong A-Level Book 3A 50 Alkyl groups stabilize the positively charged 按一下以編輯母片標題樣式 carbocation by positive inductive effect 51 New Way Chemistry for Hong Kong A-Level Book 3A Allylic CH 3 H2C H2C CH 3 按一下以編輯母片標題樣式 C H C H H H Allylic carbocation is stabilized by resonance effect. 52 New Way Chemistry for Hong Kong A-Level 3A New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 C CH3 H benzylic C C CH3 H H The +ve charge is shared by the benzene ring by resonance effect. Resonance effect > inductive effect Stability of carbocations : Benzylic > allylic > 3 > 2 > 1 > CH3+ 53 New Way Chemistry for Hong Kong A-Level 3A New Way Chemistry for Hong Kong A-Level Book 3A CH3 C H CH3 Types of halides Relative rate of SN1 reaction CH3X 1.0 CH3CH2–X (1) 1.6 (CH3)2CH–X (2) 32 (CH3)3C–X (3) 1107 按一下以編輯母片標題樣式 SN1 reactions are much more significant for 3 alkyl halides 54 New Way Chemistry for Hong Kong A-Level Book 3A 2. SN2 Reactivity depends on how easily the 按一下以編輯母片標題樣式 nucleophile can approach the reaction site. Steric effect : More alkyl groups attached to the site stronger van der Waals’ repulsions experenced by the nucleophile when approaching the site higher activation 55 lower reactivity New Way Chemistry for Hong Kong A-Level Book 3A 2. SN2 Reactivity depends on how easily the 按一下以編輯母片標題樣式 nucleophile can approach the reaction site. Steric effect : - CH3X < 1 < 2 < 3 Reactivity : - CH3X > 1 > 2 > 3 56 New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 Steric hindrance : hindrance to reaction caused by bulky groups of atoms around the reaction site 57 New Way Chemistry for Hong Kong A-Level Book 3A H H 按一下以編輯母片標題樣式 H H C H H C C H Nu C H C H H X o 1 substrate with a very bulky alkyl group H Not favours SN2 though 1 58 New Way Chemistry for Hong Kong A-Level Book 3A Types of Relative rate of Relative rate of halides SN1 reaction SN2 reaction 按一下以編輯母片標題樣式 CH3X 1.0 32 CH3CH2–X (1) 1.6 1.0 (CH3)2CH–X (2) 32 0.02 (CH3)3C–X (3) 1107 0 59 New Way Chemistry for Hong Kong A-Level Book 3A Overall reactivity : (SN1 + SN2) 按一下以編輯母片標題樣式 Benzylic, allylic, 3 >> 2, 1 CH3X (mainly SN1) (SN1 & SN2) 60 New Way Chemistry for Hong Kong A-Level Book 3A 1 benzylic and 1 allylic halides favour BOTH SN1 按一下以編輯母片標題樣式 & SN2 mechanisms. However, they proceed faster via SN1 routes. H C H H H2C C H C H 1 allylic 1 benzylic 61 New Way Chemistry for Hong Kong A-Level Book 3A Reactivity of SN reactions depends on 1. Electronic effect 按一下以編輯母片標題樣式 (a) Inductive effect (+ or -) polarization of electron cloud (b) Mesomeric effect (+ or -) polarization of electron cloud (c) Resonance effect delocalization of electron cloud 2. Steric effect 62 New Way Chemistry for Hong Kong A-Level Book 3A The Concentration of the Nucleophile SN按一下以編輯母片標題樣式 1 : no effect the nucleophile is NOT involved in the rate determining step SN2 : [nucleophile] rate of reaction 63 New Way Chemistry for Hong Kong A-Level Book 3A 63 The Reactivity of the Nucleophile (Nucleophilicity) SN1按一下以編輯母片標題樣式 : no effect the nucleophile is NOT involved in the rate determining step SN2 : Nucleophilicity rate of reaction 64 New Way Chemistry for Hong Kong A-Level Book 3A 64 Trends of Nucleophilicity : (i) Charged nucleophiles > neutral ones 按一下以編輯母片標題樣式 H2O + R – Br OH + R – Br slow fast R – OH + HBr R – OH + Br 65 New Way Chemistry for Hong Kong A-Level Book 3A 65 Trends of Nucleophilicity : (ii) For some species, 按一下以編輯母片標題樣式 the order of nucleophilicity follows the order of basicity R O Stronger base > H O Weaker base Stronger Weaker nucleophile nucleophile >> R OH Stronger base Stronger nucleophile 66 New Way Chemistry for Hong Kong A-Level Book 3A > H OH Weaker base Weaker nucleophile 66 (iii)For halide ions, the order of nucleophilicity follows the 按一下以編輯母片標題樣式 order of polarisability but not basicity Nucleophilicity / polarisability : I > Br > Cl > F Basicity : - I < Br < Cl < F Acidity : HI > HBr > HCl > HF 67 New Way Chemistry for Hong Kong A-Level Book 3A 67 Less easily polarized Cl More easily polarized Br Most easily polarized I + C Slower 按一下以編輯母片標題樣式 + C + C 68 New Way Chemistry for Hong Kong A-Level Book 3A Faster Fastest 68 R – Br + I + Na+ propanone SN2 R – I + NaBr(s) 按一下以編輯母片標題樣式 I is a stronger nucleophile than Br the forward reaction is preferred even though E(C – Br) > E(C – I) SN2 mechanism is favoured in non-polar or slightly polar solvents (e.g. propanone) (Refer to effect of solvent on p.72) 69 New Way Chemistry for Hong Kong A-Level Book 3A The Nature of the Leaving Group 1. For halogens, 按一下以編輯母片標題樣式 Bond Bond enthalpy (kJ mol-1) CF +484 C Cl +338 C Br +276 CI +238 Bond strength : C – I < C – Br < C – Cl < C – F 70 New Way Chemistry for Hong Kong A-Level Book 3A 70 The Nature of the Leaving Group 1. For halogens, 按一下以編輯母片標題樣式 Bond strength : C – I < C – Br < C – Cl < C – F The ease of leaving : I > Br > Cl > F Since the r.d.s. of both SN1 and SN2 reactions involves breaking of C – X bonds Reactivity for both SN1 and SN2 reactions : R – I > R – Br > R – Cl > R – F 71 New Way Chemistry for Hong Kong A-Level Book 3A 71 The Nature of the Leaving Group I is按一下以編輯母片標題樣式 both a good leaving group and a good nucleophile Useful in organic synthesis R–X Y Strong I nucleophile R – Y + I Y R–I Good leaving group 72 New Way Chemistry for Hong Kong A-Level Book 3A 72 The Nature of the Leaving Group 2. Neutral molecules are better leaving 按一下以編輯母片標題樣式 groups than anions Ease of leaving : Br > OH (C-O > C-Br) Nucleophilicity/basicity : OH > Br Br + R OH R 73 New Way Chemistry for Hong Kong A-Level Book 3A Br + OH 73 protonation R OH + H+ R protonation OH2 按一下以編輯母片標題樣式 Br R OH2 SN R Br + H2 O substitution betterleaving leaving Better group group Protonating the –OH group weakens the C – O bond. It causes the H2O to leave more easily than the HO 74 New Way Chemistry for Hong Kong A-Level Book 3A H H H O + O 按一下以編輯母片標題樣式 + H C O CH 3 HO H C H H Is the forward reaction preferred ? Explain No. Basicity : CH3 – O > H – O Nucleophilicity : CH3 – O > H – O Ease of leaving : CH3 – O < H – O 75 New Way Chemistry for Hong Kong A-Level Book 3A CH 3 Suggest a route for the following conversion. 按一下以編輯母片標題樣式 CH – OH CH – CN 3 3 Ease of leaving : CN > HO Nucleophilicity/basicity : CN < HO 76 New Way Chemistry for Hong Kong A-Level Book 3A Suggest a route for the following conversion. 按一下以編輯母片標題樣式 CH – OH CH – CN 3 3 PI3 reflux CN CH3 – I Acidity : HCN < HI basicity : CN > I 77 New Way Chemistry for Hong Kong A-Level Book 3A The Nature of solvents (optional) Highly polar solvents favour SN1 reactions 按一下以編輯母片標題樣式 ∵ carbocation can be stabilized by polar solvent via ion-dipole interaction Non-polar or slightly polar solvents favour SN2 reactions R – Br + I + Na+ propanone SN2 R – I + NaBr(s) 78 New Way Chemistry for Hong Kong A-Level Book 3A Unreactivity of Vinylic and Aryl Halides(at sp2 C) 按一下以編輯母片標題樣式 X X sp2 Vinylic halide Aryl halide No significant SN1 & SN2 reactions 79 New Way Chemistry for Hong Kong A-Level Book 3A 79 Interpretation : 1. The carbocation intermediates are highly 按一下以編輯母片標題樣式 unstable SN1 not favoured unstable C vinylic X allylic 80 X CH2 stabilized by resonance effect New Way Chemistry for Hong Kong A-Level Book 3A CH2 Q.45 X CH2 按一下以編輯母片標題樣式 benzylic vacant 2pz orbital c Delocalization of e among 7 C atoms 81 New Way Chemistry for Hong Kong A-Level Book 3A 81 Q.45 CH 2 按一下以編輯母片標題樣式 The positive charge is shared by the benzene ring CH2 CH2 CH2 82 New Way Chemistry for Hong Kong A-Level Book 3A CH2 Q.45 X 按一下以編輯母片標題樣式 Aryl 83 Highly unstable Each structure with TWO C atoms losing the octet New Way Chemistry for Hong Kong A-Level Book 3A Interpretation : 2. C 按一下以編輯母片標題樣式 – X bond strength : - > C X C X > H2C C X CH 2 X 84 New Way Chemistry for Hong Kong A-Level Book 3A The bonds have more 按一下以編輯母片標題樣式 s character due to C sp2 – p overlap X Better overlap C X stronger bond 85 New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 H2C C X The bonds have less s character due to sp3 – p overlap CH 2 Weaker bond X 86 New Way Chemistry for Hong Kong A-Level Book 3A C C 按一下以編輯母片標題樣式 X X Double bond character due to resonance effect Stronger bond 87 Both SN1 & SN2 not favored New Way Chemistry for Hong Kong A-Level Book 3A Double bond character due to resonance effect 按一下以編輯母片標題樣式 X X X Stronger bond Both S 1 & S 2 not favored N N 88 New Way Chemistry for Hong Kong A-Level Book 3A X X Completely-filled 按一下以編輯母片標題樣式 2pz orbital 89 New Way Chemistry for Hong Kong A-Level Book 3A 89 Interpretation : 1. The carbocation intermediates are highly 按一下以編輯母片標題樣式 unstable SN1 not favoured 2. C – X bond strength : - SN1 & SN2 not favored 3. The electron cloud of benzene tends to repel the approaching nucleophiles 90 New Way Chemistry for Hong Kong A-Level Book 3A Vinylic and aryl halides are unreactive 按一下以編輯母片標題樣式 towards SN1 and SN2 reactions as both mechanisms involve the breaking of C – X bond in the r.d.s. 91 New Way Chemistry for Hong Kong A-Level Book 3A Non-SN substitution only happens at drastic conditions 按一下以編輯母片標題樣式 OH e.g. Cl NaOH o O Na H+ 300 C, 200 atm phenol 92 New Way Chemistry for Hong Kong A-Level Book 3A Tests for halogeno-compounds Test按一下以編輯母片標題樣式 1 : Shake RX with AgNO3(aq), and observe the precipitation of AgX (Ethanol is always added as co-solvent) R – X + H2O R – OH + H+ + X X + Ag+ AgX(s) 93 New Way Chemistry for Hong Kong A-Level Book 3A Tests for halogeno-compounds Test按一下以編輯母片標題樣式 2: R – X + OH boil C2H5OH R – OH + X OH(excess) + H+(aq) otherwise OH(excess) + Ag+(aq) X + Ag+ AgX(s) 94 New Way Chemistry for Hong Kong A-Level Book 3A H2 O Ag2O black Positive results : - 按一下以編輯母片標題樣式 AgNO (aq) R – Cl R – Br R–I 3 C2H5OH AgNO3(aq) C2H5OH AgNO3(aq) C2H5OH AgCl (white ppt) AgBr (cream ppt) AgI (yellow ppt) 95 New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 AgCl(s) AgBr(s) AgI(s) Reactivity : R – I > R – Br > R – Cl 96 New Way Chemistry for Hong Kong A-Level Book 3A 96 Reactivity : - 按一下以編輯母片標題樣式 R – I > R – Br > R – Cl O R acyl > benzylic > allylic C Cl > 3 > 2 > 1 > CH3X Aryl and vinylic halides give –ve results 97 New Way Chemistry for Hong Kong A-Level Book 3A 97 O 按一下以編輯母片標題樣式 R C + Cl The carbonyl C is highly positive due to 1. –ve inductive effective of O 2. –ve inductive effective of Cl 3. –ve mesomeric effective of O 98 New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 2R’OH + 2Na 2RONa+ + H2 acidic 99 New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 2R’CCH + 2Na 2R’CCNa+ + H2 100 acidic New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 101 2R’CCH + NH2 2R’CC + NH3 acidic acetylide ion New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 Products have one more C atom than RX 102 lengthening of carbon chain New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 Lengthening of carbon chain 103 New Way Chemistry for Hong Kong A-Level Book 3A Q.46 + + OH 按一下以編輯母片標題樣式 CN OH CN OH is a better nucleophile than CN The backward reaction is favoured In acidic medium, CN loses its nucleophilicity H+ + CN HCN (weak acid) 104 New Way Chemistry for Hong Kong A-Level Book 3A Q.46 OH COOH 按一下以編輯母片標題樣式 reflux H3O+ PBr3 Br CNC2H5OH, reflux 105 New Way Chemistry for Hong Kong A-Level Book 3A reflux CN Q.46 OH COOH 按一下以編輯母片標題樣式 reflux H3O+ PI3 I CNC2H5OH, reflux 106 New Way Chemistry for Hong Kong A-Level Book 3A reflux CN Q.46 OH COOH 按一下以編輯母片標題樣式 reflux H3O+ HI I CNC2H5OH, reflux 107 New Way Chemistry for Hong Kong A-Level Book 3A reflux CN Q.46 OH COOH 按一下以編輯母片標題樣式 25C H3O+ SOCl2 Cl CNC2H5OH, reflux 108 New Way Chemistry for Hong Kong A-Level Book 3A reflux CN 按一下以編輯母片標題樣式 further substitution 109 New Way Chemistry for Hong Kong A-Level Book 3A H H H S 2 按一下以編輯母片標題樣式 CH N C Br N + Br N 3 H H H H Compete with each other for CH3Br H H H3C N H H 110 New Way Chemistry for Hong Kong A-Level Book 3A + H Nucleophilicity : - 按一下以編輯母片標題樣式 H H3C H3C H3C N H3C H3C > N H3C H > N H H > N H H Further substitution reactions occur CH3NH2 + CH3Br (CH3)2NH + CH3Br (CH3)3N + CH3Br 111 (CH3)2NH + HBr (CH3)3N + HBr (CH3)4N+Br New Way Chemistry for Hong Kong A-Level Book 3A NH3 + CH3Br CH3NH2 + HBr 1 amine 按一下以編輯母片標題樣式 CH NH + CH Br (CH ) NH + HBr 3 2 3 3 2 2 amine (CH3)2NH + CH3Br (CH3)3N + HBr 3 amine (CH3)3N + CH3Br (CH3)4N+Br 4 ammonium salt A mixture of products is obtained 112 Not suitable for preparing 1 amine New Way Chemistry for Hong Kong A-Level Book 3A NH3 + excess CH3Br CH3NH2 + HBr major 按一下以編輯母片標題樣式 To prepare 1 amino, NH3 must be used in large excess 113 New Way Chemistry for Hong Kong A-Level Book 3A 按一下以編輯母片標題樣式 114 For strong bases, elimination competes with SN New Way Chemistry for Hong Kong A-Level Book 3A Conditions in favour of Elimination 按一下以編輯母片標題樣式 1. higher temperature 2. the presence of a strong base 115 New Way Chemistry for Hong Kong A-Level Book 3A In a strongly basic medium, 按一下以編輯母片標題樣式 H C C X + OH C C + H–OH + X elimination In a strongly acidic medium, H C C X C C addition 116 New Way Chemistry for Hong Kong A-Level Book 3A ++ HH++ ++ XX- Mechanisms of Elimination Reactions(Optional) 1-step, 2nd order, bimolecular E2 按一下以編輯母片標題樣式 HO H C C C C X dehydrohalogenation 117 New Way Chemistry for Hong Kong A-Level Book 3A + X + HOH E1 2-step, 1st order, unimolecular OH 按一下以編輯母片標題樣式 H H slow C C C C + X X fast C 118 New Way Chemistry for Hong Kong A-Level Book 3A C + HOH Q.47 CH 3 H3C 按一下以編輯母片標題樣式 H2C C Br CH 3 C2H5OH, heat NaOH H3C C2H5 CH 3 C + C H CH 3 major H C C H3C H minor 119 New Way Chemistry for Hong Kong A-Level Book 3A Elimination of Dihalides 按一下以編輯母片標題樣式 H B H R H R' R Br R' Br Br H R' B more difficult R R Br 120 New Way Chemistry for Hong Kong A-Level Book 3A C C alkyne R' Elimination of Dihalides 按一下以編輯母片標題樣式 Br B H R Br R' R' Br Br R H R' H B more difficult R R H 121 New Way Chemistry for Hong Kong A-Level Book 3A C C alkyne R' Stronger bases can be used to convert dihalides to alkynes directly 按一下以編輯母片標題樣式 NH2 from NaNH2 (sodamide) C2H5O from C2H5ONa (sodium ethoxide) 122 New Way Chemistry for Hong Kong A-Level Book 3A Suggest a route for the each of the H H following conversions. 按一下以編輯母片標題樣式 H3C H3C H H C C Br Br CH 3 H3C H3C C C H H Br Br C C Br Br H Br C C H Br 123 New Way Chemistry for Hong Kong A-Level Book 3A CH 3 CH 3 CH 3 Elimination vs Substitution 按一下以編輯母片標題樣式 E1 vs S N1 When the same reactants are used, all factors (except temperature) affect E1 and SN1 reactions equally. Temperature is the ONLY factor for distinguishing between E1 and SN1 124 New Way Chemistry for Hong Kong A-Level Book 3A SN1 E1 involves the breaking of strong C–H bond H CH 按一下以編輯母片標題樣式 3 H OH C H C CH3 Only happens at higher temperatures E1 At higher temperatures, Vibrational motions of the groups around the reaction site prevent the nucleophile from approaching the reaction site. S 1 is not favoured. N 125 New Way Chemistry for Hong Kong A-Level Book 3A E1 vs SN1 : Example 按一下以編輯母片標題樣式 lower T higher T No need to distinguish between E1 & E2 126 New Way Chemistry for Hong Kong A-Level Book 3A 126 E2 vs SN2 按一下以編輯母片標題樣式 Effect of structure on reactivity SN2 : CH3X > 1 > 2 > 3 E2 : 3 > 2 > 1 CH3X Alkene 127 New Way Chemistry for Hong Kong A-Level Book 3A E2 : 3 > 2 > 1 OH 按一下以編輯母片標題樣式 H H C H H H H C C C H X H H (3) No. of reaction site : 3 > 2 > 1 128 New Way Chemistry for Hong Kong A-Level Book 3A SN2 predominates 1 favours E2 if a 按一下以編輯母片標題樣式 CH3X, 1 bulky and strong base is used 2 E2 competes with SN2 3 E2 predominates in strongly basic medium 129 New Way Chemistry for Hong Kong A-Level Book 3A SN2 H2C H2C C O H H 按一下以編輯母片標題樣式 H2C E2 Bulky and strong base H C C H H X (1) (1) SN reactions are suppressed due to great steric hindrance (2) Strong base favours E2 reactions 130 New Way Chemistry for Hong Kong A-Level Book 3A Effect of strength of base/nucleophile on reactivity 按一下以編輯母片標題樣式 Strong bases favour E2 more than E1 Strong nucleophiles favour SN2 more than SN1 However, the attacking species tends to behave as a strong base rather than a strong nucleophile. 131 New Way Chemistry for Hong Kong A-Level Book 3A Effect of strength of base/nucleophile on reactivity 按一下以編輯母片標題樣式 Reasons : - E2 reactions involve the breaking of the strong C–H bond high activation energy E2 is more favoured than SN2 when approached by a strong attacking species 132 New Way Chemistry for Hong Kong A-Level Book 3A Effect of temperature on reactivity H3C H3C H3C 按一下以編輯母片標題樣式 Same as above for SN2 C O E2 H H H C C H H E1 vs SN1 X The effect of temperature is more important for comparing E2/SN2 reactions than E1/SN1 reactions since rate determining steps are considered in the former. 133 New Way Chemistry for Hong Kong A-Level Book 3A Reaction conditions Structure Strength of Base 1o 2o 3o 按一下以編輯母片標題樣式 Temperature CH3X Strong High SN2 SN2 E2 E2 Strong Low SN2 SN2 E2/SN2 E2 Weak High SN2 SN2 SN2/E2 E1 Weak Low SN2 SN2 SN2 SN1 134 New Way Chemistry for Hong Kong A-Level Book 3A Q.49 strong Elimination (a) 按一下以編輯母片標題樣式 (CH3)2CHONa + CH3Br ? SN2 (CH3)2CH–O–CH3 + NaBr Major(100%) 135 New Way Chemistry for Hong Kong A-Level Book 3A Q.49 (b) 2 strong 按一下以編輯母片標題樣式 (CH3)2CHBr + CH3ONa SN2 E2 H (CH3)2CH–O–CH3 CH 3 C H 136 C H side-product New Way Chemistry for Hong Kong A-Level Book 3A Q.50 strong 1 (a) 按一下以編輯母片標題樣式 CH3CH2ONa + CH3CH2Br C2H5OH 55C CH3CH2 – OCH2CH3 + CH2=CH2 Major(90%) SN2 Minor(10%) E2 137 New Way Chemistry for Hong Kong A-Level Book 3A Q.50 CH 3 strong 3 C CH 3 (b) 按一下以編輯母片標題樣式 CH3CH2ONa + H3C C2H5OH 55C H CH 3 C H Br C CH 3 138 New Way Chemistry for Hong Kong A-Level Book 3A 100% E2 Q.50 CH 3 strong 3 C CH 3 CH3CH2ONa + H3C 按一下以編輯母片標題樣式 25C C2H5OH CH 3 H3C C H CH 3 OC 2H5 139 Br Minor(10%) SN1 + CH 3 C H C CH 3 Major(90%) E2 New Way Chemistry for Hong Kong A-Level Book 3A Q.48(a) 1 按一下以編輯母片標題樣式 C H O Na - + 2 5 CH2CH2Br C2H5OH, heat C H CH2 E2 is preferred to SN2 since 1. strong base and high temperature favour elimination 2. the product is stabilized by extensive delocalization of electrons 140 New Way Chemistry for Hong Kong A-Level Book 3A Q.48(b) E 按一下以編輯母片標題樣式2 C2H5O-Na+ CH2CH2Br C2H5OH, heat - + C2H5O Na CH2CH2Br C H CH2 SN2 CH2CH2OC2H5 C2H5OH, heat 141 New Way Chemistry for Hong Kong A-Level Book 3A Q.48(a) 按一下以編輯母片標題樣式 C H O Na - + 2 5 CH2CH2Br C2H5OH, heat weaker base OH- C H heat C2H5OH, 25oC Markovnikov’s addition SN2 C H2 CH 2OH (1) 142 New Way Chemistry for Hong Kong A-Level Book 3A CH2 dilute H + H C OH (2) CH 3 Q.48(c) E2 C H O Na 按一下以編輯母片標題樣式 - + 2 5 Br C2H5OH, heat (2) CH3CCl3 Br 2 transaddition Br Br 143 New Way Chemistry for Hong Kong A-Level Book 3A Q.48(d) CH3CH2CH2Br (1) H2, Pt 按一下以編輯母片標題樣式 CH2=CHCH2Br H H H NaOH H C C C H C2H5OH, heat H Br Br 144 New Way Chemistry for Hong Kong A-Level Book 3A Q.48(d) CH3CH2CH2Br (1) H2, Pt 按一下以編輯母片標題樣式 CH2=CHCH2Br H H H NaOH H C C C H C2H5OH, heat H Br CH3CH=CHBr Br 145 New Way Chemistry for Hong Kong A-Level Book 3A H2, Pt Q.48(d) CH3CH2CH2Br (1) H2, Pt 按一下以編輯母片標題樣式 CH2=CHCH2Br H H H NaOH H C C C H C2H5OH, heat H Br CH3CH=CHBr Br No recommended since a mixture of products is formed CH3CBr=CH2 H2, Pt CH3CHBrCH3 (2) 146 New Way Chemistry for Hong Kong A-Level Book 3A H2, Pt Q.48(d) H H H C H ONa 按一下以編輯母片標題樣式 2 5 H C C C H H3C C CH C2H5OH, heat H Br Br C2H5OH, heat H3C Markovnikov’s addition H (2) 147 H2 H3C C Br C-Na + C cold limited HBr H3C CH 3 H C Pt C2H5ONa Br New Way Chemistry for Hong Kong A-Level Book 3A C H Q.51(a) frontside strong attack base 按一下以編輯母片標題樣式 CH 3 (2) C H Br C2H5 E2 CH 3 C H C2H5O- heat H3C backside attack H3C + C H H C H H + C CH 3 C2H5 C H 148 New Way Chemistry for Hong Kong A-Level Book 3A C H Q.51(b) Strong (2) CH 3 CH 3 base SN2 OC H 按一下以編輯母片標題樣式 2 5 C C I C2H5 H Br inversion E2 A mixture of three alkenes H C2H5 I inversion SN2 CH3 C C2H5 H OC2H5 minor 149 New Way Chemistry for Hong Kong A-Level Book 3A CH 3 I CH 3 SN2 OH 按一下以編輯母片標題樣式 C H weaker inversion I C H Br C2H5 2 5 SN2 CH 3 C C2H5 SN2 H O-Na + 3C CH 3 CH3 C C H Br H OC2H5 C2H5 H OH major 150 New Way Chemistry for Hong Kong A-Level Book 3A base 25C CH 2 C2H5 C H 按一下以編輯母片標題樣式 151 New Way Chemistry for Hong Kong A-Level Book 3A 30.2 Nomenclature of Halogeno-compounds (SB p.210) 按一下以編輯母片標題樣式 Draw the structural formulae and give the IUPAC names of all isomers with the following molecular formulae. (a) C4H9Br Answer 152 New Way Chemistry for Hong Kong A-Level Book 3A 152 30.2 Nomenclature of Halogeno-compounds (SB p.210) (a) 按一下以編輯母片標題樣式 153 New Way Chemistry for Hong Kong A-Level Book 3A 153 30.2 Nomenclature of Halogeno-compounds (SB p.210) 按一下以編輯母片標題樣式 Draw the structural formulae and give the IUPAC names of all isomers with the following molecular formulae. (b) C4H8Br2 Answer 154 New Way Chemistry for Hong Kong A-Level Book 3A 154 30.2 Nomenclature of Halogeno-compounds (SB p.210) (b) 按一下以編輯母片標題樣式 Back 155 New Way Chemistry for Hong Kong A-Level Book 3A 155 30.2 Nomenclature of Halogeno-compounds (SB p.211) 按一下以編輯母片標題樣式 Draw the structural formulae and give the IUPAC names for all the structural isomers of C5H11Br. Answer 156 New Way Chemistry for Hong Kong A-Level Book 3A 156 30.2 Nomenclature of Halogeno-compounds (SB p.211) 按一下以編輯母片標題樣式 157 New Way Chemistry for Hong Kong A-Level Book 3A 157 30.2 Nomenclature of Halogeno-compounds (SB p.211) Back 按一下以編輯母片標題樣式 158 New Way Chemistry for Hong Kong A-Level Book 3A 158 30.4 Preparation of Halogeno-compounds (SB p.218) Back 按一下以編輯母片標題樣式 State the major products of the following reactions. (a) CH3CHOHCH2CH3 + PBr3 (b) CH3CH=CH2 + HBr (c) (d) CH3CCH + 2HBr (a) (b) (c) (d) CH3CHBrCH2CH3 CH3CHBrCH3 CH3CBr2CH3 Answer 159 New Way Chemistry for Hong Kong A-Level Book 3A 159 30.6 Nucleophilic Substitution Reactions (SB p.230) 按一下以編輯母片標題樣式 What chemical test could you use to distinguish between the members of the following pairs of compounds? (a) CH3CH2CH2CH2Cl and CH3CH2CH = CHCl (b) and Answer 160 New Way Chemistry for Hong Kong A-Level Book 3A 160 30.6 Nucleophilic Substitution Reactions (SB p.230) (a) (b) 按一下以編輯母片標題樣式 Reagent used: Br in 1,1,1-trichloroethane 2 CH3CH2CH = CHCl decolorizes the reagent while CH3CH2CH2CH2Cl shows no observable change. Reagent used: aqueous alcoholic AgNO3 reacts to give a pale yellow precipitate while does not react. Back 161 New Way Chemistry for Hong Kong A-Level Book 3A 161 30.6 Nucleophilic Substitution Reactions (SB p.232) The reactions between three bromine-containing compounds and silver nitrate solution at room conditions are summarized in the following table: 按一下以編輯母片標題樣式 Compound Sodium bromide 1-Bromobutane Bromobenzene 162 Reaction with silver nitrate solution A pale yellow precipitate appears immediately No reaction at first; a pale yellow precipitate appears after several minutes No reaction even after several hours New Way Chemistry for Hong Kong A-Level Book 3A 162 30.6 Nucleophilic Substitution Reactions (SB p.232) 按一下以編輯母片標題樣式 (a) What is the pale yellow precipitate produced in the reaction between silver nitrate and sodium bromide? (b) Write an ionic equation for the reaction. (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? (d) Briefly explain why bromobenzene does not give any precipitate with silver nitrate solution. Answer 163 New Way Chemistry for Hong Kong A-Level Book 3A 163 30.6 Nucleophilic Substitution Reactions (SB p.232) Back (a) (b) (c) (d) 按一下以編輯母片標題樣式 Silver bromide Ag+(aq) + Br–(aq) AgBr(s) The hydrolysis of 1-bromobutane takes time. Precipitation of AgBr occurs only after OH– from water has replaced Br– from 1-bromobutane. The C Br bond of bromobenzene is strengthened due to the delocalization of π 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 silver nitrate solution. 164 New Way Chemistry for Hong Kong A-Level Book 3A 164 30.6 Nucleophilic Substitution Reactions (SB p.232) Back 按一下以編輯母片標題樣式 Which is the stronger nucleophile in each of the following pairs? Explain your choice briefly. (a) OH– and H2O (b) OH– and CH3CH2O– (a) (b) Answer OH– is a stronger nucleophile than H2O because it carries a negative charge while H2O is electrically neutral. CH3CH2O– is a stronger nucleophile than OH–. It is because the ethyl group (CH3CH2 ) is an electron-releasing group. It increases the electron density on the oxygen atom. This makes CH3CH2O– a stronger nucleophile than OH–. 165 New Way Chemistry for Hong Kong A-Level Book 3A 165 30.6 Nucleophilic Substitution Reactions (SB p.233) 按一下以編輯母片標題樣式 Predict whether the following substitution reactions follow mainly SN1 or SN2 pathway. Briefly explain your answer. (a) CH3I + OH– CH3OH + I– 166 New Way Chemistry for Hong Kong A-Level Book 3A Answer 166 30.6 Nucleophilic Substitution Reactions (SB p.233) 按一下以編輯母片標題樣式 (a) The reaction follows mainly the SN2 mechanism. It is because 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. 167 New Way Chemistry for Hong Kong A-Level Book 3A 167 30.6 Nucleophilic Substitution Reactions (SB p.233) 按一下以編輯母片標題樣式 Predict whether the following substitution reactions follow mainly SN1 or SN2 pathway. Briefly explain your answer. (b) Answer 168 New Way Chemistry for Hong Kong A-Level Book 3A 168 30.6 Nucleophilic Substitution Reactions (SB p.233) 按一下以編輯母片標題樣式 (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 in the SN1 reaction is stabilized by both the inductive effect of the electron-releasing ethyl group and the resonance effect of the phenyl group. Back 169 New Way Chemistry for Hong Kong A-Level Book 3A 169 30.6 Nucleophilic Substitution Reactions (SB p.234) 按一下以編輯母片標題樣式 Can you tell why haloalkanes undergo substitution reactions more readily than alkanes, but the reaction takes place fairly slowly? Answer It is because haloalkanes possess a polar C X bond inviting the attack of nucleophiles or bases. However, the polarity of the C X bond is not so high. Back 170 New Way Chemistry for Hong Kong A-Level Book 3A 170 30.6 Nucleophilic Substitution Reactions (SB p.235) 按一下以編輯母片標題樣式 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– (a) (b) (c) Answer dilute NaOH CH3CH2O–Na+ or Na in CH3CH2OH NH3 in excess CH3I Back 171 New Way Chemistry for Hong Kong A-Level Book 3A 171 30.6 Nucleophilic Substitution Reactions (SB p.235) 按一下以編輯母片標題樣式 Give the name(s) and structural formula(e) of the major organic product(s) formed in each of the following reactions. (a) Answer (a) 172 New Way Chemistry for Hong Kong A-Level Book 3A 172 30.6 Nucleophilic Substitution Reactions (SB p.235) 按一下以編輯母片標題樣式 Give the name(s) and structural formula(e) of the major organic product(s) formed in each of the following reactions. (b) Answer (b) 173 New Way Chemistry for Hong Kong A-Level Book 3A 173 30.6 Nucleophilic Substitution Reactions (SB p.235) Back 按一下以編輯母片標題樣式 Give the name(s) and structural formula(e) of the major organic product(s) formed in each of the following reactions. (c) Answer (c) 174 New Way Chemistry for Hong Kong A-Level Book 3A 174 30.7 Elimination Reactions (SB p.239) 按一下以編輯母片標題樣式 How can the following compounds be made from bromopropane? (a) Propanol (b) Propene (c) Butanenitrile Answer (a) (b) (c) By heating in NaOH By heating in alcoholic KOH By reacting with aqueous alcoholic KCN Back 175 New Way Chemistry for Hong Kong A-Level Book 3A 175 30.7 Elimination Reactions (SB p.240) 按一下以編輯母片標題樣式 (a) Hot and concentrated alcoholic potassium hydroxide can eliminate hydrogen iodide from the compound CH3CH2CHICH3. Suggest and name two possible products. Answer (a) 176 New Way Chemistry for Hong Kong A-Level Book 3A 176 30.7 Elimination Reactions (SB p.240) Back 按一下以編輯母片標題樣式 (b) Draw the structural formulae and give the names of all possible products formed by elimination of hydrogen bromide from the dibromoalkane, CH3CHBrCHBrCH3. Answer (b) 177 New Way Chemistry for Hong Kong A-Level Book 3A 177 30.7 Elimination Reactions (SB p.240) 按一下以編輯母片標題樣式 (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 atom available on the carbon atom adjacent to the carbon atom that is directly bonded to the iodine atom. 178 New Way Chemistry for Hong Kong A-Level Book 3A 178 30.7 Elimination Reactions (SB p.240) 按一下以編輯母片標題樣式 (b) 2-Iodo-2-methylbutane gives two elimination products: one is 2-methylbut-2-ene, what is the other one? (b) 2-Methylbut-1-ene 179 New Way Chemistry for Hong Kong A-Level Book 3A Answer 179 30.7 Elimination Reactions (SB p.240) 按一下以編輯母片標題樣式 (c) Arrange the following compounds in order of increasing tendency towards elimination reactions: A: 2-bromo-2-methylbutane, B: 1-bromopentane and C: 2-bromopentane Explain your answer. 180 New Way Chemistry for Hong Kong A-Level Book 3A Answer 180 30.7 Elimination Reactions (SB p.240) 按一下以編輯母片標題樣式 (c) B <C <A The rate of elimination depends on the stability of the alkene formed. The condensed structural formulae of the alkenes formed from the elimination reactions is: A: CH3CH = CCH3CH3 B: CH3CH2CH2CH = CH2 C: CH3CH2CH = CHCH3 A more highly substituted alkene is more stable and is formed more readily. Back 181 New Way Chemistry for Hong Kong A-Level Book 3A 181
