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Source: http://www.doksinet Chapter 9 Addition Reactions of Alkenes Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 9. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary. • Addition reactions are thermodynamically favorable at temperature and disfavored at temperature. • Hydrohalogenation reactions are regioselective, because the halogen is generally placed at the substituted position, called addition. • In the presence of , addition of HBr proceeds via an antiMarkovnikov addition. • The regioselectivity of an ionic addition reaction is determined by the preference for the reaction to proceed through . • Acid-catalyzed hydration is inefficient when are possible. Dilute acid favors formation of the and while concentrated

acid favors the . • Oxymercuration-demercuration achieves hydration of an alkene without . • - can be used to achieve an anti-Markovnikov addition of water across an alkene. The reaction is stereospecific and proceeds via a addition. • Asymmetric hydrogenation can be achieved with a catalyst. • Bromination proceeds through a bridged intermediate, called a , which is opened by an SN2 process that produces an addition. • A two-step procedure for anti dihydroxylation involves conversion of an alkene to an , followed by acid-catalyzed ring opening. • Ozonolysis can be used to cleave a double bond and produce two groups. • The position of a leaving group can be changed via followed by . • The position of a π bond can be changed via followed by . Review of Skills Fill in the

blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 9. The answers appear in the section entitled SkillBuilder Review. 9.1 Drawing a Mechanism for Hydrohalogenation STEP 1 - DRAW TWO CURVED ARROWS SHOWING PROTONATION OF THE ALKENE, AND DRAW THE CARBOCATION THAT IS FORMED. STEP 2 - DRAW ONE CURVED ARROW THAT SHOWS THE HALIDE ION ATTACKING THE CARBOCATION, AND DRAW THE PRODUCT. H X + X Source: http://www.doksinet CHAPTER 9 9.2 Drawing a Mechanism for Hydrohalogenation with a Carbocation Rearrangement STEP 1 - DRAW TWO CURVED ARROWS SHOWING PROTONATION OF THE ALKENE AND DRAW THE CARBOCATION THAT IS INITIALLY FORMED. H STEP 2 - DRAW ONE CURVED ARROW SHOWING A CARBOCATION REARRANGEMENT AND DRAW THE RESULTING, MORE STABLE CARBOCATION. STEP 3 - DRAW ONE CURVED ARROW SHOWING THE HALIDE ION ATTACKING THE CARBOCATION, AND DRAW THE PRODUCT. Cl Cl + Cl 9.3 Drawing a Mechanism for an Acid-Catalyzed Hydration STEP 1 - DRAW

TWO CURVED ARROWS SHOWING PROTONATION OF THE ALKENE, AND DRAW THE RESULTING CARBOCATION. H H STEP 2 - DRAW ONE CURVED ARROW SHOWING WATER ATTACKING THE CARBOCATION, AND DRAW THE RESULTING OXONIUM ION O H H O STEP 3 - DRAW TWO CURVED ARROWS SHOWING DEPROTONATION OF THE OXONIUM ION, AND DRAW THE RESULTING PRODUCT. H H O H 9.4 Predicting the Products of Hydroboration-Oxidation DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP 1) BH3 THF + 2) H2O2, NaOH RELATIONSHIP = 9.5 Predicting the Products of Catalytic Hydrogenation DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP H2 + Pt RELATIONSHIP = 9.6 Predicting the Products of Halohydrin Formation DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP Br2 H2O + RELATIONSHIP = 173 Source: http://www.doksinet 174 CHAPTER 9 9.7 Drawing

the Products of Anti Dihydroxylation DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP H 1) MCPBA + 2) H3O+ RELATIONSHIP = 9.8 Predicting the Products of Ozonolysis DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION. 1) O3 + 2) DMS 9.9 Predicting the Products of an Addition Reaction DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION. 1) BH3 THF + 2) H2O2, NaOH 9.10 Proposing a One-Step Synthesis IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION: OH 9.11 Changing the Position of a Leaving Group IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION: Br 1) Br 2) 9.12 Changing the Position of a π Bond IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION: 1) 2) Source: http://www.doksinet 175 CHAPTER 9 Review of Reactions Identify the reagents necessary to achieve each of the following transformations. To verify that your answers are correct, look in your textbook

at the end of Chapter 9. The answers appear in the section entitled Review of Reactions. O X H O OH + En Br OH OH OH + En OH OH + En Br + En OH Br + En Br Solutions 9.1 HBr HBr Br a) b) HBr HCl Br c) d) I HI e) f) Br HBr ROOR Br ROOR Cl Source: http://www.doksinet 176 CHAPTER 9 9.2 Br Br HBr HBr a) ROOR b) 9.3 H Br Br Br + a) Cl H Cl Cl + b) H Cl c) Cl Cl + 9.4 b) a) c) d) 9.5 In this case, the less-substituted carbocation is more stable because it is resonancestabilized: Cl H O O O O + Cl Cl resonance-stabilized 9.6 HBr + a) Br Br Cl HCl b) HBr c) Br + Br Source: http://www.doksinet 177 CHAPTER 9 HI + I d) I Cl HCl e) Cl HCl Cl + f) 9.7 H Br Br Hydride Shift H a) H Br Br Br Hydride Shift H b) Br Cl H c) Cl Methyl Shift Cl 9.8 Br H Ring Expansion Br Br Source: http://www.doksinet 178 CHAPTER 9 9.9 H H Br Methyl Shift This rearrangement converts a secondary carbocation into a

more stable tertiary carbocation. Hydride Shift This rearrangement converts a tertiary carbocation into a more stable, resonancestabilized, tertiary carbocation. Br Br 9.10 a) , because the reaction proceeds via a tertiary carbocation, rather than a secondary carbocation. b) 2-methyl-2-butene, because the reaction proceeds via a tertiary carbocation, rather than a secondary carbocation. 9.11 a) To favor the alcohol, dilute sulfuric acid (mostly water) is used. Having a high concentration of water favors the alcohol according to Le Chatelier’s principle. b) To favor the alkene, concentrated sulfuric acid (which has very little water) is used. Having a low concentration of water favors the alkene according to Le Chatelier’s principle. 9.12 H H O H H O H O H H H O OH H a) H H O H H O H O H H H O OH H b) H H H O H c) H O H O H H O OH H Source: http://www.doksinet CHAPTER 9 9.13 O H O S O H O H O Me O H Me 9.14 O H O S O H O OH OH H O O H H

O (even concentrated H2SO4 has some water present) 9.15 a) OH 1) Hg(OAc)2, H2O 2) NaBH4 HO + H3O b) OH 1) Hg(OAc)2, H2O 2) NaBH4 H 3O + OH H O Me OMe 179 Source: http://www.doksinet 180 CHAPTER 9 c) 1) Hg(OAc)2, H2O OH 2) NaBH4 H 3O + 9.16 1) Hg(OAc)2 , EtOH 2) NaBH4 OEt a) 1) Hg(OAc) 2 , EtNH2 2) NaBH4 H b) N Et 9.17 1) BH3 THF a) 2) H2O2, NaOH OH 1) BH3 THF b) OH 2) H2O2, NaOH 1) BH3 THF c) OH 2) H2O2, NaOH 9.18 1) BH3 THF OH 2) H2O2, NaOH 9.19 1) BH3 THF a) 2) H2O2, NaOH + En OH Source: http://www.doksinet CHAPTER 9 181 OH 1) BH3 THF b) + En 2) H2O2, NaOH OH 1) BH3 THF c) 2) H2O2, NaOH 1) BH3 THF d) + En 2) H2O2, NaOH 1) BH3 THF e) HO HO 2) H2O2, NaOH H 1) BH3 THF f) + En 2) H2O2, NaOH OH 9.20 1) BH3 THF 2) H2O2, NaOH + En OH 9.21 Only one chirality center is formed, so both possible stereoisomers (enantiomers) are obtained, regardless of the configuration of the starting alkene: 1) BH3 THF 2) H2O2, NaOH + 1)

BH3 THF 2) H2O2, NaOH 9.22 OH OH Source: http://www.doksinet 182 CHAPTER 9 9.23 H2 Ni a) H2 Pd b) H2 + En Pt c) H2 + En Ni d) H2 Pt e) H2 Pd (meso) f) 9.24 D D2 + En Pt D 9.25 a) 1) BH3 THF b) 2) H2O2, NaOH OH Source: http://www.doksinet CHAPTER 9 9.26 Br Br2 + En Br a) Br Br2 + En Br b) Br Br Br2 + En c) Br Br2 + En Br d) 9.27 Br + En a) OH OH HO Br Br c) Br + En d) OH 9.28 Br2 OEt + En OH a) Br Et Br2 N H + En EtNH 2 b) + En + En b) Br 183 Source: http://www.doksinet 184 CHAPTER 9 9.29 The bromonium ion can open (before a bromide ion attacks), forming a resonance stabilized carbocation. This carbocation is trigonal planar and can be attacked from either side: Br Br2 Br resonance-stabilized 9.30 OH HO OH HO + En OH + En a) b) OH c) OH OH OH OH OH d) e) + En (meso) OH f) 9.31 OEt 1) MCPBA + En 2) [H2SO4] , OH a) OH O OH O b) [H2SO4] OH 9.32 a) MCPBA H3O+ O HO OH no chirality

centers MCPBA O H3O+ OH OH no chirality centers + En Source: http://www.doksinet CHAPTER 9 b) H O MCPBA H Et H Et + En H3O+ HO H Et H Et H OH meso 9.33 OH OsO4 (catalytic) + En NMO OH a) OH 1) OsO4 b) OH + En 2) NaHSO3 / H2O OH KMnO4, NaOH cold OH (meso) c) OH KMnO4, NaOH cold OH d) OH OsO4 (catalytic) OH + En OOH , NaOH e) OH OsO4 (catalytic) f) NMO + En OH 185 Source: http://www.doksinet 186 CHAPTER 9 9.34 a) H 1) O3 O O O 2) DMS O b) 1) O3 O O 2) DMS O O c) H 1) O3 O O 2) DMS H d) H 1) O3 O O 2) DMS H e) H O H 1) O3 2) DMS H meso O O H O f) 1) O3 O 2) DMS O 9.35 1) O3 a) 2) DMS O O O 1) O3 b) 2) DMS Source: http://www.doksinet CHAPTER 9 O 1) O3 c) O 2) DMS 9.36 1) BH3 THF a) OH 2) H2O2, NaOH H2 b) + En Pt HO 1) CH3CO3H c) OH 2) H3O+ OH 1) OsO4 d) + En + En 2) NaHSO3 / H2O H3O+ OH OH + En e) Br HBr f) 1) MCPBA OH + En 2) H3O+ OH g) OH 1) BH3 THF h) 2) H2O2,

NaOH OsO4 (catalytic) + En OH + En NMO i) HO 187 Source: http://www.doksinet 188 CHAPTER 9 9.37 OH KMnO4 , NaOH OH + cold OH OH Diastereomers 9.38 The products are the same: HO KMnO4 , NaOH + En cold OH (2R,3R) HO 1) MCPBA OH + En 2) H3O+ (2R,3R) 9.39 O + O H 1) O3 1) BH3 THF 2) DMS 2) H2O2, NaOH Compounds E + F Compound A Compounds B + C HBr Br Compound D 9.40 OH 1) BH3 THF + En 2) H2O2, NaOH a) Br t-BuOK b) HBr c) ROOR Br + OH OH Source: http://www.doksinet CHAPTER 9 H2 Pt d) Cl HCl e) 1) BH3 THF OH + En 2) H2O2, NaOH f) Br NaOEt g) Br HBr h) 9.41 Br HBr ROOR a) HBr Br b) HO KMnO4 , NaOH OH cold c) 1) MCPBA HO 2) H3O+ d) OH + En 9.42 Cl 1) NaOMe 2) HCl a) Cl 189 Source: http://www.doksinet 190 CHAPTER 9 1) TsCl, py OH 2) t-BuOK HO 3) BH3 THF 4) H2O2, NaOH b) Br 1) t-BuOK HO 2) BH3 THF 3) H2O2, NaOH c) OH 1) conc. H2SO4 + En 2) BH3 THF OH 3) H2O2, NaOH d) 9.43 Br 1) NaOMe 2) HBr,

ROOR a) Br 1) NaOMe 2) HBr, ROOR b) Br Br 9.44 1) NaOMe 2) HBr Br 3) t-BuOK 4) HBr, ROOR 9.45 a) 1) HBr 2) NaOMe b) 1) HBr 2) t-BuOK 1) HBr 2) NaOMe Br Source: http://www.doksinet CHAPTER 9 191 9.46 1) HBr, ROOR 2) t-BuOK a) 1) HBr 2) NaOMe b) 9.47 a) 1) HBr 2) NaOMe 1) HBr 2) NaOMe 3) HBr, ROOR 4) t-BuOK 3) HBr, ROOR 4) t-BuOK b) 9.48 A reaction is only favorable if ∆G is negative Recall that ∆G has two components: (∆H) and (-T∆S). The first term (∆H) is positive for this reaction (two sigma bonds are converted into one sigma bond and one pi bond). The second term (T∆S) is negative because ∆S is positive (one molecule is converted into two molecules) Therefore, the reaction is only favorable if the second term is greater in magnitude than the first term. This only occurs at high temperature 9.49 HO OH + En KMnO4 1) Hg(OAc)2 , H2O NaOH, cold HCl 2) NaBH4 H2 Pt Cl HO Br2 H2O HO Br + En Source: http://www.doksinet 192 CHAPTER 9 9.50

HO 1) BH3 THF 1) MCPBA OH 2) H3O + En + OH + En 2) H2O2, NaOH HBr Br2 H2 Pt Br Br Br + En 9.51 a) H H O H H O H O H H H O OH H b) H H O Hydride H H H O H O H H Shift H O OH c) H Br Br Br d) H Br Br Methyl Shift Br H Source: http://www.doksinet CHAPTER 9 9.52 Br NaOMe H2 Pt Compound A 9.53 1) HBr 2) NaOMe a) 1) HBr, ROOR 2) t-BuOK b) 9.54 OH 1) Conc. H2SO4 2) HBr, ROOR 3) t-BuOK 9.55 Two different alkenes will produce 2,4-dimethylpentane upon hydrogenation: 9.56 OH 1) MCPBA 2) H3O+ OH Compound A 9.57 a) 1) Conc. H2SO4 2) dilute H2SO4 OH 1) Conc. H2SO4 2) BH3 THF 3) H2O2, NaOH OH 193 Source: http://www.doksinet 194 CHAPTER 9 b) 1) NaOMe 2) HBr, ROOR Br Br 1) NaOMe 2) HBr c) Cl 1) NaOMe 2) H2, Pt d) OH OH OH 1) Conc. H2SO4 2) OsO4, NMO 9.58 1) HBr, ROOR 2) t-BuOK a) OH 1) Conc. H2SO4 OH 2) dilute H2SO4 b) 9.59 Br Compound A dilute H2SO4 NaOMe Compound B OH Compound C Source: http://www.doksinet CHAPTER 9

195 9.60 HBr, ROOR + En Br HBr Br 1) BH3 THF 2) H2O2, NaOH H3O+ OH + En HO 9.61 excess H2 + Pt Diastereomers 9.62 Markovnikov addition of water without carbocation rearrangements can be achieved via oxymercuration-demercuration: OH 1) Hg(OAc)2, H2O 2) NaBH4 racemic 9.63 O O H O S O H O H O H O Me OH Me O H O Me OH MeO H Source: http://www.doksinet 196 CHAPTER 9 9.64 a) H H O H Methyl Shift H H OH H O H b) Br H Br + Br 9.65 H2 (PPh3)3RhCl (meso) a) H3O+ OH b) 1) BH3 THF c) OH 2) H2O2, NaOH HO 1) MCPBA d) 2) H3O+ OH + En O H O H Source: http://www.doksinet 197 CHAPTER 9 9.66 a) Hydroboration-oxidation gives an anti-Markovnikov addition. If 1-propene is the starting material, the OH group will not be installed in the correct location. Acidcatalyzed hydration of 1-propene would give the desired product b) Hydroboration-oxidation gives a syn addition of H and OH across a double bond. This compound does not have a proton that is cis

to the OH group, and therefore, hydroboration-oxidation cannot be used to make this compound. c) Hydroboration-oxidation gives an anti-Markovnikov addition. There is no starting alkene that would yield the desired product via an anti-Markovnikov addition. 9.67 Br2 Br Br Br Br (meso) 9.68 H H H a) b) c) d) 9.69 The reaction proceeds via a resonance-stabilized carbocation, which is even lower in energy than a tertiary carbocation: O H Br O resonance-stabilized O Br Br O Source: http://www.doksinet 198 CHAPTER 9 9.70 OH Br HBr ROOR 1) BH3 THF 2) H2O2, NaOH OH OsO4, NMO Cl2 Br2 H2O HBr OH Cl OH Cl Br Br OH H2 Pt NaOMe + En Br Br2 H2O 1) O3 O 2) DMS H 1) BH3 THF 2) H2O2, NaOH O + En OH Source: http://www.doksinet CHAPTER 9 199 9.71 Br O HBr ROOR HO 1) BH3 THF 2) H2O2, NaOH 1) O3 2) DMS HO OH HO Br OsO4, NMO Br2 H3O+ HBr OH Br NaOMe H2 O H2 Pt OH OH OsO4, NMO + En 1) MCPBA 2) H3O+ OH OH + En 9.72 Addition of HBr to

2-methyl-2-pentene should be more rapid because the reaction can proceed via a tertiary carbocation. In contrast, addition of HBr to 4-methyl-1pentene proceeds via a less stable, secondary carbocation 9.73 HS Br2 + En Br H2S 9.74 Br 1) NaOMe 2) HBr 3) NaOMe 4) BH3 THF 5) H2O2, NaOH OH + En Source: http://www.doksinet 200 CHAPTER 9 9.75 H2 1) BH3 THF 2) H2 O2 , NaOH Pt Compound X OH 2,4-d imethy lpentan-1-ol H3O+ OH 9.76 Cl t-BuOK HBr Br Br + + Br 9.77 HBr, ROOR 1) O 3 2) DMS Br Compound Y C7H12 H2 Pt 9.78 H H H H H O + O H Source: http://www.doksinet CHAPTER 9 9.79 a) O H O S O H O OH OH O O H O H O S O O O HO O H HO b) O H O S O H O O O H H O H H HO HO (even concentrated H2SO4 has some water present) 9.80 Cl O 1) NaOMe 2) O3 3) DMS H O H 201 Source: http://www.doksinet 202 CHAPTER 9 9.81 a) Br Br Br Br H OH Br Br O O OH b) O H O S O H O OH OH O H O S O O O H O 9.82 I I I I O OH OH I H O O I O O O 9.83 Br

Br Br H Br + En + En + En Br Br Br + En