Organic Chemistry I

Mohammad Jafarzadeh Faculty of Chemistry, Razi University

Organic Chemistry, Structure and Function (7th edition) By P. Vollhardt and N. Schore, Elsevier, 2014 1 326 CHAPTER 9 Further Reactions of Alcohols and the Chemistry of Ethers

Figure 9-1 Four typical reaction A A modes of alcohols. In each mode, COO H Deprotonation COO H one or more of the four bonds A A marked a–d are cleaved (wavy line HHOOCOO HOCOO Ϫ denotes bond cleavage): (a) de- A a A protonation by base; (b) protonation by acid followed by uni- or Alkoxide bimolecular substitution; (b, c) 9. elimination;Further and (a, dReactions) oxidation. of Alcohols and the Chemistry of Ethers A A COO H Substitution COO H A A HHOOCOO HXOCO A b A 326 CHAPTER 9 Further ReactionsA variety of Alcoholsof reaction and themodes Chemistryavailable of Ethersto alcohols . Usually at least oneHaloalkaneof the andfour bonds marked a, b, c, or d is cleaved. other alkane derivatives

Figure 9-1 Four typical reaction A A A c H E modes of alcohols. In each mode, COO H Deprotonation COO H COO H C one or more of the four bonds A A A Elimination B marked a–d are cleaved (wavy line HHC O HOC Ϫ HHC O C OOO OO OOO E H denotes bond cleavage): (a) de- A a A A b H protonation by base; (b) protonation by acid followed by uni- or Alkoxide Alkenes bimolecular substitution; (b, c) elimination; and (a, d) oxidation. A A A A COO H Oxidation COO H COO H COO H A A Substitution HHC O C A A OOO E N HHOOCOO HXOCO d A a O A b A Aldehydes Haloalkane and and ketones other alkane derivatives 2

A c H E COO H C 9-1 REACTIONS OF ALCOHOLS WITH BASE: PREPARATION A Elimination B OF ALKOXIDES HHC O C OOO E H A b H AlkenesAs described in Section 8-3, alcohols can be both acids and bases. In this section, we shall review the methods by which the hydroxy group of alcohols is deprotonated to furnish their A A conjugate bases, the alkoxides. COO H Oxidation COO H A A Strong bases are needed to deprotonate alcohols completely HHC O C OOO E N d A a ToO remove a proton from the OH group of an alcohol (Figure 9-1, cleavage of bond a), we Aldehydesmust use a base stronger than the alkoxide. Examples include lithium diisopropylamide (Sec- and ketonestion 7-8), butyllithium (Section 8-7), and alkali metal hydrides (Section 8-6, Exercise 8-4), such as potassium hydride, KH. Such hydrides are particularly useful because the only by- product of the reaction is hydrogen gas.

Three Ways of Making Methoxide from Methanol 9-1 REACTIONS OF ALCOHOLS WITH BASE: PREPARATION OF ALKOXIDES CH(CH3)2 CH(CH3)2 Sodium reacts with water vigor- A K ϭ1024.5 A ϩϪ Ϫ ϩ ously, releasing hydrogen gas. CH3OH ϩ Li NCH(CH3)2 CH3O Li ϩ HNCH(CH3)2

40 ؍ this15.5 section,Lithium we diisopropylamideshall pKa؍ As described in Section 8-3, alcohols can be both acids and bases. pInKa review the methods by which the hydroxy group of alcohols is deprotonated to furnish their K ϭ1034.5 Ϫ ϩ conjugate bases, the alkoxides. CH3OH ϩ CH3CH2CH2CH2Li CH3O Li ϩ CH3CH2CH2CH2H

50 ؍ Butyllithium pKa 15.5؍ pKa

Strong bases are needed to deprotonate alcohols completely K ϭ1022.5 ϩ Ϫ Ϫ ϩ To remove a proton from the OH group of an alcohol (Figure 9-1, cleavage CHof bond3OH a),ϩϩ weK H CH3O K HOH 38 ؍ Potassium pKa 15.5؍ pKa must use a base stronger than the alkoxide. Examples include lithium diisopropylamide (Sec-hydride tion 7-8), butyllithium (Section 8-7), and alkali metal hydrides (Section 8-6, Exercise 8-4), such as potassium hydride, KH. Such hydrides are particularly useful because the only by- product of the reaction is hydrogen gas.

Three Ways of Making Methoxide from Methanol

CH(CH3)2 CH(CH3)2 Sodium reacts with water vigor- A K ϭ1024.5 A ϩϪ Ϫ ϩ ously, releasing hydrogen gas. CH3OH ϩ Li NCH(CH3)2 CH3O Li ϩ HNCH(CH3)2

40 ؍ Lithium diisopropylamide pKa 15.5؍ pKa

K ϭ1034.5 Ϫ ϩ CH3OH ϩ CH3CH2CH2CH2Li CH3O Li ϩ CH3CH2CH2CH2H

50 ؍ Butyllithium pKa 15.5؍ pKa

K ϭ1022.5 ϩ Ϫ Ϫ ϩ CH3OH ϩϩK H CH3O K HOH

38 ؍ Potassium pKa 15.5؍ pKa hydride 326 CHAPTER 9 Further Reactions of Alcohols and the Chemistry of Ethers

A c H E COO H C A Elimination B HHC O C OOO E H A b H Alkenes

A A COO H Oxidation COO H A A HHC O C OOO E N d A a O Aldehydes and ketones

9-1 REACTIONS OF ALCOHOLS WITH BASE: PREPARATION OF ALKOXIDES 9.1 REACTIONS OF ALCOHOLS WITH BASE: PREPARATION OF ALKOXIDES As described in Section 8-3, alcohols can be both acids and bases. In this section, we shall review the methods by which the hydroxy group of alcohols is deprotonated to furnish their The hydroxyconjugategroup bases,of alcohols the alkoxides.is deprotonated to furnish their conjugate bases, the alkoxides. Strong basesStrongare neededbases areto deprotonateneeded to deprotonatealcohols completely alcohols completely To remove Toa removeproton a fromprotonthe fromOH the OHgroup groupof ofan an alcohol,alcohol (Figurea base 9-1, strongercleavage ofthan bond athe), wealkoxide is must use a base stronger than the alkoxide. Examples include lithium diisopropylamide (Sec- used, including lithium diisopropylamide, butyllithium, and alkali metal hydrides (e.g., tion 7-8), butyllithium (Section 8-7), and alkali metal hydrides (Section 8-6, Exercise 8-4), potassium hydride,such as potassiumKH). hydride, KH. Such hydrides are particularly useful because the only by- Such hydridesproductare ofparticularly the reaction isuseful hydrogenbecause gas. the only by-product of the reaction is hydrogen gas. Three Ways of Making Methoxide from Methanol

CH(CH3)2 CH(CH3)2 Sodium reacts with water vigor- A K ϭ1024.5 A ϩϪ Ϫ ϩ ously, releasing hydrogen gas. CH3OH ϩ Li NCH(CH3)2 CH3O Li ϩ HNCH(CH3)2

40 ؍ Lithium diisopropylamide pKa 15.5؍ pKa

K ϭ1034.5 Ϫ ϩ CH3OH ϩ CH3CH2CH2CH2Li CH3O Li ϩ CH3CH2CH2CH2H

50 ؍ Butyllithium pKa 15.5؍ pKa

K ϭ1022.5 ϩ Ϫ Ϫ ϩ CH3OH ϩϩK H CH3O K HOH 3 38 ؍ Potassium pKa 15.5؍ pKa hydride 9-29-2 ReactionsReactions of Alcohols withwith StrongStrong AcidsAcids CHAPTERCHAPTER 99 327327

ExerciseExercise 9-19-1 WeWe shall shall encounterencounter laterlater manymany reactionsreactions thatthat require catalytic base, forfor example,example, catalyticcatalytic sodiumsodium methoxidemethoxide in in methanol. methanol. Assume Assume thatthat youyou wouldwould like to make such aa solutionsolution containingcontaining 1010 mmol mmol of of NaOCH3 in 1 liter of CH3OH. Would it be all right simply to add 10 mmol of NaOH to the solvent? NaOCH3 in 1 liter of CH3OH. Would it be all right simply to add 10 mmol of NaOH to the solvent? (Caution: Simply comparing pKa values (Table 2-2) is not enough. Hint: See Section 2-3.) Alkali metals(Caution:also deprotonateSimply comparingalcohols pKa values— (Tablebut by 2-2)reduction is not enough.of Hint:H+ See Section 2-3.)

1 Another commonAlkaliAlkali metalsmetalsway of obtaining alsoalso deprotonatedeprotonatealkoxides is by alcohols—butthe reaction of alcohols byby reductionreductionwith alkali ofofmetals, HH1 such as lithiumAnother. Such commonmetals way reduce of obtainingwater alkoxides—in some is cases, by the reactionviolently of— to alcoholsyield alkali with alkalimetal hydroxidesAnotherand hydrogen common gas way. of obtaining alkoxides is by the reaction of alcohols with alkali metals,metals, such such asas lithium.lithium. SuchSuch metalsmetals reducereduce water—in some cases,cases, violently—toviolently—to yieldyield alkalialkali metalmetal hydroxideshydroxides andand hydrogenhydrogen gas.gas. WhenWhen the more reactive metalsmetals (sodium,(sodium, potassium,potassium, When theandmore cesium)reactive are exposedmetals to(sodium, water in potassium,air, the hydrogenand cesiumgenerated) arecan exposedignite spontaneouslyto water in and cesium) are exposed to water in air, the hydrogen generated can ignite spontaneously air, the hydrogenor even detonate.generated can ignite spontaneously or even detonate. or even detonate. ϩϪ 2 HOOH2ϩϩ2 M (Li, Na, K, Cs) MϩϪ OH H2 2 HOOH2ϩϩ2 M (Li, Na, K, Cs) M OH H2 The alkali metals act similarly on the alcohols to give alkoxides, but the transformation is The alkali metals act similarly on the alcohols to give alkoxides, but the transformation is The alkalilessmetals vigorous.act similarly Here are ontwothe examples.alcohols to give alkoxides, but the transformation is less less vigorous. Here are two examples. vigorous. Alkoxides from Alcohols and Alkali Metals Alkoxides from Alcohols and Alkali Metals Ϫ ϩ 2 CH3CH2OH ϩϩ2 Na 2 CH3CH2OϪ Naϩ H2 2 CH3CH2OH ϩϩ2 Na 2 CH3CH2OϪNaϩ H2 2 (CH3)3COH ϩϩ2 K 2 (CH3)3COϪ Kϩ H2 2 (CH3)3COH ϩϩ2 K 2 (CH3)3CO K H2 The reactivity of the alcohols used in this process decreases with increasing substitution, The reactivity of the alcohols used in this process decreases with increasing substitution, methanol being most reactive and tertiary alcohols least reactive. 4 methanol being most reactive and tertiary alcohols least reactive. Relative Reactivity of ROH with Alkali Metals Relative Reactivity of ROH with Alkali Metals R 5 CH3 . primary . secondary . tertiary R 5 CH3 . primary . secondary . tertiary Decreasing reactivity Decreasing reactivity 2-Methyl-2-propanol reacts so slowly that it can be used to safely destroy potassium resi- dues2-Methyl-2-propanol in the laboratory. reacts so slowly that it can be used to safely destroy potassium resi- duesWhat in the are laboratory. alkoxides good for? We have already seen that they can be useful reagents in organicWhat synthesis. are alkoxides For example, good for? the We reaction have already of hindered seen that alkoxides they can with be usefulhaloalkanes reagents gives in elimination.organic synthesis. For example, the reaction of hindered alkoxides with haloalkanes gives elimination. Ϫ ϩ (CH3)3CO K , (CH3)3COH ϩ Ϫ CH3CH2CH2CH2Br Ϫ ϩ CH3CH2CHP CH2 ϩϩ(CH3)3COH K Br (CH3)3CO KE2, (CH3)3COH ϩ Ϫ CH3CH2CH2CH2Br CH3CH2CHP CH2 ϩϩ(CH3)3COH K Br E2 Less branched alkoxides attack primary haloalkanes by the SN2 reaction to give ethers. This methodLess branched is described alkoxides in Section attack primary9-6. haloalkanes by the SN2 reaction to give ethers. This method is described in Section 9-6. In Summary A strong base will convert an alcohol into an alkoxide by an acid-base reaction.In Summary The stronger A strongthe base, base the will more convert the equilibrium an alcohol isinto displaced an alkoxide to the by alkoxide an acid-base side. Alkalireac- metalstion. The react stronger with the alcohols base, the by more reduction the equilibrium to generate is hydrogendisplaced gasto the and alkoxide an alkoxide. side. Alkali This processmetals reactis retarded with alcoholsby steric byhindrance. reduction to generate hydrogen gas and an alkoxide. This process is retarded by steric hindrance. 9-2 REACTIONS OF ALCOHOLS WITH STRONG ACIDS: 9-2 ALKYLOXONIUM REACTIONS OF ALCOHOLS IONS IN SUBSTITUTION WITH STRONG AND ACIDS: ELIMINATIONALKYLOXONIUM REACTIONS IONS IN SUBSTITUTIONOF ALCOHOLS AND ELIMINATION REACTIONS OF ALCOHOLS We have seen that heterolytic cleavage of the O–H bond in alcohols is readily achieved withWe havestrong seen bases. that Can heterolytic we break cleavage the C–O of linkage the O–H (bond bond b ,in Figure alcohols 9-1) is as readily easily? achieved Yes, but nowwith we strong need bases. acid. CanRecall we (Section break the 2-3) C–O that linkage water has(bond a highb, Figure pKa (15.7): 9-1) as It easily? is a weak Yes, acid. but Consequently,now we need acid. hydroxide, Recall its(Section conjugate 2-3) that base, water is an has exceedingly a high pKa (15.7): poor leaving It is a weak group. acid. For Consequently, hydroxide, its conjugate base, is an exceedingly poor leaving group. For 9-2 Reactions of Alcohols with Strong Acids CHAPTER 9 327 9-2 Reactions of Alcohols with Strong Acids CHAPTER 9 327

Exercise 9-1 ExerciseWe shall 9-1 encounter later many reactions that require catalytic base, for example, catalytic sodium methoxide in methanol. Assume that you would like to make such a solution containing 10 mmol of We shallNaOCH encounter3 in 1 liter later of many CH3OH. reactions Would that it be require all right catalytic simply to base, add 10for mmol example, of NaOH catalytic to the sodium solvent? methoxide(Caution: in methanol. Simply comparing Assume that pK youa values would (Table like to2-2) make is not such enough. a solution Hint: containing See Section 10 2-3.)mmol of NaOCH3 in 1 liter of CH3OH. Would it be all right simply to add 10 mmol of NaOH to the solvent? (Caution: Simply comparing pKa values (Table 2-2) is not enough. Hint: See Section 2-3.) Alkali metals also deprotonate alcohols—but by reduction of H1 Another common way of obtaining alkoxides is by the reaction of alcohols with alkali1 Alkalimetals, metals such as also lithium. deprotonate Such metals reduce alcohols—but water—in some cases,by reduction violently—to ofyield H alkali Anothermetal common hydroxides way and of obtaininghydrogen gas. alkoxides When isthe by more the reactive reaction metals of alcohols (sodium, with potassium, alkali metals,and such cesium) as lithium. are exposed Such metalsto water reduce in air, water—in the hydrogen some generated cases, violently—to can ignite spontaneouslyyield alkali metalor hydroxides even detonate. and hydrogen gas. When the more reactive metals (sodium, potassium, and cesium) are exposed to water in air, the hydrogen generated can ignite spontaneously 2 H OH2ϩϩ2 M (Li, Na, K, Cs) MϩϪOH H or even detonate. O 2

The alkali metals act similarly on the alcohols to give alkoxides,ϩϪ but the transformation is 2 HOOH2ϩϩ2 M (Li, Na, K, Cs) M OH H less vigorous. Here are two examples. 2 The alkali metals act similarlyAlkoxides on the fromalcohols Alcohols to give and alkoxides, Alkali Metals but the transformation is less vigorous. Here are two examples. Ϫ ϩ 2 CH3CH2OH ϩϩ2 Na 2 CH3CH2O Na H2 Ϫ ϩ 2 (CHAlkoxides3)3COH fromϩϩ Alcohols2 K and 2Alkali (CH3 )Metals3CO K H2 Ϫ ϩ The reactivityThe reactivityof2 CHthe 3 ofCHalcohols the2O H alcoholsϩϩused 2 used Nain inthis thisprocess process2 CH 3 decreasesdecreasesCH2O Na withwith increasingincreasingH2 substitution,substitution, Ϫ ϩ methanolmethanolbeing being2most (CH most3reactive)3CO reactiveH ϩϩand and2tertiary K tertiaryalcohols alcohols2 (CH leastleast3)3C Oreactive.reactiveK . H2

The reactivity of the alcoholsRelative used Reactivity in this process of ROH decreases with Alkali with Metals increasing substitution, methanol being most reactive and tertiary alcohols least reactive. R 5 CH3 . primary . secondary . tertiary

Relative ReactivityDecreasing of ROH reactivity with Alkali Metals

R 5 CH3 . primary . secondary . tertiary 2-Methyl-2-propanol reacts so slowly that it can be used to safely destroy potassium resi- 2-Methyl-2-propanol reacts so Decreasingslowly that reactivityit can be used to safely destroy potassium residuesduesin thein thelaboratory laboratory.. What are alkoxides good for? We have already seen that they can be useful reagents in 2-Methyl-2-propanolorganic synthesis. reacts For example,so slowly the that reaction it can ofbe hinderedused to safelyalkoxides destroy with potassiumhaloalkanes resi- gives dues elimination.in the laboratory. AlkoxidesWhat canare alkoxidesbe useful goodreagents for? We havein organicalready seensynthesis that they. Forcan beexample, useful reagentsthe reaction in of (CH ) COϪKϩ, (CH ) COH hindered alkoxides with haloalkanes3 3 3 3gives elimination. Less branched alkoxides attackϩ Ϫ organicCH synthesis.3CH2CH2CH For2Br example, the reaction of hinderedCH3CH2 CHalkoxidesP CH2 withϩϩ haloalkanes(CH3)3COH gives K Br E2 primaryelimination.haloalkanes by the SN2 reaction to give ethers.

Less branched alkoxides attackϪ ϩ primary haloalkanes by the SN2 reaction to give ethers. This (CH3)3CO K , (CH3)3COH ϩ Ϫ CH3CHmethod2CH2 CHis described2Br in Section 9-6. CH3CH2CHP CH2 ϩϩ(CH3)3COH K Br E2 In Summary A strong base will convert an alcohol into an alkoxide by an acid-base reac- 5 Less tion.branched The strongeralkoxides the attack base, primarythe more haloalkanes the equilibrium by theis displaced SN2 reaction to the to alkoxide give ethers. side. This Alkali methodmetals is described react with in alcoholsSection 9-6. by reduction to generate hydrogen gas and an alkoxide. This In Summaryprocess is Aretarded strong bybase steric will hindrance. convert an alcohol into an alkoxide by an acid-base reaction. The stronger the base, the more the equilibrium is displaced to the alkoxide side. Alkali metals 9-2 react with REACTIONS alcohols by reductionOF ALCOHOLS to generate WITH hydrogen STRONG gas and ACIDS: an alkoxide. This process is retarded by steric hindrance. ALKYLOXONIUM IONS IN SUBSTITUTION AND ELIMINATION REACTIONS OF ALCOHOLS 9-2 REACTIONS OF ALCOHOLS WITH STRONG ACIDS: We haveALKYLOXONIUM seen that heterolytic IONScleavage IN of SUBSTITUTION the O–H bond in alcohols AND is readily achieved with strongELIMINATION bases. Can we REACTIONS break the C–O linkageOF ALCOHOLS (bond b, Figure 9-1) as easily? Yes, but now we need acid. Recall (Section 2-3) that water has a high pKa (15.7): It is a weak acid. Consequently, hydroxide, its conjugate base, is an exceedingly poor leaving group. For We have seen that heterolytic cleavage of the O–H bond in alcohols is readily achieved with strong bases. Can we break the C–O linkage (bond b, Figure 9-1) as easily? Yes, but now we need acid. Recall (Section 2-3) that water has a high pKa (15.7): It is a weak acid. Consequently, hydroxide, its conjugate base, is an exceedingly poor leaving group. For 9.2 REACTIONS OF ALCOHOLS WITH STRONG ACIDS: ALKYLOXONIUM IONS IN SUBSTITUTION AND ELIMINATION REACTIONS OF ALCOHOLS

The heterolytic cleavage of the O–H bond in alcohols is readily achieved328 with strongCHAPTERbases. 9 Further Reactions of Alcohols and the Chemistry of Ethers To break the C–O linkage, an acid is needed.

Water has a high pKa (15.7), is a weak acid. Hydroxide, its conjugate base, is an alcohols to undergo substitution or elimination reactions, the OH must ! rst be converted exceedingly poor leaving group. For alcohols to undergo substitution or elimination into a better leaving group. reactions, the OH must first be converted into a better leaving group. Haloalkanes from primary alcohols and HX: Water can be a Haloalkanes from primary alcohols and HX: Water can be a leaving group in S 2 N leaving group in SN2 reactions reactions Poor leaving group The simplest way of turning the hydroxy substituent in alcohols into a good leaving group is to protonate the oxygen to form an alkyloxonium ion. Recall (Section 8-3) that this process ties up one of the oxygen lone pairs by forming a bond to a proton. The positive The simplest way of turning the hydroxy substituent in alcohols into ROšOð ϩ Hϩ i charge therefore resides on the oxygen atom. Protonation changes OH from a bad leaving a good leaving group is to protonate the oxygen to form an H group into neutral water, a good leaving group. alkyloxonium ion. This reaction is reversible and, under normal conditions, the equilibrium lies on the side of unprotonated alcohol. However, this is immaterial if a nucleophile is present in the H Good mixture that is capable of trapping the oxonium species. For example, alkyloxonium ions Protonation changes OH from a bad leaving group into neutral ϩ f leaving derived from primary alcohols are subject to such nucleophilic attack. Thus, the butyloxo- ROOð water, a good leaving group. i group nium ion resulting from the treatment of 1-butanol with concentrated HBr undergoes dis- H placement by bromide to form 1-bromobutane. The originally nucleophilic (red) oxygen is 6 Alkyloxonium ion protonated by the electrophilic proton (blue) to give the alkyloxonium ion, containing an electrophilic (blue) carbon and H2O as a leaving group (green). In the subsequent SN2 reaction, bromide acts as a nucleophile. Primary Bromoalkane Synthesis from an Alcohol ϩ Ϫ CH3CH2CH2CH2"šOHCϩ H"šBrð H3CH2CH2CH2OH" 2 ϩ "šBrðð

CH3CH2CH2CH2"šBrð ϩ H2"šO

Reactivity of Oxonium Ions Does not dissociate: Dissociate:

SN2 only SN1 and E1 ϩ ϩ ϩ RprimOOH2 ӶϽRsecOOH2 RtertOOH2

Increasing ease of carbocation formation

Synthetically, this fact is exploited in the preparation of tertiary haloalkanes from tertiary alcohols in the presence of excess concentrated aqueous hydrogen halide. The product forms in minutes at room temperature. The mechanism is precisely the reverse of that of solvolysis (Section 7-2). 328 CHAPTER 9 Further Reactions of Alcohols and the Chemistry of Ethers

alcohols to undergo substitution or elimination reactions, the OH must ! rst be converted into a better leaving group.

Haloalkanes from primary alcohols and HX: Water can be a This reactionleavingis reversible group inand, SN2under reactionsnormal conditions, the equilibrium lies on the side of Poor leaving group unprotonatedThe simplestalcohol way. However, of turning thethis hydroxyis immaterial substituentif ina alcoholsnucleophile into a isgoodpresent leavingin groupthe mixture that is capableis to protonateof trapping the oxygenthe oxonium to form species an alkyloxonium. ion. Recall (Section 8-3) that this process ties up one of the oxygen lone pairs by forming a bond to a proton. The positive ROšOð ϩ Hϩ i charge therefore resides on the oxygen atom. Protonation changes OH from a bad leaving H For example,group intoalkyloxonium neutral water, ionsa goodderived leaving group.from primary alcohols are subject to such nucleophilicThisattack reaction. Thus, is reversiblethe butyloxonium and, under normalion conditions,resulting thefrom equilibriumthe treatment lies on the ofside1 -butanol with concentratedof unprotonatedHBr undergoes alcohol. However,displacement this is immaterialby bromide if a nucleophileto form 1 - isbromobutane present in the. H Good mixture that is capable of trapping the oxonium species. For example, alkyloxonium ions ϩ f leaving derived from primary alcohols are subject to such nucleophilic attack. Thus, the butyloxo- ROOð The originally nucleophilic (red) oxygen is protonated by the electrophilic proton (blue) to i groupgive the niumalkyloxonium ion resultingion, fromcontaining the treatmentan ofelectrophilic 1-butanol with(blue) concentratedcarbon HBrand undergoesH O as dis-a leaving H 2 group (green)placement. by bromide to form 1-bromobutane. The originally nucleophilic (red) oxygen is Alkyloxonium ion protonated by the electrophilic proton (blue) to give the alkyloxonium ion, containing an electrophilic (blue) carbon and H2O as a leaving group (green). In the subsequent SN2 reac- In the subsequenttion, bromideS Nacts2 reaction, as a nucleophile.bromide acts as a nucleophile. Primary Bromoalkane Synthesis from an Alcohol ϩ Ϫ CH3CH2CH2CH2"šOHCϩ H"šBrð H3CH2CH2CH2OH" 2 ϩ "šBrðð

CH3CH2CH2CH2"šBrð ϩ H2"šO

Iodoalkane Synthesis The reaction of primary alcohols with concentrated HI proceeds in similar fashion to 7 furnish primary iodoalkanes (margin). On the other hand, concentrated HCl is sluggish in HO(CH2)6OHϩ 2 HI turning primary alcohols into the corresponding chloroalkanes, because chloride ion is a 1,6-Hexanediol relatively weak nucleophile under protic conditions (Section 6-8). Therefore, this conversion, while possible, is usually carried out by other reagents (Section 9-4). In general, the acid- catalyzed SN2 reaction of primary alcohols with HBr or HI is a good way of preparing I(CH2)6Iϩ 2 H2O simple primary haloalkanes. What about secondary and tertiary alcohols? 85% 1,6-Diiodo- hexane Secondary and tertiary alcohols and HX: Water can be a leaving group to form carbocations in SN1 and E1 reactions Alkyloxonium ions derived from secondary and tertiary alcohols, in contrast with their primary counterparts, lose water with increasing ease to give the corresponding carbocations. The reason for this difference in behavior is the difference in carbocation stability (Section 7-5). Primary carbocations are too high in energy to be accessible under ordinary laboratory conditions, whereas secondary and tertiary carbocations are generated with increasing ease. Thus, primary alkyloxonium ions undergo only SN2 reactions, whereas their secondary and tertiary relatives enter into SN1 and E1 processes. When good nucleophiles are present, we observe SN1 products.

Reactivity of Oxonium Ions Does not dissociate: Dissociate:

SN2 only SN1 and E1 ϩ ϩ ϩ RprimOOH2 ӶϽRsecOOH2 RtertOOH2

Increasing ease of carbocation formation

alcohols to undergo substitution or elimination reactions, the OH must ! rst be converted into a better leaving group.

Haloalkanes from primary alcohols and HX: Water can be a leaving group in SN2 reactions Poor leaving group The simplest way of turning the hydroxy substituent in alcohols into a good leaving group is to protonate the oxygen to form an alkyloxonium ion. Recall (Section 8-3) that this process ties up one of the oxygen lone pairs by forming a bond to a proton. The positive ROšOð ϩ Hϩ i charge therefore resides on the oxygen atom. Protonation changes OH from a bad leaving H group into neutral water, a good leaving group. This reaction is reversible and, under normal conditions, the equilibrium lies on the side of unprotonated alcohol. However, this is immaterial if a nucleophile is present in the H Good mixture that is capable of trapping the oxonium species. For example, alkyloxonium ions ϩ f leaving derived from primary alcohols are subject to such nucleophilic attack. Thus, the butyloxo- ROOð i group nium ion resulting from the treatment of 1-butanol with concentrated HBr undergoes dis- The reaction of primary alcohols with concentrated HI proceedsH in similarplacementfashion byto bromidefurnish to form 1-bromobutane. The originally nucleophilic (red) oxygen is primary iodoalkanes. Alkyloxonium ion protonated by the electrophilic proton (blue) to give the alkyloxonium ion, containing an electrophilic (blue) carbon and H2O as a leaving group (green). In the subsequent SN2 reac- Concentrated HCl is sluggish in turning primary alcohols into thetion, bromidecorresponding acts as a nucleophile. chloroalkanes, because chloride ion is a relatively weak nucleophile under protic conditionsPrimary. Bromoalkane Synthesis from an Alcohol Therefore, this conversion, while possible, is usually carried out by other reagents. ϩ Ϫ CH3CH2CH2CH2"šOHCϩ H"šBrð H3CH2CH2CH2OH" 2 ϩ "šBrðð

In general, the acid-catalyzed SN2 reaction of primary alcohols with HBr or HI is a good way CH3CH2CH2CH2"šBrð ϩ H2"šO of preparing simple primary haloalkanes. Iodoalkane Synthesis The reaction of primary alcohols with concentrated HI proceeds in similar fashion to furnish primary iodoalkanes (margin). On the other hand, concentrated HCl is sluggish in HO(CH2)6OHϩ 2 HI turning primary alcohols into the corresponding chloroalkanes, because chloride ion is a 1,6-Hexanediol relatively weak nucleophile under protic conditions (Section 6-8). Therefore, this conversion, while possible, is usually carried out by other reagents (Section 9-4). In general, the acid- catalyzed SN2 reaction of primary alcohols with HBr or HI is a good way of preparing I(CH2)6Iϩ 2 H2O simple primary haloalkanes. What about secondary and tertiary alcohols? 85% 1,6-Diiodo- hexane Secondary and tertiary alcohols and HX: Water can be a leaving group to form carbocations in S 1 and E1 reactions 8 N Alkyloxonium ions derived from secondary and tertiary alcohols, in contrast with their primary counterparts, lose water with increasing ease to give the corresponding carbocations. The reason for this difference in behavior is the difference in carbocation stability (Section 7-5). Primary carbocations are too high in energy to be accessible under ordinary laboratory conditions, whereas secondary and tertiary carbocations are generated with increasing ease. Thus, primary alkyloxonium ions undergo only SN2 reactions, whereas their secondary and tertiary relatives enter into SN1 and E1 processes. When good nucleophiles are present, we observe SN1 products.

Reactivity of Oxonium Ions Does not dissociate: Dissociate:

SN2 only SN1 and E1 ϩ ϩ ϩ RprimOOH2 ӶϽRsecOOH2 RtertOOH2

Increasing ease of carbocation formation

alcohols to undergo substitution or elimination reactions, the OH must ! rst be converted into a better leaving group.

Haloalkanes from primary alcohols and HX: Water can be a leaving group in SN2 reactions Poor leaving group The simplest way of turning the hydroxy substituent in alcohols into a good leaving group is to protonate the oxygen to form an alkyloxonium ion. Recall (Section 8-3) that this process ties up one of the oxygen lone pairs by forming a bond to a proton. The positive ROšOð ϩ Hϩ i charge therefore resides on the oxygen atom. Protonation changes OH from a bad leaving H group into neutral water, a good leaving group. This reaction is reversible and, under normal conditions, the equilibrium lies on the side of unprotonated alcohol. However, this is immaterial if a nucleophile is present in the H Good mixture that is capable of trapping the oxonium species. For example, alkyloxonium ions ϩ f leaving derived from primary alcohols are subject to such nucleophilic attack. Thus, the butyloxo- ROOð i group nium ion resulting from the treatment of 1-butanol with concentrated HBr undergoes dis- H placement by bromide to form 1-bromobutane. The originally nucleophilic (red) oxygen is Alkyloxonium ion protonated by the electrophilic proton (blue) to give the alkyloxonium ion, containing an electrophilic (blue) carbon and H2O as a leaving group (green). In the subsequent SN2 reaction, bromide acts as a nucleophile. Primary Bromoalkane Synthesis from an Alcohol ϩ Ϫ CH3CH2CH2CH2"šOHCϩ H"šBrð H3CH2CH2CH2OH" 2 ϩ "šBrðð

CH3CH2CH2CH2"šBrð ϩ H2"šO

Iodoalkane Synthesis The reaction of primary alcohols with concentrated HI proceeds in similar fashion to furnish primary iodoalkanes (margin). On the other hand, concentrated HCl is sluggish in HO(CH2)6OHϩ 2 HI turning primary alcohols into the corresponding chloroalkanes, because chloride ion is a 1,6-Hexanediol relatively weak nucleophile under protic conditions (Section 6-8). Therefore, this conversion, while possible, is usually carried out by other reagents (Section 9-4). In general, the acid- Secondarycatalyzedand StertiaryN2 reactionalcohols of primaryand alcoholsHX with: Water HBr orcan HI isbe a gooda leaving way of preparinggroup to form I(CH2)6Iϩ 2 H2O simple primary haloalkanes. What about secondary and tertiary alcohols? 85% carbocations in SN1 and E1 reactions 1,6-Diiodo- hexane AlkyloxoniumSecondaryions derived and fromtertiarysecondary alcoholsand andtertiary HX:alcohols, Water incancontrast be a with their primary counterparts,leavinglose groupwater withto formincreasing carbocationsease to give in SNthe1 andcorresponding E1 reactionscarbocations. The reason forAlkyloxoniumthis difference ions in derivedbehavior from is secondarythe difference and tertiaryin carbocation alcohols, in contraststability with. their primary counterparts, lose water with increasing ease to give the corresponding carbocations. The reason for this difference in behavior is the difference in carbocation stability Primary carbocations(Section 7-5). Primaryare too carbocationshigh in areenergy too highto in beenergyaccessible to be accessibleunder underordinary ordinarylaboratory conditions,laboratorywhereas conditions,secondary whereasand tertiary secondarycarbocations and tertiary carbocationsare generated are generatedwith increasing with ease. increasing ease. Thus, primary alkyloxonium ions undergo only SN2 reactions, whereas their Thus, primarysecondaryalkyloxonium and tertiary relativesions undergo enter intoonly SN1 andSN2 E1reactions, processes. Whenwhereas good theirnucleophilessecondary and tertiary relativesare present,enter we observeinto S NS1N1and products.E1 processes. When good nucleophiles are present, we observe SN1 products. Reactivity of Oxonium Ions Does not dissociate: Dissociate:

SN2 only SN1 and E1 ϩ ϩ ϩ RprimOOH2 ӶϽRsecOOH2 RtertOOH2

Increasing ease of carbocation formation 9 Synthetically, this fact is exploited in the preparation of tertiary haloalkanes from tertiary alcohols in the presence of excess concentrated aqueous hydrogen halide. The product forms in minutes at room temperature. The mechanism is precisely the reverse of that of solvolysis (Section 7-2). Synthetically, this fact is exploited in the preparation of tertiary haloalkanes from tertiary alcohols in the presence of excess concentrated9-29-2 Reactions Reactionsaqueous of of Alcohols Alcoholshydrogen withwith StrongStronghalide .AcidsAcidsThe productCHAPTERCHAPTER 99 329 forms in minutes at room temperature. The mechanism is precisely the reverse of that of solvolysis. ConversionConversion of of 2-Methyl-2-propanol 2-Methyl-2-propanol into into 2-Bromo-2-methylpropane 2-Bromo-2-methylpropane

(CH(CH3)33)COH3COH ϩϩ HBrHBr (CH(CH3)33)CBr3CBr ϩϩ HH22OO ExcessExcess ReactionRReactionR

MechanismMechanism of of the the S SN1N 1Reaction Reaction of of Tertiary Tertiary Alcohols Alcohols with with Hydrogen Hydrogen Halides Halides ϩϩ ϪϪ

š š " " š " (CH ) C OH ϩ H Br" (CH ) C OH ϩ " Br" (CH3)3C3 OOOO"šO"H ϩ H Br"š" (CH3)33C3 OO"OH" 22 ϩ Br"š" ϩϩ ϪϪ

š " š " š š "

H O ϩ (CH ) C ϩ " Br" H O ϩ (CH ) COBr " H2"šO2" ϩ (CH3)33C3 ϩ Br"š" H2"šO2" ϩ (CH33)33COBr"š" MechanismMechanism

The reasonTheThefor reason thereasonsuccess for for the the success successof this of ofprocess this this process processis thatis is that thatrelatively relatively relatively low lowlow temperatures temperaturestemperatures areareare suf-suf-sufficient to generate! cient! cientthe to to generatetertiary generate carbocation,the the tertiary tertiary carbocation, carbocation,thus largely thus thus largely preventinglargely preventing preventingcompeting competingcompetingE 1 E1E1reactions reactionsreactions . (Section(Section 7-6). 7-6). Indeed, Indeed, at at higher higher temperatures temperatures (or (or in in the the absence absence of of good good nucleophiles), nucleophiles), At highereliminationeliminationtemperatures becomes becomes dominant.*(or dominant.*in the This Thisabsence explains explains whyof whygood protonated protonatednucleophiles), secondary secondary alcohols alcoholselimination showshow thethebecomes most complex behavior in the presence of HX, following S 2, S 1, and E1 pathways: They dominantmost. complex behavior in the presence of HX, following SNN2, SNN1, and E1 pathways: They are relatively hindered, compared with their primary counterparts (thus reduced SN2 reactivity), 10 are relatively hindered, compared with their primary counterparts (thus reduced SN2 reactivity), andand relatively relatively slow slow in in forming forming carbocations, carbocations, compared compared with with their their tertiary tertiary counterparts counterparts (e.g., retarded SN1 reactivity); review Section 7-9. (e.g., retarded SN1 reactivity); review Section 7-9. TheThe E1 E1 reaction, reaction, here here called called dehydration dehydration because because it it results results in in the the loss loss of of a a molecule molecule of water (Figure 9-1, breaking bonds b and c; see also Sections 9-3 and 9-7), is one of the of water (Figure 9-1, breaking bonds b and c; see also Sections 9-3 and 9-7), is one of the AnimationAnimation methodsmethods for for the the synthesis synthesis of of alkenes alkenes (Section (Section 11-7). 11-7). Rather Rather than than the the “nucleophilic” “nucleophilic” acids acids HBrHBr and and HI, HI, so so called called because because the the conjugate conjugate base base is is a a good good nucleophile, nucleophile, “nonnucleophilic”“nonnucleophilic” ANIMATEDANIMATED MECHANISM:MECHANISM: acids, such as H3PO4 or H2SO4, are employed. Alcohol dehydration acids, such as H3PO4 or H2SO4, are employed. Alcohol dehydration Alcohol Dehydration by the E1 Mechanism Alcohol Dehydration by the E1 Mechanism OH OH ReactionRReactionR H2SO4, 130 –140ЊC H2SO4, 130 –140ЊC ϪH2O ϪH2O 87% Cyclohexanol Cyclohexene87% Cyclohexanol Cyclohexene

Mechanism of the Dehydration of Cyclohexanol Mechanism of the Dehydration of Cyclohexanol H HHϩ Mechanism HHHEϩ Mechanism ðšO OH HEšO ðšO O šO ϩ ϩ H ϩ H ϩ ϩϩH H2šO ϩ H3O # ϩ # ϩ ϩϩH H2#šO ϩ H3O #

2 In the example, the conjugate base of the acid is the poor nucleophile HSO4 , and proton 2 Inloss the fromexample, the intermediate the conjugate carbocation base of the is observed.acid is the Dehydration poor nucleophile of tertiary HSO alcohols4 , and protonis even losseasier, from often the intermediateoccurring at carbocationslightly above is observed.room temperature. Dehydration of tertiary alcohols is even easier, often occurring at slightly above room temperature.

Exercise 9-2 Exercise 9-2 Write the structure of the product that you expect from the reaction of 4-methyl-1-pentanol with Writeconcentrated the structure aqueous of the HI. product Give the that mechanism you expect of from its formation.the reaction of 4-methyl-1-pentanol with concentrated aqueous HI. Give the mechanism of its formation.

*The preference for elimination at high temperatures has its origin in its relatively large positive entropy *The(Section preference 2-1), as for two elimination molecules (alkeneat high 1temperatures H2O) result hasfrom its one origin (alcohol), in its andrelatively the DS large8 term positive in the expression entropy DG8 5 DH8 2 TDS8 increases with temperature (see also Exercise 7-17). (Section 2-1), as two molecules (alkene 1 H2O) result from one (alcohol), and the DS8 term in the expression DG8 5 DH8 2 TDS8 increases with temperature (see also Exercise 7-17). This explains why protonated secondary alcohols show the most complex behavior in the presence of HX, following SN2, SN1, and E1 pathways:

They are relatively hindered, compared with their primary counterparts (thus reduced SN2 reactivity), and relatively slow in forming carbocations, compared with their tertiary counterparts (e.g., retarded SN1 reactivity).

The E1 reaction, here called dehydration, because it results in the loss of a molecule of water, is one of the methods for the synthesis of alkenes. Rather than the “nucleophilic” acids HBr and HI (called because the conjugate base is a good nucleophile),

“nonnucleophilic” acids, such as H3PO4 or H2SO4, are employed.

– In the example, the conjugate base of the acid is the poor nucleophile HSO4 , and proton loss from the intermediate carbocation is observed. Dehydration of tertiary alcohols is even easier, often occurring at slightly above room temperature.

11 9-2 Reactions9-2 Reactions of Alcohols of Alcohols with withStrong Strong Acids AcidsCHAPTERCHAPTER 9 9 329 329

ConversionConversion of 2-Methyl-2-propanol of 2-Methyl-2-propanol into 2-Bromo-2-methylpropane into 2-Bromo-2-methylpropane

(CH3)(CH3COH3)3COHϩ HBrϩ HBr (CH3)3(CHCBr3)3CBrϩ Hϩ2O H2O Excess Excess ReactionR ReactionR

MechanismMechanism of the of S Nthe1 Reaction SN1 Reaction of Tertiary of Tertiary Alcohols Alcohols with Hydrogen with Hydrogen Halides Halides ϩ ϩ Ϫ Ϫ

š š " " š " ϩ " ϩ " " (CH3)(CH3COO3)"O3CHOO"šOH Hϩ Br"H Br"š (CH3)3(CHCO3)"OH3CO2 "OH2 Brϩ" Br"š ϩ ϩ Ϫ Ϫ

š " " š " ϩ ϩ š " " ϩ š " H2"O H2"šO (CHϩ 3)(CH3C 3)3C Br"ϩ Br"š H2"O H2"šO (CHϩ 3)3(CHCO3Br)"3COBr"š MechanismMechanism

The reasonThe reason for the for success the success of this of process this process is that is relatively that relatively low temperatures low temperatures are suf- are suf- ! cient! cientto generate to generate the tertiary the tertiary carbocation, carbocation, thus largely thus largely preventing preventing competing competing E1 reactions E1 reactions (Section(Section 7-6). 7-6). Indeed, Indeed, at higher at higher temperatures temperatures (or in (or the in absence the absence of good of goodnucleophiles), nucleophiles), eliminationelimination becomes becomes dominant.* dominant.* This explains This explains why protonated why protonated secondary secondary alcohols alcohols show theshow the most complexmost complex behavior behavior in the inpresence the presence of HX, of following HX, following SN2, S SNN1,2, and SN1, E1 and pathways: E1 pathways: They They are relativelyare relatively hindered, hindered, compared compared with their with primary their primary counterparts counterparts (thus reduced(thus reduced SN2 reactivity), SN2 reactivity), and relativelyand relatively slow slowin forming in forming carbocations, carbocations, compared compared with their with tertiary their tertiary counterparts counterparts (e.g., (e.g.,retarded retarded SN1 reactivity); SN1 reactivity); review review Section Section 7-9. 7-9. 330 CHAPTER 9 FurtherThe E1The reaction, Reactions E1 reaction, here called here of calleddehydration Alcohols dehydration because and because itthe results it Chemistry results in the in loss the of loss aof molecule of Ethers a molecule of waterof water(Figure (Figure 9-1, breaking 9-1, breaking bonds bonds b and bc ;and see calso; see Sections also Sections 9-3 and 9-3 9-7), and is9-7), one isof one the of the AnimationAnimation methodsmethods for the for synthesis the synthesis of alkenes of alkenes (Section (Section 11-7). 11-7). Rather Rather than the than “nucleophilic” the “nucleophilic” acids acids HBr andHBr HI, and so HI, called so called because because the conjugate the conjugate base isbase a good is a nucleophile,good nucleophile, “nonnucleophilic” “nonnucleophilic” ANIMATED ANIMATED MECHANISM: MECHANISM: acids,acids, such assuch H3 POas H4 or3PO H42 SOor 4H, 2areSO 4employed., are employed. Alcohol dehydrationAlcohol dehydration Exercise 9-3 AlcoholAlcohol Dehydration Dehydration by the by E1 the Mechanism E1 Mechanism Write the structure of the products that you expect from the reaction of 1-methylcyclohexanol with OH (a) concentrated HCl and (b) OHconcentrated H2SO4. Compare and contrast the mechanisms of the 2 2 ReactionR ReactionR two processes. (Hint: Compare the relative nucleophilicity of Cl and HSO4 . Caution: When H SO , 130 –140ЊC writing mechanisms, use “arrow 2pushing”4 H2SO4, 130 to –140depictЊC electron ! ow; write out every step separately; ϪH2O ϪH O formulate complete structures, including charges2 and relevant electron pairs; and draw explicit reaction arrows to connect starting materials or intermediates87% 87% with their respective products. Don’t use shortcuts, and don’tCyclohexanol beCyclohexanol sloppy!) CyclohexeneCyclohexene

MechanismMechanism of the of Dehydration the Dehydration of Cyclohexanol of Cyclohexanol

HHϩHHϩ MechanismMechanism In SummaryH H TreatmentHE of HE alcohols with strong acid leads to protonation to give alkyl- ðšO OðšO O šO šO oxonium ions, which, when primary, undergoϩ SϩN2 reactions in the presence of good nucleophiles. Alkyloxonium ions from secondary or H tertiaryH alcohols convert into carbocations, ϩ ϩ ϩ ϩ # ϩϩH H šO š ϩ H O # which furnishϩϩ productsH of substitution and elimination (dehydration).2# H2#O ϩ3 H3O

Ϫ 2 2 In theIn example, the example, the conjugate the conjugate base ofbase the of acid the is acid theX is,poor S Nthe2 nucleophilepoor nucleophile HSO4 HSO, and4 proton, and proton RX ϩ H2šO loss fromloss thefrom intermediate the intermediate carbocation carbocation is observed. is observed.R Dehydration ϭ prim Dehydration of tertiary of tertiary alcohols alcohols" is even is even easier,easier, often oftenoccurring occurring at slightly at slightly above aboveroomH roomtemperature. temperature. Hϩ ROHš ROϩOð " Ϫ X , SN1 ExerciseExercise 9-2 9-2 H ϪH O RX 2 ϩ R ϩ Write Writethe structure the structure of the productof the product that you that expect you expect from the from reaction the reaction of 4-methyl-1-pentanol of 4-methyl-1-pentanolϪH , E1 with with R ϭ sec, tert 12 concentratedconcentrated aqueous aqueous HI. Give HI. the Give mechanism the mechanism of its formation. of its formation. Alkene

*The preference*The preference for elimination for elimination at high at temperatures high temperatures has its hasorigin its inorigin its relatively in its relatively large positive large positive entropy entropy (Section 2-1), as two molecules (alkene 1 H O) result from one (alcohol), and the DS8 term in the expression 9-3(Section 2-1),CARBOCATION as two molecules (alkene2 1 REARRANGEMENTSH2O) result from one (alcohol), and the DS8 term in the expression DG8 5D DGH8 85 2 D THD8 S28 increasesTDS8 increases with temperature with temperature (see also (see Exercise also Exercise 7-17). 7-17).

When alcohols are transformed into carbocations, the carbocations themselves are subject to rearrangements. The two rearrangements, known as hydride shifts and alkyl shifts, can occur with most types of carbocations. The rearranged molecules can then undergo further SN1 or E1 reactions. The result is likely to be a complex mixture, unless we can establish a thermodynamic driving force toward one speci" c product. Hydride shifts give new S 1 products ReactionR N Treatment of 2-propanol with concentrated hydrogen bromide gives 2-bromopropane, as expected. However, exposure of the more highly substituted secondary alcohol 3-methyl- 2-butanol to the same reaction conditions produces a surprising result. The expected SN1 product, 2-bromo-3-methylbutane, is only a minor component of the reaction mixture; Normal SN1 Reaction the major product is 2-bromo-2-methylbutane. of an Alcohol (No Rearrangement) Hydride Shift in the SN1 Reaction of an Alcohol with HBr OH H OH H Br Br H A A A HBr, 0ЊC A A A A CH3COCCH3 CH3COCCH3 ϩ CH3COCCH3 ϩ HOOH CH3CHCH3 ϩ HBr A A A A A A 0ЊC H3C H H3C H H3C H Minor product Major product Br 3-Methyl-2- 2-Bromo-3- 2-Bromo-2- A butanol methylbutane methylbutane CH3CHCH3 ϩ HOOH (Normal product) (Rearranged product) What is the mechanism of this transformation? The answer is that carbocations can undergo rearrangement by hydride shifts, in which the hydrogen (yellow) moves with both electrons Animation from its original position to the neighboring carbon atom. Initially, protonation of the alcohol followed by loss of water gives the expected secondary carbocation. A shift of the tertiary ANIMATED MECHANISM: hydrogen to the electron-de" cient neighbor then generates a tertiary cation, which is more Carbocation rearrangement stable. This species is " nally trapped by bromide ion to give the rearranged SN1 product. 9.3 CARBOCATION REARRANGEMENTS

The rearranged molecules can then undergo further SN1 or E1 reactions.

The result is likely to be a complex mixture, unless we can establish a thermodynamic driving force toward one specific product.

Hydride shifts give new SN1 products

13 330 CHAPTER 9 Further Reactions of Alcohols and the Chemistry of Ethers 330 CHAPTER 9 Further Reactions of Alcohols and the Chemistry of Ethers

Exercise 9-3 Exercise 9-3 Write the structure of the products that you expect from the reaction of 1-methylcyclohexanol with Write(a) concentrated the structure HClof the and products (b) concentrated that you expect H2SO from4. Compare the reaction and contrastof 1-methylcyclohexanol the mechanisms ofwith the 2 2 (a)two concentrated processes. HCl(Hint: and Compare (b) concentrated the relative H2 SOnucleophilicity4. Compare and of Clcontrast and HSOthe mechanisms4 . Caution: of When the 2 2 twowriting processes. mechanisms, (Hint: Compareuse “arrow the pushing” relative to nucleophilicity depict electron of ! ow;Cl write and HSOout every4 . Caution: step separately; When writingformulate mechanisms, complete use structures, “arrow pushing”including to charges depict electronand relevant ! ow; electron write out pairs; every and step draw separately; explicit formulatereaction arrowscomplete to connectstructures, starting including materials charges or intermediates and relevant with electron their respectivepairs; and products. draw explicit Don’t reactionuse shortcuts, arrows toand connect don’t be starting sloppy!) materials or intermediates with their respective products. Don’t use shortcuts, and don’t be sloppy!)

In Summary Treatment of alcohols with strong acid leads to protonation to give alkyl- Inoxonium Summary ions, Treatment which, when of alcoholsprimary, undergo with strong SN2 acidreactions leads in to the protonation presence of to good give nucleo- alkyl- oxoniumphiles. Alkyloxoniumions, which, when ions primary, from secondary undergo S orN 2 tertiary reactions alcohols in the presence convert into of good carbocations, nucleophiles.which Alkyloxonium furnish products ions of fromsubstitution secondary and orelimination tertiary alcohols(dehydration). convert into carbocations, which furnish products of substitution and elimination (dehydration).

Ϫ X , SN2

Ϫ RX ϩ H2šO R ϭ prim " H X , SN2 ϩ RX ϩ H2šO H R ϭ prim " ROHš R ϩOH " ϩ O ð H XϪ, S 1 ROHš ROϩOð N RX " H ϪH2O Ϫ ϩ X , SN1 R ϪHϩ, E1 H RϪ ϭH sec,O tert RX 2 ϩ Alkene R ϪHϩ, E1 R ϭ sec, tert Alkene

9-3 CARBOCATION REARRANGEMENTS 9-3 CARBOCATION REARRANGEMENTS When alcohols are transformed into carbocations, the carbocations themselves are subject Whento rearrangements. alcohols are transformed The two rearrangements, into carbocations, known the as carbocations hydride shifts themselves and alkyl are shifts, subject can Treatment of 2-propanol towithoccur rearrangements.concentrated with most typesThe twoofhydrogen carbocations. rearrangements,bromide The knownrearrangedgives as hydridemolecules2-bromopropane, shifts can and then alkyl undergo shifts,as further can expected. occurSN1 withor E1 most reactions. types Theof carbocations. result is likely The to rearranged be a complex molecules mixture, can unless then weundergo can establish further a thermodynamic driving force toward one speci" c product. SN1 or E1 reactions. The result is likely to be a complex mixture, unless we can establish Exposure of the more highlya Hydridethermodynamicsubstituted shifts drivingsecondary give force new towardalcohol S 1one products speci3-methyl" c product.-2-butanol to the same ReactionR N reaction conditions producesTreatmenta surprising of 2-propanolresult with. concentrated hydrogen bromide gives 2-bromopropane, as Hydride shifts give new SN1 products ReactionR expected. However, exposure of the more highly substituted secondary alcohol 3-methyl- Treatment of 2-propanol with concentrated hydrogen bromide gives 2-bromopropane, as The expected S 1 product2-butanol(2-bromo to the-3 - samemethylbutane) reaction conditionsis only producesa minor a surprisingcomponent result. Theof expectedthe N expected. However, exposure of the more highly substituted secondary alcohol 3-methyl- S 1 product, 2-bromo-3-methylbutane, is only a minor component of the reaction mixture; reaction mixture, while the2-butanolmajorN product to the sameis 2 reaction-bromo conditions-2-methylbutane produces a. surprising result. The expected Normal S 1 Reaction the major product is 2-bromo-2-methylbutane. N S 1 product, 2-bromo-3-methylbutane, is only a minor component of the reaction mixture; of an Alcohol N Normal S 1 Reaction the major product is 2-bromo-2-methylbutane. (No Rearrangement)N Hydride Shift in the SN1 Reaction of an Alcohol with HBr of an Alcohol (No Rearrangement)OH H HydrideOH Shift in the SN1H ReactionBr of an AlcoholBr Hwith HBr A A A HBr, 0ЊC A A A A CHH3COOHCCH3 CHH3COBrCCH3 ϩ CHBr3COHCCH3 ϩ HOOH CHOH3CHCH3 ϩ HBr A A A A A HBr, 0ЊC AA AA AA AA 0ЊC CHH3C3COCCHH 3 CHH3C3COCHCH3 ϩ CHH3C3COCHCH3 ϩ HOOH CH3CHCH3 ϩ HBr A A MinorA productA MajorA productA Br 0ЊC H33-Methyl-2-C H H32-Bromo-3-C H H32-Bromo-2-C H A butanol Minormethylbutane product Majormethylbutane product CHBr3CHCH3 ϩ HOOH 3-Methyl-2- (Normal2-Bromo-3- product) (Rearranged2-Bromo-2- product) A butanol methylbutane methylbutane CH3CHCH3 ϩ HOOH What is the mechanism of this transformation?(Normal product) The answer (Rearranged is that product) carbocations can undergo rearrangement by hydride shifts, in which the hydrogen (yellow) moves with both electrons What is the mechanism of this transformation? The answer is that carbocations can undergo Animation from its original position to the neighboring carbon atom. Initially, protonation of the alcohol14 rearrangementfollowed by lossby hydride of water shifts, gives in the which expected the hydrogen secondary (yellow) carbocation. moves A with shift both of the electrons tertiary ANIMATEDAnimation MECHANISM: fromhydrogen its original to the position electron-de to the" cient neighboring neighbor carbon then generatesatom. Initially, a tertiary protonation cation, ofwhich the alcohol is more followed by loss of water gives the expected secondary carbocation. A shift of the tertiary Carbocation rearrangement stable. This species is " nally trapped by bromide ion to give the rearranged SN1 product. ANIMATED MECHANISM: hydrogen to the electron-de" cient neighbor then generates a tertiary cation, which is more Carbocation rearrangement stable. This species is " nally trapped by bromide ion to give the rearranged SN1 product. The mechanism of this transformation: carbocations can undergo rearrangement by hydride shifts, in which the hydrogen (yellow) moves with both electrons from its original position to the neighboring carbon atom. A shift of9-3the Carbocationtertiary hydrogen Rearrangementsto the electronCHAPTER- 9 331 deficient neighbor then generates a tertiary cation, which is more stable. Mechanism of Carbocation Rearrangement Step 1. Protonation Step 2. Loss of water

HH HH Mechanism HEϩ HEϩ H ðšOH H OÂ H OÂ H A A A A A A ϩ ϩ CH3CO CCH3 ϩ H CH3CO CCH3 CH3CO CCH3 CH3COCCH3 ϩ H2$šO A A A A A A A A H3C H H3C H H3C H H3C H

Step 3. Hydride shift Step 4. Trapping by bromide

H H H )Br>) H A ϩϩA ϩ A Ϫ A A CH3COCCH3 CH3COCCH3 CH3COCCH3 ϩ )Brp>) CH3COCCH3 A A A A A A A A H3C H H3C H H3C H H3C H Secondary Tertiary carbocation carbocation (Less stable) (More stable) 15

The details of the transition state of the observed hydride shift are shown schematically Note: Color is used to indicate the electrophilic (blue), in Figure 9-2. A simple rule to remember when executing hydride shifts in carbocations is nucleophilic (red), and that the hydrogen and the positive charge formally exchange places between the two neigh- leaving-group (green) boring carbon atoms participating in the reaction. character of the reacting Hydride shifts of carbocations are generally very fast—faster than SN1 or E1 reactions. centers. Therefore, a color In part, this rapidity is due to hyperconjugation, which weakens the C–H bond (Section 7-5 may “switch” from one group and Figure 9-2B). They are particularly favored when the new carbocation is more stable or atom to another as the than the original one, as in the example depicted in Figure 9-2. reaction proceeds.

Migrating hydride ‡ H H H H + δ + δ + H3C + C C C C C C

H3C CH3 H3C H H3C H H3C H3C CH3 CH3 sp3 sp2 sp2 sp3 Both carbons are rehybridizing A

Hyperconjugation Migrating hydride Hyperconjugation

‡ H H H + + H H3C + + C C δ C C δ C C

H3C CH3 H3C H H3C H H3C H3C CH3 CH3

p back lobe p back lobe B increasing decreasing Figure 9-2 The rearrangement of a carbocation by a hydride shift: (A) dotted-line notation; (B) orbital picture. Note that the migrating hydrogen and the positive charge exchange places. In addition, you can see how hyperconjugation weakens the C–H bond by effecting some electron transfer into the empty neighboring p orbital. 9-3 Carbocation Rearrangements CHAPTER 9 331

9-3 Carbocation Rearrangements CHAPTER 9 331 Mechanism of Carbocation Rearrangement

Step 1. Protonation Mechanism of Carbocation RearrangementStep 2. Loss of water HH HH Mechanism Step 1. Protonation HEϩ HEStepϩ 2. Loss of water H ðšOH H OÂ H OÂ H A A HHA A HHA A ϩ Mechanism ϩ HEϩ HEϩ CHH3CðOšO CCHH 3 ϩ H CHH3COO CCH3 CHH3COO CCH3 CHH3COCCH3 ϩ H2$šO A A A AÂ A AÂ A A A A A A A A ϩ H C H ϩ H C H H C H H C H CH3C3 O CCH3 ϩ H CH3C3 O CCH3 CH3C3 O CCH3 CH3C3 OCCH3 ϩ H2$šO A A A A A A A A H C H H C H H C H H C H Step3 3. Hydride shift 3 3 Step 4. Trapping3 by bromide

Step H3. Hydride shift H StepH 4. Trapping by bromide)Br>) H A ϩϩA ϩ A Ϫ A A CHH3COCCH3 CH3COHCCH3 CH3COHCCH3 ϩ )Brp>) CH)Br>3C)OHCCH3 AA A A AA A AA AA AA H C ϩϩH H C H HϩC H Ϫ H C H CH3C3 OCCH3 CH3C3 OCCH3 CH3C3 OCCH3 ϩ )Brp>) CH3C3 OCCH3 SecondaryA A ATertiaryA A A A A H3carbocationC H H3carbocationC H H3C H H3C H Secondary(Less stable) (MoreTertiary stable) carbocation carbocation Note: Color is used to indicate (LessThe stable) details of the transition(More stable) state of the observed hydride shift are shown schematically the electrophilic (blue), in Figure 9-2. A simple rule to remember when executing hydride shifts in carbocations is The details of the transition state of the observed hydride shift are shown schematically Note:nucleophilic Color is (red),used toand indicate that the hydrogen and the positive charge formally exchange places between the two neigh- the electrophilic (blue), in Figure 9-2. A simple rule to remember when executing hydride shifts in carbocations is leaving-group (green) boring carbon atoms participating in the reaction. nucleophiliccharacter of (red), the reacting and that the hydrogen and the positive charge formally exchange places between the two neigh- Hydride shifts of carbocations are generally very fast—faster than SN1 or E1 reactions. leaving-groupcenters. Therefore, (green) a color The hydrogenboringIn part, carbon thisand rapidity atomsthe participatingis duepositive to hyperconjugation, in the reaction.charge which formallyweakens the C–Hexchange bond (Section 7-5places charactermay between“switch” of the fromreacting one thegroup two andHydride Figure shifts 9-2B). of They carbocations are particularly are generally favored very when fast—faster the new carbocationthan SN1 or isE1 more reactions. stable centers.or atom Therefore, to another a as color the neighboringInthan part,carbon the this original rapidityatoms one, is dueas intoparticipating thehyperconjugation, example depicted inwhich inthe Figure weakensreaction 9-2. the C–H. bond (Section 7-5 mayreaction “switch” proceeds. from one group and Figure 9-2B). They are particularly favored when the new carbocation is more stable or atom to another as the than the original one, as in the example depicted in Figure 9-2. reaction proceeds. Hydride shifts of carbocations are generally very fast—faster than SN1 or E1 reactions. This rapidity is due to hyperconjugation, which weakensMigrating hydridethe C‡–H bond. H Migrating hydride H H ‡ H H + δ + δ + H3C + H C C C H C C C H H C H3C + CH3 H3C δ + δ + H H3 3C + H C C C C C C H3C 3 H3C CH3 CH3 H C sp sp2CH H C H H C sp2 sp3 H 3 3 3 Both carbons are rehybridizing 3 AH3C H3C CH3 CH3 sp3 sp2 sp2 sp3 Both carbons are rehybridizing A Hyperconjugation Migrating hydride Hyperconjugation

Hyperconjugation Migrating hydride ‡ Hyperconjugation H H H ‡ H + + H H H H3C + + C +C δ C C δ +C C H H3C H3C CH3 H3C + + H H3C H C C δ C C δ C C H3C H3C CH3 CH3 H3C CH3 H3C H H3C H H3C H3C CH3 CH3 p back lobe p back lobe 16 B increasing decreasing p back lobe p back lobe BFigure 9-2 The rearrangement of a carbocation byincreasing a hydride shift:decreasing (A) dotted-line notation; (B) orbital picture. Note that the migrating hydrogen and the positive charge exchange places. In Figureaddition, 9-2 you The can rearrangement see how hyperconjugation of a carbocation weakens by a thehydride C–H shift:bond (A)by dotted-lineeffecting some notation; electron (B)transfer orbital intopicture. the emptyNote thatneighboring the migrating p orbital. hydrogen and the positive charge exchange places. In addition, you can see how hyperconjugation weakens the C–H bond by effecting some electron transfer into the empty neighboring p orbital. Primary carbocations are too unstable to be formed by rearrangement. 9-3 Carbocation Rearrangements CHAPTER 9 333 Carbocations of comparable stability, for example, the pairs secondary–secondary or 9-3 Carbocation Rearrangements CHAPTER 9 333 tertiary–tertiary, equilibrate readily. Any added nucleophile will trap all carbocations present, furnishing mixturesOHof products. Br Br A HBr, 0ЊC A A CH3CCHOH2CH2CH3 CH3CCHBr 2CH2CH3 ϩ CH3CH2CCHBr2CH3 AA HBr, 0ЊC A A A A CHH3CCH2CH2CH3 CHH3CCH2CH2CH3 ϩ CH3CHH 2CCH2CH3 A A A Carbocation Hrearrangements take place regardlessH of the nature of the precursorH to the carbocation: alcohols (this section), haloalkanes (Chapter 7), and alkyl sulfonates (Section 6-7). CarbocationForCarbocation example,rearrangements solvolysisrearrangements of take2-bromo-3-ethyl-2-methylpentane takeplace place regardlessregardless of theof nature thein ethanol ofnature the (ethanolysis) precursorof the to gives precursorthe car- to the carbocationthebocation:: twoalcohols, possible alcohols haloalkanes,tertiary (this ethers.section),and haloalkanesalkyl sulfonates (Chapter 7), .and alkyl sulfonates (Section 6-7). For example, solvolysis of 2-bromo-3-ethyl-2-methylpentane in ethanol (ethanolysis) gives the two possible tertiaryRearrangement ethers. in Solvolysis of a Haloalkane

Br H CH3CH2 O H H CHO 2CH3 A A RearrangementA in SolvolysisA of a HaloalkaneA A CH3CH2OH CH3COOCCH2CH3 CH3C CCH2CH3 ϩϩCH3CO 23CHCCH HOBr Br H CH3CH2 O H H CHO 2CH3 AA AA A A A A A AA A HC3 CH2CH3 CH3CH2OH HC3 CH2CH3 HC3 CH2CH3 CH3COOCCH2CH3 CH3C CCH2CH3 ϩϩCH3CO 23CHCCH HOBr 2-Bromo-3-ethyl-A A 2-Ethoxy-3-ethyl-A A 3-Ethoxy-3-ethyl-A A 2-methylpentane 2-methylpentane 2-methylpentane HC3 CH2CH3 HC3 CH2CH3 HC3 CH2CH3 (Normal product) (Rearranged product) 2-Bromo-3-ethyl- 2-Ethoxy-3-ethyl- 3-Ethoxy-3-ethyl- 17 2-methylpentane 2-methylpentane 2-methylpentane (Normal product) (Rearranged product) Exercise 9-6 Give a mechanism for the preceding reaction. Then predict the outcome of the reaction of 2-chloro- 4-methylpentaneExercise 9-6 with methanol. (Hint: Try two successive hydride shifts to the most stable carbocation.) Give a mechanism for the preceding reaction. Then predict the outcome of the reaction of 2-chloro- 4-methylpentane with methanol. (Hint: Try two successive hydride shifts to the most stable carbocation.) Carbocation rearrangements also give new E1 products How does the rearrangement of intermediates affect the outcome of reactions under conditions thatCarbocation favor elimination? rearrangements At elevated temperatures also and ingive relatively new nonnucleophilic E1 products media, rearranged carbocations yield alkenes by the E1 mechanism (Sections 7-6 and 9-2). For example,How treatment does the ofrearrangement 2-methyl-2-pentanol of intermediates with aqueous affect the sulfuric outcome acid of at reactions 808C gives under the conditions same majorthat favor alkene elimination? product as Atthat elevated formed temperatures when the starting and in material relatively is 4-methyl-2-pentanol.nonnucleophilic media, The rear- conversionranged carbocations of the latter yield alcohol alkenes includes by the a hydrideE1 mechanism shift of (Sectionsthe initial 7-6carbocation and 9-2). followed For exam- byple, deprotonation. treatment of 2-methyl-2-pentanol with aqueous sulfuric acid at 808C gives the same major alkene product as that formed when the starting material is 4-methyl-2-pentanol. The conversion of the latter alcoholRearrangement includes a hydride in E1 Elimination shift of the initial carbocation followed Note: Carbocations can by deprotonation. always give SN1 and E1 OH H C CH CH H OH products. Their relative ratios A 3 2 3 A A H2SO4, 80ЊC i f H2SO4, 80ЊC dependNote: on Carbocationsthe structure of can CH3COCH2CH2CH3 RearrangementCCP in E1 Elimination CH3COCH2CCH3 ϪH O ϪH O the cation as well as the A 2 f i 2 A A always give SN1 and E1 OH H C H H OH nucleophilicity of potential CH3 3H3C CH2CH3 With CH3 H products. Their relative ratios A Major product rearrangement A A trapping agents and the H2SO4, 80ЊC i f H2SO4, 80ЊC depend on the structure of CH2-Methyl-2-pentanol3COCH2CH2CH3 2-Methyl-2-penteneCCP 4-Methyl-2-pentanolCH3COCH2CCH3 temperature.the cation as well as the A ϪH2O f i ϪH2O A A nucleophilicity of potential CH3 H3C H With CH3 H Major product rearrangement trapping agents and the Exercise2-Methyl-2-pentanol 9-7 2-Methyl-2-pentene 4-Methyl-2-pentanol temperature. (a) Give mechanisms for the preceding E1 reactions. (b) Treatment of 4-methylcyclohexanol with hot acid gives 1-methylcyclohexene. Explain by a mechanism. (Hint: Consider several sequential HExercise shifts.) 9-7 (a) Give mechanisms for the preceding E1 reactions. (b) Treatment of 4-methylcyclohexanol with hot acid gives 1-methylcyclohexene. Explain by a mechanism. (Hint: Consider several sequential OtherH shifts.) carbocation rearrangements are due to alkyl shifts Carbocations, particularly when lacking suitable (secondary and tertiary) hydrogens next to the positively charged carbon, can undergo another mode of rearrangement, known as alkyl groupOther migration carbocation or alkyl shift. rearrangements are due to alkyl shifts Carbocations, particularly when lacking suitable (secondary and tertiary) hydrogens next to the positively charged carbon, can undergo another mode of rearrangement, known as alkyl group migration or alkyl shift. 9-3 Carbocation Rearrangements CHAPTER 9 333

OH Br Br A HBr, 0ЊC A A CH3CCH2CH2CH3 CH3CCH2CH2CH3 ϩ CH3CH2CCH2CH3 A A A H H H

Carbocation rearrangements take place regardless of the nature of the precursor to the carbocation: alcohols (this section), haloalkanes (Chapter 7), and alkyl sulfonates (Section 6-7). For example, solvolysis of 2-bromo-3-ethyl-2-methylpentane in ethanol (ethanolysis) gives the two possible tertiary ethers.

Rearrangement in Solvolysis of a Haloalkane

Br H CH3CH2 O H H CHO 2CH3 A A A A A A CH3CH2OH CH3COOCCH2CH3 CH3C CCH2CH3 ϩϩCH3CO 23CHCCH HOBr A A A A A A HC3 CH2CH3 HC3 CH2CH3 HC3 CH2CH3 2-Bromo-3-ethyl- 2-Ethoxy-3-ethyl- 3-Ethoxy-3-ethyl- 2-methylpentane 2-methylpentane 2-methylpentane (Normal product) (Rearranged product)

Exercise 9-6 Give a mechanism for the preceding reaction. Then predict the outcome of the reaction of 2-chloro- 4-methylpentane with methanol. (Hint: Try two successive hydride shifts to the most stable carbocation.)

Carbocation rearrangements also give new E1 products How does the rearrangement of intermediates affect the outcome of reactions under conditions that favor elimination? At elevated temperatures and in relatively nonnucleophilic media, rearranged carbocations yield alkenes by the E1 mechanism (Sections 7-6 and 9-2). For example, treatment of 2-methyl-2-pentanol with aqueous sulfuric acid at 808C gives the same Carbocation rearrangementsmajor alkene alsoproductgive as thatnew formedE 1whenproducts the starting material is 4-methyl-2-pentanol. The conversion of the latter alcohol includes a hydride shift of the initial carbocation followed At elevated temperaturesby deprotonation.and in relatively nonnucleophilic media, rearranged carbocations yield alkenes by the E1 mechanism. Rearrangement in E1 Elimination Note: Carbocations can always give SN1 and E1 OH H C CH CH H OH products. Their relative ratios A 3 2 3 A A H2SO4, 80ЊC i f H2SO4, 80ЊC depend on the structure of CH3COCH2CH2CH3 CCP CH3COCH2CCH3 the cation as well as the A ϪH2O f i ϪH2O A A nucleophilicity of potential CH3 H3C H With CH3 H Major product rearrangement trapping agents and the 2-Methyl-2-pentanol 2-Methyl-2-pentene 4-Methyl-2-pentanol temperature.

Other carbocation rearrangements are due to alkyl shifts 334 CHAPTER 9 FurtherExercise Reactions9-7 of Alcohols and the Chemistry of Ethers Carbocations, particularly(a) Givewhen mechanismslacking for thesuitable preceding (secondaryE1 reactions. (b) Treatmentand tertiary) of 4-methylcyclohexanolhydrogens withnext to the positively charged carbon,hot acid givescan 1-methylcyclohexene.undergo another Explainmode by a mechanism.of rearrangement, (Hint: Consider severalknown sequentialas alkyl H shifts.) group migration or alkyl shift. Rearrangement by Alkyl Shift in SN1 Reaction

H3C CH3 Br CH3 ReactionR A A HBr A A Other carbocationCH3 CrearrangementsOCOH areCH due3CO CtoCH alkyl3 shifts A A ϪHOH A A Carbocations, particularly when lacking suitable (secondary and tertiary) hydrogens next to H3C H H3C H the positively charged carbon, can undergo another mode of rearrangement, known as alkyl 94% group migration or alkyl shift. 18 3,3-Dimethyl-2-butanol 2-Bromo-2,3-dimethylbutane

As in the hydride shift, the migrating group takes its electron pair with it to form a bond to the neighboring positively charged carbon. The moving alkyl group and the positive charge formally exchange places.

Mechanism of Alkyl Shift

H3C CH3 H3C CH3 " Brš " CH3 Ϫ

ϪH šO ϩ " Brš " A A 2" A ϩϩCH shift A " A A Mechanism ϩ 3 CH3CCO "šOH ϩ H CH3COCCH3 CH3COCCH3 CH3COCCH3 Ϫ

A A š A A A A " " A A ϩH2"O Ϫ Br"š H3C H H3C H H3C H H3C H

The rates of alkyl and hydride shifts are comparable when leading to carbocations of similar stability. However, either type of migration is faster when furnishing tertiary carbocations relative to those ending in their secondary counterparts. This explains why, in the preceding discussion of hydride shifts, alkyl group rearrangement was not observed: The less substituted cation would have been the result. Exceptions to this observation are found only if there are other compelling reasons for the preference of alkyl migration, such as electronic stabilization or steric relief (see Worked Example 26 and Problem 61).

Working with the Concepts: Formulating a More Complex Solved Exercise 9-8 Carbocation Rearrangement Molecules of the type A have been shown to dehydrate to B. Formulate a mechanism.

H3C % ≥ H2SO4 ϪH O H 2 ≥ 0 OH CH3 A B

Remember WHIP Strategy First we need to look carefully at the function(s) in A and B, as well as the reagent: We recognize What the presence of a tertiary alcohol that is being treated with acid. These conditions are suggestive How of the formation of a carbocation. Moreover, the net reaction is dehydration, indicating an E1 Information process. Second, it is important to inspect the carbon skeleton in A and compare it with that in B: Proceed We can see that the methyl group has migrated to its neighboring carbon. Conclusion: We are dealing with a carbocation rearrangement featuring a methyl shift. Solution • The ! rst steps are protonation followed by loss of water to generate the carbocation.

H3C H3CH3C % ≥ ϩ Hϩ % ≥ % ≥ H HHϪH2O ≥ ≥ ϩ ðOH ðOH2 " ϩ 334 CHAPTER 9 Further Reactions of Alcohols and the Chemistry of Ethers

Rearrangement by Alkyl Shift in SN1 Reaction

H3C CH3 Br CH3 ReactionR A A HBr A A CH3COCOH CH3COCCH3 A A ϪHOH A A H3C H H3C H 94% 3,3-Dimethyl-2-butanol 2-Bromo-2,3-dimethylbutane As in the hydride shift, the migrating group takes its electron pair with it to form a bond to the neighboringAspositively in the hydridecharged shift, thecarbon migrating. The groupmoving takes its electronalkyl group pair withand it tothe formpositive a bond charge to the neighboring positively charged carbon. The moving alkyl group and the positive formally exchange places. charge formally exchange places.

Mechanism of Alkyl Shift

H3C CH3 H3C CH3 " Brš " CH3 Ϫ

ϪH šO ϩ " Brš " A A 2" A ϩϩCH shift A " A A Mechanism ϩ 3 CH3CCO "šOH ϩ H CH3COCCH3 CH3COCCH3 CH3COCCH3 Ϫ

A A š A A A A " " A A ϩH2"O Ϫ Br"š H3C H H3C H H3C H H3C H

The rates of alkyl and hydride shifts are comparable when leading to carbocations of similar stability. However, either type of migration is faster when furnishing tertiary carbocations The rates ofrelativealkyl toand thosehydride ending shiftsin their aresecondarycomparable counterparts.when Thisleading explains why,to carbocations in the precedingof similar stability. discussion of hydride shifts, alkyl group rearrangement was not observed: The less substituted cation would have been the result. Exceptions to this observation are found only if Either type thereof migration are other compellingis faster reasonswhen for furnishingthe preferencetertiary of alkyl migration,carbocations such asrelative electronicto those ending in theirstabilizationsecondary or stericcounterparts relief (see Worked. Example 26 and Problem 61).

Working with the Concepts: Formulating a More Complex Solved Exercise 9-8 Carbocation Rearrangement 19 Molecules of the type A have been shown to dehydrate to B. Formulate a mechanism.

H3C % ≥ H2SO4 ϪH O H 2 ≥ 0 OH CH3 A B

H3C H3CH3C % ≥ ϩ Hϩ % ≥ % ≥ H HHϪH2O ≥ ≥ ϩ ðOH ðOH2 " ϩ 9-3 Carbocation Rearrangements CHAPTER 9 335

2 • This cation has, in principle, various options. For example, it could be trapped by HSO4 2 as a nucleophile (SN1). However, HSO4 is very weak in this respect (it is a poor base; Sec- tion 6-8). Moreover, even if it did form a bond to the cationic center, this would be readily 2 reversible (HSO4 is an excellent leaving group; Section 6-7). It could also lose a proton in two possible ways. This is also reversible and not observed. Instead, this cation provides an opportunity to move the methyl group to the neighboring position, as in B, by rearrangement. 2 • Finally, deprotonation by HSO4 (written as “proton loss”) provides the product.

H3C % ϩ ≥ ≥ ϩ B H H ϪH ϩ 0 CH3

Exercise 9-9 Try It Yourself At 1008C, 3,3-dimethyl-2-butanol gives three products of E1 reaction: one derived from the carbocation present prior to rearrangement, the other two from that formed after an alkyl shift has taken place. Give the structures of these elimination products.

Primary alcohols may undergo rearrangement Primary alcohols may undergo rearrangement TreatmentTreatmentof ofa aprimary primary alcoholalcohol withwith HBrHBr or HIor normallyHI normally produces producesthe correspondingthe corresponding haloalkane through S 2 reaction of the alkyloxonium ion (Section 9-2). However, it is possible haloalkane through NSN2 reaction of the alkyloxonium ion. in some cases to observe alkyl and hydride shifts to primary carbons bearing leaving It is possiblegroups, evenin some thoughcases primaryto carbocationsobserve alkyl are notand formedhydride in solution.shifts toForprimary example,carbons treating bearing 2,2-dimethyl-1-propanol (neopentyl alcohol) with strong acid causes rearrangement, despite leaving groups, even though primary carbocations are not formed in solution. the fact that a primary carbocation cannot be an intermediate.

Rearrangement in a Primary Substrate

CH3 Br A A HBr, ∆ ReactionR CH3CCH2OH CH3CCH2CH3 A ϪHOOH A CH3 CH3 2,2-Dimethyl-1-propanol 2-Bromo-2-methylbutane (Neopentyl alcohol)

In this case, after protonation to form the alkyloxonium ion, steric hindrance interferes with Nucleophile: red direct displacement by bromide (Section 6-10). Instead, water leaves at the same time as a Electrophile: blue methyl group migrates from the neighboring carbon, thus bypassing formation of a primary Leaving20 group: green carbocation.

Mechanism of Concerted Alkyl Shift

CH3 CH3 Ϫ ðBršð ϩ ϩ š ϩ Bršðð A H A Ϫ H2"O ϩ " A Mechanism CH3CCH2šOH CH3COCH2OšOH2 CH3CCH2CH3 CH3CCH2CH3 " Ϫ A A A Ϫ Br"šðð A CH3 CH3 CH3 CH3

Rearrangements of primary substrates are relatively dif! cult processes, usually requiring elevated temperatures and long reaction times. 9-3 Carbocation Rearrangements CHAPTER 9 335

H3C % ϩ ≥ ≥ ϩ B H H ϪH ϩ 0 CH3

Primary alcohols may undergo rearrangement Treatment of a primary alcohol with HBr or HI normally produces the corresponding haloalkane through SN2 reaction of the alkyloxonium ion (Section 9-2). However, it is possible in some cases to observe alkyl and hydride shifts to primary carbons bearing leaving groups, even though primary carbocations are not formed in solution. For example, treating 2,2-dimethyl-1-propanol (neopentyl alcohol) with strong acid causes rearrangement, despite the fact that a primary carbocation cannot be an intermediate.

Rearrangement in a Primary Substrate

CH3 Br In this case, after protonation toA form the alkyloxonium ion,A steric hindrance interferes with HBr, ∆ ReactionR direct displacement by bromideCH3C.CH2OH CH3CCH2CH3 A ϪHOOH A CH3 CH3 Instead, water leaves at2,2-Dimethyl-1-propanolthe same time as a methyl2-Bromo-2-methylbutanegroup migrates from the neighboring carbon, thus bypassing formation(Neopentyl alcohol)of a primary carbocation.

In this case, after protonation to form the alkyloxonium ion, steric hindrance interferes with Nucleophile: red direct displacement by bromide (Section 6-10). Instead, water leaves at the same time as a Electrophile: blue Rearrangementsmethyl groupof migratesprimary fromsubstrates the neighboringare relativelycarbon, thus difficultbypassingprocesses, formation of ausually primary requiringLeaving group: green elevated carbocation.temperatures and long reaction times.

Mechanism of Concerted Alkyl Shift

CH3 CH3 Ϫ ðBršð ϩ ϩ š ϩ Bršðð A H A Ϫ H2"O ϩ " A Mechanism CH3CCH2šOH CH3COCH2OšOH2 CH3CCH2CH3 CH3CCH2CH3 " Ϫ A A A Ϫ Br"šðð A CH3 CH3 CH3 CH3

Rearrangements of primary substrates are relatively dif! cult processes, usually requiring elevated temperatures and long reaction times. 21