Ch 8 - Reactions of Alkenes
61 Pages

Ch 8 - Reactions of Alkenes

Course Number: CHEM 140, Spring 2010

College/University: Whitman

Word Count: 20469

Rating:

Document Preview

8 Reactions of Alkenes A ll al ke nes have a commo n feature : a carbo n-carbo n double bo nd. The reac tio ns of al ke nes arise from the reactivity of the carbon-carbo n double bo nd. O nce agai n, the co ncept of the fu nctio nal group helps to organize a nd sim plify the study of chemical reactio ns. By studyi ng the characteristic reactio ns of the double bo nd, we can predict the reactio ns of al ke nes...

Unformatted Document Excerpt
Coursehero >> Washington >> Whitman >> CHEM 140

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

of 8 Reactions Alkenes A ll al ke nes have a commo n feature : a carbo n-carbo n double bo nd. The reac tio ns of al ke nes arise from the reactivity of the carbon-carbo n double bo nd. O nce agai n, the co ncept of the fu nctio nal group helps to organize a nd sim plify the study of chemical reactio ns. By studyi ng the characteristic reactio ns of the double bo nd, we can predict the reactio ns of al ke nes we have never see n before. Because si ngle bo nds (sigma bo nds) are more stable than pi bonds, the most co mmon re actio ns of double bo nds transform t he pi bo nd i nto a sigma bo nd. For example, catalytic hydrogenation converts the C C p i bo nd and the H - H sigma bo nd i nto two C - H sig ma bo nds (Sectio n 7-7). The reaction is exothermic (i.H0 about -80 to -120 k limol or about -20 to -30 kcal jmol), showi ng that the product is more stable than the reactants. = = 8-1 Reactivity of the Ca rbon-Ca rbon Double Bond '" / C=C / '" + H -H catalyst ) I I -C-CI I HH + e nergy Hydroge natio n of an al ke ne is a n example of an addition, o ne of the three major reactio n types we have studied: addition, elimi nation, and substitutio n. I n an additio n, two molecules combi ne to form o ne product molecule. Whe n an al ke ne u ndergoes addition, two groups add to the carbo n atoms of the double bo nd and the carbo ns become saturated. I n ma ny ways, additio n is the reverse of elimination, i n which o ne molecule splits i nto two fragme nt molecules. I n a substitution, o ne fragme nt replaces another fragme nt i n a molecule. Addition '" / C=C / '" + X-Y Elimination -C-CX I I Y I I Substitution I I -C-CI I X y / '" C=C / '" + X-Y - C -X I I + y- -C I I - y + X321 322 C hapter 8: Reactio ns of Alkenes Addition is the most commo n reactio n of alke nes, a nd i n this chapter we co nsid er additio ns to alke nes i n detail. A wide variety of fu nctio nal groups can be formed by addi ng suitable reage nts to the double bo nds of alke nes. In pri nciple, many differe nt reage nts could add to a double bo nd to form more stable products ; that is , the reactio ns are e nergetically favorable. Not all of these reactio ns have co nve nie nt rates, howe ver. For example, the reactio n of ethyle ne with hydroge n (to gi ve ethane) is stro ngly exothermic, but the rate is very slow. A mixture of ethyle ne a nd hydroge n ca n remai n for years without appreciable reactio n. Addi ng a catalyst such as p lati num, palladium, or nickel allows the reactio n to take place at a rapid rate. Some reage nts react with carbo n-carbo n double bo nds without the aid of a cat alyst. To u nderstand what types of reage nts react with double bo nds, co nsider the structure of the pi bo nd . Although the electro ns i n the sigma bo nd framework are tightly held, the pi bo nd is delocalized abo ve a nd below the sigma bo nd (Figure 8-1 ). The pi-bo ndi ng electro ns are spread farther from the carbon nuclei, a nd they are more loosely held. A stro ng electrophile has a n affinity for these loosely held electro ns; it can pull them away to form a new bo nd (Figure 8-2), l eavi ng o ne of the carbon atoms with o nly three bo nds a nd a positive charge: a carbocatio n. I n effect, the double bo nd has reacted as a nucleophile, do nati ng a pair of electro ns to the electrophile. Most additio n reactio ns i nvolve a seco nd step i n which a nucleophile attacks the carbocatio n (as i n the seco nd step of the S N 1 reactio n), formi ng a stable addition prod uct. I n the product, both the electrophile a nd the nucleophile are bo nded to the carbo n atoms that were co nnected by the double bo nd. This reactio n is outli ned i n Key Mech a nism Box 8-1 , i de ntifyi ng the electrophile as E+ a nd the nucleophile as Nuc: -. This type of reactio n requires a stro ng electrophile to attract the electro ns of the pi bo nd a nd ge nerate a carbocatio n i n the rate-limiti ng step . Most al ke ne reactio ns fall i nto this large class of electrophilic additions to alke nes. 8-2 El ectrop h i l i c Add ition to Al kenes ... Figure 8-1 The elec trons i n the pi bo nd are s pread farther from the carbon nucle i and are more loosel y hel d than the si gma electro ns. Figure 8-2 The pi bo nd as a nucleophile. A strong electro phile attracts the electrons out of the pi bo nd to form a ne w si gma bo nd, ge nerati ng a carbocatio n. The (re d) c urve d arro w sho ws the mo vement of e lectro ns, fro m the e lectro n-rich pi bo nd to the elec tron- poor elec tro ph ile. '1/"'1/ ..... C+-- C "1/,,,. '" e mpt y p orbital ."'"t-:"- -- KEY MECHANISM 8-1 A wide variety of electrophilic additio ns i nvol ve similar mechanisms. First, a stro ng electrophile attracts the loosely hel d electro ns from the pi bo nd of an alke ne. The e1ec trophile forms a sigma bo nd to o ne of the carbo ns of the (former) double bo nd, while the other carbo n becomes a carbocation. The carbocation (a stro ng electro phile) reacts with a nucleo phile (o ften a weak nucleophile) to form another sigma bo nd. Attack of the pi bond on the electrophile for ms a car bocation. E on the more substituted carbon Q o . S tep 1: - C - C+ I I / "- + 8-2 Electrophilic Addition to Alkenes Step 2: Attack by a nuc1eophile gives the addition product. 323 -C-C+ '" I E I /_ Nuc: + -----0> II -C-CII E Nuc EXAMPLE: Ionic addition of HBr to 2-butene This example shows what happens when gaseous HEr adds to 2-butene. The proton in HEr is electrophilic; it reacts with the alkene to form a carbocation. Bromide ion reacts rapidly with the carbocation to give a stable product in which the elements of HBr have added to the ends of the double bond. Step 1: Protonation of the double bond forms a carbocation. Step 2: Bromide ion attacks the carbocation. HH II CH -C=C-CH 3 \H 3 u!: . HH II CH -C-C-CH 3 3I+ H + :Br: HH II .. CH -C-C-CH + :Br: .. 3I+ 3 H ) HH II CH -C-C-CH 3II 3 H :Br: PROBLEM: Explain why the + charge of the carbocation always appears at the carbon of the (former) double bond that has NOT bonded to the electrophile. We will consider several types of additions to alkenes, using a wide variety of reagents: water, borane, hydrogen, carbenes, halogens, oxidizing agents, and even other alkenes. Most, but not all, of these will be electrophilic additions. Table TABLE 8-1 8 -1 summarizes Types of Additions to Alkenes / c=c " " / Type of Addition ) [Elements Added] a OH Product x H hydration ) rH 20] -C-C- I I halogenation I I I I I I I [X2], an oxidation ) -C-C- I X I I I I I I H hydrogenation H h alohydrin formation X [HOX], an oxidation OH [H2], a reduction ) -C-C- ) -C-C- I I I OH OH hydroxylation [HOOHJ, an oxidation oxidative cleavage -C-C- I H HX a ddition [HX] X I I -C-C- I I [02], an oxidation ) / " C=O / O=C " H cyclopropanation ) e poxidation [0], an oxidation ) -C-C- /\ "/ C [CH2] -C-C- /\ H I I I I "These are not the reagents used but simply the groups that appear in the product. 324 Chapter 8: Reactio ns of Alkenes the classes of additions we will cover. Note that the table shows what ele me nts have added across the double bond i n the fi nal product, but it says nothi ng about reage nts or mechanis ms. As we study these reactio ns, you should note the regiochemistry of each reactio n, also called the orientation of addition, m eaning which part of the reage nt adds to which e nd of the double bo nd. Also note the stereochemistry i f the reaction is stereospecific. 8-3 Add ition of Hyd rogen H a l i des to Al kenes S-3A Orientation of Addition: Markovnikov's Rule The si mple mechanism show n for additio n of HBr to 2-butene applies to a large nu mber of electrophilic additio ns. We can use this mechanism to predict the outcome of some fairly complicated reactio ns. For example, the addition of RBr to 2- methyl-2-butene could lead to either of two products , yet only one is observed. CH3 + H - Br C H3 - C -CH-CH3 Br H observed I C H3 I I or C H3 - C - CH -CH3 H I I Br I not observed The first step is proto natio n of the double bo nd. If the proton adds to the seco ndary carbo n, the product will be diffe re nt from the o ne formed if the proto n adds to the tertiary carbo n. C H3 CH3 - C - CH -CH3 add H+ to secondary carbon ) I + H I Be tertiary carbocation C H3 C H3 - C I H - Br '-..:;. )CH- CH3 add H+ to tertiary carbon ) C H3 CH3 - C - CH -CH3 H secondary carbocation I I + Br - PROBLEM-SOLVING Stability of carbocations: Hi-ltv 3 > 2 > 1 > +CH3 W hen the proto n adds to the seco ndary carbo n, a tertiary carbocatio n results. Whe n the proto n adds to the tertiary carbo n atom , a secondary carbocatio n results. The tertiary carbocatio n is more stable (see Sectio n 4- 1 6A) , so the first reactio n is favored. The seco nd half of the mechanism produces the fi nal product of the additio n of HBr to 2- methyl-2-butene. CH3 C H3 - C - CH - CH3 Br : I C H3 H J + I C H3 - C -CH - CH3 Br H I I I Note that proto nation of o ne carbo n ato m of a double bo nd gives a carbocatio n o n the carbo n atom that was not proto nated. Therefore , the proto n adds to the e nd of the dou ble bond that is less substituted to give the m ore substituted carbocation (the more sta ble carbocatio n). 8-3 A ddition of Hydrogen Halides to Alkenes 325 MECHANISM 8-2 Step 1: Ionic Addition of HX to an Alkene Protonatio n of the pi bond forms a c arbo catio n. '" / Step 2: . '"C =C + H .X: \ '" I C- C / I H + :X:- + on the more substituted carbon Attack by the h alide io n gives the additio n product . I '" :X:C - C .. / I .. + H :X: H -C-C- I I I I EXAMPLE: The ionic additio n o f HBr to prope ne shows proto nation of the less substituted c ar bon to give the more substituted c arboc atio n. Reaction with bromide ion completes the additio n. CH3 H H-C-C-H :1;I.r= H product I I I I *- I /H H-C-C I +"H H CH3 Positive charge on less substituted carbon. Less stable; 1101 formed. There are m any ex amples o f reactio ns where the proto n adds to the less substituted c arbo n atom o f the double bo nd i n order to produce the more substituted c arbocatio n. The additio n of HBr ( and other hydroge n h alides) is s aid to be regioselective bec ause i n e ach c ase, o ne of the two possible orie nt atio ns of additio n results prefere nti ally over the other. M arkovnikov's Rule A R ussi an chemist, Vl adimir M arkov nikov, first showed the orie nt atio n o f additio n o f HBr to alke nes i n 186 9. M arkov nikov stated: MARKO VN IKO V'S RULE: The additio n of a proto n acid to the double bo nd o f an alke ne re sults i n a product with the acid proto n bo nded to the c arbo n atom th at alre ady holds the gre ater number o f hydroge n atoms. This is the origi nal stateme nt o f Markovnikov's rule. Re actions th at fo llow this ru le are s aid to follow Markovnikov orientation and give the Markovnikov product. We are o fte n i nterested i n addi ng electrophiles other th an proto n acids to the double bonds o f alke nes. M arkov nikov ' s rule c an be exte nded to i nclude a wide variety o f other additio ns, b ased o n the additio n o f the electrophile i n such a w ay as to produce the most stable c arboc atio n. MARKO VNIKO V' S RULE (exte nded) : I n an electrophilic additio n to an alke ne , the electrophile add s i n such a w ay as to ge nerate the most st able i ntermedi ate. 326 C hapter 8: Reactions of Alkenes Figure 8-3 An electrophile adds to the less substituted end of the double bond to give the more substituted (and therefore more stable) carbocation. a H 3 : QCH3 .. : r: \ H I r: -----7 '-...A ' H H H . -----7 aC product H H . Positive charge Less stable; on less substituted carbon. notformed. Figure 8-3 shows how HBr adds to I-methylcyclohexene to give the product with an additional hydrogen bonded to the carbon that already had the most bonds to hydrogen (one) in the alkene. Note that this orientation results from addition of the proton in the way that generates the more stable carbocation. Like HBr, both HCI and HI add to the double bonds of alkenes, and they also fol low Markovnikov's rule; for example, CH -C-CH-CH CH3 3 HCl P ROBLEM 8-1 HI 0 H H'-..., / C ;'H H CH3 Cl H I I I 2 CQ (c) Predict the major products of the following reactions. (a) CH3-CH=CH2 + I-methylcyclohexene + HBr HI (b) 2-methylpropene + HCl (d) 4-methylcyclohexene + HBr P ROBLEM 8 - 2 When 1,3-butadiene reacts with 1 mol of HBr, both 3-bromo-l -butene and J-bromo-2-butene are formed. Propose a mechanism to account for this mixture of products. ' :' .. R - O - O -R . .. In 1 933, M. S. Kharasch and F. W. Mayo showed that anti-Markovnikov products result from addition of HBr (but not HCl or HI) in the presence of peroxides. Perox ides give rise to free radicals that act as catalysts to accelerate the addition, causing it to occur by a different mechanism. The oxygen-oxygen bond in peroxides is rather weak. It can break to give two radicals. 8-3B Free-Radical Addition of HBr: Anti-Markovnikov Addition -----7 heat R-O ' .. + ' O-R .. + 1 50 kJ (+ 3 6 kcal) Alkoxy radicals (R - 0) catalyze the anti-Markovnikov addition of HBr. The mechanism of this free-radical chain reaction is shown next. 8-3 Addition of Hydrogen Halides to Alkenes 327 MECHANISM 8-3 Initiation: Radicals are formed. Free-Radical Addition of HBr to Alkenes Step 1 : Propagation: A radical reacts to generate another radical. / '" C=C '" / A bromine radical adds to the double bond to generate an alkyl radical on the more substituted carbon atom. Br ----7 R-O-O-R R-O + O-R R-O + H-Br R-O-H + Br Step 2: The bromine radical generated in Step 2 goes on to react in Step 1, continuing the chain. I '" I I Initiation: Radicals are formed. The alkyl radical abstracts a hydrogen atom from HEr to generate the product and a bromine radical. Br H Br I I I / -C-C- + Br - C-C + H -Br R -O- O-R R-O + O-R R-O + H-Br R-O-H + Br C-C-H / I H 3C H "'. I + EXAMPLE: Free-radical addition of HBr to propene. Step 1: Propagation: A radical reacts to generate another radical. A bromine radical adds to the double bond to generate an alkyl radical on the secondary carbon atom. H Br Step 2: The bromine radical generated in Step 2 goes on to react in Step C -C - H / I H 3C H "' . I The alkyl radical abstracts a hydrogen atom from HBr to generate the product and a bromine radical. . on the 2 carbon H Br + H-Br H-C-C-H CH3 H I I H I 1, continuing the chain. Br I + B r 328 Chapter 8: Reactions of Alkenes G .. R-ol + H-Br: .. .. '" The bromine radical lacks an octet of electrons in its valence shell, making it electron It adds to a double bond, forming a neW free radical with deficient and electrophilic. the odd electron on a carbon atom. = Let's consider the individual steps. In the initiation step, free radicals generated from the peroxide react with HBr to form bromine radicals. .. .. R-O-H + :Br jjHO -63 kJ (-15 kcal) kcal) This free radical reacts with an HBr molecule to form a C-H bond and generate another bromine radical. The regenerated bromine radical reacts with another molecule of the alkene, continuing the chain reaction. Note that each propagation step starts with one free radical and ends with another free radical. The number of free radicals is constant, until free radicals come together and terminate the chain reaction. CH3 -C-CH-CH3 CH3 but not CH3 -C-CH-CH3 Br CH3 .. II -C-C- + :Br Br H . . :Br + .. C=C/ '" -C-C :Br: I I / '" jjHD = -12 kJ ( -3 I I jjHO = -25 kJ (-6 kcal) Radical Addition of HBr to Unsymmetrical Alkenes Now we must explain the anti-Markovnikov orientation found in the products of the peroxide-catalyzed reaction. When the alkene is unsymmetrical, adding the bromine radical to the second ary end of the double bond forms a tertiary radical. I A s we saw in the protonation of an alkene, the electrophile (in this case, Br ) adds to the less substituted end of the double bond, and the unpaired electron appears on the more substituted carbon to give the more stable free radical. This intermediate reacts with HBr to give the anti-Markovnikov product, in which H has added to the more substituted end of the double bond: the end that started with fewer hydrogens. tertiary radical (more stable) secondary radical (less stable) Br I I I ' PROBLEM-SOLVING Stability of radicals: Hi-ltv . CH3 30 > 20 > 10 > Note that both mechanisms for the addition of HBr to an alkene (with and with out peroxides) follow our extended statement of Markovnikov's rule: In both cases, the electrophile adds to the less substituted end of the double bond to give the more stable carbocation or free radical. In the ionic reaction, the electrophile is H+. In the peroxide-catalyzed free-radical reaction, Br is the electrophile. Many students wonder why the reaction with Markovnikov orientation does not take place in the presence of peroxides, together with the free-radical chain reaction. It actually does take place, but the peroxide-catalyzed reaction is faster. If just a tiny anti-Markovnikov product CH3 I CH3-C-CH-CH3 + H-Br . Br I CH3-C-CH-CH3 + Br H Br CH3 I I I 8-3 Addition of H ydrogen Halides t o Alkenes 329 b it of peroxide is present, a mixture of Markovnikov and anti-Markovnikov products results. If an appreciable amount of peroxide is present, the radical chain reaction is so much faster than the uncatalyzed ionic reaction that only the anti-Markovnikov product is observed. The reversal of orientation in the presence of peroxides is called the peroxide effect. I t occurs only with the addition of HBr to alkenes. The reaction of an alkyl rad ical with HCl is strongly endothermic, so the free-radical chain reaction is not effec tive for the addition of HC\. I / C] - C - C ' I + H - Cl C] - C - C - H I I I I + CI' t1HO = + 4 kJ ( + 10 kcal) 2 S imilarly, the reaction of an iodine atom with an alkene is strongly endothermic, and the free-radical addition of HI is not observed. Only HBr has just the right reactiv ity for each step of the free-radical chain reaction to take place. I + / C =C / I - C - C' I I / t1HO = + 54kJ ( +13kcal) PROBLEM-SOLVING P ROBLEM 8-3 Predict the major products of the following reactions, and propose mechanisms to support your predictions. Remember to write out complete Hi-ltv structures, including all bonds and charges, when writing a mechanism or determining the course of a reaction. o (a) 2-methylpropene (b) I-methylcyclopentene + HBr + CH3 - C -O-O-C - CH3 + HBr + CH3CH2 - O - O-CH2CH3 I 0 I (c) I-phenylpropene + H Br + d i-t-butyl peroxide ( PhenYl = Ph = 0-) SOLVED PROBLEM 8-1 Show how you would accomplish the following synthetic conversions. (a) Convert I-methylcyclohexene to I-bromo- l -methylcyclohexane. This synthesis requires the addition of HBr to an alkene with Markovnikov orientation. Ionic addition of HBr gives the correct product. SOLUTION + I-methylcyclohexene HBr I-bromo-I-methylcyclohexane (b) Convert I-methylcyclohexanol to I-bromo-2-methylcyclohexane. SOLUTION This synthesis requires the conversion of an alcohol to an alkyl bromide with the bromine atom at the neighboring carbon atom. This is the anti-Markovnikov product, which could be formed by the radical-catalyzed addition of HBr to I-methylcyclohexene. + I-methylcyclohexene H Br R-O-O-R ) heat (X CH3 Br ( Continued ) l-bromo-2-methylcyclohexane 330 Chapter 8: Reactions of Alkenes l-Methylcyclohexene is easily synthesized by the dehydration of I-methylcyclohexa nol. The most substituted alkene is the desired product. + l -methylcyc1ohexanol l-methylcyc1ohexene The two-step synthesis is summarized as follows: HEr ROaR ) (XCH3 I-methylcyclohexanol I-methy lcyc10hexene Br l-bromo-2-methylcyclohexane P ROBLEM 8-4 (a) I-butene I-bromobutane (b) I-butene 2-bromobutane (c) 2-l11ethylcyclohexanol I-bromo-l-methylcyclohexane Show how you would accomplish the following synthetic conversions. (d) 2-l11ethyl-2-butanol 2-broI110-3-l11ethylbutane 8-4 Add ition of Water: Hyd rati on of A l ke n es A n alkene may react with water in the presence of a strongly acidic catalyst to form an alcohol. Formally, this reaction is a hydration (the addition of water), with a hydrogen atom adding to one carbon and a hydroxyl group adding to the other. Hydration of an alkene is the reverse of the dehydration of alcohols we studied in Section 7- 10. Hydration of an alkene H I - c OH I - I c - I alken e alcohol ( Markovnikov orientation) Dehydration of an alcohol H OH I I - C - CI I alcohol alkene For dehydrating alcohols, a concentrated dehydrating acid (such as H2S04 or H3P04) is used to drive the equilibrium to favor the alkene. Hydration of an alkene, on the other hand, is accomplished by adding excess water to drive the equilibrium toward the alcohol. 8-4A Mechanism of Hydration The principle of microscopic reversibility states that a forward reaction and a reverse reac tion taking place under the same conditions (as in an equi]jbrium) must follow the same re action pathway in microscopic detail. The hydration and dehydration reactions are the two complementary reactions in an equi]jblium; therefore, they must follow the same reaction pathway. It makes sense that the lowest-energy transition states and intermediates for the reverse reaction are the same as those for the forward reaction, except in reverse order. According to the principle of microscopic reversibility, we can write the hydra tion mechanism by reversing the order of the steps of the dehydration (Section 7-10). 8-4 Addition of Water: Hydration of Alkenes 3 31 Protonation of the double bond forms a carbocation. Nucleophilic attack by water, fol lowed by loss of a proton, gives the alcohol. MECHANISM 8-4 Step 1: Acid-Catalyzed Hydration of an Alkene Protonation of the double bond forms a carbocation. / Step 2: Cb L H "' .. <-='- "c-L I / + on the more substituted carbon Nucleophilic attack by water gives a protonated alcohol. H .. "'+ I H20 : ' + " C-C Step 3: / I H I H- O:t- H II - C-CI I Deprotonation gives the alcohol. ' HII - C -CI I Step 1: b 0.. + H20: ( ) H -O: H II - C-CI I + H30+ EXAMPLE: Acid-catalyzed hydration of propene. Protonation of the double bond forms a secondary carbocation. H" C=C " / H3C H propene Step 2: fOLH H+ .. ----i> H,, f +- C-H C / I H3C H + H20 '----___ Nucleophilic attack by water gives a protonated alcohol. Step 3: H I H- O:t- H II H-C-C -H II CH3 H Deprotonation gives the alcohol. H' II H-C- C-H II CH3H b :-;\ + H20: .. ----i> H-O: H II H-C-C-H II CH3H 2-propanol + H3 0+ 332 C hapter 8: Reactions of Alkenes 8 -48 C H3 -C=CH-CH3 I CH3 H + Step 1 of the hydration mechanism is similar to the first step in the addition of HBr. The proton adds to the less substituted end of the double bond to form the m ore substituted carbocation. Water attacks the carbocation to give (after loss of a proton) the alcohol with the -OH group on the more substituted carbon. Like the addition of hydrogen halides, hydration is regioselective: It follows Markovnikov's rule, giving a product in which the new hydrogen has added to the less substituted end of the double bond. Consider the hydration of 2-methyl-2-butene: Orientation of Hydration The proton adds to the less substituted end of the double bond, so the positive charge appears at the more substituted end. Water attacks the carbocation to give the proto nated alcohol. 3 , more stable 2, less stable ) H -O-H .. I+ I I C H3-C-CH-CH 3 but not CH3-C-CH-CH 3 + + I I C H3 H H CH 3 The reaction follows Markovnikov's rule. The proton has added to the end of the dou ble bond that already had more hydrogens (that is, the less substituted end), and the -OH group has added to the more substituted end. Like other reactions that involve carbocation intermediates, hydration may take place with rearrangement. For example, when 3,3-dimethyl-l -butene undergoes acid catalyzed hydration, the major product results from rearrangement of the carbocation intermediate. CH3 -C-CH=CH2 3,3-dimethyl- l -butene CO : H .. -;/ "H20 :,----"" H H I I CH 3-C- CH-CH3 I CH3 C H3 -C-CH-CH3 OH H I I I CH3 CH3 CH3 I I 2 ,3-dimethyl-2-butanol (major product) P ROBLEM 8- 5 PROBLEM-SOLVING Propose a mechanism to show how 3,3-dimethyl- l -butene reacts with dilute aqueous H2S04 to give 2,3-dimethyl-2-butanol and a small amount of 2,3-dimethyl-2-butene. When predicting products for Htnl/ P ROBLEM 8- 6 electrophilic additions, first draw the structure of the carbocation (or other intermediate) that results from electrophilic attack. Predict the products of the following hydration reactions. ( a) I -methylcyclopentene + dilute acid (b) 2-phenylpropene + dilute acid (e) I-phenylcyclohexene + dilute acid Many alkenes do not easily undergo hydration in aqueous acid. Some alkenes are near ly insoluble in aqueous acid, and others undergo side reactions such as rearrangement, polymerization, or charring under these strongly acidic conditions. In some cases, the overall equilibrium favors the alkene rather than the alcohol. No amount of catalysis can cause a reaction to occur if the energetics are unfavorable. Oxymercuration-demercuration is another method for converting alkenes to alcohols with Markovnikov orientation. Oxymercuration-demercuration works with many alkenes that do not easily undergo direct hydration, and it takes place under milder conditions. No free carbocation is formed, so there is no opportunity for rearrangements or polymerization. OJ.. }'mercuration-Demercuration 8-5 H ydration by Oxymercuration-Demercuration 333 8- 5 Hyd ration by Oxymercu rati on Demercu ration '" / C=C / '" + H g(OAch HoQ -) I I -C -C I HO I CH -C-O-Hg-O-C-CH 3 o II Hg(OAc)2 The reagent for mercuration is mercuric acetate, Hg(OCOCH3 h abbreviated Hg( OAc h There are several theories as to how this reagent acts as an electrophile; the simplest one is that mercuric acetate dissociates slightly to form a positively charged mercury species, +Hg( OAc ). 0 3 CH 3-C-O - Hg+ o I +Hg(OAc) + I -c-c- HgOAc (Markovnikov orientation) II II HO H O xymercuration involves an electrophilic attack on the double bond by the positively charged mercury species. The product is a mercurinium ion, an organometallic cation containing a three-membered ring. In the second step, water from the solvent attacks the mercurinium ion to give (after deprotonation) an organomercurial alcohol. A subsequent reaction is demercuration, to remove the mercury. Sodium borohydride (NaBH4, a reducing agent) replaces the mercuric acetate fragment with a hydrogen atom. MECHANISM 8-5 Step 1 : Electrophilic attack forms a mercurinium ion. '" / Oxymercuration of an Alkene Step 2: Water opens the ring to give an organomercurial alcohol. H20 : .. )1 -C-CI\ c:,Hg(OAC) I C =C ;;- . (C '" + Hg(OAc) . H P : 0H .. (I H20 : II -C-CI I Hg(O Ac) -C-C/\ I I OAc H g+ I mercurinium ion ------? II --: OH I organomercurial alcohol C C 1 H g(OAc) + H3 0+ ( Continued ) 334 Chapter 8: Reactions of Alkenes Demercuration replaces the mercuric fragment with hydrogen to give the alcohol. 4 - C - C - + NaBH4 + 4 - OH OH 1 Hg(OAc) 1 1 1 4 - C - C - + NaB (OH)4 + 4 Hg OH alcohol 1 H 1 1 1 + 4 - OAc o rganomercurial alcohol Step 1: EXAMPLE: Oxymercuration-demercuration of propene. Electrophilic attack forms a mercurinium ion. OAc ----7 mercurinium ion ) ", / \ C-C-H / 1 HC H 3 H H g+ 1 Step 2: propene Water opens the ring to give an organomercurial alcohol. OAc H H3 C ", l \ J C - C- H / H g+ 1 H ----7 H ij: )k H C - C- C - H 3 11 H - O:r H H20: H 1 Hg(OAc) 1 H .. (I H C - C - C - H + H 0+ 3 3 11 OH H 1 Hg(OAc) 1 s ubstituted carbon. Demercuration replaces the mercuric fragment with hydrogen to give the alcohol. H H C-C-C-H 3 OH H 1 Hg(OAc) 1 H 1 1 H C -C-C-H 3 OH H 2-propanol 1 1 H 1 1 O xymercuration-demercuration of an unsymmetrical alkene generally gives Markovnikov orientation of addition, as shown by the oxymercuration of propene in the preceding example. In this unsymmetrical case, the mercurinium ion has a considerable amount of positive charge on the more substituted carbon atom. Attack by water occurs on this more electrophilic carbon, giving Markovnikov orientation. The electrophile, +Hg( OAc), remains bonded to the less substituted end of the double bond. Reduction of the organomercurial alcohol gives the Markovnikov alcohol: 2propanol. Similarly, oxymercuration-demercuration of 3,3-dimethyl- l -butene gives the Markovnikov product, 3,3-dimethyl-2-butanol, in excellent yield. Contrast this unre arranged product with the reaITanged product formed in the acid-catalyzed hydration of the same alkene in Section 8-4 . Oxymercuration-demercuration reliably adds water across B the double bond of an alkene with Markovnikov orientation and without reanangement. 8-6 Alkox ymercurati on-Demercurati o n ( C H h C - C - C- H 3 H H g(OAc) OH H I 3 ,3-dimethyl- l -butene mercurinium ion OH H ( CH hC- C-C- H 3 I I H H g(OAc) M arkovniko v product II Markovnikov product I I 335 I 3,3-dimethyl-2-butanol (94% overall) Of the methods we have seen for Markovnikov hydration of alkenes, oxymercu ration-demercuration is most commonly used in the laboratory. It gives better yields than direct acid-catalyzed hydration, it avoids the possibility of reanangements, and it does not involve harsh conditions. There are also disadvantages, however. Organomer curial compounds are highly toxic. They must be used with great care and then must be disposed of properly. W hen mercuration takes place in an alcohol solvent, the alcohol serves as a nucle ophile to attack the mercurinium ion. The resulting product contains an alkoxy ( - 0 - R ) group. In effect, alkoxymercuration-demercuration converts alkenes to ethers by adding an alcohol across the double bond of the alkene. " / C =C 8- 6 AI koxyme rcu ration Demercu ration / " + H g(OAc)2 ---7 ROH -C-C- -C-C- RO As we have seen, an alkene reacts to form a mercurinium ion that is attacked by the nucleophilic solvent. Attack by an alcohol solvent gives an organomercurial ether that can be reduced to the ether. \ - C-C- I I HgOAc I I RO II I HgOAc I -C-C- I RO I H (M arko vnikov orientation) I I C/ + Hg(OAc) R- J : I I I H R ---:O + l ) H ",----, : Q - R -C- C - II Hg(OAc) - C-C R-O : The solvent attacks the mercurinium ion at the more substituted end of the double bond (where there is more 0+ charge), giving Markovnikov orientation of addition. The Hg(OAc) group appears at the less substituted end of the double bond. Reduction gives the Markovnikov product, with hydrogen at the less substituted end of the double bond. organomercurial ether I II II - R-O : II -c-cI organomercurial ether I Hg(OAc) Hg(OAc) NaBH. ) - C- CR-O I H I I M ercury and its compounds were used for centuries as ingredients i n antibacterial and drugs, skin creams, com antiseptics. Mercury I an ether pounds are q u ite toxic, however. In the body, mercury combines with the thiol groups of critical en zymes, i nactivating them. Mercury poisoning causes b ra i n and kidney damage, often leading to death. 336 Chapter 8 : Reac tions of Alkenes SOLVED PROBLEM 8- 2 Show the intermediates and products that result from alkoxymercuration-demercuration of I -methylcyclopentene, using methanol as the solvent. S OLUT ION Mercuric acetate adds to I -methylcyclopentene to give the cyclic mercurinium ion. This ion has a considerable amount of positive charge on the more substituted tertiary carbon atom. Methanol attacks this carbon. a H + Hg(OAc) CH3 ) W, ; r H I -methylcyclopentene CH3 - O - H CH3 fI (OAC) N Hg(OAc) \.-\: CH3 H OCH3 mercurinium ion trans intermediate (product of anti addition) Reduction of the intermediate gives the Markovnikov product, I -methoxy - l -methylcyclopentane. N' Hg(OAc) \'-\:' CH3 H i ntermediate OCH3 I -methoxy- l -methylcyclopentane C( H' OCH3 P ROBL E M 8- 7 (a) Propose a mechanism for the following reaction. CH3 -C=CH-CH3 CH3 I Hg(OAc)2' CH3 CH20H ) CH3CHzO CH3 -C-CH-CH3 CH3 I I (90%) Hg(OAc) I P ROBLEM 8-8 (b) Give the structure of the product that results when this intermediate is reduced by sodium borohydride. P ROBLEM 8- 9 Predict the major products of the following reactions. ( a) I -methylcyclohexene + aqueous Hg( OAc h (e) 4-chlorocycloheptene + Hg ( OAc h in CH30H (b) the product from part (a), treated with NaBH4 (d) the product from part (c), treated with NaBH4 Show how you would accomplish the following synthetic conversions. (b) l -iodo-2-methylcyclopentane --> I -methylcyclopentanol (a) I -butene --> 2-methoxybutane (e) 3-methyl-l -pentene --> 3-methyl-2-pentanol Explain why acid-catalyzed hydration would be a poor choice for the reaction in (c). 8-7 Hyd roboration of Al kenes We have seen two methods for hydrating an alkene with Markovnikov orientation. What if we need to convert an alkene to the anti-Markovnikov alcohol? For example, the following transformation cannot be accomplished using the hydration procedures covered thus far. ? CH 3 C H3 - C - CH - CH3 H OH (anti -Markovnikov) I I I 2-methyl-2-butene 3-methyl-2-butanol 8-7 Hydroboration of Alkenes 337 S uch an anti-Markovnikov hydration was i mpossible u ntil H. C. Brown, of Purdue University, discovered that diborane ( B2H6) adds to alkenes with anti Markovnikov orientation to form alkylboranes, which can be oxidized to give anti-Markovnikov alcohols. This discovery led to the development of a large field of borane chemistry, for which B rown received the Nobel Prize in chemistry in 1 979. CH3 I C H3 - C- CH - CH3 I I H B H2 2-methyl-2-butene an alkylborane o xidize C H3 , C H3 - C - CH -CH3 I I H OH 3 -methyl-2-butanol (>90%) D iborane ( B2H6) is a dimer composed of two molecules of borane (BH 3 ) ' The bonding in diborane is unconventional, using three-centered (banana-shaped) bonds with protons in the middle of therp. Diborane is in equilibrium with a small amount of borane (BH 3 ) ' a strong Lewis acid with only six valence electrons. three-centered bond 2 diborane H" , H / B -H borane D iborane is an inconvenient reagent. It is a toxic, flammable, and explosive gas. It is more easily u sed as a complex with tetrahydrofuran (THF), a cyclic ether. This complex reacts like diborane, yet the solution is easily measured and transferred. 2 tetrahydrofuran (THF) I /. CH2- CH2 CH2 -CH? - '"0: + C H2 -C 2 2 ) O =- H CH 2 -CH2 I H diborane borane- THF complex B H3 " THF The BH 3 ' THF reagent is the form of borane commonly used in organic reac tions. BH 3 adds to the double bond of an alkene to give an alkylborane. B asic hydrogen peroxide oxidizes the alkylborane to an alcohol. In effect, hydroboration-oxidation converts alkenes to alcohols by adding water across the double bond, with anti Markovnikov orientation. Hydrohoration - oxidation: '" / C=C '" / + B H3 " THF I I -C-C I I H B-H I H I I -C -CI I H OR a nti-Markovnikov orientation (syn stereochemistry) 338 Chapter 8: Reactions of Alkenes 8-7A Mechanism of Hydroboration Borane is an electron-deficient compound. It has only six valence electrons, so the boron atom cannot have an octet. Acquiring an octet is the driving force for the unusu al bonding structures ("banana" bonds, for example) found in boron compounds. As an electron-deficient compound, B H 3 is a strong electrophile, capable of adding to a double bond. This hydroboration of the double bond is thought to occur in one step, with the boron atom adding to the less substituted end of the double bond, as shown in Mechanism 8-6. In the transition state, the electrophilic boron atom withdraws electrons from the pi bond, and the carbon at the other end of the double bond acquires a partial positive charge. This partial charge is more stable on the more substituted carbon atom. The product shows boron bonded to the less substituted end of the double bond and hydro gen bonded to the more substituted end. Also, steric hindrance favors boron adding to the less hindered, less substituted end of the double bond. MECHANISM 8-6 H ydroboration of an Al kene Borane adds to the double bond in a single step. Boron adds to the less hindered, less substituted carbon, and hydrogen adds to the more substituted carbon. CH3 H 1 0+ 1 CH3- <;= == <;= - CH3 H - - - BH? 0- more stable transition state / CH3 H 1 1 C H3 - C - C - CH 3 1 1 H BH2 C H3 H 1 1 0+ CH3- <;= == <;= - CH3 , , , , H2B - - -H 0less stable transition state The boron atom is removed by oxidation, using aqueous sodium hydroxide and hy drogen peroxide (HOOH or H202) to replace the boron atom with a hydroxyl ( - OH ) group. CH3 C H3 -C-C-CH3 1 1 B H2 H I H I CH3 I C H3 -C-CH -CH3 1 1 H OH This hydration of an alkene by hydroboration-oxidation is another example of a reaction that does not follow the original statement of Markovnikov's rule (the product is anti-Markovnikov), but still follows our understanding of the reasoning behind Markovnikov's rule. The electrophilic boron atom adds to the less substituted end of the double bond, placing the positive charge (and the hydrogen atom) at the more sub stituted end. 8-7 SO L VED PROB L E M 8-3 Show how you would convert I-methylcyclopentanol to 2-methylcyclopentanol. H ydroboration of A lkenes 339 PROBLEM-SOLVING syntheses. Work backward on multistep HtltP SO L U T ION I-methylcyclopentene. The use of Working back ward, use hydroboration-oxidation to form 2-methyl-cyclopentanol from and (2) above and below the reaction arrow indicates individual steps in a two-step seq uence. (1) (I) B H 3 . THF OH (2) H P2 ' I-methylcyclopentene tral1s-2-methylcyclopentanol U + CH 3 ' '' OH The 2-methylcyclopentanol that results from this synthesis is the pure trans isomer. This stere ochemical result is discussed in Section 8 -7C. I -M ethylcyclopentene is the most substituted alkene that results from dehydration of I-methylcyclopentanol. Dehydration of the alcohol would give the correct alkene. I-methylcyclopentanol I -methylcyclopentene PROB L E M 8- 1 0 Predict the ma or products of the following reactions. j (c) (e) (a) propene + BH3 ' THF 2-methyl-2-pentene + BH 3 . THF I-methylcyclohex ene + BH3 ' THF (d) (f) (b) the product from part ( a) + H202/OHthe product from part ( c) the product from part ( e) + + H 20 2/OHH2 02/OH- PROB L E M 8- 1 1 (c) Show how you would accomplish the following synthetic conv ersions. (a) I -butene ---7 I -butanol (b) I-butene ---7 2-butanol 2-bromo-2,4-dimethylpentane ---7 2,4-dimethyI-3-pentanol 8-78 Stoichiometry of Hydroboration For simplicity, we have neglected the fact that 3moles of an alkene react with each mole of BH 3 . Each B bond in BH 3 can add across the double bond of an alkene. The first addition forms an alkylborane, the second a dialkylborane, and the third a Ilialkylborane. H Summar), H -H B / '" '-.. / c=c '-.. / -H + ) II/ -C-C-B I I '" alkylborane H HH '-.. / c=c / '-.. ) -i-?tB-H ( dialkylborane '-.. / c =c '-.. / ( -i-?rB trialkylborane 3 '" / C =C / '" 3 I I -C -C I I Trialkylboranes react exactly as we have discussed, and they oxidize to give anti Markovnikov alcohols. Trialkylboranes are quite bulky, further reinforcing the prefer ence for boron to add to the less hindered carbon atom of the double bond. Boranes are H OH 340 Chapter 8 : Reactions of Alkenes often drawn as the 1 : 1 monoalky Iboranes to simplify their structure and emphasize the organic part of the molecule. 8 -7C Stereochemistry of Hydroboration The simultaneous addition of boron and hydrogen to the double bond (as shown in Mechanism 8 -7) leads to a syn addition: Boron and hydrogen add across the double bond on the same side of the molecule. (If they added to opposite sides of the mole cule, the process would be an anti addition.) T he stereochemistry of the hydroboration-oxidation of I -methylcyclopentene is shown next. Boron and hydrogen add to the same face of the double bond (syn) to form a trialkylborane. Oxidation of the trialkylborane replaces boron with a hydroxyl group in the same stereochemical position. The product is trans-2-methylcyclopentanol. A racemic mixture is expected because a dural product is formed from achiral reagents. H ..." " B H H transition state trans-2-methylcyclopentanol (85% overall ) (racemic mixture o f enantiomers) The second step (oxidation of the borane to the alcohol) takes place with reten tion of configuration. Hydroperoxide ion adds to the borane, causing the alkyl group to migrate from boron to oxygen. The alkyl group migrates with retention of configura tion because it moves with its electron pair and does not alter the tetrahedral structure of the migrating carbon atom. Hydrolysis of the borate ester gives the alcohol. Formation of hydroperoxide ion F) +:O H -O-O-H H " " ' ( ) H-O-O: + H20 : A ddition of hydroperoxide and migration of the alkyl group R R-B I " I :O -O-H " R -B -O - O - H " R I R hyclroperoxicle R /' I ,, , _ _ " R ----i> 0; " R -B -O : R b orate ester I .. I + : OH l R m igrates ) Twice more to oxidize the other two alkyl groups R R -B -O R I O- R I O - B -O R I I R I t rialkyl borate ester 8-7 Hydroboration of Alkenes Hydrolysis of the borate ester 3 41 : O-B - O : R : OH R .. : O-R I C .. I .. I O-R ? --vT R : OH R I .. O -R O -B R : OH I O-R I I :0: R I O- B O-H + R- OH + - OH R I I I (The other two O R groups hydrolyze similarly) Hydroboration of alkenes is another example of a stereospecific reaction, i n which different stereoisomers of the starting compound react to give different stereoisomers of the product. Problem 8- 14considers the different products formed by the hydroboration-oxidation of two acyclic diastereomers. SOLVED PROBLE M 8-4 A norbornene molecule labeled with deuterium is subjected to hydroboration-oxidation. Give the structures of the intermediates and products. J): ( D "' B H; T HF elida ( inside) face deuterium- labeled norbornene a lkylborane alcohol (racemic mixture) SOLUT ION The syn addition of BH3 across the double bond of norbornene takes place mostly from the more accessible outside (exo) face of the double bond. Oxidation gives a product with both the hydrogen atom and the hydroxyl group in exo positions. (The less accessible inner face of the double bond is called the endo face.) PROBLE M 8-1 2 In the hydroboration of I-methylcyclopentene shown in Solved Problem 8-3, the reagents are achi.ral, and the products are chiral. The product is a racemic mixture of trans-2-methylcy clopentanol, but only one enantiomer is shown. Show how the other enantiomer is formed. P ROBLE M 8-1 3 (a) I -methylcycloheptene + BH3 . THF, then H202> OH(b) trans-4,4-dimethyl-2-pentene + B H3 ' THF, then H202, OH- Predict the major products of the following reactions. Include stereochemistry where applicable. (c) /, H + B H3 ' THF, then HP2' OH- CH3 PROBLE M 8-1 4 (a) When (Z)-3-methyl-3-hexene undergoes hydroboration-oxidation, two isomeric prod ucts are formed. Give their structures, and label each asymmetric carbon atom as (R) or ( S). What is the relationship between these isomers? (b) Repeat part (a) for (E)-3-methyl-3-hexene. What is the relationship between the products formed from (Z)-3-methyl-3-hexene and those formed from (E)-3-methyl-3-hexene? 342 C hapter 8: Reactions of Alkenes P ROBLEM 8- 1 5 Show how you would accomplish the following transformations. (. l (b l OH (c) I -methylcycloheptanol --7 2-methylcycloheptanol P ROBLEM 8- 1 6 eo eo CQ CO OH When HBr adds across the double bond of 1 ,2-dimethylcyclopentene, the product is a mix ture of the cis and trans isomers. Show why this addition is not stereospecific. 8-8 Add ition of H a logens to Al kenes H alogens add to alkenes to form vicinal dihalides. X "-.... C=C '-... /' /' + X2 --7 -C-CX I I I I u sually anti addition S-SA Mechanism of Halogen Addition A halogen molecule ( Br2, C12, or 12 ) i s electrophilic; a nucleophile can react with a halogen, displacing a halide ion: Nuc : + : Br-Br: .. .. .. N uc- Br: .. + : Br : - .. In this example, the nucleophile attacks the electrophilic nucleus of one bromine atom, and the other bromine serves as the leaving group, departing as bromide ion. Many re actions fit this general pattern; for example: . HO - Br: + : Br : HO : - + : Br - Br: .. . . \:;: .. .. .. . .. + : Cl :- .. C =C "-.... /' "-.... . . + : Br- Br: .. \:;: .. : Br /+ \ - C- C I I bromoniulll ion + : Br : - .. In the last reaction, the pi electrons of an alkene attack the bromine molecule, expelling bromide ion. A bromonium ion results, containing a three-membered ring with a positive charge on the bromine atom. This bromonium ion is similar in structure to the mercurinium ion discussed in Section 8-5. Similar reactions with other halogens form other halonium ions. The structures of a chloronium ion, a bromonium ion, and an iodonium ion are shown next. 8 -8 Addition of Halogens to Alkenes Examples 343 '" Cl " /+ \ - C -C c hloronium ion I I '" Br" /+ \ - C -C bromonium ion I I ." ( 1+ \ - C- C iodonium ion I I U nlike a nonnal carbocation, all the atoms in a halonium ion have filled octets. The three-membered ring has considerable ring strain, however, which, combined with a positive charge on an electronegative halogen atom, makes the halonium ion strongly electrophilic. Attack by a nucleophile, such as a halide ion, opens the halonj um ion to give a stable product. MECHANISM 8-7 Step 1: A ddition of Halogens to Alkenes Electrophilic attack forms a halonium ion. /' ,, C = C + : X. - X : . .. " .. \l.: '" x " /+\ -C-Chal onium ion I I + : X:- Step 2 : The halide ion opens the halonium ion. :X: .. -C -C :X: I I I I X- attacks from the back side Step 1 : EXAMPLE: Addition of Br2 to propene. Electrophilic attack forms a bromonium ion. H ", :Br- Br: C =C . . . "" / H3 C H propene f:)l\ .. .. H H3C I ' Br' I\ C -C . . + \H H bromonium ion W hen a solution of bromine (red-brown) is added to cyclohexene, the bromine color quickly disappears because bromine adds across the double bond. When bromine is added to cyclohexane (at right), the color persists. Step 2 : Bromide ion opens the bromonium ion H3C C - C/ / \H Br H H Br 1 ,2-dibromopropane Chlorine and bromine commonly add to alkenes by the halonium ion mechanism. Iodination is used less frequently because diiodide products decompose easily. Any solvents used must be inert to the halogens; methylene chloride CH2CI2 , chlorofonn CHCI3 , and carbon tetrachloride CCI4 are the most frequent choices. ( ) ( ) ( ) 344 Chapter 8: Reactions of Alkenes 8 -88 Stereochemistry of Halogen Addition The addition of bromine to cyclopentene is a stereospecific anti addition. cyc10pentene trans- l ,2-dibromocyc1opentane (92%) d H H but not i?H Br Br cis- I ,2-dibromocyclopentane (not formed) Anti stereochemistry results from the bromonium ion mechanism. When a nucleophile attacks a halonium ion, it must do so from the back side, in a manner similar to the S N 2 displacement. This back-side attack assures anti stereochemistry of addition. H Br H trans + enantiomer Halogen addition is another example of a stereospecific reaction, in which different stereoisomers of the starting material give different stereoisomers of the prod uct. Figure 8-4 shows additional examples of the anti addition of halogens to alkenes. The addition of bromine has been used as a simple chemical test for the pres ence of olefinic double bonds. A s olution of bromine in carbon tetrachloride is a clear, deep red color. When this red solution is added to an alkene, the red bromine color disappears (we say it is "decolorized"), and the solution becomes clear and (JC H H + Cl 2 CI CI + CI Cl -f------J H H racemic trans- l ,2-dichlorocyc1ohexane H H cyc10hexene Br H3 C H / ,%0 C-C ' H / CH3 Br (+ enantiomer) + Br + H CH3 C H3 H Br + CH3 Br H CH3 + H + Br cis-2-butene F igure 8-4 ( )-2,3-dibromobutane Ex amples of the anti addition of halogens to alkenes. The stereospecific anti addition gives predictable stereoisomers of the products. trans-2-butene meso-2,3-dibromobutane = :f:: C H3 8-9 Formation of Halohydrins 345 colorless. (Although there are other functional groups that decolorize bromine, few do it as quickly as alkenes.) P ROBLEM 8-1 7 Give mechanisms to account for the stereochemistry of the products observed from the addi tion of bromine to cis- and trans-2-butene (Figure 8-4). Why are two products formed from the cis isomer but only one from the trans? (Making models will be helpful.) P ROBLEM 8-1 8 Propose mechanisms and predict the major products of the following reactions. Include stere ochemistry where appropriate. (a) cycloheptene + Br2 in CH2Cl2 ( E)-3-decene + Br2 in CCI4 (b) (0) CD PROBLEM-SOLVING + C I, ;n CHCl, (d) ---0---< M odels may be helpful whenever com plete structures, including all HinZ; stereochemistry is involved. Write + 2 C I2 in CCI4 bonds and charges, when writing mechanisms. A halohydrin is an alcohol with a halogen on the adjacent carbon atom. In the pres ence of water, halogens add to alkenes to form halohydrins. The halogen adds to the alkene to give a hal onium ion, which is strongly electrophilic. Water acts as a nucle ophile to open the hal onium ion and form the halohydrin. 8- 9 Fo rmation of H a l ohyd rins M ECHANISM 8-8 Step 1 : Formation of Halohydrins Electrophilic attack forms a hal onium iOIl. / ,,C =C + :X. \""" {(; : . .. " .. -C-C/ \ halonium ion . X '+/\ :X:- (X Step 2: = Cl, Br, or I ) Water opens the halonium ion; deprotonation gives the halohydrin. ----7 :0 Hi):\..J -<JI .... H H+ back-side attack II - C-CII : x: ----7 - c -c - :x: I I :Q I H I + Hp + X- halohydrin Markovnikov orientation anti stereochemistry EXAMPLE: Addition of (12 to propene in water. Step 1: E lectrophilic attack forms a chloronium ion. H ()i\ .. .. ",C =C :Cl-Cl: . . . / " H3C H propene + Cl I C-\ R" 9 \'H H H3C . . ' ' + : r: ( Continued) chloronium ion 346 C hapter 8: Reactions of A lkenes Step 2: Back-side attack by water opens the chloronium ion. H" ,C, c;'\" H H3C ) . H H20 : + r C' c\ i ---7 Step 3: Water removes a proton to give the chlorohydrin. H :CI: H3C / C-C / \" H H-Q : H c hlorohydrin When halogenation takes place with no solvent or with an inert solvent such as carbon tetrachloride ( CCI4) or chloroform ( CHCI3), only the halide ion is available as a nucleophile to attack the halonium ion. A d ihalide results. But when an alkene reacts with a halogen in the presence of a nucleophilic solvent such as water, a solvent molecule is the most likely nucleophile to attack the halonium ion. When a water mol ecule attacks the hal onium ion, the final product is a halohydrin, with a halogen on one carbon atom and a hydroxyl group on the adjacent carbon. The product may be a chlorohydrin, a bromohydrin, or an iodohydrin, depending on the halogen. I I I I I I - C-C- I I C l OH -C-CBr OH - I I C I - I C - OH I chlorohydrin bromohydrin iodohydrin S tereochemistry of Halohydrin Formation Because the mechanism involves a halonium ion, the stereochemistry of addition is anti, as in halogenation. For example, the addition of bromine water to cyclopentene gives trans-2-bromocyclopentanol, the product of anti addition across the double bond. H / Br + enantiomer a " H OH cyclopentene trans-2-bromocyclopentanol (cyclopentene bromohydrin) P ROBLEM 8-1 9 Propose a mechanism for the addition of bromine water to cyclopentene, being careful to show why the trans product results and how both enantiomers are formed. O rientation of Halohydrin Formation Even though a h alonium ion is involved, rather than a carbocation, the extended version of Markovnikov ' s rule applies to halo hydrin formation. When propene reacts with chlorine water, the major product has the electrophile (the chlorine atom) bonded to the less substituted carbon of the double bond. The nucleophile (the hydroxyl group) is bonded to the more substituted carbon. H2C -CH-CH 3 I I Cl OH + HCl 8-9 F ormation of Halohydrins 347 The Markovnikov orientation observed in halohydrin formation is explained by the structure of the halonium ion intermediate. The two carbon atoms bonded to the halogen have partial positive charges, with a larger charge (and a weaker bond to the halogen) on the more substituted carbon atom (Figure 8-5). The nucleophile (water) attacks this more substituted, more electrophilic carbon atom. The result is both anti stereochemistry and Markovnikov orientation. H- H -; C --;:8 /ci 8 <;\ ' H .. .... Figure 8-5 laraer 8 + on the b more substituted carbon : O-H H ------.- -- I C H3 I Orientation of halohydrin formation. The more substituted carbon of the chloronium ion bears more positive charge than the less substituted carbon. Attack by water occurs on the more substituted carbon to give the Markovnikov product. This hal onium ion mechanism can be used to explain and predict a wide variety of reactions in both nucleophilic and non-nucleophilic solvents. The halonium ion mechanism is similar to the mercurinium ion mechanism for oxymercuration of an alkene with Markovnikov orientation (Section 8-5). SOLVED PROBLEM 8- 5 Propose a mechanism for the reaction of I -methylcyclopentene with bromine water. S OLUTION I -Methylcyclopentene reacts with bromine to give a bromonium ion. Attack by water could occur at either the secondary carbon or the tertiary carbon of the bromonium ion. Attack actually occurs at the more substituted carbon, which bears more of the positive charge. The product is formed as a racemic mixture. H '+ H :OH O CH3 " "" H 3 "''''' CH3 " ,,2' + H20: Hi): . .. ) .. Br : .. . \ 11,, . B r : "" "Br : : Br-Br: . .. 20 H : iir:H H ry . V . o. " " , N \ ) Q. + enantiomer SOLVED PROBLEM 8 - 6 When cyclohexene is treated with bromine in saturated aqueous sodium chloride, a mixture of trans-2-bromocyclohexanol and trans I-bromo-2-chlorocyclohexane results. Propose a mechanism to account for these two products. S OLUT ION Cyclohexene reacts with bromine to give a bromonium ion, which will react with any available nucleophile. The most abundant nucleophiles in saturated aqueous sodium chloride solution are water and chloride ions. Attack by water gives the bromohydrin, and attack by chloride gives the dihalide. Either of these attacks gives anti stereochemistry. cyclohexene o a I" , " H :::: Br + a" r o. : I : " "" Br H OH " H + enantiomer trans- l -bromo-2-chlorocyclohexane H bromonium 20 : H iOIl ("f, "Br H + enantiomer trans-2-bromocyclohexanol 348 Ch apter 8 : R eactions o f Al kene s P ROBLEM 8- 2 0 The solutions to Solved Problem 8 -5 and Solved Problem 8 -6 showed only how one enan tiomer of the product is formed. For each product, show how an equally probable reaction forms the other enantiorner. Predict the major product(s) for each reaction. Include stereochemistry where appropriate. (b) 2 -methyl-2-butene + B r2/H 20 (a) I -methyIcyclopentene + CI2/H20 (d) trans-2-butene + CI2/H 20 ( c) c is-2-butene + CI2/H20 (e) I -methyIcyclopentene + Br2 in saturated aqueous NaCl P ROBLEM 8- 2 1 PROBLEM-SOLVING The opening of a halonium ion is Htl1v P ROB L E M 8- 2 2 driven by its electrophilic nature. The weak nucleophile attacks the carbon bearing more positive charge. Show how you would accomplish the following synthetic conversions. ( a) 3 -methyl-2-pentene 2 -chloro-3-methyl-3-pentanol ( b) c hlorocyclohexane trans-2-chlorocyclohexanol (c) I -methyIcyc1opentanol 2 -chloro- l -methyIcyc1opentanol 8- 1 0 Cata lytic Hyd rogenation of A l ke n es A lthough we have mentioned catalytic hydrogenation before (Sections 7 -7 and 8- 1 ), we now consider the mechanism and stereochemistry in more detail. Hydrogenation of an alkene is formally a reduction, with H2 adding across the double bond to give an alkane. The process usually requires a catalyst containing Pt, Pd, or Ni. / C =C / + H2 catalyst ) - C -C I I HH I I Example C H3-CH =CH-CH3 + H2 Pt For most alkenes, hydrogenation takes place at room temperature, using hydrogen gas at atmospheric pressure. The alkene is usually dissolved in an alcohol, an alkane, or acetic acid. A small amount of platinum, palladium, or nickel catalyst is added, and the container is shaken or stirred while the reaction proceeds. Hydrogenation actually takes place at the surface of the metal, where the liquid solution of the alkene comes into contact with hydrogen and the catalyst. Hydrogen gas is adsorbed onto the surface of these metal catalysts, and the catalyst weakens the H - H bond. In fact, if H 2 and O2 are mixed in the presence of a platinum catalyst, the two isotopes quickly scramble to produce a random mixture of HO, H2, and O2. (No scrambling occurs in the absence of the catalyst.) Hydrogenation is an example of heterogeneous catalysis, because the (solid) catalyst is in a different phase from the reactant solution. In contrast, homogeneous catalysis involves reactants and catalyst in the same phase, as in the acid-catalyzed dehydration of an alcohol. Because the two hydrogen atoms add from a solid surface, they add with syn stere ochemistry. For example, when 1 ,2-dideuteriocyclohexene is treated with hydrogen gas over a catalyst, the product is the cis isomer resulting from syn addition (Figure 8-6). The Parr hydrogenation apparatus shakes the reaction vessel (containing the alkene and the solid catalyst), while a pressurized cylinder supplies hydrogen. One face of the alkene pi bond binds to the catalyst, which has hydrogen adsorbed on its surface. Hydrogen inserts into the pi bond, and the product is freed from the catalyst. Both hydrogen atoms add to the face of the double bond that is complexed with the catalyst. C( D o O H 8- 1 0 C atalytic Hydrogenation of Alkenes 349 HH Pt Pt Pt II catalyst with hydrogen adsorbed .A. Syn stereochemistry in catalytic hydrogenation . A s olid heterogeneous catalyst adds two hydrogen atoms to the same face of the pi bond (syn stereochemistry) . F igure 8-6 catalyst with hydrogen and alkene adsorbed hydrogen inserted into C = C alkane product released from catalyst S oluble homogeneous catalysts, such as Wilkinson 's catalyst, also catalyze the hydrogenation of carbon-carbon double bonds. PPh3 Ph3P....... Rh / / ....... PPh CI 3 (Wilkinson's catalyst) -C-C- I I I H I H Wilkinson's catalyst is not chiral, but its triphenylphosphine (PPh3) ligands can be replaced by chiral ligands to give chiral catalysts that are capable of converting opti cally inactive starting materials to optically active products. Such a process is called asymmetric induction o r enantioselective synthesis. F or example, Figure 8-7 shows a chiral ruthenium complex catalyzing an enantioselective hydrogenation of a car bon-carbon double bond to give a large excess of one enantiomer. Because the catalyst is chiral, the transition states leading to the two enantiomers of product are diastere omeric. They have different energies, and the transition state leading to the (R) enan tiomer is favored. Ryoji Noyori and William Knowles shared the 200 1 Nobel Prize in chemistry for their work on chirally catalyzed hydrogenation reactions. Enantioselective synthesis is particularly important in the pharmaceutical indus try, because only one enantiomer of a chiral drug is likely to have the desired effect. For example, levodopa [ ( ) -dopa or I-dopa] is used in patients with Parkinson's disease to counteract a deficiency of dopamine, one of the neurotransmitters in the brain. Dopamine itself is useless as a drug because it cannot cross the "blood-brain barrier"; that is, it cannot get into the cerebrospinal fluid from the bloodstream. ( )-Dopa, on the other hand, is an amino acid related to tyrosine. It crosses the blood-brain barrier - F ig ure 8-7 96% e.e. (R) Ru(BINAP)CI2 = Chiral hydrogenation catalysts. Rhodium and ruthenium phosphines are effective homogeneous catalysts for hydrogenation. Chiral ligands can be attached to accomplish asymmetric induction, the creation of a new asymmetric carbon as mostly one enantiomer. 3 50 C hapter 8: R eactions of Alkenes () into the cerebrospinal fluid, where it undergoes enzymatic conversion to dopamine. ( Only the - enantiomer of dopa can be transformed into dopamine; the other enan tiomer, + ) -dopa, is toxic to the patient. The correct enantiomer can be synthesized from an achiral starting material by catalytic hydrogenation using a complex of rhodium with a chiral ligand called mop. S uch an enantioselective synthesis is more than efficient making a racemic mixture, resolving it into enantiomers, and discarding the unwanted enantiomer. Rh(DIOP)CI2 H0 -Q _ HO (S)-( - )-dopa H 2N 1 CH2 '' " ' C H ' " """"' brain enzymes ) COOH dopamine Rh(DIOP)CI 2 T he enzymatic reduction of a dou ble bond is a key step i n the forma tion cell of wall a fatty acid that is ultimately incorporated of the into the that P ROBLE M 8-23 bacterium Give the expected major product for each reaction, including stereochemistry where applicable. (a) I -butene + H 2/Pt (b) cis- 2 -bu tene + R 2/Ni causes tuberculosis. The antituber culosis drug isoniazid blocks this enzyme, preventing reduction of the double bond. Without an i ntact cell wall, the bacteria die. (e) P ROBLE M 8- 24 W + H,IPt (d) ---0-< + """ H,1Pt P ROBLE M 8- 2 5 One of the principal components of lemon grass oil is limonene, C l OR 1 6. When limonene is treated with excess hydrogen and a platinum catalyst, the product is an alkane of formula C I oR 20. What can you conclude about the structure of limonene? isoniazid The chiral BINAP ligand shown in Figure 8-7 contains no asymmetric carbon atoms. Explain how this ligand is chiral. ( ) 8 -1 1 Add ition of Ca rbenes to Al ke n es Methylene :CH2 is the simplest of the carbenes: uncharged, reactive intermediates ( ) that have a carbon atom with two bonds and two nonbonding electrons. Like borane B H 3 ' m ethylene is a potent electrophile because it has an unfilled octet. It adds to the electron-rich pi bond of an alkene to form a cyclopropane. methylene 8 - 1 1 Addition of Carbenes to Alkenes 35 1 Heating or photolysis of diazomethane gives nitrogen gas and methylene: [- : N=N=CH2 .. + diazomethane heat or ultraviolet light ) + :C / '" H H methylene Two difficulties arise when using diazomethane to cyclopropanate double bonds. First, it is extremely toxic and explosive. A safer reagent would be more con venient for routine use. Second, methylene generated from diazomethane is so reac tive that it inserts into C - H bonds as well as C = C bonds. In the reaction of propene with diazomethane-generated methylene, for example, several side products are formed. C H2 - N = N : , ;-; + hv propene H - CH2 - H 2C + H '" / C =C / '" H H P ROBLEM 8 - 2 6 ( a) I -methylcyclohexene Show how the insertion of methylene into a bond of cyclohexene can produce the following. (b) 3-methylcyclohexene (e) 0 0",'''0', ( 8 -1 1 A The Simmons-Smith Reaction Two DuPont chemists discovered a reagent that converts alkenes to cyclopropanes in bet ter yields than diazomethane, with fewer side reactions. The Simmons-Smith reaction, named in their honor, is one of the best ways of making cyclopropanes. The Simmons-Smith reagent is made by adding methylene iodide to the "zinc-copper couple," zinc dust that has been activated with an impurity of copper. The reagent probably resembles iodomethyl zinc iodide, ICH2ZnI. This kind of reagent is called a carbenoid b ecause it reacts much like a carbene, but it does not actually contain a divalent carbon atom. ICH2ZnI S immons - Smith reagent (a carbenoid) o (59%) 352 Chapter 8: Reactions of Alkenes 8 -1 1 B Formation of Carbenes by A l pha Elimination Carbenes are also formed by reactions of halogenated compounds with bases. If a carbon atom has bonds to at least one hydrogen and to enough halogen atoms to make the hydrogen slightly acidic, it may be possible to form a carbene. For example, bromoform (CHBr3) reacts with a 50% aqueous solution of potassium hydroxide to form dibromocarbene. CHBr3 bromoform + K+ -OH (---7 ( ) : CBr3 : CBr2 K+ + + H2O : Bc B r-C=-Br Br This dehydrohalogenation is called an alpha elimination (a e limination) because the hydrogen and the halogen are lost from the same carbon atom. The more common dehydrohalogenations (to form alkenes) are called beta eliminations because the hydrogen and the halogen are lost from adjacent carbon atoms. Dibromocarbene formed from CHBr3 can add to a double bond to form a dibromocyclopropane. IV dibromocarbene The products of these cyclopropanations retain any cis or trans stereochemistry of the reactants. H N aOH, Hp o CHB r3 KOHlHP ) /i'vBr Br Ph CHBr3 ) NaOH, Hp H CH3CH2 CH2CH3 H Br Cl Ph H P ROBLEM 8- 2 7 ( a) c yclohexene Predict the major products of the following reactions. + C HCl3, 5 0% N aOH/H20 CHBro (e) ~ + 50% NaOH/Hp 8- 1 3 Acid-Catalyzed Opening of Epoxides 3 55 M ost epoxides are easily isolated as stable products if the solution is not too acidic. Any moderately strong acid protonates the epoxide, however. Water attacks the proto nated epoxide, opening the ring and forming a 1,2-diol, commonly called a glycol. Acid-Cata Iyzed Opening of Epoxides 8 -13 MECHANISM 8-10 Step 1: Acid-Catalyzed Opening of Epoxides The crucial step is a back-side attack by the solvent on the protonated epoxide. Protonation of the epoxide activates it toward nucleophilic attack . H o Hp+ /\ -C-Cepoxide I I :0+ /\ -C-C/ \ protonated epoxide I Step 2: Back-side attack by the solvent (water) opens the ring. H protonated epoxide '0+ :OH II i \ .) -C-C-C-C / )\ I I+ :0 Hi) back-side attack /\ HH I .. ) Step 3: Deprotonation gives the diol product. :OH .. -C-C:0+ / H I I I I .. :OH () .. OH 2 -C-C- I I : OH I I + H30+ U a glycol (anti orientation) Step EXAMPLE: Acid-catalyzed hydrolysis of propylene oxide (epoxypropane). 1: Protonation of the epoxide. H I '0' :0+ /\ H 0+ /\ +H-C-C-CH3 ---'- H-C-C-CH3 / \ II H H HH epoxypropane (propylene oxide) protonated epoxide T he body oxidizes the alkene com ponents of drugs and other sub stances epoxide to epoxides, which are The then hydrolyzed to diols by an hydrolase enzyme. more reactive epoxides are rapidly converted to water-soluble diols and eliminated in the urine. Epox ide hydrolase enzymes are some times used in organic synthesis to ( Continued) produce chiral diols. 3 56 Chap ter 8: Reactions of A l ken es . Steps 2 and 3: B ack-side attack by water, then deprotonation of the p roduct H- C - C - CH H / J\ H2 ' ' .. 1 Oj ) H protonated epoxide O 3 ---- back-side attack J 11 H - C-C-CH3 1 1+ H :0 /\ HH :O H H :O H H ( ) H - C - C - CH3 H : QH propane- l ,2-diol (propylene glycol) 1 1 1 1 + H30 + 0 .. 0H2 B ecause glycol formation involves a back-side attack on a protonated epoxide, the result is anti orientation of the hydroxyl groups on the double bond. For example, when 1 ,2-epoxycyclopentane ("cyclopentene oxide") is treated with dilute mineral acid, the product is pure trans- l ,2-cyclopentanediol. cyclopentene oxide + O.J HI PROBLEM 8-30 H : OH 0 6{ HO O H 0 H OH H trans- l,2-cyclopentanediol (racemic) (a) Propose a detailed mechanism for the conversion of cis-3-hexene to the epoxide (3,4-epoxyhexane) and the ring-opening reaction to give the glycol, 3,4-hexanediol. In your mechanism, pay particular attention to the stereochemistry of the intermedi ates and products. ( b) R epeat part (a) for trans-3-hexene. Compare the products obtained from cis- and trans3-hexene. Is this reaction sequence stereospecific? Epoxidation reagents can be chosen to favor either the epoxide or the glycol. Per oxyacetic acid i s used in strongly acidic water solutions. The acidic solution protonates the epoxide and converts it to the glycol. Peroxybenzoic acids are weak acids that can be used in nonnucleophilic solvents such as carbon tetrachloride. m-Chloroperoxyben zoic acid in CCl4 generally gives good yields of epoxides. Figure 8-8 compares the uses of these reagents. PROBLEM 8-31 M agnesium monoperoxyphthalate (MMPP) epoxidizes alkenes much like MCPBA. MMPP is more stable, however, and it may be safer to use for large-scale and industrial reactions. Propose a mechanism for the reaction of trans-2-methyl-3-heptene with M MPP, and predict the structure of the product(s). 8- 1 3 Acid-Catalyzed Opening of Epoxides Magnesium monoperoxyphthalate, MMPP 3 57 2 PROBLEM 8-32 Predict the major products of the following reactions. (a) cis-2-hexene + M CPBA in chloroform (b) trans-3-hexene + peroxyacetic acid (CH3C03H ) in water (c) I -methylcyclohexene + MMPP in ethanol (d) trans-cyclodecene + peroxyacetic acid in water (e) cis-cyclodecene + M CPBA in CH2CI2, then dilute aqueous acid PROBLEM 8-33 When 1 ,2-epoxycyclohexane (cyclohexene oxide) is treated with anhydrous HCI in methanol, the principal product is trans-2-methoxycyclohexanol. Propose a mechanism to account for the formation of this product. o o 0 Ph - C - OOH CCl 4 II 0:0 ( 1 00%) 0 CH3 -C- OOH H3 0+ II 0 [O:oJ not isolated CH3 - C - OH Hp+ stronger acid, nucleophilic solvent II H dOH H OH -:0 + enantiomer ( 75%) Figure 8-8 Reagents for epoxidation. Peroxyacetic acid is used in strongly acidic aqueous solutions. Alkenes are epoxidized, then opened to glycols in one step. Weakly acidic peroxyacids, such as peroxybenzoic acid, can be used in nonaqueous solutions to give good yields of epoxides. 3 58 Chapter 8: Reactions of Alkenes Converting an alkene to a glycol requires adding a hydroxyl group to each end of the double bond. This addition is called hydroxylation o f the double bond. We have seen that epoxidation of an alkene, followed by acidic hydrolysis, gives anti hydroxylation of the double bond. Reagents are also available for the hydroxylation of alkenes with s yn stereochemistry. The two most common reagents for this purpose are osmium tetroxide and potassium permanganate. - C-COR OR (or KMn04, -OH) syn addition Syn Hydroxylation of Alkenes 8 -14 I I I I 8-14A Osm i u m Tetroxide Hyd roxyl ation O smium tetroxide (OS04, sometimes called osmic acid) reacts with alkenes in a con certed step to form a cyclic osmate ester. Hydrogen peroxide hydrolyzes the osmate ester and reoxidizes osmium to osmium tetroxide. The regenerated osmium tetroxide catalyst continues to hydroxylate more molecules of the alkene. '" / C P -I /" alkene 1 0q 0 C osmic acid ----? ,,/ C- O /\ s IC- O/OO osmate ester / "' 0 H 20 2 ,,/ C -OH /\ I C-OH glycol + OS04 B ecause the two carbon-oxygen bonds are formed simultaneously with the cyclic osmate ester, the oxygen atoms add to the same face of the double bond; that is, they add with syn stereochemistry. The following reactions show the use of OS04 and H202 for the syn hydroxylation of alkenes. cis-glycol ( 65%) concerted formation of osmate ester H OH H H CH2CH3 ", / C C / '" CH2CH3 H cis-3-hexene II 0,0, ",0, H H -r: I OH ' CH2CH3 meso-3,4-hexanediol 8-14 Syn Hydroxylation of Alkenes 8- 1 4 B Permanganate Hydroxylation 359 Osmium tetroxide is expensive, highly toxic, and volatile. A cold, dilute solution of potassium permanganate ( KMn04 ) also hydroxylates alkenes with syn stereochem istry, with slightly reduced yields in most cases. Like osmium tetroxide, permanganate adds to the alkene double bond to form a cyclic ester: a manganate ester in this case. The basic solution hydrolyzes the manganate ester, liberating the glycol and producing a brown precipitate of manganese dioxide, Mn02' H W OH cis-glycol H OH (49%) concerted formation of manganate ester I n addition to its synthetic value, the permanganate oxidation of alkenes pro vides a simple chemical test for the presence of an alkene. When an alkene is added to a clear, deep purple aqueous solution of potassium permanganate, the solution loses its purple color and becomes the murky, opaque brown color of Mn02. (Although there are other functional groups that decolorize permanganate, few do it as quickly as alkenes.) 8- 14C Choosi n g a Reagent To hydroxylate an alkene with syn stereochemistry, which is the better reagent: osmium tetroxide or potassium permanganate? Osmium tetroxide gives better yields, but per manganate is cheaper and safer to use. The answer depends on the circumstances. If the starting material is only 2 mg of a compound 15 steps along in a difficult syn thesis, we use osmium tetroxide. The better yield is crucial because the starting material is precious and expensive, and little osmic acid is needed. If the hydroxylation is the first step in a synthesis and involves 5 kg of the starting material, we use potassium perman ganate. The cost of buying enough osmium tetroxide would be prohibitive, and dealing with such a large amount of a volatile, toxic reagent would be inconvenient. On such a large scale, we can accept the lower yield of the permanganate oxidation. PROBLEM 8 -3 4 Predict the major products of the following reactions, i ncl udi ng stereochemistry. (a) cyclohexene + KMn04/H20 ( b) cyclohexene + peroxyacetic acid in water (c) cis-2-pentene + O s04/H202 (d) cis-2-pentene + p eroxyacetic acid in water (e) trans-2-pentene + O s04/H202 (f) trans-2-pentene + peroxyacetic acid in water PROBLEM 8-35 Show how you would accomplish the following conversions. (a) cis-3-hexene to meso-3,4-hexanediol (b) cis-3-hexene to (d,l)-3,4-hexanediol (c) trans-3-hexene to meso-3,4-hexanediol (d) trans-3-hexene to (d,I)-3,4-hexanediol 3 60 C hapter 8: Reactions of Alkenes 8- 1 5A Cleavage by Perma nganate Oxidative Cleavage of Alkenes 8 -1 5 In a potassium permanganate hydroxylation, if the solution is warm or acidic or too concentrated, oxidative cleavage of the glycol may occur. Mixtures of ketones and carboxylic acids are formed, depending on whether there are any oxidizable alde hydes in the initial fragments. A terminal =CH2 group is oxidized to CO2 and water. Figure 8-9 shows the oxidative cleavage of a double bond by warm or concen trated permanganate. 8- 1 58 Ozonolysis R' R'" / C=C'" / H R + Like permanganate, ozone cleaves double bonds to give ketones and aldehydes. However, ozonolysis is milder, and both ketones and aldehydes can be recovered with out further oxidation. R' R'" R", O /R' (CH3)2S / / C C=O + O=C'" 03 /C'" / / H R 0-0 H R ozonide Ozone is a powerful lung irritant, causing a cough, sore throat, and tiredness. It can also increase a per son's sensitivity to allergens. The mechanism may involve peroxida tion of the double bonds of the fatty acids that make up the surfac tants and the membranes of the cells lining the bronchial airways and lungs. Ozone (03) is a high-energy form of oxygen produced when ultraviolet light or an electrical discharge passes through oxygen gas. Ultraviolet light from the sun con verts oxygen to ozone in the upper atmosphere. This "ozone layer" shields the earth from some of the high-energy ultraviolet radiation it would otherwise receive. 02 + 142 kJ (34 kcal) 03 Ozone has 1 42 kllmol of excess energy over oxygen, and it is much more reactive. A Lewis structure of ozone shows that the central oxygen atom bears a positive charge, and each of the outer oxygen atoms bears half a negative charge. :0=9-9:] [:Q-9=0: '" ketone aldehyde (warm, concd.) l .. + .. .. + .. R R' R - - -H OH OH ii glycol Examples J ----;> R ""C=O R/ ketone (stable) + :J C ( aldehyde (oxidizable) R' O=C/ ""OH acid OH O C I (warm, concd.) (warm, concd.) iYoOH Yci5g o +C O2 .... Figure 8-9 Permanganate cleavage of alkenes. Warm, concentrated KMn04 oxidizes alkenes to glycols, then cleaves the glycols. The products are initially ketones and aldehydes, but aldehydes are oxidized to carboxylic acids under these conditions. 8-12 Epoxidation of Alkenes PROBLEM 8-2 8 3 53 (a) trans-2-butene -7 trans- I ,2-dimethylcyclopropane Show how you would accomplish each of the following synthetic conversions. (b) cyclopentene (c) cyclohexanol O><H C l Cl Br Some of the most important reactions of alkenes involve oxidation. When we speak 8 -12 of oxidation, we usually mean reactions that form carbon-oxygen bonds. (Halo gens are oxidizing agents, and the addition of a halogen molecule across a double Epoxidation bond is formally an oxidation as well.) Oxidations are particularly important of Alkenes because many common functional groups contain oxygen, and alkene oxidations are some of the best methods for introducing oxygen into organic molecules. We will consider methods for epoxidation, hydroxylation, and oxidative cleavage of alkene double bonds. An epoxide is a three-membered cyclic ether, also called an oxirane. Epoxides are valuable synthetic intermediates used for converting alkenes to a variety of other functional groups. An alkene is converted to an epoxide by a peroxyacid, a carboxylic acid that has an extra oxygen atom in a - 0 - 0- (peroxy) linkage. " / c=c / " alkene + R-C-O-O-H peroxyacid 0 II 0 " /\ / c-c " / epoxide (oxirane) + R-C-O-H acid 0 II The epoxidation of an alkene is clearly an oxidation, since an oxygen atom is added. Peroxyacids are highly selective oxidizing agents. Some simple peroxy acids (sometimes called peracids) and their corresponding carboxylic acids are shown next. R-C-O-H a carboxylic acid 0 II CH3-C-0-H acetic acid 0 II G j 0 C-O-H 0 benzoic acid, PhC02H il R-C-O-O-H a peroxyacid A 0 II C H 3-C-0 -0 -H peroxyacetic acid 0 II < 1--O-O-H peroxybenzoic acid, PhC03H Epoxides are often found as com ponents of natural products that plants use as defense mechanisms against insects. Typically, the epox ide from killed. reacts with critical their cellular normal enzymes or DNA, preventing them carrying out functions. As a result, the insect is peroxyacid epoxidizes an alkene by a concerted electrophilic reaction where sever al bonds are broken and several are formed at the same time. Starting with the alkene and the peroxyacid, a one-step reaction gives the epoxide and the acid directly, without any intermediates. 3 54 C hapter 8: R ea tio c ns of Al kenes Epoxidation of Alkenes ----- MECHANISM 8-9 Peroxyacids epoxidize alkenes in a one-step (concerted) process. O H alkene peroxyacid transition state epoxide EXAMPLE: Epoxidation of propene by peroxyacetic acid. /R C I 0 / acid /H C" / C - CH3 ll 1 11 0 C ",Hr H3 C / H propene (propylene) peroxyacetic acid H Because the epoxidation takes place in one step, there is no opportunity for the alkene molecule to rotate and change its cis or trans geometry. The epoxide retains whatever stereochemistry is present in the alkene. The following examples use m-chloroperoxybenzoic acid (MCPBA), a common epoxidizing reagent, to convert alkenes to epoxides having the same cis or trans stere ochemistry. MCPBA is used for its desirable solubility properties: The peroxyacid dis solves, then the spent acid precipitates out of solution. cis cis trans trans PROBLEM 8-29 Predict the products, including stereochemistry where appropriate, for the m-chloroperoxy benzoic acid epoxidations of the following alkenes. (a) cis-2-hexene (b) trans-2-hexene (c) cis-cyclodecene (d) transldcn cyc o e e e 8- 1 5 Oxidative Cleavage of Alkenes 3 61 Ozone reacts with an alkene to form a cyclic compound called a p rimary ozonide or molozonide (because 1 mole of ozone has been added). The molozonide has two peroxy ( - 0 - 0 - ) linkages, so it is quite unstable. It rearranges rapidly, even at low temperatures, to form an ozonide. \/ / \ .9: C",-"O" II" :0+ C \ /.:J ' . ' . mol ozonide (primary ozonide) ozonide Ozonides are not very stable, and they are rarely isolated. In most cases, they are im mediately reduced by a mild reducing agent such as zinc or (more recently) dimethyl sulfide. The products of this reduction are ketones and aldehydes. R dimethyl sulfide ozonide "" R C =O / + O=C"" / R' + CH3-S-CH3 dimethyl sulfoxide (DMSO) o II H ketones, aldehydes ""C=O O=C/R' + C =C"" / ""H / R R H The following reactions show the products obtained from ozonolysis of some representative alkenes. Note how ( 1) and ( 2) are used with a single reaction arrow to denote the steps in a two-step sequence. R "" / R' R CH3CH2CHO 3-nonene + CH3(CH2)4CHO ( 65%) cD H H K-!o H C=O I H "0 + H2C=O CHOl1 o One of the most common uses of ozonolysis has been for determining the posi tions of double bonds in alkenes. For example, if we were uncertain of the position of the methyl group in a methylcyclopentene, the products of ozonolysis-reduction would confirm the structure of the original alkene. 3 62 C hapter 8: Reactions of Alkenes l-methylcyclopentene 3-methylcyclopentene PROBLEM-SOLVING T o predict the products from ozonolysis of an a l kene, erase the double bond and add two oxygen atoms as carbonyl ((=0) groups where the double bond used to be. Hi-It;; SOLVED PROBLEM 8-7 O zonolysis-reduction of an unknown alkene gives an equimolar mixture of cyclohex anecarbaldehyde and 2-butanone. Determine the structure of the original alkene. cyclohexanecarbaJdehyde SOLUTION 2-butanone We can reconstruct the alkene by removing the two oxygen atoms of the carbonyl groups ( C = O ) a nd connecting the remaining carbon atoms with a double bond. One uncertainty remains, however: The original alkene might be either of two possible geometric isomers. Ozone is a powerful oxidizing agent that is sometimes used instead of chlorine to disinfect the water in swimming pools. Ozone oxidizes organic matter, and it kills bacteria and algae. Ozone is used instead of chlorine because it can be generated on-site (rather than storing and using highly toxic chemicals such as chlorine gas or sodium hypochlo rite) and because it doesn't produce as many harmful byproducts. PROBLEM 8-36 Give structures of the alkenes that would give the following products upon ozonolysis-reduction. II II (a) CH3-C-CHz-CHz-CHz-C-CHz-CH3 ( b) cyclohexanone o 0 and II CH3-CH2-CHz-C-H o 8 - 1SC Compari son of Perma nganate Cleavage and Ozonolysis Both permanganate and ozonolysis break the carbon-carbon double bond and replace it with carbonyl (C = 0) groups. In the permanganate cleavage, any aldehyde prod ucts are further oxidized to carboxylic acids. In the ozonolysis-reduction procedure, the aldehyde products are generated in the dimethyl sulfide reduction step (and not in the presence of ozone), and they are not oxjdized. 8-16 Polymerization of Alkenes 3 63 coned . KMn04 OH ) (not isolated) PROBLEM 8-37 (,) monomers) (a) (E)-3-methyl-3-octene + ozone, then (CH3hS (b) (Z)-3-methyl-3-octene + warm, concentrated KMn04 Predict the major products of the following reactions. + PROBLEM-SOLVING 0,. theo (CH,J,S (d) l-ethylcycloheptene + ozone, then (CH3hS (e) l-ethylcycloheptene + warm, concentrated KMn04 (0 l-ethylcycloheptene + cold, dilute KMn04 Osmium tetroxide, cold, dilute KMn04, and epoxidation oxidize the p i bond of an alkene but leave the sigma bond intact. Ozone and warm, concentrated KMn04 break the double bond entirely to give carbonyl compounds. Hi-ltv A polymer is a large molecule composed of many smaller repeating units (the 8 -16 bonded together. Alkenes serve as monomers for some of the most common polymers, such as polyethylene, polypropylene, polystyrene, poly(vinyl Polymerization chloride), and many others. Alkenes generally undergo addition polymerization, the of Alkenes rapid addition of one molecule at a time to a growing polymer chain. There is general ly a reactive intermediate (cation, anion, or radical) at the growing end of the chain; for that reason, addition polymers are also called chain-growth polymers. Many alkenes undergo addition polymerization under the right conditions. The chain-growth mechanism involves addition of the reactive end of the growing chain across the double bond of the alkene monomer. Depending on the structure of the monomer, the reactive intermediates may be carbocations, free radicals, or carbanions. 8 -1 6A tion, Cationic Po lymerization Alkenes that easily form carbocations are good candidates for cationic polymeriza which is just another example of electrophilic addition to an alkene. Consider what happens when pure isobutylene is treated with a trace of concentrated sulfuric acid. Protonation of the alkene forms a carbocation. If a large concentration of isobutylene is available, another molecule of the alkene may act as the nuc1eophile and attack the carbocation to form the dimer (two monomers joined together) and give an other carbocation. If the conditions are right, the growing cationic end of the chain will keep adding across more molecules of the monomer. The polymer of isobutylene is polyisobutylene, one of the constituents of butyl rubber used in inner tubes and other synthetic rubber products. 3 64 Cha p t r 8: e Reactions of Alkenes Attack by the second molecule of isobutylene Profonation H2S04 + /CH3 H2C=C"'CH3 isobutylene / /CH3 CHo-C+ "'-CH3 H2C=C"'-CH3 CH3 I /CH3 CH3-C-CH2-c+ "'I CH3 CH3 dimer Attack by a third molecule to give a trimer CH3 CH I / /CH3 CH3-C-CH2-C+ "'-CH3 H2C=C"'-CH3 I CH3 ' dimer third monomer CH3 CH3 I I /CH3 CH3-C-CH2-C-CH2-C+ "'I I CH3 CH3 CH3 trimer polymer Loss of a proton is the most common side reaction that terminates chain growth: HS04 CH3 H) CH3 CH3 I I h/ CH3-C-CHo-C-C-C+ I I I "'CH3 - CH3 H CH3 CH3 CH3 I I -H+ CH3-C-CHo-C-CH=C/CH3 "'-CH3 I I CH3 - CH3 Boron trifluoride (BF3 ) is an excellent catalyst for cationic polymerization because it leaves no good nucleophile that might attack a carbocation intermediate and end the polymerization. Boron trifluoride is electron-deficient and a strong Lewis acid. It usually contains a trace of water that acts as a co-catalyst by adding to BF3 and then protonating the monomer. Protonation occurs at the less substituted end of an alkene double bond to give the more stable carbocation. Each additional monomer molecule adds with the same orientation, always giving the more stable carbocation. The following reaction shows the polymerization of styrene (vinylbenzene) using BF3 as the catalyst. F I + O-H / B'1 "'-F I-I F First chain-lengthening ste p ----7 F HH H 1 +/ _/- ) / F-B =- Q "'-c=c H "'-Ph 1 F styrene ----7 F H 1 /H 1 /H F-B =- O : H-C-C+ + 1 .. F "Ph H /H C H3-C{ "''" H2C=C Ph -----7 Ph H I /H CH3-C-CH2-c+ "'Ph I Ph After many steps the p olymerization continues H / C/H -CH2-CH-CH?-C+ - "'2C "'I Ph Ph Ph - growing polymer chain = H -CH2-CH-CH2-CH-CH2-C{ "'I I Ph Ph Ph 8-16 Po lymeri ation of Alkenes z 3 65 The most likely ending of this BFrcatalyzed polymerization is the loss of a proton from the carbocation at the end of the chain. This side reaction terminates one chain, but it also protonates another molecule of styrene, initiating a new chain. Termination of a polymer chain H I _.(D -:t C+ + H2C\ /H '\ - CH -CH-C =C" I I "Ph Ph Ph H __ _ 2 H H I / H2C-C+ " Ph polystyrene starts another chai n The product of this polymerization is polystyrene: a clear, brittle plastic that is often used for inexpensive lenses, transparent containers, and styrofoam insulation. Polystyrene is also the major component of the resin beads that are used to make syn thetic proteins. (See Section 24-11.) PROBLEM 8-38 Propose a mechanjsm for the following reaction. 2(CH3hC=CH - CH3 + cat. H + ----+ 2,3,4,4-tetramethyl-2-hexene PROBLEM 8-39 Show the fIrst three steps (as far as the tetramer) in the BF3-catalyzed polymerization of propylene to form polypropylene. PROBLEM 8-40 When cyclohexanol is dehydrated to cyclohexene, a gummy green substance forms on the bottom of the flask. Suggest what this residue might be, and propose mechanism for its for mation (as far as the dimer). a 8-1 6B Free-Rad ical Polymeri zation Many alkenes undergo free-radical polymerization when they are heated with radical initiators. For example, styrene polymerizes to polystyrene when it is heat ed to lOODC with a peroxide initiator. A radical adds to styrene to give a resonance stabilized radical, which then attacks another molecule of styrene to give an elongated radical. Initiation step Propagation step ROOR -----7 heat 2 RO ' H H rr I I RO-C-C-C - C' I I I H H H "H Q : styrene stabilized radical styrene growing chain 366 C hapter 8: Reactions of Alkenes Each propagation step adds another molecule of styrene to the radical end of the growing chain. This addition always takes place with the orientation that gives anoth er resonance-stabilized benzylic (next to a benzene ring) radical. Propagation step RO-C-C-C-C H growing chain lY )0 I H I H I \ -----7 -----7 -----7 -----7 add many more styrene molecules H C-C H y I H I n styrene e longated chain n = polystyrene about 100 to 10,000 C hain growth may continue with addition of several hundred or several thousand styrene units. Eventually, the chain reaction stops, either by the coupling of two chains or by reaction with an impurity (such as oxygen) or simply by running out of monomer. PROBLEM 8-41 Show the intermediate that would result if the growing chain added to the other end of the styrene double bond. Explain why the final polymer has phenyl groups substituted on alter nate carbon atoms rather than randomly distributed. Ethylene is also polymerized by free-radical chain-growth polymerization. With ethylene, the free-radical intermediates are less stable, so stronger reaction conditions are required. Ethylene is commonly polymerized by free-radical initiators at pressures around 3000 atm and temperatures of about 200C. The product, called low-density polyethylene, i s the material commonly used in polyethylene bags. PROBLEM 8-42 Propose a mechanism for reaction of the first three ethylene units in the polymerization of ethylene in the presence of a peroxide. n H2C=CH2 ethylene ROOR h igh pressure polye thy l ene 8 - 16C Anionic Polymerization L ike cationic polymerization, anionic polymerization depends on the presence of a stabi lizing group. To stabilize anions, the double bond should have a strong elec tron-withdrawing group such as a carbonyl group, a cyano group, or a nitro group. Methyl a-cyanoacrylate contains two powerful electron-withdrawing groups, and it undergoes nucleophilic additions very easily. If this liquid monomer is spread in a thin film between two surfaces, traces of basic impurities (metal oxides, etc.) can catalyze its rapid polymerization. The solidified polymer joins the two surfaces. The chemists who first made this monomer noticed how easily it polymerizes and real ized that it could serve as a fast-setting glue. Methyl a-cyanoacrylate is sold com mercially as Super Glue. 8- 1 6 P olymerization of Alkenes Initiation step 3 67 t race 0f base Super Glue H COOCH3 HO-C-C:- -CN 'H I .... I highly stabilized anion C hain lengthening step ------> __ _ growing chain PROBLEM 8-43 monomer COOCH3 {_ 1 _ _ c:::-COOCH3 '--CN H CN H I I I elongated chain m CH}1{_r lH CN 11 polymer Draw a mechanism for a base-catalyzed polymerization of methyl a-methacrylate to give the Plexiglas polymer. H---... C=C...... COOCH3 H...... ---""'CH3 methyl a-methacrylate P R O BL E M -S OLV I N G ST R AT E GY Organic Synthesis Alkyl halides and alkenes are readily made from other compounds, and they are easily con verted to other functional groups. This flexibility makes them useful as reagents and inter mediates for organic synthesis. Alkenes are particularly important for industrial syntheses because they are inexpensive and available in large quantities from cracking and dehydro genation of petroleum fractions. Organic synthesis i s the preparation of desired compounds from readily available materials. Synthesis is one of the major areas of organic chemistry, and nearly every chapter of this book involves organic synthesis in some way. A synthesis may be a simple one-step reaction, or it may involve many steps and incorporate a subtle strategy for assembling the correct carbon skeleton with all the functional groups in the right positions. Many of the problems in this book are synthesis problems. In some synthesis problems, you are asked to show how to convert a given starting material to the desired product. There are obvious one-step answers to some of these problems, but others may require several steps and there may be many correct answers. In solving multistep synthetic problems, it is often helpful to analyze the problem backward: Begin with the desired product (called the target compound) and see how it might be mentally changed or broken down to give the starting materials. This backward approach to synthesis is called a retrosynthetic analysis. Some problems allow you to begin with any compounds that meet a certain restric tion. For example, you might be allowed to use any alcohols containing no more than four carbon atoms. A retrosynthetic analysis can be used to break down the target compound into fragments no larger than four carbon atoms; then those fragments could be formed from the appropriate alcohols by functional group chemistry. The following suggestions should help you solve synthesis problems: 1. D o not guess a starting material and try every possible reaction to convert it to the target compound. Rather, begin with the target compound and use a retrosynthetic analysis to simplify it. (Continued) 3 68 Chapter 8: Reactions of Alkenes Use simple equations, with reagents written above and below the arrows, to show the reactions. The equations do not have to be balanced, but they should include all the reagents and conditions that are important to the success of the reaction. Br2, light H+, H20 NaOH, alcohol )B C )D A heat 3. Focus on the functional groups, since that is generally where reactions occur. Do not use any reagents that react with a functional group that you don't intend to modify. In solving multistep synthesis problems, you will rarely be able to "see" the solution immediately. These problems are best approached systematically, working backward and considering alternative routes. To illustrate a systematic approach that can guide you in solv ing synthesis problems, we will work through the synthesis of a complex ether starting from alkenes. The problem-solving method described here will be extended in future chapters to multistep syntheses based on the reactions of additional functional groups. A systematic retrosynthetic analysis begins with an examination of the structure of the product. We will consider the synthesis of the following compound from alkenes containing up to five carbon atoms. 2. 1. Review the functional groups and carbon skeleton of the target compound. The target compound is an ether. One alkyl group is a five-carbon cyclopentane ring with two oxygen atoms situated trans. The other group has three carbons containing a reactive epoxide ring. 2. Review the functional groups and carbon skeletons of the starting materials (if specified), and see how their skeletons might fit together in the target compound. The synthesis is to begin with alkenes containing up to five carbon atoms, so all the func tional groups in the product must be derived from alkenes. Most likely, we will start with cyclopentene to give the five-carbon ring and propene to give the three-carbon chain. 3. Compare methods for synthesizing the functional groups in the target compound, and select the reactions that are most likely to give the correct product. This step may require writing several possible reactions and evaluating them. Ethers can be synthesized by nucleophilic reactions between alkyl halides and alkoxides (Section 6-9). The target compound might be formed by SN2 attack of an alkoxide ion on an alkyl halide in either of two ways: o +Br a \ \,\\' .O The first reaction is better because the SN2 attack is on a primary alkyl halide, while the second is on a secondary halide. Also, in the second reaction the alkoxide might simply deprotonate the alcohol on the left and cause the reaction to fail. 4. In general, reactive functional groups are best put into place toward the end of a synthesis. C( +-O a'"\' Br o OH /1 o /1 o .. 0 OH OH The target compound contains a reactive epoxide ring. Epoxides react with acids and bases, and the epoxide might not survive the crucial ether-forming reaction just shown. 8-16 Polymerization of Alkenes Perhaps the epoxide is best added after formation of the ether. That gives us the following final two steps in the synthesis: 3 69 /1"" .. . v: OH 5. Working backward through as many steps as necessary, compare methods for synthesizing the reactants needed for the final step. 0 , C( OH o This process may require writing several possible reaction sequences and evaluating them, keeping in mind the specified starting materials. Two reactants are needed to form the ether: an allylic halide and an alkoxide ion. Alkoxide ions are commonly formed by the reaction of an alcohol with sodium metal: The alkoxide needed to make the ether is formed by adding sodium to a trans diol. Trans diols are formed by epoxidation and hydrolysis of alkenes (Section 8- 1 3). o " /I\"O, H The other piece we need is an allylic bromide. Allylic bromides are formed by al lylic bromination of alkenes (Section 6-6B). v: .. Na OH a \"" .. OH 0 Na + + H2 i Br 6. Summarize the complete synthesis in the forward direction, including all steps and all reagents, and check it for errors and omissions. This summary is left to you (Problem 8-44), as a review of both the chemistry involved in the synthesis and the method used to develop multistep syntheses. PROBLEM 8-44 Summarize the synthesis outlined in the problem-solving strategy. This summary should be in the synthetic (forward) direction, showing each step and all reagents. Problem 8-45 requires devising several multistep syntheses. As practice in working such problems, we suggest that you proceed in order through the five steps just outlined. PROBLEM 8-45 Show how you would synthesize each compound, starting with alkenes or cycloalkenes that contain no more than six carbon atoms. You may use any additional reagents you need. OH (b) aC=N (0) aJy 3 70 Chapter 8: Reactions of Alkenes SUMMARY Reactions of Alkenes I 1 . Electrophilic Additions a. Addition of hydrogen halides (Section 8-3) - C-C H X '" / C=C / '" + H -X I I I I (HX = HCI, HBr, or HI) Markovnikov orientation (anti-Markovnikov with HBr and peroxides) Example no p roXides CH3 CH3 -C=CH2 2-methylpropene I + HBr I I I I CH3 CH3 - C -CH3 Br I I t-butyl bromide (Markovnikov orientation) CH3 CH3-CH -CH2Br peroxides I isobutyl bromide (anti-Markovnikov orientation) b. Acid-catalyzed hydration (Section 8-4) -C-CH OH \ / C=C / \ + H20 ------? + H (Markovnikov orientation) Example OH CH3- CH=CH2 propene + H20 CH3-CH-CH3 2-propanol I c.Oxymercuration-demercuration (Section 8-5) \ / C=C \ / + Hg(OAc)z HO - C-CHO H g OA c I I I I -C-CHO H I I I I (Markovnikov orientation) Example OH CH3 -CHCH2CH3 2-butanol d. Alkoxymercuration-demercuration I (Section 8-6) \ / C=C / \ + Hg(OAc)z ROH -C-CRO HgOAc I I I I -C-CRO H I I I I (Markovnikov orientation) 8-16 Polymerization of Alkenes Example 37 1 H2 C= CH-CH2 -CH3 I-butene ( I) Hg(OAch, CH30H ) (2) NaBH4 H3C-CH- CH? -CH3 OCH3 I 2-methoxybutane e. Hydroboration-oxidation (Section 8-7) ---i> / " C =C " / + B H3' T HF -C-C- H I I B H2 I I -C-C- H I I OH I I anti-Markovnikov orientation (syn stereochemistry) Example (2) H2 02, OH (1) B H3 . THF f. Polymerization (Section 8 -16) " / Example C= C / " I / R- C - C+ "I '\.. ----' C=C'\.. / / III / R - C-C -C-C+ I I I " ---;. pol y mer ( also radical and anionic polymerization) n C H3 -CH= CH2 p ropylene BF3 ) l -l lk t J H3 p olypropylene ) n 2. Reduction: Catalytic Hydrogenation (Section 8- 10) - C -CH H ( syn addition) " / C =C / " + H2 Pt, Pd, or Ni I I I I 3. Addition ofCarbenes: Cyclopropanation (Section 8-11) " / C =C / "= -C-CH, Cl, Br, I, or -COOEt) y I /\ \/ I C X (X,Y Example cyclohexene o + C HBr, B r Br (Continued) 372 Chapter 8: Reactions of Alkenes 4. Oxidative Additions a. Addition of halogens (Section 8 -8) / C=C "/ Example "- x - C-C- I I I X (anti addition) I cyclohexene " ilr, Br trans-l,2-dibromocyclohexane NBS provides a trace of Br2 that (with light as initiator) allows radical substitution to proceed faster than the ionic addition. (Section 6-6B) NBS, hv ) (trace Br2) cyclohexene b. 3-bromocyclohexene (Section 8-9) Halohydrin formation + anti addition (Markovnikov orientation) c. Epoxidation Q if "\ B r CH3 ""'OH "- / C=C alkene / (Section 8-12) R- C-O-O-H "- + 0 II -C - C - I \/ peroxyacid syn addition 0 I + R -C-O-H 0 II Example cyclohexene 0 O-0 " - C-OOH II Cl epoxycyclohexane Cl (cyclohexene oxide) 00 O>-r + OH d. / "- Anti hydroxylation (Section 8- 1 3) -C-C\/ o C=C "- / + R-C-O-O-H II o I I OH I I -C-CI I OH 8-16 Polymerization of Alkenes Example o 373 cycIohexene o H-C-OOH, H30+ II ) trans-cyclohexane-l,2-diol Q Ji / OH '\ H em e. Syn hydroxylation (Section 8-14) + "-C=C / / "Example + -OH,Hp fI "' OH '\ OH II -C-CII OH OH ( syn addition) I I cycIohexene o C=C cis-cycIohexane- l,2-diol Q 5. Oxidative Cleavage ofAlkenes (Section 8-15) a. R R Ozonolysis " / / "- R' + H 03 -4 + ozonide O =C / R' " H ketones and aldehydes Example H CH 3 - C = C - CH 3 2-methyl-2-butene acetaldehyde acetone I CH3 I b. Potassium permanganate R "-C=C / R' ,,/ R H + KMn04 warm R R "- / C=O + / R' O =C" OH ketones and acids (aldehydes are oxidized) Example H CH3 II CH3-C=C-CH3 + KMn04 2-methyl-2-butene warm acetic acid acetone 3 74 C hapter 8: R eactions of Alkenes addition A reaction involving an increase in the number of groups attached to the alkene and a decrease in the number of elements of unsaturation. (p. 3 21) anti addition: A n addition in which two groups add to opposite faces of the double bond (as in addition of Br2). (p. 344) electrophilic addition: A n addition in which the electrophile (electron-pair acceptor) bonds to one of the double-bonded carbons first, followed by the nucleophile. (p. 322) syn addition: A n addition in which two groups add to the same face of the double bond (as in osmium tetroxide hydroxylation). (p. 348) addition polymer (chain-growth polymer) A polymer that results from rapid addition of one molecule at a time to a growing polymer chain, usually with a reactive intermediate (cation, radical, or anion) at the growing end of the chain. (p. 363) alkoxy group (alkoxyl group) ( - 0 - R) An alkyl group bonded through an oxygen atom, as in an ether. alkoxymercuration The addition of mercuric acetate to an alkene in an alcohol solution, forming an aLkoxymercurial intermediate. Demercuration gives an ether. (p. 335) " / C hapte r 8 G l ossa ry / C=C " + Hg(OAc)2 R -OH R-O alpha elimination (a e limination) The elimination of two atoms or groups from the same car bon atom. Alpha eliminations are frequently used to form carbenes. (p. 352) II -C -C II R- O -C - C- I H gOAc I H I I CHBr3 + KOH ----'> : CBr2 + H 20 + KEr anionic polymerization T he process of forming an addition polymer by chain-growth poly merization involving an anion at the end of the growing chain. (p. 3 66) asymmetric induction (enantioselective synthesis) The formation of an optically active product from an optically inactive starting material. Such a process requires the use of an opti cally active reagent or catalyst. (p. 3 49) beta elimination ( f3 elimination) The elimination of two atoms or groups from adjacent carbon atoms. This is the most common type of elimination. (p. 352) H - C-C - I Br I I I + KOH / " C=C / " + HzO + KEf carbene A reactive intermediate with a neutral carbon atom having only two bonds and two non bonding electrons. Methylene ( :CH2 ) is the simplest carbene. (p. 350) cationic polymerization The process of forming an addition polymer by chain-growth poly merization involving a cation at the end of the growing chain. (p. 363) chain-growth polymer See addition polymer. ( p. 363) demercuration The removal of a mercury species from a molecule. Demercuration of the products of oxymercuration and alkoxymercuration is usually accomplished using sodium borohydride. (p. 333) epoxide (oxirane) A three-membered cyclic ether. (p. 353) epoxidation: Formation of an epoxide, usually from an alkene. A peroxyacid is generally used for alkene epoxidations. free-radical polymerization The process of forming an addition polymer by chain-growth polymerization involving a free radical at the end of the growing chain. (p. 365) glycol A 1 ,2-diol. (p. 355) halogenation The addition of a halogen (X2) to a molecule, or the free-radical substitution of an X for an H (p. 344) halohydrin A beta-haloalcohol, with a halogen and a hydroxyl group on adjacent carbon atoms. (p. 345) " / C=C / " - C -C CI OH I I I I + H CI a chlorohydrin Chapter 8 G lossary 3 75 halonium ion A reactive, cationic intermediate with a three-membered ring containing a halo gen atom; usually, a c hloronium ion, a b romonium ion, or an i odonium ion. ( p. 342) heterogeneous catalysis U se of a catalyst that is in a separate phase from the reactants. For example, a platinum hydrogenation catalyst is a solid, a separate phase from the liquid alkene. (p. 3 48) homogeneous catalysis U se of a catalyst that is in the same phase as the reactants. For exam ple, the acid catalyst in hydration is in the liquid phase with the alkene. (p. 3 48) hydration The addition of water to a molecule. Hydration of an alkene forms an alcohol. (p. 3 30) II -C -C II hydroboration The addition of borane (BH3 ) or one of its derivatives (BH3 ' THF, for exam ple) to a molecule. (p. 3 36) hydrogenation T he addition of hydrogen to a molecule. The most common hydrogenation is the addition of H2 across a double bond in the presence of a catalyst (catalytic hydrogenation or catalytic reduction). (p. 3 48) hydroxylation The addition of two hydroxyl groups, one at each carbon of the double bond; formally, an oxidation. (p. 3 58) H OH II - C-C II Markovnikov's rule (original statement) When a proton acid adds to the double bond of an alkene, the proton bonds to the carbon atom that already has more hydrogen atoms. (extended statement) In an electrophilic addition to an alkene, the electrophile adds in such a way as to generate the most stable intermediate. (p. 3 24) HO OH + H CI H Markovnikov product c(:' Markovnikov orientation: An orientation of addition that obeys the original statement of Markovnikov's rule; one that gives the Markovnikov product. ( p. 3 25) anti-Markovnikov orientation: An orientation of addition that is the opposite of that pre dicted by the original statement of Markovnikov's rule; one that gives the anti-Markovnikov product. (p. 3 26) monomer One of the small molecules that bond together to form a polymer. (p. 3 63) organic synthesis T he preparation of desired organic compounds from readily available materials. (p. 3 67) oxidative cleavage T he cleavage of a carbon-carbon bond through oxidation. Carbon-carbon double bonds are commonly cleaved by ozonolysis/reduction or by warm, concentrated per manganate. (p. 3 60) oxymercuration T he addition of aqueous mercuric acetate to an alkene. (p. 3 33) "" / C=C / "" HO + H g(OAc)2 -C -C HgOAc I I I I + HOAc ozonolysis T he use of ozone, usually followed by reduction, to cleave a double bond. (p. 3 66) peroxide effect The reversal of orientation of HBr addition to aLkenes in the presence of per oxides. A free-radical mechanism is responsible for the peroxide effect. (p. 3 29) peroxyacid (peracid) A carboxylic acid with an extra oxygen atom and a peroxy ( - 0 - 0 - ) linkage. The general formula is RC03H. (p. 3 53) polymer A h igh-molecular-weight compound composed of many molecules of a smaller, simpler compound called the monomer. (p. 3 63) polymerization: T he reaction of monomer molecules to form a polymer. 3 76 C hapter 8: R eactions of Alkenes regioselective reaction A reaction in which one direction of bond making or bond breaking occurs preferentially over all other directions. For example, the addition of HCl is regioselec tive, predicted by Markovnikov's rule. Hydroboration-oxidation is regioselective because it consistently gives anti-Markovnikov orientation. (p. 325) retrosynthetic analysis A method of working backward to solve multistep synthetic prob lems. (p. 367) Simmons-Smith reaction A cyclopropanation of an alkene using the carbenoid reagent generated from diiodomethane and the zinc-copper couple. (p. 35 1 ) o CH212, Zn(Cu)) Simmons-Smith reaction (J> stereospecific reaction A reaction that converts different stereoisomers of the starting material into different stereoisomers of the product. (p. 341 ) I E ssentia l ProblemSolving Skil l s i n Chapter 8 1 . Predict the products of additions, oxidations, reductions, and cleavages of alkenes, including ( a) orientation of reaction (regiochemistry), (b) stereochemistry. 2. Propose logical mechanisms to explain the observed products of alkene reactions, including regiochemistry and stereochemistry. 3. Use retrosynthetic analysis to solve multistep synthesis problems with alkenes as reagents, intermediates, or products. 4. W hen more than one method is usable for a chemical transformation, choose the better method and explain its advantages. 5. Use clues provided by products of reactions such as ozonolysis to determine the struc ture of an unknown alkene. In studying these reaction-intensive chapters, students ask whether they should "memorize" all the reactions. Doing organic chemistry is like speaking a foreign language, and the reactions are our vocabulary. Without knowing the words, how can you construct sentences? Making flash cards often helps. In organic chemistry, the mechanisms, regiochemistry, and stereochemistry are our grammar. You must develop facility w ith the reactions, as you develop facility with the words and grammar you use in speaking. Problems and multistep syntheses are the sen tences of organic chemistry. You must practice c ombining all aspects of your vocabulary in solving these problems. Students who fail exams often do so because they have memorized the vocabulary, but they have not practiced doing problems. Others fail because they think they can do problems, but they lack the vocabulary. If you understand the reactions and can do the end-of-chapter problems without looking back, you should do well on your exams. S tudy Problems 8-46 Define each term, and give an example. ( a) d imerization (d) stereospecific addition ( g) Markovnikov addition (j) hydrogenation (m) heterogeneous catalysis ( p) hydroxylation (s) hydroboration (v) oxymercuration-demercuration (y) alkoxymercuration-demercuration (b) polymerization (e) syn addition (h) anti-Markovnikov addition (k) hydration (n) halogenation (q) epoxidation (t) alpha elimination (w) carbene addition (z) monomer (c) electrophilic addition (f) a nti addition ( i) peroxide effect (I) (0) ( r) (u) h omogeneous catalysis halohydrin oxidative cleavage beta elimination (x) cationic polymerization (aa) addition polymer S tudy Problems 8-47 3 77 Predict the major products of the fol lowing reactions, and give the structures of any intermediates. Include stereochem istry where appropriate. (oj Y'" ( I ) 03 (b) 0-0-- ------0> Br2 CCI4 (c) (d) (2) ( CH3)2S ( ' J V 0-- ( I ) BH3 . THF (2) H202, -OH R H Br ( g) PhC03H ;. ( h) -----? 054 HP2 GJ ( m J (OJ U (:( eo C H3C03H H +, HP ) ( kJ ----? H2 Pt ( ( I ) Hg(OAc)2' Hp nJ (2) N aBH4 U eo KMoO,, -OH , H', H,O (O V (;l U U ( IJ ------0> ) HCI ) ROOR KMn04' -OH (cold, dil) " (2) (CH3)2S ( I ) 03 (P J eo (b) C l2 H20 8-48 Propose mechanisms consistent with the following reactions. (a) Y'" ( c) H Br ROOR " ~ Br H2S04 Hp " ------;> H Br ( e) H Cl C H30H ) ~ + Br (d) OCH) + CI + () I + Br C HEr3 NaOH ) ()1 B' Br (f) B r2 LiCI in CH30H 3 Br OCH3 Jy Jy CI - (g) 2 H H c=c '\.. H 0 H+ H CH -c-c( C HI I H Q I -o , 3 78 8-49 C hapter 8 : Reactions of Alkenes S how how you would synthesize each compound using methylenecyclohexane as your starting material . (a) (d) (g) 8-50 Limonene is one of the compounds that give lemons their tangy odor. Show the structures of the products expected when limonene reacts with an excess of each of these reagents. (fo U et O H ( b) (,) ( h) () B' (c) V (fo Ck' CH; methylenecyclohexane OH OH (n OH Cl 0) l imonene 8 -51 ( a) borane in tetrahydrofuran, followed by basic hydrogen peroxide ( b) m -chloroperoxybenzoic acid (c) ozone, then dimethyl sulfide (d) a m ixture of osmic acid and hydrogen peroxide ( e) hot, concentrated potassium permanganate (1) peroxyacetic acid in water hydrogen and a platinum catalyst (g) (h) hydrogen bromide gas (i) h ydrogen bromide gas in a solution containing dimethyl peroxide (j) bromine water (k) c hlorine gas (I) mercuric acetate in methanol, followed by sodium borohydride (m) methylene iodide pretreated with the zinc-copper couple The structures of three monomers are shown. In each case, show the structure of the polymer that would result from polymerization of the monomer. Vinyl chloride is polymerized to "vinyl" plastics and PVC pipe. Tetrafluoroethylene polymerizes to Teflon , used as non-stick coatings and PTFE valves and gaskets. Acrylonitrile is polymerized to Orlon, used in sweaters and carpets. F" F / C = C" /F F acrylonitrile tetraftuoroethylene *8-52 8-53 8-54 W hen styrene (vinylbenzene) is commercially polymerized, about 1-3% of 1 ,4-divinylbenzene is often added to the styrene. The incorporation of some divinylbenzene gives a polymer with more strength and better resistance to organ ic solvents. Explain how a very small amount of divinylbenzene has a marked effect on the properties of the polymer. The cationic polymerization of isobutylene (2-methylpropene) is shown in Section 8- 1 6A. Isobutylene is often polymer ized under free-radical conditions. Propose a mechanism for the free-radical polymerization of isobutylene. Poly(ethyl acrylate) has the formula Give the structure of the ethyl acrylate monomer. S tudy Problems 8 -55 3 79 8 -56 8-57 8-58 D raw the structures of the following compounds, and determine which member of each pair is more reactive toward the addition of HBr. (a) propene or 2-methylpropene ( b) cyclohexene or I -methylcyclohexene (e) I -butene or 1 ,3-butadiene C yclohexene is dissolved in a solution of l ithium chloride in chloroform. To this solution is added one equivalent of bromine. The material isolated from this reaction contains primarily a mixture of trans- l ,2-dibromocyclohexane and trans- l -bromo-2-chlorocyclohexane. Propose a mechanism to show how these compounds are formed. Draw a reaction-energy diagram for the propagation steps of the free-radical addition of HBr to isobutylene. Draw curves representing the reactions leading to both the Markovnikov and the anti-Markovnikov products. Compare the values of fl eD and Ea for the rate-limiting steps, and explain why only one of these products is observed. Give the products expected when the following compounds are ozonized and reduced. ca) ef a a .. ( b) (b) Cd) JS) (e) 8-59 S how how you would make the following compounds from a suitable cyclic alkene. ( a) CH3 OH OH H C1 OH (d) ( e) 8 -60 8 -61 U nknown X , C SH9Br, does not react with bromine or with dilute KMn04. Upon treatment with potassium t-butoxide, X gives only one product, Y , C SH8. Unlike X, Y decolorizes bromine and changes KMn04 from purple to brown. Catalytic hydrogenation of Y g ives methylcyclobutane. Ozonolysis-reduction of Y gives dialdehyde Z , C SH802. Propose consistent structures for X, Y, and Z . I s there any aspect of the structure of X that is still unknown? O ne of the constituents of turpentine is a-pinene, formula CJOH 1 6. The fol lowing scheme (called a "road map") gives some reactions of a-pinene. Determine the structure of a-pinene and of the reaction products A through E . E O CQ OH A OH OH (f) c::( CO Br OCH3 C JOH 1S02 H 3 O' PhC0 3 H D I C JOH l6Brz Be, CC14 Br2 Hp C 1oH l 6O 8-62 The sex attractant of the housefly has the formula C23H46. When treated with warm potassium permanganate, this pheromone gives two products: CH3 ( CH2 ) 1 2COOH and CH3 (CH2hCOOH. Suggest a structure for this sex attractant. Explain which PaIt of the structure is uncertain. j -flr a-pinene C 1 oH l6 1 B ( I ) 03 (2) (CH3)2S C J OH 1 7OBr H ,SO, heat C IOH l SBr c 1 CH3 3 80 8-63 C hapter 8: R eactions of Alkenes I n contact with a platinum catalyst, an unknown alkene reacts with 3 equivalents of hydrogen gas to give l -isopropyl4-methylcyclohexane. When the unknown alkene is ozonized and reduced, the products are the following: o H-C-H II 0 H - C - CH2 -C-C-CH3 I 0 II 0 I o C H3- C - CH2- C - H II 0 I Deduce the structure of the unknown alkene. * 8-64 Propose a mechanism for the following reaction. o 8 -65 The two butenedioic acids are called fumaric acid (trans) and maleic acid (cis). 2,3-Dihydroxybutanedioic acid is called tartaric acid. H HOOC / " C=C / " COOH H HOOC H / " C=C " / COOH H HOOC-CH-CH-COOH OH I OH I fumaric acid maleic acid tartaric acid S how how you would convert (a) fumaric acid to ( ) -tartaric acid. ( b) fumaric acid to meso-tartaric acid. ( c) maleic acid to ( ) -tartaric a cid. (d) m aleic acid to meso-tartaric acid. 8 -66 8-67 T he compound BD3 is a deuterated form of borane. Predict the product formed when l -methylcyc1ohexene reacts with BD3 . THF, followed by basic hydrogen peroxide. A routine addition of HBr across the double bond of a vinylcyclopentane gave an unexpected rearranged product. Propose a mechanism for the formation of this product, and explain why the rearrangement occurs. H Br 8 -68 A n unknown compound decolorizes bromine in carbon tetrachloride, and it undergoes catalytic reduction to give decalin. When treated with warm, concentrated potassium permanganate, this compound gives cis-cyclohexane- I ,2-dicarboxylic acid and oxalic acid. Propose a structure for the unknown compound. u nknown compound (1 1J J, Co /JCd.) " /J o COOH cis cycl ohexane l 2 d icarboxyl ic acid co (X - -,decal in COOH + o H O - C - C - OH oxalic acid II 0 II ( further oxidation ) S tudy Problems * 8-69 3 81 Many enZymes catalyze reactions that are similar to reactions we might use for organic synthesis. Enzymes tend to be stereospecific in their reactions, and asymmetric induction is common. The following reaction, part of the tricarboxylic acid cycle of cell respiration, resembles a reaction we might use in the laboratory ; however, the enzyme-catalyzed reac tion gives only the (S) enantiomer of the product, malic acid. H "/ C C COOH fumarase /" HOOC H fumaric acid II HO COOH H / C C H2COOH (S)-malic acid COOD product in D20 I (a) W hat type of reaction does fumarase catalyze? (b) Is fumaric acid crural? Is malic acid crural? In the enzyme-catalyzed reaction, is the product (malic acid) optically active? "'8-70 (c) If we could run the preceding reaction in the laboratory using sulfuric acid as the catalyst, would the product (malic acid) be optically active? (d) Do you expect the fumarase enzyme to be a chiral molecule? (e) When the enzyme-catalyzed reaction takes place in D20, the only product is the stereoisomer just pictured. No enan tiomer or diastereomer of this compound is formed. Is the enzyme-catalyzed reaction a syn or anti addition? (0 Assume we found conditions to convert fumaric acid to deuterated malic acid using hydroboration with BD3 . THF, followed by oxidation with D202 and NaOD. Use Fischer projections to show the stereoisomer(s) of deuterated malic acid you would expect to be formed. (a) The following cyclization has been observed in the oxymercuration-demercuration of this unsaturated alcohol. Propose a mechanism for this reaction. ( I) Hg(OAc)2 (2) N aBH4 (b) Predict the product of formula C7HI3BrO from the reaction of this same unsaturated alcohol with bromine. Propose *8-71 8-72 a mechanism to support your prediction. An inexperienced graduate student treated 5-decene with borane in THF, placed the flask in a refrigerator, and left for a party. When he returned from the party, he discovered that the refrigerator was broken, and it had gotten quite warm inside. Although all the THF had evaporated from the flask, he treated the residue with basic hydrogen peroxide. To his surprise, he recovered a fair yield of I -decanol. Use a mechanism to show how this reaction might have occurred. (Hint: The addition of BH3 is reversible.) We have seen many examples where halogens add to alkenes with anti stereochemistry via the halonium ion mechanism. However, when l -phenylcyclohexene reacts with chlorine in carbon tetrachloride, a mixture of the cis and trans isomers of the product is recovered. Propose a mechanism, and explain this lack of stereospecificity. + I -phenylcycJohexene transand cis1 ,2-dichloro- l -phenylcycJohexane

Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Whitman - CHEM - 140
A l kynes9HH9- 1AlkynesIntroductionare hydrocarbons that contain carbon-carbon triple bonds. Alkynes are also called acetylenes because they are derivatives of acetylene, the simplest alkyne. H-C=C-Hacetylene ethyneC H3CH 2 - C - C - Hethyl acet
Whitman - CHEM - 140
Structure and Synthesis of Alcohols101 0-1IntroductionAlcohols are organic compounds containing hydroxyl ( - OH ) groups. They are some of the most common and useful compounds in nature, in industry, and around the house. The word alcohol is one of th
Whitman - CHEM - 140
R eacti ons of Alcohols11TABLE 1 1 -1Alcohols are important organic compounds because the hydroxyl group is easily converted to almost any other functional group. In Chapter 10, we studied reac tions that form alcohols. In this chapter, we seek to und
Whitman - CHEM - 140
I nfra red S pectrosco py a n d M a ss S pectrometry12fixed mirrorsamplelaser calibration beam 12-1detectorI n trod ucti o nOne of the most important tasks of organic chemistry is the determination of organic structures. When an interesting compou
Whitman - CHEM - 140
13N uclea r M ag n etic Reso n a n ce S pectrosco pyH( \ \ Ji nduced magnetic fieldcirculationHNuclear magnetic resonance spectroscopy (NMR) i s the most powerful tool available for organic structure determination. Like IR spectroscopy, NMR can be
Whitman - CHEM - 140
E th ers, E poxi d es, a nd S u lfidesl S-crown-6 with K+ s olvated14or aryl (benzene ring) groups. Like alcohols, ethers are related to water, with alkyl groups replacing the hydrogen atoms. In an alcohol, one hydrogen atom of water is replaced by an
Whitman - CHEM - 140
Conjugated Systems, Orbital Symmetry, and Ultraviolet SpectroscopyDouble bonds can interact with each other if they are separated by just one single bond. Such interacting double bonds are said to be conjugated. D ouble bonds with two or more single bond
Whitman - CHEM - 140
Aromatic Compounds16H -,. .In 1 825, Michael Faraday isolated a pure compound of boiling point 80C from the oily mixture that condensed from illuminating gas, the fuel burned in gaslights. Elemental analysis showed an unusually small hydrogen-to-carbon
Whitman - CHEM - 140
Reactions of Aromatic CompoundsEPM17of anisoleWith an understanding of the properties that make a compound aromatic, we now consider the reactions of aromatic compounds. A large part of this chapter is devoted to e lectrophilic aromatic substitution,
Whitman - CHEM - 140
18Ketones and AldehydesWheywiarle stofucentcompoundsance taoinorganithce chemistry, biochemistry, andin biololgy.because dy i cont ing C tofethe common rtyalpesmportcarbonyl compounds are l isted in Table detai Some of Csarbonylare constituentareofevery
Whitman - CHEM - 140
A m i n eselectrostatic potential map of trimethylamine191 9- 1 Introd uction( S)-coni inebonded toarecompounds. atom. As serve many functions orsome ofalkyl mostarylsuch as theorganic derivatives of class, arnines includein more organisms,important
Whitman - CHEM - 140
20C a rboxyl ic Aci d sThe combination of a carbonyl group and a hydroxyl on the same carbon atom is called a carboxyl group. Compounds containing the carboxyl group are distinctly acidic and are called c arboxylic acids. -C -O- Hcarboxyl group2 0- 1
Whitman - CHEM - 140
Whitman - CHEM - 140
nH aJHUa1d-NOS \f3d9-2Lh -E -O N8SI
Whitman - CHEM - 140
LUT ONS MANUALCalifornia Polytechnic State UniversityJan William SimekORC CHEMISTRYSIXTH EDITIONL. G. Wade, Jr.&quot;.JL.-':'Prentice HallUpper Saddle River, NJ07458
Whitman - CHEM - 140
Assistant Editor: Carole Snyder Project Manager: Kristen Kaiser Executive Editor: Nicole Folchetti Executive Managing Editor: Kathleen Schiaparelli Assistant Managing Editor: Becca Richter Production Editor: Kathryn O'Neill Supplement Cover Manager: Paul
Whitman - CHEM - 140
PrefaceTABLE OF CONTENTS.vSymbols and Abbreviations . viiChapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chap
Whitman - CHEM - 140
PREFACE Hints for Passing Organic ChemistryDo you want to pass your course in organic chemistry? Here is my best advice, based on over thirty years of observing students learning organic chemistry: Hint #1: Do the problems. It seems straightforward, but
Whitman - CHEM - 140
Some Web Stuff Prentice-Hall maintains a w eb site dedicated to the Wade text: try www.prenhall.com/wade. Two essential web sites providing spectra are listed on the bottom of p. 270. Acknowledgments No project of this scope is ever done alone. These are
Whitman - CHEM - 140
SYMBOLS AND ABBREVIA TIONSB elow is a list of symbols and abbreviations used in this Solutions Manual, consistent with those used in the textbook by Wade. (Do not expect all of these to make sense to you now. You will learn them throughout your study of
Whitman - CHEM - 140
Symbols and Abbreviations, continued SUBSTITUENT GROUPS, continued Ac Cy Ts an acetyl group: a cyclohexyl group: CH3 - C oII0-o II B oc a t-butoxycarbonyl group (amino acid and peptide chemistry): (CH3)3C -0 - C IItosyl, or p-toluenesulfonyl group: CH
Whitman - CHEM - 140
Symbols and Abbreviations, c ontinued REAGENTS AND SOLVENTS, c ontinued pyridinium chlorochromate, Cr03 HCI N PCCS ia2BHdisiamylboraneC H3 H H H CH 3 I I I I I H-C-C-B-C-C-H I I I I C H3 CH 3 CH3 CH3 ')THFtetrahydrofuranS PECTROSCOPY infrared spect
Whitman - CHEM - 140
C HAPTER I-INTRODUCTION A ND REVIEW1-11-2P 1 s 2 2s 2 2p6 3s 2 3px 1 3py 1 3pz 1 1 s 2 2s2 2p6 3s 1 S 1 s 2 2s 2 2p6 3s 2 3px 2 3py 13pz 1 Mg 1 s 2 2s2 2p6 3s 2 Is 2 2s2 2p6 3s 2 3px 1 CI 1 s 2 2s2 2p6 3s 2 3px2 3py2 3pz 1 AI Ar 1 s 2 2s 2 2p6 3s 2 3px
Whitman - CHEM - 140
1-5The symbols &quot; 8+&quot; and &quot; 8-&quot; indicate bond polarity by showing partial charge. (In the arrow symbolism, the arrow should point to the partial negative charge.) (a) C -C 1(b) (g)C- O8+(c) (h)C-N N-S8-N-O8-8+(i) N - B8-8+1-6Non-zero forma
Whitman - CHEM - 140
1-7continued. .(c) :O-N=O (d) (e)I I+O=N-O: .+.-H-C=C-C-HHHHI. H-C-C=C-HHHHIIIH - C = C - C - H . . H - C - C=C - H I I I I I I HHH HHH (f) Sulfur can have up to 12 e lectrons around it because it has d orbitals accessible . :0 : : 0: :
Whitman - CHEM - 140
1 -8 continued H H H + + + \+ \ \ C - C = N - O: C = C - N - O: C=C - N=O (b) / / / 1 1 1 11 1 1 H H H H :0 : H :0 : H :0 : minor major major These two forms have equivalent energy and are major because they have full octets, more bonds, and less charge s
Whitman - CHEM - 140
1 -8 continued :0 : II (h) H - C - N - HI.II-I . . .. :0 : + I H - C =N H-I(no charge separation)majorHm inorH1 -9 Your Lewis structures may appear different from these. As long as the atoms are connected in the same order and by the same type
Whitman - CHEM - 140
1 -10 continued(e )(g)H H :0 : H H H (h) :0 : H I II I I II I I I H ' /C:-., /,C-C - H H-C - C-C-C-C-H c &quot;c I I I I I I H II H HHH C C, H/ , C 'i H I H 1 -1 1 There is often more than one correct way to write condensed structural formulas. You must oft
Whitman - CHEM - 140
1 - 1 2 continued (b) 32.0 g C 1 2.0 g/mole 2 .67 moles C 1.34 moles 1.99 2 C 6.67 g H 5H 4 .93 6 .60 m oles H -:- 1.34 m oles l.01 g/mole IS.7 g N 1 4.0 g/mole 1 .34 moles N -:- 1.34 moles 1 N 4 2.6 g a . 1 6.0 g/mole - 2.66 moles a -;- 1.34 moles 1 .99
Whitman - CHEM - 140
1 -13 1 mole HEr (a) 5 .00 g HBr x 80.9 g HBr 0.06 1 8 moles HEr0.0618 moles=0 .06 1 8 moles HBr0.06 1 8 moles H 30 + (100% dissociated) x=1 00 mL-H30 +10001LmL=0.6 1 8 moles H30 + 1 L solution=pH=10gIO [H30+]-logl o (0.6 1 8)(b)0 .03
Whitman - CHEM - 140
1-15 (a) HCOOH stronger acid pKa 3.76 (b) CH3COOweaker base (c) CH30H stronger acid pKa 1 5.5+-CN stronger base+-.-HCOOweaker base+H CN FAVORS weaker PRODUCTS acid pK a 9.22+C H30H weaker acid pK a 1 5 .5 NaNH2 stronger base.-CH 3 COOH strong
Whitman - CHEM - 140
1 -l7 In Solved Problem 1-4, the structures of ethanol and methylamine are shown to be similar to methanol and ammonia, respectively. We must infer that their acid-base properties are also similar. (a) This problem can be viewed in two ways. 1 ) Quantitat
Whitman - CHEM - 140
1 -1 8 continued (d) - . 0 Na+ :O - H + H-S-H . stronger acid stronger base (e) H /' I CH3 - H + C H3-O: stronger base H stronger acid+ &quot; _ equilibrium favors PRODUCTS _-. H - O - H + Na+ : S-H conjugate acid conjugate base weaker base weaker acid CH3
Whitman - CHEM - 140
1 - 19 continued (e) Bronsted-Lowry-c-proton transferC H - C - C - H +/.- H = H -. - C - H .- H - b = b - H :O - ) k base acid. . . :(f)C H3 - - H H base.+C H3 - S=!: acid()i:H :0:+:Cl :..-_+H - O-H.1 -20 Learning organic chemistry
Whitman - CHEM - 140
1 -24Cl: N (b) P (a) N . . . p . : C l I C l: I I c : . .!'&quot; I C l: :Cl: :Cl: :Cl: :CI: C ANNOT EXIST NCls violates the octet rule; nitrogen can have no more than eight electrons (or four atoms) around it. Phosphorus, a third-row element, can have more t
Whitman - CHEM - 140
1-26 continued H - O: H :0:. .( c)H-C = C-C-C-C-O-H HIIIIIHIHHII1 -27 In each set below, the second structure is a more correct line formula. Since chemists are human (surprise!), they will take shortcuts where possible; the first structure i
Whitman - CHEM - 140
1 -28 continued (c) There are several other possibilities as well. Your answer may be correct even if it does not appear here. Check with others in your study group. HH H HHH HHH I I I I I I I I I :O - C-C - O - C - H :O - C - C-C - O: :O-C-C - C - H I I
Whitman - CHEM - 140
HH (b) H . _ _H \ HI H . C ,0 0/ C &quot; / N . H H I HH H H :0: III&quot; H - N - C-C-C-C-O - H IIII HHHH(e)H H (g) I \ C-C \ :0: H, &quot; II \ H - C-C C-S=O / \ / I H C = C :0: I \ I HH HHH \I /C, ,.H C, :N II H I H H\ C I H / H - C , C - H H HH I I,. I \ CH H -
Whitman - CHEM - 140
1 -34 Non-zero formal charges are shown by the atoms. + + (a) H - C = N = N : H - C - N - N:H ,H/ H 1 (b) H C 1 1 1 , H H H - C -+N - O: / H H I H H C I 1 + + 1 (c) H - C = C-C - H (d) H - C-N=O (e) H - C-O - C - H H /1' H H 1 I +1 111 11 HCH HHH H : 0:
Whitman - CHEM - 140
1 -37 continued (c)&lt;&gt;C H24(d) (e)0+ . 0-0 &lt;&gt;+0 /; :4+ O. .CH,-+C H2(f)(g) (h)(i) C H 3 - C = C - C = C - C - CH3 I I I I I HHHHHQ:0 :&lt; &gt; : -o + (&gt; CN- - CN+-H + 0 . 0 0+0. .0/ +CH2+()C H,0 .-/--:0=0 .-H4H.0+. o ._
Whitman - CHEM - 140
1-38 . (a) O=S-O:+.:O-S=O++...O=S=O.(b) 0=0-0:+.(c) The last resonance form of S0 2 has no equivalent form in 03. Sulfur, a third row element, can have more than eight electrons around it because of d orbitals, whereas oxygen, a second row e
Whitman - CHEM - 140
1 -40 continued (c) :0 : :0: II -. . II C H3 - C - ? - C - CH3 minorH. : 0:. .1-1. . .C H3 - C = C - C - CH3I:0:\ -majorHIII:0:.1-1. . ..C H3 - C - C = C - CH3II. :0:n egative charge on electronegative atoms-equal energy (d)+ + -+ C H3
Whitman - CHEM - 140
1 -4 1 continued (c) H - C - CH3 \ Hno resonance stabi li zati onH - C - C - N: \ H..H - C = C = N: \ H ./ CH c' 2 \H C .:/+more stable-reson ance stabi l i zedmore stable-resonance stabilized (e) CH3 - N - CH 3 1 CH3 - C - CH 3+ .I-I . . - .CC
Whitman - CHEM - 140
1 -45 The newly formed bond is shown in bold. . ('. . CH 3 - - CH3 + (a) CH 3 - .O: CH 3-CI. : . . electrophile nucleophile Lewis acid Lewis base (b) CH3 - O - CH 3 + H - O - H +1 ) H 3 c ucleoPhile Lewis base e I ectrophIO Ie Lewis acid (c) CH 3 - ? - CH
Whitman - CHEM - 140
(h)1 -45 continu F-B - F I F e lectrophile Lewis acidCH2 = CH2 nucleophile Lewis baseF + -I F - f - CH2 - CH2 F+( i) B F3 - CH2 - CH2 electrophile Lewis acid 1 -46 (a) H2S04++C H2 = CH2 nucleophile Lewis base. B F3 - CH2 - CH2 - CH2 - CH2+C H3C
Whitman - CHEM - 140
1 -48 From the amounts of CO2 and H20 generated, the milligrams of C and H in the original sample can be determined, thus giving by difference the amount of oxygen in the 5.00 mg sample. From these values, the empirical formula and empirical weight can be
Whitman - CHEM - 140
C HAPTER 2-STRUCTURE AND PROPERTIES OF ORGANIC MOLECULES2 - 1 The fundamental principle of organic chemistry is that a molecule's chemical and physical properties depend on the molecule's structure: the structure-function or structure-reactivity correlat
Whitman - CHEM - 140
2 -5 (a) linear, bond angle 1 80 (b) all atoms are sp3 ; tetrahedral geometry and bond angles of 1 09 around each atom not a bqnd-;-shows not a bond-s how s '. H H lone Rair coming \ , lbon pair gomg ehmd paper &quot;'- / out 01 paper I I H , . 0 , . H H-C-O
Whitman - CHEM - 140
2 -6 Carbon-2 is sp hybridized. If the p orbitals making the pi bond between C-l and C-2 are in the plane of the paper (putting the hydrogens in front of and behind the paper), then the other p orbital on C-2 must be perpendicular to the plane of the pape
Whitman - CHEM - 140
2-7 continued (c) the nitrogen and the carbon bonded to it are sp hybridized; the left carbon is sp2H \ 1 200 c /H( .N.:1 800HII H /c-cN:oil(d) the boron and the oxygens bonded to it are sp2 hybridizedH - O: H \ I B-O: I H - O. : .II.PH-
Whitman - CHEM - 140
2-8 Very commonly in organic chemistry, we have to determine whether two structures are the same or different, and if they are different, what structural features are different. In order for two structures to be the same, all bonding connections have to b
Whitman - CHEM - 140
2-1 1 M odels will be helpful here. (a) cis-trans isomers-the first is trans, the second is cis (b) constitutional isomers-the carbon skeleton is different (c) constitutional isomers-the bromines are on different carbons in the first structure, on the sam
Whitman - CHEM - 140
2 -14 continued (c)L &quot; c / n et dipole 0 F =F(d)( e)0, . 0 'x o &quot; &quot; o ( g)( i)ro or 0; 13 1 1 C /H ' CH3 n etn et/ N ,,/Q1 oeach end oxygen has one-half negative charge as it is the composite of two resonance forms; see n et solution to 1 -
Whitman - CHEM - 140
( hydrogen bonds shown as wavy bond)2- 1 7 ( a) (CH3hCHCH2CH2CH(CH3h has less branching and boils at a higher temperature than (CH3hCC(CH3h . (b) CH3(CH2)sCH2 0 H can form hydrogen bonds and will boil at a much higher temperature than CH3(CH2) 6CH3 which
Whitman - CHEM - 140
( a)2-19HHHHH I I I I I H-C-C-C-C-C-H I I I I I HHHHH( d)a lkane (Usually, we use the tenn &quot;alkane&quot; only when no other groups are present.)H I H-C-C H-C H'( b)H HH I I I H-C-C=C-C-C-H I I I I I HHHHHa lkene( c)HHH I I I H - C - C : C - C - C -
Whitman - CHEM - 140
2 -2 1( a)H1 1H1 1011 1H1 1HH-C-C-C-N-C-H H H H H( b)H-C-C-N-C-C-H H11H1 1 1H1 1H1 1)H1 1H1 1011H1 1H-C-C-C-O-C-H H H'HHHHa mideHa mineHI IHH1 1C'HHe ster( e)H-C-C-O-C-C-H HIIHI 1HI I(f)H1 1H1 1
Whitman - CHEM - 140
2-22(d)c ontinuedII H C - NHa mide( this also looks like an aldeh yde, b u t an amide h as higher &quot;priority &quot; as y o u wi l l s e e l ater)g( e)CH6 c:3)Ha mineS uggested by student R i c h ard King: R i s the symbol that organic chemists use t
Whitman - CHEM - 140
2 -24 c ontinued(g) aldehyde: contains a carbonyl group with a hydrogen on one side HHa H-C-C-C-H H HIIIIII( h) aromatic hydrocarbon : a cyclic hydrocarbon with alternating double and single bonds( i ) carboxylic acid: contains a c arbonyl group w
Whitman - CHEM - 140
2 -27 c ontinued (h)i n front o f the p lane of the p aperbeh i nd the p l ane of the pape;(i) (j cfw_)H/a&quot;'&quot;Cangles around sp 3 atom angles around sp 2 atoms1 09 1 20HI, ,HH.-/2 -28For cl ari ty in these picture s , b o n d s between
Whitman - CHEM - 140
( i)s p 3 _sp 3I OJ - &lt; o, f&quot;' ;H_s p 3 _sp 2sp 2 -ssp 2 _ sp 3 2 -29 The second resonance form of formamide is a minor but significant resonance contri butor. It shows that the nitrogen-carbon bond h as some double bond character, requiring that
Whitman - CHEM - 140
2 -32( a) cis (b) The cop l an ar atoms i n the structures to the left and below are marked w i th asterisks.(c)HH JC*C - C ,&quot; ,&quot;T here are sti l l six copl anar atoms.2 -33QD M(d), ,eH,CH]H*transC ollinear atoms are marked with asteri s