401 CHAPTER 16 ETHERS, EPOXIDES, AND SULFIDES SOLUTIONS TO TEXT PROBLEMS 16.1 (b) Oxirane is the IUPAC name for ethylene oxide. A chloromethyl group (ClCH 2 @) is attached to position 2 of the ring in 2-(chloromethyl)oxirane. This compound is more commonly known as epichlorohydrin. (c) Epoxides may be named by adding the prefix epoxy to the IUPAC name of a parent compound, specifying by number both atoms to which the oxygen is attached. 16.2 1,2-Epoxybutane and tetrahydrofuran both have the molecular formula C 4 H 8 O—that is, they are constitutional isomers—and so it is appropriate to compare their heats of combustion directly. Angle strain from the three-membered ring of 1,2-epoxybutane causes it to have more internal energy than tetrahydrofuran, and its combustion is more exothermic. 1,2-Epoxybutane; heat of combustion 2546 kJ/mol (609.1 kcal/mol) H 2 C CHCH 2 CH 3 Tetrahydrofuran; heat of combustion 2499 kJ/mol (597.8 kcal/mol) O O 1-Butene CH 2 CH 3 CH 2 CH 3,4-Epoxy-1-butene CHCH CH 2 H 2 C O H 2 C CH 2 Oxirane H 2 C CHCH 2 Cl 2-(Chloromethyl)oxirane O O 32 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 402 ETHERS, EPOXIDES, AND SULFIDES 16.3 An ether can function only as a proton acceptor in a hydrogen bond, but an alcohol can be either a proton acceptor or a donor. The only hydrogen bond possible between an ether and an alcohol is therefore the one shown: 16.4 The compound is 1,4-dioxane; it has a six-membered ring and two oxygens separated by CH 2 —CH 2 units. 16.5 Protonation of the carbon–carbon double bond leads to the more stable carbocation. Methanol acts as a nucleophile to capture tert-butyl cation. Deprotonation of the alkyloxonium ion leads to formation of tert-butyl methyl ether. 16.6 Both alkyl groups in benzyl ethyl ether are primary, thus either may come from the alkyl halide in a Williamson ether synthesis. The two routes to benzyl ethyl ether are 16.7 (b) A primary carbon and a secondary carbon are attached to the ether oxygen. The secondary car- bon can only be derived from the alkoxide, because secondary alkyl halides cannot be used in the preparation of ethers by the Williamson method. The only effective method uses an allyl halide and sodium isopropoxide. Elimination will be the major reaction of an isopropyl halide with an alkoxide base. H11001(CH 3 ) 2 CHONa Sodium isopropoxide Allyl bromide H 2 C CHCH 2 Br CHCH 2 OCH(CH 3 ) 2 NaBrH11001 Allyl isopropyl ether Sodium bromide H 2 C C 6 H 5 CH 2 Br CH 3 CH 2 ONaH11001 Benzyl bromide Sodium ethoxide C 6 H 5 CH 2 OCH 2 CH 3 NaBrH11001 Benzyl ethyl ether Sodium bromide C 6 H 5 CH 2 ONa CH 3 CH 2 BrH11001 Sodium benzyloxide Bromoethane C 6 H 5 CH 2 OCH 2 CH 3 NaBrH11001 Benzyl ethyl ether Sodium bromide H11001H11001(CH 3 ) 3 C OCH 3 H11001 H O H CH 3 H 2 OCH 3 H11001 (CH 3 ) 3 COCH 3 tert-Butyl methyl ether H11001(CH 3 ) 2 CCH 3 H11001 (CH 3 ) 3 C OCH 3 H H11001 O H CH 3 H11001 H H11001 (CH 3 ) 2 CCH 2 2-Methylpropene (CH 3 ) 2 CCH 3 H11001 tert-Butyl cation OO 1,4-dioxane (‘‘6-crown-2”) O R R Ether OH R Alcohol Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) Here the ether is a mixed primary–tertiary one. The best combination is the one that uses the primary alkyl halide. The reaction between (CH 3 ) 3 CBr and C 6 H 5 CH 2 O H11002 is elimination, not substitution. 16.8 16.9 (b) If benzyl bromide is the only organic product from reaction of a dialkyl ether with hydrogen bromide, then both alkyl groups attached to oxygen must be benzyl. (c) Since 1 mole of a dihalide, rather than 2 moles of a monohalide, is produced per mole of ether, the ether must be cyclic. 16.10 As outlined in text Figure 16.4, the first step is protonation of the ether oxygen to give a dialkylox- onium ion. In the second step, nucleophilic attack of the halide ion on carbon of the oxonium ion gives 4-iodo-1-butanol. The remaining two steps of the mechanism correspond to those in which an alcohol is converted to an alkyl halide, as discussed in Chapter 4. H11001 Water H 2 OI H11002 H11001 H11001I OH 2 1,4-Diiodobutane I I H11001 I H11002 Hydrogen iodide H I H11001 4-Iodo-1-butanol I OH H11001 OH 2 I H11001I Iodide ion Dialkyloxonium ion HO H11001 H11002 4-Iodo-1-butanol I OH H11001H11001O Tetrahydrofuran I Iodide ion H I Hydrogen iodide Dialkyloxonium ion HO H11001 H11002 2HBr heat H11001 O Tetrahydropyran BrCH 2 CH 2 CH 2 CH 2 CH 2 Br 1,5-Dibromopentane H 2 O Water Dibenzyl ether C 6 H 5 CH 2 OCH 2 C 6 H 5 2C 6 H 5 CH 2 Br H11001 Benzyl bromide Water HBr heat H 2 O Diethyl ether CH 3 CH 2 OCH 2 CH 3 H11001H11001 Oxygen 6O 2 Carbon dioxide 4CO 2 Water 5H 2 O KBrH11001(CH 3 ) 3 COCH 2 C 6 H 5 Benzyl tert-butyl ether Potassium bromide H11001(CH 3 ) 3 COK Potassium tert-butoxide C 6 H 5 CH 2 Br Benzyl bromide ETHERS, EPOXIDES, AND SULFIDES 403 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 16.11 The cis epoxide is achiral. It is a meso form containing a plane of symmetry. The trans isomer is chiral; its two mirror-image representations are not superposable. Neither the cis nor the trans epoxide is optically active when formed from the alkene. The cis epox- ide is achiral; it cannot be optically active. The trans epoxide is capable of optical activity but is formed as a racemic mixture because achiral starting materials are used. 16.12 (b) Azide ion is a good nucleophile, reacting readily with ethylene oxide to yield 2-azidoethanol. (c) Ethylene oxide is hydrolyzed to ethylene glycol in the presence of aqueous base. (d) Phenyllithium reacts with ethylene oxide in a manner similar to that of a Grignard reagent. (e) The nucleophilic species here is the acetylenic anion CH 3 CH 2 C>C: H11002 , which attacks a carbon atom of ethylene oxide to give 3-hexyn-1-ol. 16.13 Nucleophilic attack at C-2 of the starting epoxide will be faster than attack at C-1, because C-1 is more sterically hindered. Compound A, corresponding to attack at C-1, is not as likely as com- pound B. Compound B not only arises by methoxide ion attack at C-2 but also satisfies the stereo- chemical requirement that epoxide ring opening take place with inversion of configuration at the site of substitution. Compound B is correct. Compound C, although it is formed by methoxide substitution at the less crowded carbon of the epoxide, is wrong stereochemically. It requires Ethylene oxide H 2 C CH 2 NH 3 CCH 2 CH 3 NaC 3-Hexyn-1-ol (48%) CH 3 CH 2 C CCH 2 CH 2 OH O C 6 H 5 CH 2 CH 2 OH 2-Phenylethanol 1. C 6 H 5 Li, diethyl ether 2. H 3 O H11001 Ethylene oxide H 2 C CH 2 O HOCH 2 CH 2 OH Ethylene glycol NaOH H 2 O Ethylene oxide H 2 C CH 2 O Ethylene oxide N 3 CH 2 CH 2 OH 2-Azidoethanol NaN 3 ethanol–water H 2 C CH 2 O N N ] H11002 [ N CH 3 H 3 C HH O cis-2,3-Epoxybutane (Plane of symmetry passes through oxygen and midpoint of carbon–carbon bond.) CH 3 H H 3 CH O H H 3 C HC 3 O Nonsuperposable mirror-image (enantiomeric) forms of trans-2,3-epoxybutane Plane of symmetry 404 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website substitution with retention of configuration, which is not the normal mode of epoxide ring opening. 16.14 Acid-catalyzed nucleophilic ring opening proceeds by attack of methanol at the more substituted carbon of the protonated epoxide. Inversion of configuration is observed at the site of attack. The correct product is compound A. The nucleophilic ring openings in both this problem and Problem 16.13 occur by inversion of configuration. Attack under basic conditions by methoxide ion, however, occurs at the less hindered carbon of the epoxide ring, whereas attack by methanol under acid-catalyzed conditions occurs at the more substituted carbon. 16.15 Begin by drawing meso-2,3-butanediol, recalling that a meso form is achiral. The eclipsed confor- mation has a plane of symmetry. Epoxidation followed by acid-catalyzed hydrolysis results in anti addition of hydroxyl groups to the double bond. trans-2-Butene is the required starting material. Osmium tetraoxide hydroxylation is a method of achieving syn hydroxylation. The necessary start- ing material is cis-2-butene. 16.16 Reaction of (R)-2-octanol with p-toluenesulfonyl chloride yields a p-toluenesulfonate ester (tosylate) having the same configuration; the stereogenic center is not involved in this step. Reaction OO C H 3 C CH 3 HH C cis-2-Butene via meso-2,3-Butanediol (CH 3 ) 3 COOH, OsO 4 (cat) (CH 3 ) 3 COH, HO H11002 C H HO OH CH 3 H CH 3 H CH 3 C H CH 3 O CC O Os trans-2-Butene CC H 3 C H H CH 3 trans-2,3-Epoxybutane O CC meso-2,3-Butanediol H 3 O H11001 C H CH 3 CH 3 H C H 3 C H OH H HO CH 3 CH 3 COOH O meso-2,3-Butanediol C H HO OH CH 3 H CH 3 C CH 3 OH H11002H H11001 O H H11001 CH 3 HOO CH 3 H11001 H HH Protonated form of 1-methyl-1,2- epoxycyclopentane Compound A OCH 3 H CH 3 CH 3 HO ETHERS, EPOXIDES, AND SULFIDES 405 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website of the tosylate with a nucleophile proceeds by inversion of configuration in an S N 2 process. The product has the S configuration. 16.17 Phenyl vinyl sulfoxide lacks a plane of symmetry and is chiral. Phenyl vinyl sulfone is achiral; a plane of symmetry passes through the phenyl and vinyl groups and the central sulfur atom. 16.18 As shown in the text, dodecyldimethylsulfonium iodide may be prepared by reaction of dodecyl methyl sulfide with methyl iodide. An alternative method is the reaction of dodecyl iodide with dimethyl sulfide. The reaction of a sulfide with an alkyl halide is an S N 2 process. The faster reaction will be the one that uses the less sterically hindered alkyl halide. The method presented in the text will proceed faster. 16.19 The molecular ion from sec-butyl ethyl ether can also fragment by cleavage of a carbon–carbon bond in its ethyl group to give an oxygen-stabilized cation of mH20862z 87. 16.20 All the constitutionally isomeric ethers of molecular formula C 5 H 12 O belong to one of two general groups: CH 3 OC 4 H 9 and CH 3 CH 2 OC 3 H 7 . Thus, we have CH 3 CH 2 OCH 2 CH 2 CH 3 and CH 3 CH 2 OCH(CH 3 ) 2 Ethyl propyl ether Ethyl isopropyl ether CH 3 OCH 2 CH(CH 3 ) 2 Isobutyl methyl ether CH 3 OC(CH 3 ) 3 tert-Butyl methyl ether CH 3 OCH 2 CH 2 CH 2 CH 3 Butyl methyl ether CH 3 OCHCH 2 CH 3 CH 3 sec-Butyl methyl ether CH 3 CH 2 CHCH 2 CH 3 CH 3 O H11001 CH 3 CH 2 H11001 CHCH 2 CH 3 CH 3 O H11001 m/z 87 Dimethyl sulfide Dodecyl iodide Dodecyldimethylsulfonium iodide (CH 3 ) 2 SCH 3 (CH 2 ) 10 CH 2 IH11001 CH 3 (CH 2 ) 10 CH 2 S(CH 3 ) 2 I H11002 H11001 Phenyl vinyl sulfoxide (chiral) S H11001 C 6 H 5 O H11002 CHH 2 C Phenyl vinyl sulfone (achiral) S C 6 H 5 O H11002 O H11002 CHH 2 C 2H11001 H11001 C 6 H 5 S H11002 Na H11001 Sodium benzenethiolate CC 6 H 5 S (S)-1-Methylheptyl phenyl sulfide(R)-1-Methylheptyl tosylate O SCH 3 O O H CH 3 C H H 3 C CH 3 (CH 2 ) 4 CH 2 CH 2 (CH 2 ) 4 CH 3 H11001OHC H H 3 C CH 3 (CH 2 ) 4 CH 2 (R)-2-Octanol Cl CH 3 O S O p-Toluenesulfonyl chloride OC H H 3 C CH 3 (CH 2 ) 4 CH 2 CH 3 O S O (R)-1-Methylheptyl tosylate 406 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website These ethers could also have been named as “alkoxyalkanes.” Thus, sec-butyl methyl ether would become 2-methoxybutane. 16.21 Isoflurane and enflurane are both halogenated derivatives of ethyl methyl ether. Isoflurane: Enflurane: 16.22 (a) The parent compound is cyclopropane. It has a three-membered epoxide function, and thus a reasonable name is epoxycyclopropane. Numbers locating positions of attachment (as in “1,2-epoxycyclopropane”) are not necessary, because no other structures (1,3 or 2,3) are pos- sible here. (b) The longest continuous carbon chain has seven carbons, and so the compound is named as a derivative of heptane. The epoxy function bridges C-2 and C-4. Therefore is 2-methyl-2,4-epoxyheptane. (c) The oxygen atom bridges the C-1 and C-4 atoms of a cyclohexane ring. (d) Eight carbon atoms are continuously linked and bridged by an oxygen. We name the com- pound as an epoxy derivative of cyclooctane. 1 5 2 3 4 6 7 8 O 1,5-Epoxycyclooctane O 1 2 3 5 4 6 1,4-Epoxycyclohexane CH 2 CH 2 CH 3 H 3 C H 3 C 1 2 3 4 567 O O Epoxycyclopropane Cl C F H F F C O CHF 2 2-Chloro-1,1,2-trifluoroethyl difluoromethyl ether FC F Cl F CH O CHF 2 1-Chloro-2,2,2-trifluoroethyl difluoromethyl ether ETHERS, EPOXIDES, AND SULFIDES 407 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 16.23 (a) There are three methyl-substituted thianes, two of which are chiral. (b) The locants in the name indicate the positions of the sulfur atoms in 1,4-dithiane and 1,3,5- trithiane. (c) Disulfides possess two adjacent sulfur atoms. 1,2-Dithiane is a disulfide. (d) Two chair conformations of the sulfoxide derived from thiane are possible; the oxygen atom may be either equatorial or axial. 16.24 Intramolecular hydrogen bonding between the hydroxyl group and the ring oxygens is possible when the hydroxyl group is axial but not when it is equatorial. 16.25 The ethers that are to be prepared are First examine the preparation of each ether by the Williamson method. Methyl propyl ether can be prepared in two ways: CH 3 Br Methyl bromide H11001 CH 3 CH 2 CH 2 ONa Sodium propoxide CH 3 OCH 2 CH 2 CH 3 Methyl propyl ether CH 3 ONa Sodium methoxide H11001 CH 3 CH 2 CH 2 Br 1-Bromopropane CH 3 OCH 2 CH 2 CH 3 Methyl propyl ether andCH 3 OCH 2 CH 2 CH 3 Methyl propyl ether CH 3 OCH(CH 3 ) 2 Isopropyl methyl ether CH 3 CH 2 OCH 2 CH 3 Diethyl ether O O HO Less stable conformation; no intramolecular hydrogen bonding O O O H More stable conformation; stabilized by hydrogen bonding S O H11002 S O H11001 H11002 H11001 S S 1,2-Dithiane S S 1,4-Dithiane S SS 1,3,5-Trithiane S CH 3 2-Methylthiane (chiral) S CH 3 3-Methylthiane (chiral) S CH 3 4-Methylthiane (achiral) 408 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Either combination is satisfactory. The necessary reagents are prepared as shown. Isopropyl methyl ether is best prepared by the reaction The reaction of sodium methoxide with isopropyl bromide will proceed mainly by elimination. Methyl bromide is prepared as shown previously; sodium isopropoxide can be prepared by adding sodium to isopropyl alcohol. Diethyl ether may be prepared as outlined: 16.26 (a) Secondary alkyl halides react with alkoxide bases by E2 elimination as the major pathway. The Williamson ether synthesis is not a useful reaction with secondary alkyl halides. (b) Sodium alkoxide acts as a nucleophile toward iodoethane to yield an alkyl ethyl ether. (R)-2-Ethoxybutane O H11002 Na H11001 H11001 CH 3 CH 2 I C OCH 2 CH 3 C CH 3CH 3 CH 2 H CH 3CH 3 CH 2 H H11001H11001H11001CH 3 CH 2 CHCH 3 ONa Sodium 2-butanolate Br Bromocyclohexane CH 3 CH 2 CHCH 3 OH 2-Butanol Cyclohexene NaBr Sodium bromide CH 3 CH 2 ONa H11001 CH 3 CH 2 Br Sodium ethoxide Ethyl bromide CH 3 CH 2 OCH 2 CH 3 H11001 Diethyl ether NaBr Sodium bromide CH 3 CH 2 OH Ethanol CH 3 CH 2 Br Ethyl bromide PBr 3 (or HBr) CH 3 CH 2 OH Ethanol CH 3 CH 2 ONa Sodium ethoxide Na H11001CH 3 Br Methyl bromide (CH 3 ) 2 CHONa Sodium isopropoxide CH 3 OCH(CH 3 ) 2 Isopropyl methyl ether CH 3 CH 2 CH 2 OH 1-Propanol CH 3 CH 2 CH 2 ONa Sodium propoxide Na CH 3 OH Methanol CH 3 Br Methyl bromide PBr 3 (or HBr) CH 3 CH 2 CH 2 OH 1-Propanol CH 3 CH 2 CH 2 Br 1-Bromopropane PBr 3 (or HBr) CH 3 OH Methanol CH 3 ONa Sodium methoxide Na ETHERS, EPOXIDES, AND SULFIDES 409 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The ether product has the same absolute configuration as the starting alkoxide because no bonds to the stereogenic center are made or broken in the reaction. (c) Vicinal halohydrins are converted to epoxides on being treated with base. (d) The reactants, an alkene plus a peroxy acid, are customary ones for epoxide preparation. The reaction is a stereospecific syn addition of oxygen to the double bond. (e) Azide ion is a good nucleophile and attacks the epoxide function. Substitution occurs at carbon with inversion of configuration. The product is trans-2-azidocyclohexanol. ( f ) Ammonia is a nucleophile capable of reacting with epoxides. It attacks the less hindered car- bon of the epoxide function. Aryl halides do not react with nucleophiles under these conditions, and so the bromine sub- stituent on the ring is unaffected. (g) Methoxide ion attacks the less substituted carbon of the epoxide ring with inversion of configuration. 1-Benzyl-1,2- epoxycyclohexane 1-Benzyl-trans-2- methoxycyclohexanol (98%) CH 2 C 6 H 5 O OCH 3 H11002 OCH 3 CH 2 C 6 H 5 OH 2-(o-Bromophenyl)-2- methyloxirane 1-Amino-2-(o-bromophenyl)- 2-propanol C H 3 C OH CH 2 NH 2 Br H 3 C Br O NH 3 methanol 1,2-Epoxycyclohexane trans-2- Azidocyclohexanol (61%) H H O OH H N 3 H NaN 3 dioxane–water H11001H11001COH O Benzoic acid(Z)-1-Phenylpropene CC HH CH 3 cis-2-Methyl-3- phenyloxirane O HH CH 3 Peroxybenzoic acid COOH O CH 3 CH 2 CHCH 2 Br OH CH 3 CH 2 CH CH 2 NaOH Br O CH 3 CH 2 CH CH 2 H11002 1-Bromo-2-butanol 1,2-Epoxybutane O 410 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (h) Under acidic conditions, substitution is favored at the carbon that can better support a positive charge. Aryl substituents stabilize carbocations, making the benzylic position the one that is attacked in an aryl substituted epoxide. (i) Tosylate esters undergo substitution with nucleophiles such as sodium butanethiolate. ( j) Nucleophilic substitution proceeds with inversion of configuration. 16.27 Oxidation of 4-tert-butylthiane yields two sulfoxides that are diastereomers of each other. Oxidation of both stereoisomeric sulfoxides yields the same sulfone. 16.28 Protonation of oxygen to form an alkyloxonium ion is followed by loss of water. The resulting car- bocation has a plane of symmetry and is achiral. Capture of the carbocation by methanol yields both enantiomers of 2-methoxy-2-phenylbutane. The product is racemic. (Achiral carbocation) H H11001 H11002H 2 O OH 2 H11001 CH 3 OH (H11002H H11001 ) (R)-(H11001)-2-Phenyl-2-butanol C C 6 H 5 CH 3 CH 2 CH 3 OH 2-Methoxy-2-phenylbutane (racemic) C C 6 H 5 CH 2 CH 3 H 3 C CH 3 OH11001OCH 3 C C 6 H 5 CH 3 CH 2 CH 3 C C 6 H 5 CH 3 CH 2 CH 3 H11001 C C 6 H 5 CH 3 CH 2CH 3 S O H11002 (CH 3 ) 3 C H11001 H11001 S H11001 O H11002 (CH 3 ) 3 C S O H11002 (CH 3 ) 3 C 2H11001 O H11002 NaIO 4 H 2 O 2 S (CH 3 ) 3 C 4-tert-Butylthiane C 6 H 5 S H11002 Na H11001 HCH 3 Cl H C 6 H 5 C 6 H 5 HCH 3 HSC 6 H 5 C 6 H 5 C 6 H 5 H11001CH 3 (CH 2 ) 16 CH 2 OTs Octadecyl tosylate CH 3 CH 2 CH 2 CH 2 SNa Sodium butanethiolate CH 3 CH 2 CH 2 CH 2 SCH 2 (CH 2 ) 16 CH 3 Butyl octadecyl sulfide CH CH 2 HCl CHCl 3 2-Phenyloxirane 2-Chloro-2-phenylethanol (71%) CHCH 2 OH Cl O ETHERS, EPOXIDES, AND SULFIDES 411 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 16.29 The proper approach to this problem is to first write the equations in full stereochemical detail. (a) It now becomes clear that the arrangement of groups around the stereogenic center remains un- changed in going from starting materials to products. Therefore, choose conditions such that the nu- cleophile attacks the CH 2 group of the epoxide rather than the stereogenic center. Base-catalyzed hydrolysis is required; aqueous sodium hydroxide is appropriate. The nucleophile (hydroxide ion) attacks the less hindered carbon of the epoxide ring. (b) Inversion of configuration at the stereogenic center is required. The nucleophile must therefore at- tack the stereogenic center, and acid-catalyzed hydrolysis should be chosen. Dilute sulfuric acid would be satisfactory. The nucleophile (a water molecule) attacks that carbon atom of the ring that can better support a positive charge. Carbocation character develops at the transition state and is better supported by the carbon atom that is more highly substituted. 16.30 The key intermediate in the preparation of bis(2-chloroethyl) ether from ethylene is 2-chloroethanol, formed from ethylene by reaction with chlorine in water. Heating 2-chloroethanol in acid gives the desired ether. 16.31 (a) There is a temptation to try to do this transformation in a single step by using a reducing agent to convert the carbonyl to a methylene group. No reagent is available that reduces esters in this way! The Clemmensen and Wolff–Kishner reduction methods are suitable only for aldehydes and ketones. The best way to approach this problem is by reasoning backward. The desired Cl 2 , H 2 O H H11001 , heat ClCH 2 CH 2 OH Ethylene 2-Chloroethanol H 2 CCH 2 ClCH 2 CH 2 OCH 2 CH 2 Cl Bis(2-chloroethyl) ether HOCH 2 H 2 O H 2 SO 4 H 2 C C CH 3 H H H11001 OH 2 OH H CH 3 C O HOCH 2 O H CH 3 (S)-1,2-Propanediol OH H CH 3 C H 2 O NaOH H 2 C CH 3 HO H11002 C H O HOCH 2 CH 3 C H OH O H CH 3 (R)-1,2-Propanediol(R)-1,2-Epoxypropane HOCH 2 OH H C CH 3 412 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website product is an ether. Ethers can be prepared by the Williamson ether synthesis involving an alkyl halide and an alkoxide ion. or Both the alkyl halide and the alkoxide ion are prepared from alcohols. The problem then be- comes one of preparing the appropriate alcohol (or alcohols) from the starting ester. This is readily done using lithium aluminum hydride. Then and The following sequence is also appropriate once methanol and benzyl alcohol are obtained by reduction of methyl benzoate: and Benzyl methyl ether CH 2 OCH 3 Sodium benzyloxide CH 2 ONa H11001 Bromomethane CH 3 Br H11001 NaBr CH 3 OH Methanol Bromomethane CH 3 Br PBr 3 or HBr CH 2 OH Benzyl alcohol Sodium benzyloxide CH 2 ONa Na CH 2 Br NaOCH 3 Benzyl bromide Sodium methoxide Benzyl methyl ether CH 2 OCH 3 NaBrH11001H11001 CH 2 OH Benzyl alcohol Benzyl bromide CH 2 Br HBr or PBr 3 CH 3 OH Methanol Sodium methoxide CH 3 ONa Na 1. LiAlH 4 2. H 2 O COCH 3 O Methyl benzoate Benzyl alcohol Methanol CH 2 OH H11001 CH 3 OH CH 2 OCH 3 CH 3 XH11001 H11002 CH 2 O CH 2 OCH 3 CH 2 X H11001 H11002 CH 3 O ETHERS, EPOXIDES, AND SULFIDES 413 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) All the methods that we have so far discussed for the preparation of epoxides are based on alkenes as starting materials. This leads us to consider the partial retrosynthesis shown. The key intermediate, 1-phenylcyclohexene, is both a proper precursor to the desired epoxide and readily available from the given starting materials. A reasonable synthesis is Preparation of the required tertiary alcohol, 1-phenylcyclohexanol, completes the synthesis. (c) The necessary carbon skeleton can be assembled through the reaction of a Grignard reagent with 1,2-epoxypropane. The reaction sequence is therefore The epoxide required in the first step, 1,2-epoxypropane, is prepared as follows from iso- propyl alcohol: 2-Propanol (isopropyl alcohol) Propene H 2 SO 4 heat CH 3 COOH O CH 3 CH CH 2 CH 3 CHCH 3 OH 1,2-Epoxypropane CH 3 CH CH 2 O H11001C 6 H 5 MgBr Phenylmagnesium bromide (from bromobenzene and magnesium) C 6 H 5 CH 2 CHCH 3 OH 1-Phenyl-2-propanol1,2-Epoxypropane H 2 C CHCH 3 O C 6 H 5 C 6 H 5 CH 2 CHCH 3 OH H 2 C CHCH 3 H11002 H11001 O 1. Mg, diethyl ether 3. H 3 O H11001 2. cyclohexanone C 6 H 5 Br Bromobenzene OH C 6 H 5 1-Phenylcyclohexanol K 2 Cr 2 O 7 H 2 SO 4 , H 2 O OH Cyclohexanol O Cyclohexanone O C 6 H 5 C 6 H 5 H H11001 heat CH 3 COOH OH C 6 H 5 O OH C 6 H 5 O C 6 H 5 Target molecule C 6 H 5 Key intermediate 414 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Because the target molecule is an ether, it ultimately derives from two alcohols. Our first task is to assemble 3-phenyl-1-propanol from the designated starting material benzyl alcohol. This requires formation of a primary alcohol with the original carbon chain extended by two carbons. The standard method for this transformation involves reaction of a Grignard reagent with ethylene oxide. After 3-phenyl-1-propanol has been prepared, its conversion to the corresponding ethyl ether can be accomplished in either of two ways: or alternatively The reagents in each step are prepared from ethanol. (e) The target epoxide can be prepared in a single step from the corresponding alkene. Disconnections show that this alkene is available through a Diels–Alder reaction. H11001 H 2 CCH 2 CH 3 COOH O Bicyclo[2.2.2]oct-2-ene O 2,3-Epoxybicyclo[2.2.2]octane CH 3 CH 2 Br PBr 3 Ethanol Ethyl bromide CH 3 CH 2 OH or HBr CH 3 CH 2 ONa Na Ethanol Sodium ethoxide CH 3 CH 2 OH C 6 H 5 CH 2 CH 2 CH 2 OCH 2 CH 3 1. Na 3-Phenyl-1-propanol Ethyl 3-phenylpropyl ether C 6 H 5 CH 2 CH 2 CH 2 OH 2. CH 3 CH 2 Br C 6 H 5 CH 2 CH 2 CH 2 Br PBr 3 3-Phenyl-1-propanol 1-Bromo-3-phenylpropane Ethyl 3-phenylpropyl ether (1-ethoxy-3-phenylpropane) C 6 H 5 CH 2 CH 2 CH 2 OH C 6 H 5 CH 2 CH 2 CH 2 OCH 2 CH 3 or HBr NaOCH 2 CH 3 ethanol C 6 H 5 CH 2 Br PBr 3 Benzyl alcohol Benzyl bromide 3-Phenyl-1-propanol C 6 H 5 CH 2 OH C 6 H 5 CH 2 CH 2 CH 2 OH or HBr 1. Mg, diethyl ether 3. H 3 O H11001 2. H 2 C CH 2 O C 6 H 5 CH 2 CH 2 CH 2 OCH 2 CH 3 C 6 H 5 CH 2 CH 2 CH 2 OH CH 3 CH 2 OHH11001 ETHERS, EPOXIDES, AND SULFIDES 415 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The reaction of 1,3-cyclohexadiene with ethylene gives the desired substance. 1,3-Cyclohexadiene is one of the given starting materials. Ethylene is prepared from ethanol. ( f ) Retrosynthetic analysis reveals that the desired target molecule may be prepared by reaction of an epoxide with an ethanethiolate ion. Styrene oxide may be prepared by reaction of styrene with peroxyacetic acid. The necessary thiolate anion is prepared from ethanol by way of the corresponding thiol. Reaction of styrene oxide with sodium ethanethiolate completes the synthesis. 16.32 (a) A reasonable mechanism is one that parallels the usual one for acid-catalyzed ether formation from alcohols, modified to accommodate these particular starting materials and products. Begin with protonation of one of the oxygen atoms of ethylene glycol. The protonated alcohol then reacts in the usual way with another molecule of alcohol to give an ether. (This ether is known as diethylene glycol.) HOCH 2 CH 2 OH H11001 H11002H 2 O H11002H H11001 HOCH 2 CH 2 O H H HOCH 2 CH 2 HOCH 2 CH 2 OH H11001 HOCH 2 CH 2 OCH 2 CH 2 OH Diethylene glycol Ethylene glycol H 2 SO 4 HSO 4 H11002 HOCH 2 CH 2 HOCH 2 CH 2 OH H HH11001H11001O H11001 Sodium ethanethiolate CH 3 CH 2 SNa C 6 H 5 CHCH 2 SCH 2 CH 3 CH 3 CH 2 OH OH Styrene oxide H11001C 6 H 5 CH CH 2 O Ethanol Ethanethiol CH 3 CH 2 OH CH 3 CH 2 SH Sodium ethanethiolate CH 3 CH 2 SNa 1. HBr 3. NaOH S2. (H 2 N) 2 C NaOH CH 3 CO 2 HC 6 H 5 CH CH 2 O H11001CH 3 COOHH11001 Styrene Peroxyacetic acid Styrene oxide Acetic acid C 6 H 5 CH CH 2 O H11002 SCH 2 CH 3 H11001C 6 H 5 CHCH 2 SCH 2 CH 3 OH C 6 H 5 CH CH 2 O Ethanol CH 3 CH 2 OH H 2 SO 4 heat Ethylene H 2 CCH 2 H11001 H 2 CCH 2 1,3-Cyclohexadiene Ethylene Bicyclo[2.2.2]oct-2-ene 416 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Diethylene glycol then undergoes intramolecular ether formation to yield 1,4-dioxane. (b) The substrate is a primary alkyl halide and reacts with aqueous sodium hydroxide by nucle- ophilic substitution. The product of this reaction now has an alcohol function and a primary chloride built into the same molecule. It contains the requisite functionality to undergo an intramolecular Williamson reaction. 16.33 (a) The first step is a standard Grignard synthesis of a primary alcohol using formaldehyde. Compound A is 3-buten-1-ol. Addition of bromine to the carbon–carbon double bond of 3-buten-1-ol takes place readily to yield the vicinal dibromide. When compound B is treated with potassium hydroxide, it loses the elements of HBr to give compound C. Because further treatment of compound C with potassium hydroxide converts it to D by a second dehydrobromination, a reasonable candidate for C is 3-bromotetrahydrofuran. Compound D KOH 25H11034C KOH heat 3,4-Dibromo-1-butanol (compound B) BrCH 2 CHCH 2 CH 2 OH Br O 3-Bromotetrahydrofuran (compound C) Br O BrCH 2 CHCH 2 CH 2 OH Br 3,4-Dibromo-1-butanol (compound B) H 2 C CHCH 2 CH 2 OH 3-Buten-1-ol Br 2 1. Mg 2. H 2 C 3. H 3 O H11001 O H 2 C CHCH 2 Br Allyl bromide H 2 C CHCH 2 CH 2 OH 3-Buten-1-ol (compound A) O O 1,4-Dioxane H11001 Cl H11002 Cl H11002 H 2 C H 2 C O CH 2 CH 2 O HOCH 2 CH 2 OCH 2 CH 2 Cl H11002 OCH 2 CH 2 OCH 2 CH 2 Cl H 2 OHO H11002 H11001H11001 Bis(2-chloroethyl) ether HO H11002 Cl H11002 HOCH 2 CH 2 OCH 2 CH 2 ClClCH 2 CH 2 OCH 2 CH 2 Cl H11001H11001 H 2 O O CH 2 CH 2 CH 2 CH 2 O HOH H H11001 O CH 2 CH 2 CH 2 CH 2 OH H11001 O O H11002H 2 O H11002H H11001 H 2 SO 4 HOCH 2 CH 2 OCH 2 CH 2 HOCH 2 CH 2 OCH 2 CH 2 OH H11001 O HSO 4 H11002 H H H11001 H11001 ETHERS, EPOXIDES, AND SULFIDES 417 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Ring closure occurs by an intramolecular Williamson reaction. Dehydrohalogenation of compound C converts it to the final product, D. The alternative series of events, in which double-bond formation proceeds ring closure, is unlikely, because it requires nucleophilic attack by the alkoxide on a vinyl bromide. (b) Lithium aluminum hydride reduces the carboxylic acid to the corresponding primary alcohol, compound E. Treatment of the vicinal chlorohydrin with base results in formation of an epox- ide, compound F. As actually carried out, the first step proceeded in 56–58% yield, the second step in 65–70% yield. (c) Treatment of the vicinal chlorohydrin with base results in ring closure to form an epoxide (compound G). Recall that attack occurs on the side opposite that of the carbon–chlorine bond. Compound G undergoes ring opening on reaction with sodium methanethiolate to give compound H. NaSCH 3 Compound G H HH 3 C CH 3 CC O Compound H H H HO SCH 3 H 3 C H 3 C CC H Cl HOH H11013 CH 3 CH 3 CC NaOH (2R,3S)-3-Chloro-2-butanol trans-2,3-Epoxybutane (compound G) H HH 3 C CH 3 CC O H 3 C H Cl H HO CH 3 CO 2 H CH 3 HCl 1. LiAlH 4 2. H 2 O KOH H 2 O CH 2 OH CH 3 HCl (S)-2-Chloro-1-propanol (compound E) (R)-1,2-Epoxypropane (compound F) H H 3 C CH 2 C O CH 2 OHC H H 3 C Cl H11013 KOH BrCH 2 CHCH 2 CH 2 OH Br BrCH CHCH 2 CH 2 OH KOH (Cyclization of this intermediate does not occur.) Br H11002 O BrCH 2 CHCH 2 CH 2 OH Br KOH Compound B O Br Compound C CH CH 2 Br Br CH 2 CH 2 O H11002 418 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Because it gives an epoxide on treatment with a peroxy acid, compound I must be an alkene; more specifically, it is 1,2-dimethylcyclopentene. Compounds J and L have the same molecular formula, C 7 H 14 O 2 , but J is a liquid and L is a crys- talline solid. Their molecular formulas correspond to the addition of two OH groups to compound I. Osmium tetraoxide brings about syn hydroxylation of an alkene; therefore com- pound J must be the cis diol. Acid-catalyzed hydrolysis of an epoxide yields a trans diol (compound L): 16.34 Cineole contains no double or triple bonds and therefore must be bicyclic, on the basis of its molecu- lar formula (C 10 H 18 O, index of hydrogen deficiency H11005 2). When cineole reacts with hydrogen chloride, one of the rings is broken and water is formed. The reaction that takes place is hydrogen halide-promoted ether cleavage. In such a reaction with excess hydrogen halide, the C—O—C unit is cleaved and two carbon–halogen bonds are formed. This suggests that cineole is a cyclic ether because the product contains both newly formed carbon–halogen bonds. A reasonable structure consistent with these facts is CH 3 H 3 CCH 3 O Cineole (C 10 H 18 O) Cineole 2HClH11001 H 2 OH11001 (C 10 H 18 Cl 2 ) Cl Cl CH 3 H 3 CCCH 3 CH 3 H 3 C O H 2 SO 4 H 2 O trans-1,2-Dimethylcyclopentane- 1,2-diol (compound L) CH 3 H 3 C HO OH OsO 4 (CH 3 ) 3 COOH (CH 3 ) 3 COH, HO H11002 1,2-Dimethylcyclopentene (compound I) H 3 C CH 3 cis-1,2-Dimethylcyclopentane- 1,2-diol (compound J) HO OH H 3 CCH 3 H 3 CCH 3 1,2-Dimethylcyclopentene (compound I) 1,2-Dimethyl-1,2- epoxycyclopentane (compound K) O H 3 CCH 3 C 6 H 5 CO 2 OH ETHERS, EPOXIDES, AND SULFIDES 419 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 16.35 Recall that p-toluenesulfonate (tosylate) is a good leaving group in nucleophilic substitution reac- tions. The nucleophile that displaces tosylate from carbon is the alkoxide ion derived from the hy- droxyl group within the molecule. The product is a cyclic ether, and the nature of the union of the two rings is that they are spirocyclic. 16.36 (a) Because all the peaks in the 1 H NMR spectrum of this ether are singlets, none of the protons can be vicinal to any other nonequivalent proton. The only C 5 H 12 O ether that satisfies this re- quirement is tert-butyl methyl ether. (b) A doublet–septet pattern is characteristic of an isopropyl group. Two isomeric C 5 H 12 O ethers contain an isopropyl group: ethyl isopropyl ether and isobutyl methyl ether. The signal of the methine proton in isobutyl methyl ether will be split into more than a septet, however, because in addition to being split by two methyl groups, it is coupled to the two pro- tons in the methylene group. Thus, isobutyl methyl ether does not have the correct splitting pattern to be the answer. The correct answer is ethyl isopropyl ether. (c) The low-field signals are due to the protons on the carbon atoms of the C—O—C linkage. Because one gives a doublet, it must be vicinal to only one other proton. We can therefore specify the partial structure: This partial structure contains all the carbon atoms in the molecule. Fill in the remaining va- lences with hydrogen atoms to reveal isobutyl methyl ether as the correct choice. (d) Here again, signals at low field arise from protons on the carbons of the C—O—C unit. One of these signals is a quartet and so corresponds to a proton on a carbon bearing a methyl group. H 3 CC H OC Quartet H 3 CCH 2 O CH(CH 3 ) 2 Low-field singlet Low-field doublet C C CC H OC H Low-field doublet (CH 3 ) 2 CHOCH 2 CH 3 Ethyl isopropyl ether (CH 3 ) 2 CHCH 2 OCH 3 Isobutyl methyl ether H 3 C O C(CH 3 ) 3 Singlet at H9254 3.2 ppm Singlet at H9254 1.2 ppm OH CH 2 CH 2 CH 2 OTs C 6 H 5 C 15 H 20 O O C 6 H 5 O CH 2 CH 2 OTs H 2 C C 6 H 5 H11002 base intramolecular nucleophilic substitution 420 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The other carbon of the C—O—C unit has a hydrogen whose signal is split into a triplet. This hydrogen must therefore be attached to a carbon that bears a methylene group. These data permit us to complete the structure by adding an additional carbon and the requisite number of hydrogens in such a way that the signals of the protons attached to the carbons of the ether linkage are not split further. The correct structure is ethyl propyl ether. 16.37 A good way to address this problem is to consider the dibromide derived by treatment of compound A with hydrogen bromide. The presence of an NMR signal equivalent to four protons in the aro- matic region at H9254 7.3 ppm indicates that this dibromide contains a disubstituted aromatic ring. The four remaining protons appear as a sharp singlet at H9254 4.7 ppm and are most reasonably contained in two equivalent methylene groups of the type ArCH 2 Br. Because the dibromide contains all the carbons and hydrogens of the starting material and is derived from it by treatment with hydrogen bromide, it is likely that compound A is a cyclic ether in which a CH 2 OCH 2 unit spans two of the carbons of a benzene ring. This can occur only when the positions involved are ortho to each other. Therefore 16.38 The molecular formula of a compound (C 10 H 13 BrO) indicates an index of hydrogen deficiency of 4. One of the products obtained on treatment of the compound with HBr is benzyl bromide (C 6 H 5 CH 2 Br), which accounts for seven of its ten carbons and all the double bonds and rings. Thus, the compound is a benzyl ether having the formula C 6 H 5 CH 2 OC 3 H 6 Br. The 1 H NMR spectrum in- cludes a five-proton signal at H9254 7.4 ppm for a monosubstituted benzene ring and a two-proton singlet at H9254 4.6 ppm for the benzylic protons. This singlet appears at low field because the benzylic protons are bonded to oxygen. The 6 remaining protons appear as two overlapping 2-proton triplets at H9254 3.6 and 3.7 ppm, along with a 2-proton pentet at H9254 2.2 ppm, consistent with the unit @OCH 2 CH 2 CH 2 Br. The compound is C 6 H 5 CH 2 OCH 2 CH 2 CH 2 Br. 16.39 The high index of hydrogen deficiency (5) of the unknown compound C 9 H 10 O and the presence of six signals in the 120–140-ppm region of the 13 C NMR spectrum suggests the presence of an aro- matic ring. The problem states that the compound is a cyclic ether, thus the oxygen atom is con- tained in a second ring fused to the benzene ring. As oxidation yields 1,2-benzenedicarboxylic acid, the second ring must be attached to the benzene ring by carbon atoms. C C (C 9 H 10 O) CO 2 H CO 2 H 1,2-Benzenedicarboxylic acid Na 2 Cr 2 O 7 H 2 SO 4 , H 2 O, heat C 6 H 5 CH 2 OC 3 H 6 Br H11001C 6 H 5 CH 2 Br C 3 H 6 Br 2 HBr heat 2HBrH11001 CH 2 O CH 2 H 2 OH11001 CH 2 Br H9254 5.1 ppm Compound A CH 2 Br H9254 4.7 ppm CH 3 CH 2 OCH 2 CH 2 CH 3 Quartet Triplet H 3 CCH 2 C H OC H Quartet Triplet ETHERS, EPOXIDES, AND SULFIDES 421 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Two structures are possible with this information; however, only one of them is consistent with the presence of three CH 2 groups in the 13 C NMR spectrum. The compound is 16.40–16.45 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for these exercises. SELF-TEST PART A A-1. Write the structures of all the isomeric ethers of molecular formula C 4 H 10 O, and give the cor- rect name for each. A-2. Give the structure of the product obtained from each of the following reactions. Show stereo- chemistry where it is important. A-3. Outline a scheme for the preparation of cyclohexyl ethyl ether using the Williamson method. A-4. Outline a synthesis of 2-ethoxyethanol, CH 3 CH 2 OCH 2 CH 2 OH, using ethanol as the source of all the carbon atoms. A-5. Provide the reagents necessary to complete each of the following conversions. In each case give the structure of the intermediate product. A-6. Provide structures for compounds A and B in the following reaction scheme: COCH 3 A (C 7 H 8 O) CH 3 OHH11001 O 1. LiAlH 4 2. H 2 O B (C 9 H 12 O)A 1. Na 2. CH 3 CH 2 I ?(b) COH CH 3 CH 3 CH 3 CO CH 2 HO Br HO SCH 3 ?(a) ? NaIO 4 ( f ) Product of part (e)C 6 H 5 CH CH 2 (c)? HI O (e) ? CH 3 CH 2 Br C 6 H 5 SNa(Z)-2-butene(b 1. CH 3 COOH 2. H 3 O H11001 O (d)? CH 3 CH 2 SNa CH 3 CH 2 OH O CH 3 CH 3 OH(a)? 1. Na 2. CH 3 I C CH 3 CH 2 H CH 3 O not H9254 68 ppm H9254 28 ppm H9254 65 ppm O CH 3 422 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-7. Using any necessary organic or inorganic reagents, provide the steps to carry out the follow- ing synthetic conversion: A-8. Give the final product, including stereochemistry, of the following reaction sequence: PART B B-1. An acceptable IUPAC name of the compound shown is (a) 1-Benzyl-3-methylpentyl ethyl ether (b) Ethyl 3-methyl-1-methylphenyl-2-hexyl ether (c) Ethyl 4-methyl-1-phenyl-2-hexyl ether (d) 5-Ethoxy-3-methyl-6-phenylhexane B-2. The most effective pair of reagents for the preparation of tert-butyl ethyl ether is (a) Potassium tert-butoxide and ethyl bromide (b) Potassium tert-butoxide and ethanol (c) Sodium ethoxide and tert-butyl bromide (d) tert-Butyl alcohol and ethyl bromide B-3. The best choice of reactant(s) for the following conversion is B-4. For which of the following ethers would the 1 H NMR spectrum consist of only two singlets? (b)CH 3 OCH 2 CH 2 OCH 3 (d) All these O (c) CH 3 CH 3 CH 3 OCCH 3 (a) CH 3 CH 3 CH 3 OH OH (a) CH 3 OH OH (b) CH 3 Br OH (d ) CH 3 Br OH (c) O? CH 3 HO H11002 CH 3 CH 2 CHCH 2 CHCH 2 OCH 2 CH 3 CH 3 H H 3 C Br 2 H 2 O NaOH H 2 O NH 3 H 2 O ? O OH OHH 3 C ? ETHERS, EPOXIDES, AND SULFIDES 423 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-5. Heating a particular ether with HBr yielded a single organic product. Which of the follow- ing conclusions may be reached? (a) The reactant was a methyl ether. (b) The reactant was a symmetric ether. (c) The reactant was a cyclic ether. (d) Both (b) and (c) are correct. B-6. Treating anisole (C 6 H 5 OCH 3 ) with the following reagents will give, as the major product, 1. (CH 3 ) 3 CCl, AlCl 3 ; 2. Cl 2 , FeCl 3 ; 3. HBr, heat B-7. What is the product of the following reaction? B-8. Identify product Z in the following reaction sequence: (a)(b)(c) (d)(e) CC H CH 3 COCH 2 H O CH 2 CH 2 OCCH 3 O CC H CH 3 COCH 2 H O CH 2 CH 2 OCCH 3 O O O O OCCH 3 O O HO OH HOCH 2 CH 2 CH 2 OH XCC H 2 Lindlar Pd H 2 SO 4 heat YZ CH 3 COOH O (CH 3 ) 3 CCH 2 CHSCH 3 (d) OH (CH 3 ) 3 CCHCH 2 OH(b) SCH 3 CH 3 SCH 2 CHC(CH 3 ) 3 (c) OCH 2 CH 3 CH 3 SCH 2 CHC(CH 3 ) 3 (a) OH O H11001 NaSCH 3 ? (CH 3 ) 3 C ethanol Br Cl C(CH 3 ) 3 (d) Br Br C(CH 3 ) 3 (b) OH Cl C(CH 3 ) 3 (e) OCH 3 Br C(CH 3 ) 3 (c) OH Br C(CH 3 ) 3 (a) 424 ETHERS, EPOXIDES, AND SULFIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-9. The major product of the following sequence is B-10. Which of the following best represents the rate-determining transition state for the reaction shown? (c) OC H H H H9254H11001 H9254H11002 (e) C H H H O H9254H11002 H9254H11002 Br (b) Na H9254H11001 Br H9254H11002 C H H H H9254H11001 (d) Br H9254H11002 O H9254H11002 C HH H (a) CBr H H H H9254H11001 H9254H11002 ONa CH 3 BrH11001 Br OCH 2 CH 3 (c) CH 2 OCH 2 CH 3 OCH 2 CH 3 (e) CH 2 Br OCH 2 CH 3 (b) Br CH 2 OCH 2 CH 3 (d) CH 3 OCH 2 CH 3 (a) CH 3 Br Br 2 light CH 3 CH 2 OK ? ETHERS, EPOXIDES, AND SULFIDES 425 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website