CHAPTER 15 ALCOHOLS, DIOLS, AND THIOLS SOLUTIONS TO TEXT PROBLEMS 15.1 The two primary alcohols, 1-butanol and 2-methyl-1-propanol, can be prepared by hydrogenation of the corresponding aldehydes. The secondary alcohol 2-butanol arises by hydrogenation of a ketone. Tertiary alcohols such as 2-methyl-2-propanol, (CH 3 ) 3 COH, cannot be prepared by hydrogenation of a carbonyl compound. 15.2 (b) A deuterium atom is transferred from NaBD 4 to the carbonyl group of acetone. 3(CH 3 ) 2 CO CH 3 C CH 3 O BD 3 D CH 3 CO CH 3 D B H11002 4 H11002 CH 3 C CH 3 D OBD 3 H11002 H20898 H20898 H 2 , Ni 2-Butanone CH 3 CCH 2 CH 3 O 2-Butanol CH 3 CHCH 2 CH 3 OH H 2 , Ni (CH 3 ) 2 CHCH 2 OH 2-Methyl-1-propanol2-Methylpropanal (CH 3 ) 2 CHCH O H 2 , Ni CH 3 CH 2 CH 2 CH 2 OH 1-ButanolButanal CH 3 CH 2 CH 2 CH O 364 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 365 On reaction with CH 3 OD, deuterium is transferred from the alcohol to the oxygen of [(CH 3 ) 2 CDO] 4 B H11002 . Overall: (c) In this case NaBD 4 serves as a deuterium donor to carbon, and CD 3 OH is a proton (not deu- terium) donor to oxygen. (d) Lithium aluminum deuteride is a deuterium donor to the carbonyl carbon of formaldehyde. On hydrolysis with D 2 O, the oxygen–aluminum bond is cleaved and DCH 2 OD is formed. 15.3 The acyl portion of the ester gives a primary alcohol on reduction. The alkyl group bonded to oxy- gen may be primary, secondary, or tertiary and gives the corresponding alcohol. 15.4 (b) Reaction with ethylene oxide results in the addition of a @CH 2 CH 2 OH unit to the Grignard reagent. Cyclohexylmagnesium bromide (or chloride) is the appropriate reagent. 15.5 Lithium aluminum hydride is the appropriate reagent for reducing carboxylic acids or esters to alcohols. 1. LiAlH 4 2. H 2 O 3-Methyl-1,5-pentanedioic acid 3-Methyl-1,5-pentanediol HOCCH 2 CHCH 2 COH O O CH 3 HOCH 2 CH 2 CHCH 2 CH 2 OH CH 3 1. diethyl ether 2. H 3 O H11001 H11001 Cyclohexylmagnesium bromide MgBr 2-Cyclohexylethanol CH 2 CH 2 OH Ethylene oxide H 2 C CH 2 O 1. LiAlH 4 2. H 2 O CH 3 CH 2 CH 2 OH 1-PropanolIsopropyl propanoate 2-Propanol H11001 HOCH(CH 3 ) 2 CH 3 CH 2 COCH(CH 3 ) 2 O 4D 2 O Methanol-d-O-d 4DCH 2 OD H11001 H11002 Al(OD) 4 H11002 Al(OCH 2 D) 4 HC H O AlD 3 D H11002 HC H D OAlD 3 H11002 (DCH 2 O) 4 Al H110023HCH O NaBD 4 CD 3 OH Benzaldehyde C 6 H 5 CH O Benzyl alcohol-1-d C 6 H 5 CHOH D NaBD 4 CH 3 OD (CH 3 ) 2 C (CH 3 ) 2 COD 2-Propanol-2-d-O-dAcetone O D H11001CH 3 COD CH 3 D CH 3 C CH 3 D O B[OCD(CH 3 ) 2 ] 3 OCH 3 D OCH 3 B[OCD(CH 3 ) 2 ] 3 H11002H11002 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 366 ALCOHOLS, DIOLS, AND THIOLS Any alkyl group may be attached to the oxygen of the ester function. In the following example, it is a methyl group. 15.6 Hydroxylation of alkenes using osmium tetraoxide is a syn addition of hydroxyl groups to the dou- ble bond. cis-2-Butene yields the meso diol. trans-2-Butene yields a racemic mixture of the two enantiomeric forms of the chiral diol. The Fischer projection formulas of the three stereoisomers are 15.7 The first step is proton transfer to 1,5-pentanediol to form the corresponding alkyloxonium ion. Rewriting the alkyloxonium ion gives HO is equivalent toCH 2 CH 2 CH 2 CH 2 CH 2 O H H H11001 O H O HH H11001 OH H11001 H11001HOCH 2 CH 2 CH 2 CH 2 CH 2 1,5-Pentanediol H OSO 2 OH Sulfuric acid OSO 2 OH H11002 Hydrogen sulfateConjugate acid of 1,5-pentanediol HOCH 2 CH 2 CH 2 CH 2 CH 2 O H H H11001 CH 3 CH 3 H HHO OH (2S,3S)-2,3-Butanediol CH 3 CH 3 HO OHH H (2R,3R)-2,3-Butanediol CH 3 CH 3 H OHH OH meso-2,3-Butanediol CC H 3 C H H CH 3 trans-2-Butene OsO 4 , (CH 3 ) 3 COOH (CH 3 ) 3 COH, HO H11002 H11001 (2R,3R)-2,3-Butanediol (2S,3S)-2,3-Butanediol C HO OH C HH 3 C H CH 3 HO OH C H H 3 C H 3 C H C CC H 3 C H CH 3 H cis-2-Butene OsO 4 , (CH 3 ) 3 COOH (CH 3 ) 3 COH, HO H11002 meso-2,3-Butanediol C CH 3 H 3 C H H HO OH C 1. LiAlH 4 2. H 2 O HOCH 2 CH 2 CHCH 2 CH 2 OH 3-Methyl-1,5-pentanediolDimethyl 3-methyl-1,5-pentanedioate CH 3 OCCH 2 CHCH 2 COCH 3 O O CH 3 CH 3 H11001 2CH 3 OH Methanol Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 367 The oxonium ion undergoes cyclization by intramolecular nucleophilic attack of its alcohol function on the carbon that bears the leaving group. Loss of a proton gives oxane. 15.8 (b) The relationship of the molecular formula of the ester (C 10 H 10 O 4 ) to that of the starting dicar- boxylic acid (C 8 H 6 O 4 ) indicates that the diacid reacted with 2 moles of methanol to form a diester. 15.9 While neither cis- nor trans-4-tert-butylcyclohexanol is a chiral molecule, the stereochemical course of their reactions with acetic anhydride becomes evident when the relative stereochemistry of the ester function is examined for each case. The cis alcohol yields the cis acetate. The trans alcohol yields the trans acetate. 15.10 Glycerol has three hydroxyl groups, each of which is converted to a nitrate ester function in nitro- glycerin. CH 2 ONO 2 CHONO 2 CH 2 ONO 2 Nitroglycerin CH 2 OH CHOH CH 2 OH Glycerol 3HNO 3 H 2 SO 4 H11001(CH 3 ) 3 C OH trans-4-tert-Butylcyclohexanol CH 3 COCCH 3 O O Acetic anhydride (CH 3 ) 3 C OCCH 3 O trans-4-tert-Butylcyclohexyl acetate H11001(CH 3 ) 3 C OH cis-4-tert-Butylcyclohexanol CH 3 COCCH 3 O O Acetic anhydride (CH 3 ) 3 C OCCH 3 O cis-4-tert-Butylcyclohexyl acetate H11001 COHHOC O O H H11001 2CH 3 OH Methanol 1,4-Benzenedicarboxylic acid COCH 3 CH 3 OC O O Dimethyl 1,4-benzenedicarboxylate H11001 O Oxane O H H11001 Conjugate acid of oxane H11001 OSO 2 OH H11002 Hydrogen sulfate Sulfuric acid H OSO 2 OH H11001 H 2 O O H H11001 O H O H H11001 Conjugate acid of 1,5-pentanediol Conjugate acid of oxane Water H Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 15.11 (b) The substrate is a secondary alcohol and so gives a ketone on oxidation with sodium dichro- mate. 2-Octanone has been prepared in 92–96% yield under these reaction conditions. (c) The alcohol is primary, and so oxidation can produce either an aldehyde or a carboxylic acid, depending on the reaction conditions. Here the oxidation is carried out under anhydrous condi- tions using pyridinium chlorochromate (PCC), and the product is the corresponding aldehyde. 15.12 (b) Biological oxidation of CH 3 CD 2 OH leads to loss of one of the C-1 deuterium atoms to NAD H11001 . The dihydropyridine ring of the reduced form of the coenzyme will bear a single deuterium. (c) The deuterium atom of CH 3 CH 2 OD is lost as D H11001 . The reduced form of the coenzyme contains no deuterium. 15.13 (b) Oxidation of the carbon–oxygen bonds to carbonyl groups accompanies their cleavage. (c) The CH 2 OH group is cleaved from the ring as formaldehyde to leave cyclopentanone. HIO 4 H11001 OH CH 2 OH 1-(Hydroxymethyl)- cyclopentanol O Cyclopentanone O HCH Formaldehyde (CH 3 ) 2 CHCH 2 CH CHCH 2 C 6 H 5 OH OH 1-Phenyl-5-methyl-2,3-hexanediol (CH 3 ) 2 CHCH 2 CH H11001 O HCCH 2 C 6 H 5 O 3-Methylbutanal 2-Phenylethanal HIO 4 CH 3 CH 2 OD D H11001 CH 3 CH O H11001H11001H11001 alcohol dehydrogenase CNH 2 N R O H11001 CNH 2 HH N R O Ethanol-O-d Ethanal NADHNAD H11001 CH 3 CD 2 OH H H11001 CH 3 CD O H11001H11001H11001 alcohol dehydrogenase CNH 2 N R O H11001 CNH 2 HD N R O 1,1-Dideuterio- ethanol 1-Deuterio- ethanal NADDNAD H11001 CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 OH 1-Heptanol Heptanal CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH O PCC CH 2 Cl 2 CH 3 CH(CH 2 ) 5 CH 3 OH 2-Octanol CH 3 C(CH 2 ) 5 CH 3 O 2-Octanone Na 2 Cr 2 O 7 H 2 SO 4 , H 2 O 368 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 369 15.14 Thiols may be prepared from the corresponding alkyl halide by reaction with thiourea followed by treatment of the isothiouronium salt with base. Thus, an acceptable synthesis of 1-hexanethiol from 1-hexanol would be 15.15 The three main components of “essence of skunk” are 15.16 The molecular weight of 2-methyl-2-butanol is 88. A peak in its mass spectrum at mH20862z 70 corre- sponds to loss of water from the molecular ion. The peaks at mH20862z 73 and mH20862z 59 represent stable cations corresponding to the cleavages shown in the equation. 15.17 (a) The appropriate alkene for the preparation of 1-butanol by a hydroboration–oxidation sequence is 1-butene. Remember, hydroboration–oxidation leads to hydration of alkenes with a regioselectivity opposite to that seen in acid-catalyzed hydration. (b) 1-Butanol can be prepared by reaction of a Grignard reagent with formaldehyde. An appropriate Grignard reagent is propylmagnesium bromide. 1. diethyl ether 2. H 3 O H11001 CH 3 CH 2 CH 2 MgBr CH 3 CH 2 CH 2 CH 2 OH 1-Butanol H11001 HCH O Mg diethyl ether CH 3 CH 2 CH 2 Br CH 3 CH 2 CH 2 MgBr Propylmagnesium bromide1-Bromopropane CH 3 CH 2 CH 2 H11001CH 3 CH 2 CH 2 CH 2 OH HCH O H11002 CH 3 CH 2 CH 2 CH 2 OHCH 3 CH 2 CH CH 2 1. B 2 H 6 2. H 2 O 2 , HO H11002 1-Butene 1-Butanol CH 3 CH 2 CH 3 CH 3 OH C H11001 CH 3 H11001 CH 3 CCH 2 CH 3 OH H11001 H11001CH 3 CCH 3 OH CH 2 CH 3 H11001 m/z 73 m/z 59 CC H 3 C CH 2 SHH H trans-2-Butene-1-thiol HH CC H 3 C CH 2 SH cis-2-Butene-1-thiol3-Methyl-1-butanethiol CH 3 CHCH 2 CH 2 SH CH 3 CH 3 (CH 2 ) 4 CH 2 OH 1-Hexanol CH 3 (CH 2 ) 4 CH 2 Br 1-Bromohexane CH 3 (CH 2 ) 4 CH 2 SH 1-Hexanethiol PBr 3 HBr, heat 1. (H 2 N) 2 CS 2. NaOH Isothiouronium salt (not isolated) H11001 NaOH RBr Alkyl bromide (H 2 N) 2 CS Thiourea RSH Thiol Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) Alternatively, 1-butanol may be prepared by the reaction of a Grignard reagent with ethylene oxide. In this case, ethylmagnesium bromide would be used. (d) Primary alcohols may be prepared by reduction of the carboxylic acid having the same num- ber of carbons. Among the reagents we have discussed, the only one that is effective in the reduction of carboxylic acids is lithium aluminum hydride. The four-carbon carboxylic acid butanoic acid is the proper substrate. (e) Reduction of esters can be accomplished using lithium aluminum hydride. The correct methyl ester is methyl butanoate. ( f ) A butyl ester such as butyl acetate may be reduced with lithium aluminum hydride to prepare 1-butanol. (g) Because 1-butanol is a primary alcohol having four carbons, butanal must be the aldehyde that is hydrogenated. Suitable catalysts are nickel, palladium, platinum, and ruthenium. (h) Sodium borohydride reduces aldehydes and ketones efficiently. It does not reduce carboxylic acids, and its reaction with esters is too slow to be of synthetic value. CH 3 CH 2 CH 2 CH 2 OH 1-ButanolButanal CH 3 CH 2 CH 2 CH O NaBH 4 water, ethanol, or methanol H 2 , Pt CH 3 CH 2 CH 2 CH 2 OH 1-ButanolButanal CH 3 CH 2 CH 2 CH O 1. LiAlH 4 2. H 2 O CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 OH 1-Butanol EthanolButyl acetate CH 3 COCH 2 CH 2 CH 2 CH 3 O H11001 1. LiAlH 4 2. H 2 O CH 3 CH 2 CH 2 COCH 3 CH 3 CH 2 CH 2 CH 2 OH CH 3 OH 1-Butanol MethanolMethyl butanoate O H11001 1. LiAlH 4 , diethyl ether 2. H 2 O CH 3 CH 2 CH 2 COH CH 3 CH 2 CH 2 CH 2 OH 1-ButanolButanoic acid O CH 3 CH 2 MgBr CH 3 CH 2 CH 2 CH 2 OHH11001 1-ButanolEthylene oxide 1. diethyl ether 2. H 3 O H11001 H 2 C CH 2 O Mg diethyl ether CH 3 CH 2 Br CH 3 CH 2 MgBr Ethylmagnesium bromideEthyl bromide CH 3 CH 2 CH 2 CH 2 OH H11001 H11002 CH 3 CH 2 H 2 C CH 2 O 370 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 371 15.18 (a) Both (Z)- and (E)-2-butene yield 2-butanol on hydroboration–oxidation. (b) Disconnection of one of the bonds to the carbon that bears the hydroxyl group reveals a feasi- ble route using a Grignard reagent and propanal. The synthetic sequence is (c) Another disconnection is related to a synthetic route using a Grignard reagent and acetaldehyde. (d–f ) Because 2-butanol is a secondary alcohol, it can be prepared by reduction of a ketone having the same carbon skeleton, in this case 2-butanone. All three reducing agents indicated in the equations are satisfactory. OH CH 3 CHCH 2 CH 3 1. LiAlH 4 2. H 2 O 2-Butanol2-Butanone CH 3 CCH 2 CH 3 O OH CH 3 CHCH 2 CH 3 NaBH 4 CH 3 OH 2-Butanol2-Butanone CH 3 CCH 2 CH 3 O OH CH 3 CHCH 2 CH 3 H 2 , Pd (or Pt, Ni, Ru) 2-Butanol2-Butanone CH 3 CCH 2 CH 3 O CH 3 CH 2 Br Ethylmagnesium bromide OH CH 3 CH 2 MgBr CH 3 CH 2 CHCH 3 Mg diethyl ether 1. CH 3 CH 2. H 3 O H11001 O 2-ButanolEthyl bromide CH 2 CH 3 CH 3 CH Acetaldehyde OH CH 3 CH H11001 O Disconnect this bond. CH 3 CH 2 C H11002 CH 3 MgBr Methylmagnesium bromide Methyl bromide CH 3 Br Mg diethyl ether 2-Butanol CH 3 CHCH 2 CH 3 OH 1. CH 3 CH 2 CH 2. H 3 O H11001 O CH 3 H11002 H11001 Propanal HCCH 2 CH 3 O CHCH 2 CH 3 H 3 C OH Disconnect this bond. 1. B 2 H 6 2. H 2 O 2 , HO H11002 (Z)- or (E)-2-butene CH 3 CH CHCH 3 2-Butanol CH 3 CHCH 2 CH 3 OH Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 15.19 (a) All the carbon–carbon disconnections are equivalent. The synthesis via a Grignard reagent and acetone is (b) An alternative route to tert-butyl alcohol is addition of a Grignard reagent to an ester. Esters react with 2 moles of Grignard reagent. Thus, tert-butyl alcohol may be formed by reacting methyl acetate with 2 moles of methylmagnesium iodide. Methyl alcohol is formed as a by- product of the reaction. 15.20 (a) All of the primary alcohols having the molecular formula C 5 H 12 O may be prepared by reduc- tion of aldehydes. The appropriate equations are 1. LiAlH 4 , diethyl ether 2. H 2 O 2,2-Dimethylpropanal O (CH 3 ) 3 CCH 2,2-Dimethyl-1-propanol (CH 3 ) 3 CCH 2 OH 1. LiAlH 4 , diethyl ether 2. H 2 O 3-Methylbutanal O (CH 3 ) 2 CHCH 2 CH 3-Methyl-1-butanol (CH 3 ) 2 CHCH 2 CH 2 OH 2-Methyl-1-butanol 1. LiAlH 4 , diethyl ether 2. H 2 O 2-Methylbutanal CH 3 CH 2 CHCH O CH 3 CH 3 CH 2 CHCH 2 OH CH 3 CH 3 CH 2 CH 2 CH 2 CH 2 OH 1-Pentanol 1. LiAlH 4 , diethyl ether 2. H 2 O Pentanal CH 3 CH 2 CH 2 CH 2 CH O Methyl acetate tert-Butyl alcoholMethylmagnesium iodide C CH 3 CH 3 OH CH 3 CH 3 COCH 3 O H11001 Methyl alcohol CH 3 OHH110012CH 3 MgI 1. diethyl ether 2. H 3 O H11001 CH 3 Br Methylmagnesium bromide CH 3 MgBr (CH 3 ) 3 COH Mg diethyl ether 1. CH 3 CCH 3 2. H 3 O H11001 O tert-Butyl alcoholMethyl bromide CH 3 CCH 3 Acetone C CH 3 H 3 COH CH 3 O H11001CH 3 H11002 372 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 373 (b) The secondary alcohols having the molecular formula C 5 H 12 O may be prepared by reduction of ketones. (c) As with the reduction of aldehydes in part (a), reduction of carboxylic acids yields primary alcohols. For example, 1-pentanol may be prepared by reduction of pentanoic acid. The remaining primary alcohols, 2-methyl-1-butanol, 3-methyl-1-butanol, and 2,2-dimethyl- 1-propanol, may be prepared in the same way. (d) As with carboxylic acids, esters may be reduced using lithium aluminum hydride to give primary alcohols. For example, 2,2-dimethyl-1-propanol may be prepared by reduction of methyl 2,2-dimethylpropanoate. 15.21 (a) The suggested synthesis is a poor one because bromination of butane yields a mixture of 1-bromobutane and 2-bromobutane, 2-bromobutane being the major product. Br 2 light or heat CH 3 CH 2 CH 2 CH 3 Butane CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane (minor product) 2-Bromobutane (major product) H11001 CH 3 CHCH 2 CH 3 Br Br 2 light or heat KOH CH 3 CH 2 CH 2 CH 3 Butane CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane CH 3 CH 2 CH 2 CH 2 OH 1-Butanol 1. LiAlH 4 , diethyl ether 2. H 2 O Methyl 2,2-dimethylpropanoate O (CH 3 ) 3 CCOCH 3 2,2-Dimethyl-1-propanol (CH 3 ) 3 CCH 2 OH 1. LiAlH 4 , diethyl ether 2. H 2 O Pentanoic acid O CH 3 CH 2 CH 2 CH 2 COH 1-Pentanol CH 3 CH 2 CH 2 CH 2 CH 2 OH 1. LiAlH 4 , diethyl ether 2. H 2 O 3-Methyl-2-butanone O (CH 3 ) 2 CHCCH 3 3-Methyl-2-butanol OH (CH 3 ) 2 CHCHCH 3 1. LiAlH 4 , diethyl ether 2. H 2 O 3-Pentanone O CH 3 CH 2 CCH 2 CH 3 3-Pentanol OH CH 3 CH 2 CHCH 2 CH 3 1. LiAlH 4 , diethyl ether 2. H 2 O 2-Pentanone O CH 3 CH 2 CH 2 CCH 3 2-Pentanol OH CH 3 CH 2 CH 2 CHCH 3 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) The suggested synthesis will fail because the reaction of 2-bromo-2-methylpropane with potassium hydroxide will proceed by elimination rather than by substitution. The first step in the process, selective bromination of 2-methylpropane to 2-bromo-2-methylpropane, is satisfactory because bromi- nation is selective for substitution of tertiary hydrogens in the presence of secondary and pri- mary ones. (c) Benzyl alcohol, unlike 1-butanol and 2-methyl-2-propanol, can be prepared effectively by this method. Free-radical bromination of toluene is selective for the benzylic position. Benzyl bromide can- not undergo elimination, and so nucleophilic substitution of bromide by hydroxide will work well. (d) The desired transformation fails because it produces more than one enantiomer. The reactant ethylbenzene is achiral and although its bromination will be highly regioselective for the benzylic position, the product will be a racemic mixture of (R) and (S)-1-bromo-1-phenylethane. The alcohol produced by hydrolysis will also be racemic. Furthermore, the hydrolysis step will give mostly styrene by an E2 elimination, rather than 1-phenylethanol by nucleophilic substitution. 15.22 Glucose contains five hydroxyl groups and an aldehyde functional group. Its hydrogenation will not affect the hydroxyl groups but will reduce the aldehyde to a primary alcohol. 15.23 (a) 1-Phenylethanol is a secondary alcohol and so can be prepared by the reaction of a Grignard reagent with an aldehyde. One combination is phenylmagnesium bromide and ethanal (acetaldehyde). C 6 H 5 MgBr H11001C 6 H 5 CHCH 3 OH 1-Phenylethanol Phenylmagnesium bromide HCCH 3 O Ethanal (acetaldehyde) H 2 (120 atm) Ni, 140H11034C HO OH OH O OH OH H Glucose HO OH OH OH OH OH Sorbitol Br 2 light or heat KOH CH 2 CH 3 Ethylbenzene CHCH 3 Br 1-Bromo-1-phenylethane CHCH 3 OH 1-Phenylethanol Br 2 light or heat KOH CH 3 Toluene CH 2 Br Benzyl bromide CH 2 OH Benzyl alcohol Br 2 light or heat KOH (CH 3 ) 3 CH 2-Methylpropane (CH 3 ) 3 CBr 2-Bromo-2- methylpropane (CH 3 ) 3 COH 2-Methyl-2- propanol 374 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 375 Grignard reagents—phenylmagnesium bromide in this case—are always prepared by reaction of magnesium metal and the corresponding halide. Starting with bromobenzene, a suitable synthesis is described by the sequence (b) An alternative disconnection of 1-phenylethanol reveals a second route using benzaldehyde and a methyl Grignard reagent. Equations representing this approach are (c) Aldehydes are, in general, obtainable by oxidation of the corresponding primary alcohol. By recognizing that benzaldehyde can be obtained by oxidation of benzyl alcohol with PCC, we write (d) The conversion of acetophenone to 1-phenylethanol is a reduction. Any of a number of reducing agents could be used. These include 1. NaBH 4 , CH 3 OH 2. LiAlH 4 in diethyl ether, then H 2 O 3. H 2 and a Pt, Pd, Ni, or Ru catalyst reducing agent Acetophenone O C 6 H 5 CCH 3 1-Phenylethanol C 6 H 5 CHCH 3 OH PCC CH 2 Cl 2 1-PhenylethanolBenzyl alcohol C 6 H 5 CH 2 OH C 6 H 5 CHCH 3 OH 2. H 3 O H11001 1. CH 3 MgI, diethyl ether Benzaldehyde C 6 H 5 CH O CH 3 I Iodomethane CH 3 MgI Methylmagnesium iodide Mg diethyl ether 1-Phenylethanol C 6 H 5 CHCH 3 OH 2. H 3 O H11001 1. C 6 H 5 CH O H11001C 6 H 5 CHCH 3 OH 1-Phenylethanol Benzaldehyde C 6 H 5 CH O CH 3 MgI Methylmagnesium iodide C 6 H 5 MgBrC 6 H 5 Br Bromobenzene Phenylmagnesium bromide 1-Phenylethanol Mg diethyl ether 1. CH 3 CH 2. H 3 O H11001 C 6 H 5 CHCH 3 OH O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) Benzene can be employed as the ultimate starting material in a synthesis of 1-phenylethanol. Friedel–Crafts acylation of benzene gives acetophenone, which can then be reduced as in part (d). Acetic anhydride can be used in place of acetyl chloride. 15.24 2-Phenylethanol is an ingredient in many perfumes, to which it imparts a rose-like fragrance. Numerous methods have been employed for its synthesis. (a) As a primary alcohol having two more carbon atoms than bromobenzene, it can be formed by reaction of a Grignard reagent, phenylmagnesium bromide, with ethylene oxide. The desired reaction sequence is therefore (b) Hydration of sytrene with a regioselectivity contrary to that of Markovnikov’s rule is required. This is accomplished readily by hydroboration–oxidation. (c) Reduction of aldehydes yields primary alcohols. Among the reducing agents that could be (and have been) used are 1. NaBH 4 , CH 3 OH 2. LiAlH 4 in diethyl ether, then H 2 O 3. H 2 and a Pt, Pd, Ni, or Ru catalyst (d) Esters are readily reduced to primary alcohols with lithium aluminum hydride. 1. LiAlH 4 , diethyl ether 2. H 2 O C 6 H 5 CH 2 COCH 2 CH 3 C 6 H 5 CH 2 CH 2 OH 2-PhenylethanolEthyl 2-phenylethanoate O reducing agent 2-Phenylethanol C 6 H 5 CH 2 CH 2 OH 2-Phenylethanal C 6 H 5 CH 2 CH O 1. B 2 H 6 , diglyme 2. H 2 O 2 , HO H11002 2-Phenylethanol C 6 H 5 CH 2 CH 2 OH Styrene C 6 H 5 CH CH 2 Mg diethyl ether CH 2 H 2 C O 1. 2. H 3 O H11001 C 6 H 5 CH 2 CH 2 OH 2-Phenylethanol C 6 H 5 MgBr Phenylmagnesium bromide C 6 H 5 Br Bromobenzene C 6 H 5 CH 2 CH 2 OH H11001C 6 H 5 MgBr H 2 CCH 2 O O O (CH 3 COCCH 3 ) H11001 O CH 3 CCl Acetyl chloride AlCl 3 Benzene Acetophenone O CCH 3 376 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 377 (e) The only reagent that is suitable for the direct reduction of carboxylic acids to primary alco- hols is lithium aluminum hydride. Alternatively, the carboxylic acid could be esterified with ethanol and the resulting ethyl 2-phenylethanoate reduced. 15.25 (a) Thiols are made from alkyl halides by reaction with thiourea, followed by hydrolysis of the isothiouronium salt in base. The first step must therefore be a conversion of the alcohol to an alkyl bromide. (b) To obtain 1-hexanol from alcohols having four carbons or fewer, a two-carbon chain exten- sion must be carried out. This suggests reaction of a Grignard reagent with ethylene oxide. The retrosynthetic path for this approach is The reaction sequence therefore becomes Given the constraints of the problem, we prepare ethylene oxide by the sequence (c) The target molecule 2-hexanol may be mentally disconnected as shown to a four-carbon unit and a two-carbon unit. CH 3 CH CH 2 CH 2 CH 2 CH 3 H11001CH 3 CH CH 2 CH 2 CH 2 CH 3 H11002 OH O Ethylene H 2 C H 2 SO 4 heat CH 3 COOH O CH 3 CH 2 OH Ethanol CH 2 H 2 C CH 2 O Mg diethyl ether CH 3 CH 2 CH 2 CH 2 Br 1-Bromobutane from part (a) CH 3 CH 2 CH 2 CH 2 MgBr Butylmagnesium bromide CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OH 1-Hexanol 2. H 3 O H11001 1. H 2 C CH 2 O CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OH H11001CH 3 CH 2 CH 2 CH 2 MgBr H 2 C CH 2 O CH 3 CH 2 CH 2 CH 2 SH 1-Butanethiol1-Bromobutane CH 3 CH 2 CH 2 CH 2 Br 1-Butanol CH 3 CH 2 CH 2 CH 2 OH 2. NaOH 1. (H 2 N) 2 C S or PBr 3 HBr H H11001 C 6 H 5 CH 2 COCH 2 CH 3 Ethyl 2-phenylethanoate C 6 H 5 CH 2 CH 2 OH 2-Phenylethanol2-Phenylethanoic acid Ethanol CH 3 CH 2 OHC 6 H 5 CH 2 COH O O reduce as in part (d) H11001 1. LiAlH 4 , diethyl ether 2. H 2 O C 6 H 5 CH 2 COH C 6 H 5 CH 2 CH 2 OH 2-Phenylethanol2-Phenylethanoic acid O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The alternative disconnection to H11002 :CH 3 and reveals a plausible approach to 2-hexanol but is inconsistent with the requirement of the problem that limits starting mate- rials to four carbons or fewer. The five-carbon aldehyde would have to be prepared first, mak- ing for a lengthy overall synthetic scheme. An appropriate synthesis based on alcohols as starting materials is (d) Hexanal may be obtained from 1-hexanol [prepared in part (b)] by oxidation in dichloromethane using pyridinium chlorochromate (PCC) or pyridinium dichromate (PDC). (e) Oxidation of 2-hexanol from part (c) yields 2-hexanone. PCC or PDC can also be used for this transformation. ( f ) Oxidation of 1-hexanol with chromic acid (sodium or potassium dichromate in aqueous sul- furic acid) yields hexanoic acid. Use of PDC or PCC in dichloromethane is not acceptable because those reagents yield aldehydes on reaction with primary alcohols. (g) Fischer esterification of hexanoic acid with ethanol produces ethyl hexanoate. (h) Vicinal diols are normally prepared by hydroxylation of alkenes with osmium tetraoxide and tert-butyl hydroperoxide. 2-Methylpropene 2-Methyl-1,2- propanediol (CH 3 ) 2 CCH 2 OH OH (CH 3 ) 2 CCH 2 OsO 4 (CH 3 ) 3 COOH, HO H11002 (CH 3 ) 3 COH CH 3 CH 2 OHH11001CH 3 (CH 2 ) 4 CO 2 H Hexanoic acid from part ( f ) Ethyl hexanoateEthanol H H11001 CH 3 (CH 2 ) 4 COCH 2 CH 3 O CH 3 (CH 2 ) 4 CH 2 OH CH 3 (CH 2 ) 4 CO 2 H 1-Hexanol from part (b) Hexanoic acid K 2 Cr 2 O 7 H 2 SO 4 , H 2 O CH 3 CHCH 2 CH 2 CH 2 CH 3 OH CH 3 CCH 2 CH 2 CH 2 CH 3 O 2-Hexanol 2-Hexanone Na 2 Cr 2 O 7 H 2 SO 4 , H 2 O CH 3 (CH 2 ) 4 CH 2 OH CH 3 (CH 2 ) 4 CH O PCC or PDC CH 2 Cl 2 1-Hexanol from part (b) Hexanal CH 3 CHCH 2 CH 2 CH 2 CH 3 2-HexanolButylmagnesium bromide from part (b) CH 3 CH 2 CH 2 CH 2 MgBr H11001 2. H 3 O H11001 1. diethyl ether CH 3 CH O Ethanal OH CH 3 CH 2 OH CH 3 CH O PCC CH 2 Cl Ethanol Ethanal O HCCH 2 CH 2 CH 2 CH 3 378 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 379 The required alkene is available by dehydration of 2-methyl-2-propanol. (i) The desired aldehyde can be prepared by oxidation of the corresponding primary alcohol with PCC or PDC. The necessary alcohol is available through reaction of a tert-butyl Grignard reagent with formaldehyde, as shown by the disconnection 15.26 (a) The simplest route to this primary chloride from benzene is through the corresponding alco- hol. The first step is the two-carbon chain extension used in Problem 15.24a. The preparation of ethylene oxide is shown in Problem 15.25b. 2-Phenylethanol CH 2 CH 2 OH 1-Chloro-2-phenylethane CH 2 CH 2 Cl SOCl 2 Benzene Bromobenzene Br 2-Phenylethanol CH 2 CH 2 OH Br 2 FeBr 3 CH 2 1. Mg, diethyl ether 3. H 3 O H11001 2. H 2 C O (CH 3 ) 3 CMgCl 1,1-Dimethylethylmagnesium chloride (tert-butylmagnesium chloride) (CH 3 ) 3 CCH 2 OH 2,2-Dimethyl-1-propanol 1. H 2 C 2. H 3 O H11001 O, diethyl ether (CH 3 ) 3 COH 2-Methyl-2-propanol (tert-butyl alcohol) (CH 3 ) 3 CCl 2-Chloro- 2-methylpropane (tert-butyl chloride) (CH 3 ) 3 CMgCl 1,1-Dimethylethyl- magnesium chloride (tert-butylmagnesium chloride) HCl Mg diethyl ether CH 3 OH Methanol Formaldehyde HCH O PCC or PDC CH 2 Cl 2 H11001(CH 3 ) 3 CCH 2 OH (CH 3 ) 3 CMgCl H 2 CO 2,2-Dimethyl-1-propanol (CH 3 ) 3 CCH 2 OH 2,2-Dimethylpropanal PCC or PDC CH 2 Cl 2 (CH 3 ) 3 CCH O 2-Methyl-2-propanol 2-Methylpropene (CH 3 ) 3 COH H 3 PO 4 heat (CH 3 ) 2 CCH 2 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) A Friedel–Crafts acylation is the best approach to the target ketone. Because carboxylic acid chlorides are prepared from the corresponding acids, we write (c) Wolff–Kishner or Clemmensen reduction of the ketone just prepared in part (b) affords isobutylbenzene. A less direct approach requires three steps: 15.27 (a) Because 1-phenylcyclopentanol is a tertiary alcohol, a likely synthesis would involve reaction of a ketone and a Grignard reagent. Thus, a reasonable last step is treatment of cyclopentanone with phenylmagnesium bromide. Cyclopentanone is prepared by oxidation of cyclopentanol. Any one of a number of oxidizing agents would be suitable. These include PDC or PCC in CH 2 Cl 2 or chromic acid (H 2 CrO 4 ) generated from Na 2 Cr 2 O 7 in aqueous sulfuric acid. (b) Acid-catalyzed dehydration of 1-phenylcyclopentanol gives 1-phenylcyclopentene. 1-Phenylcyclopentene H 2 SO 4 , heat or H 3 PO 4 , heat C 6 H 5 1-Phenylcyclopentanol OH C 6 H 5 O CyclopentanoneCyclopentanol OH H oxidize O Cyclopentanone 1-Phenylcyclopentanol OH C 6 H 5 1. C 6 H 5 MgBr, diethyl ether 2. H 3 O H11001 2-Methyl-1-phenyl-1- propanone 2-Methyl-1-phenyl-1- propanol NaBH 4 CH 3 OH C 6 H 5 CHCH(CH 3 ) 2 C 6 H 5 CCH(CH 3 ) 2 O OH H 2 SO 4 H 2 Pt 2-Methyl-1- phenylpropene C 6 H 5 CH C(CH 3 ) 2 Isobutylbenzene C 6 H 5 CH 2 CH(CH 3 ) 2 heat 2-Methyl-1-phenyl-1-propanone H 2 NNH 2 , HO H11002 triethylene glycol, heat or Zn(Hg), HCl Isobutylbenzene C 6 H 5 CH 2 CH(CH 3 ) 2 C 6 H 5 CCH(CH 3 ) 2 O (CH 3 ) 2 CHCH 2 OH 2-Methyl-1-propanol 2-Methylpropanoyl chloride K 2 Cr 2 O 7 H 2 SO 4 , heat SOCl 2 2-Methylpropanoic acid (CH 3 ) 2 CHCOH O (CH 3 ) 2 CHCCl O Benzene 2-Methylpropanoyl chloride 2-Methyl-1-phenyl-1- propanone AlCl 3 H11001 (CH 3 ) 2 CHCCl O CCH(CH 3 ) 2 O 380 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 381 (c) Hydroboration–oxidation of 1-phenylcyclopentene gives trans-2-phenylcyclopentanol. The elements of water (H and OH) are added across the double bond opposite to Markovnikov’s rule and syn to each other. (d) Oxidation of trans-2-phenylcyclopentanol converts this secondary alcohol to the desired ketone. Any of the Cr(VI)-derived oxidizing agents mentioned in part (a) for oxidation of cyclopentanol to cyclopentanone is satisfactory. (e) The standard procedure for preparing cis-1,2-diols is by hydroxylation of alkenes with osmium tetraoxide. ( f ) The desired compound is available either by ozonolysis of 1-phenylcyclopentene: or by periodic acid cleavage of the diol in part (e): (g) Reduction of both carbonyl groups in the product of part ( f ) gives the desired diol. H 2 , Pt (or Pd, Ni, Ru) or NaBH 4 , H 2 O or 1. LiAlH 4 , diethyl ether 2. H 2 O 5-Oxo-1-phenyl-1-pentanone C 6 H 5 CCH 2 CH 2 CH 2 CH O O 1-Phenyl-1,5-pentanediol C 6 H 5 CHCH 2 CH 2 CH 2 CH 2 OH OH HIO 4 5-Oxo-1-phenyl-1-pentanone C 6 H 5 CCH 2 CH 2 CH 2 CH O O 1-Phenyl-cis-1,2- cyclopentanediol HO OH C 6 H 5 H H C 6 H 5 1. O 3 2. H 2 O, Zn 1-Phenylcyclopentene 5-Oxo-1-phenyl-1-pentanone C 6 H 5 CCH 2 CH 2 CH 2 CH O O H C 6 H 5 OsO 4 , (CH 3 ) 3 COOH (CH 3 ) 3 COH, HO H11002 1-Phenylcyclopentene 1-Phenyl-cis-1,2-cyclopentanediol HO H OH C 6 H 5 Cr(VI) oxidation 2-Phenylcyclopentanonetrans-2-Phenylcyclo- pentanol HO H H C 6 H 5 O H C 6 H 5 H C 6 H 5 1. B 2 H 6 , diglyme 2. H 2 O 2 , HO H11002 1-Phenylcyclopentene trans-2-Phenylcyclopentanol HO H H C 6 H 5 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 15.28 (a, b) Primary alcohols react in two different ways on being heated with acid catalysts: they can condense to form dialkyl ethers or undergo dehydration to yield alkenes. Ether formation is favored at lower temperature, and alkene formation is favored at higher temperature. (c) Nitrate esters are formed by the reaction of alcohols with nitric acid in the presence of a sulfuric acid catalyst. (d) Pyridinium chlorochromate (PCC) oxidizes primary alcohols to aldehydes. (e) Potassium dichromate in aqueous sulfuric acid oxidizes primary alcohols to carboxylic acids. ( f ) Amide ion, a strong base, abstracts a proton from 1-propanol to form ammonia and 1-propanolate ion. This is an acid–base reaction. (g) With acetic acid and in the presence of an acid catalyst, 1-propanol is converted to its acetate ester. This is an equilibrium process that slightly favors products. (h) Alcohols react with p-toluenesulfonyl chloride to give p-toluenesulfonate esters. 1-Propanol CH 3 CH 2 CH 2 OH H11001H11001HCl p-Toluenesulfonyl chloride CH 3 SO 2 Cl pyridine Propyl p-toluenesulfonate O O CH 3 CH 2 CH 2 OS CH 3 CH 3 CH 2 CH 2 OH 1-Propanol Propyl acetate HCl Acetic acid CH 3 COHH11001 Water H 2 OH11001 O CH 3 COCH 2 CH 2 CH 3 O NaNH 2 H11001 Sodium amide CH 3 CH 2 CH 2 OH 1-Propanol CH 3 CH 2 CH 2 ONa Sodium 1-propanolate NH 3 H11001 Ammonia CH 3 CH 2 CH 2 OH 1-Propanol Propanoic acid K 2 Cr 2 O 7 H 2 SO 4 , H 2 O heat CH 3 CH 2 COH O CH 3 CH 2 CH 2 OH 1-Propanol Propanal PCC CH 2 Cl 2 CH 3 CH 2 CH O HONO 2 H11001 Nitric acid CH 3 CH 2 CH 2 OH 1-Propanol CH 3 CH 2 CH 2 ONO 2 Propyl nitrate H 2 OH11001 Water H 2 SO 4 (cat) 2CH 3 CH 2 CH 2 OH 1-Propanol CH 3 CH 2 CH 2 OCH 2 CH 2 CH 3 Dipropyl ether H 2 OH11001 Water H 2 SO 4 140°C CH 3 CH 2 CH 2 OH 1-Propanol H 2 OH11001 Water H 2 SO 4 200°C Propene CH 3 CH CH 2 382 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 383 (i) Acyl chlorides convert alcohols to esters. ( j) The reagent is benzoic anhydride. Carboxylic acid anhydrides react with alcohols to give esters. (k) The reagent is succinic anhydride, a cyclic anhydride. Esterification occurs, but in this case the resulting ester and carboxylic acid functions remain part of the same molecule. 15.29 (a) On being heated in the presence of sulfuric acid, tertiary alcohols undergo elimination. (b) The combination of reagents specified converts alkenes to vicinal diols. (c) Hydroboration–oxidation of the double bond takes place with a regioselectivity that is opposite to Markovnikov’s rule. The elements of water are added in a stereospecific syn fashion. (d) Lithium aluminum hydride reduces carboxylic acids to primary alcohols, but does not reduce carbon–carbon double bonds. CO 2 H Cyclopentene-4- carboxylic acid CH 2 OH (3-Cyclopentenyl)- methanol 1. LiAlH 4 , diethyl ether 2. H 2 O C 6 H 5 1-Phenylcyclobutene C 6 H 5 HO H trans-2-Phenylcyclobutanol (82%) 1. B 2 H 6 , diglyme 2. H 2 O 2 , HO H11002 (CH 3 ) 3 COOH, OsO 4 (cat) (CH 3 ) 3 COH, HO H11002 2,3-Dimethyl-2-butene (CH 3 ) 2 C C(CH 3 ) 2 2,3-Dimethyl-2,3-butanediol (72%) (CH 3 ) 2 C C(CH 3 ) 2 HO OH H 2 SO 4 heat H 3 CC 6 H 5 4-Methyl-1- phenylcyclohexene (81%) 4-Methyl-1- phenylcyclohexanol H 3 C C 6 H 5 OH pyridine Hydrogen propyl succinate CH 3 CH 2 CH 2 OCCH 2 CH 2 COH O O H11001CH 3 CH 2 CH 2 OH 1-Propanol Succinic anhydride O O O 1-Propanol CH 3 CH 2 CH 2 OH H11001 H11001 pyridine Benzoic anhydride C 6 H 5 COCC 6 H 5 O O Propyl benzoate CH 3 CH 2 CH 2 OCC 6 H 5 O Benzoic acid C 6 H 5 COH O 1-Propanol CH 3 CH 2 CH 2 OH H11001H11001HCl pyridine p-Methoxybenzoyl chloride O CH 3 O CCl Propyl p-methoxybenzoate CH 3 CH 2 CH 2 OC OCH 3 O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) Chromic acid oxidizes the secondary alcohol to the corresponding ketone but does not affect the triple bond. (f ) Lithium aluminum hydride reduces carbonyl groups efficiently but does not normally react with double bonds. (g) Alcohols react with acyl chlorides to yield esters. The O@H bond is broken in this reaction; the C@O bond of the alcohol remains intact on ester formation. (h) Carboxylic acid anhydrides react with alcohols to give esters. Here, too, the spatial orientation of the C@O bond remains intact. (i) The substrate is a carboxylic acid and undergoes Fischer esterification with methanol. CH 3 OH H 2 SO 4 4-Chloro-3,5- dinitrobenzoic acid COH O 2 N O 2 N Cl O Methyl 4-chloro-3,5- dinitrobenzoate (96%) O 2 N O 2 N Cl COCH 3 O H11001H11001 exo-Bicyclo[2.2.1]- heptan-2-ol OH H Acetic anhydride CH 3 COCCH 3 O O exo-Bicyclo[2.2.1]hept- 2-yl acetate (90%) OCCH 3 H O Acetic acid CH 3 COH O pyridine 3,5-Dinitrobenzoyl chloride CCl O 2 N O 2 N O trans-3-Methylcyclohexanol OH H 3 C trans-3-Methylcyclohexyl-3,5- dinitrobenzoate (74%) NO 2 NO 2 O OC H 3 CH11001 1. LiAlH 4 , diethyl ether 2. H 2 O CH 3 CCH 2 CH CHCH 2 CCH 3 4-Octen-2,7-dione OO OH CH 3 CHCH 2 CH CHCH 2 CHCH 3 4-Octen-2,7-diol (75%) OH H 2 CrO 4 H 2 SO 4 , H 2 O acetone 3-Octyn-2-one (80%) CH 3 CC C(CH 2 ) 3 CH 3 O 3-Octyn-2-ol C(CH 2 ) 3 CH 3 CH 3 CHC OH 384 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 385 ( j) Both ester functions are cleaved by reduction with lithium aluminum hydride. The product is a diol. (k) Treatment of the diol obtained in part ( j) with periodic acid brings about its cleavage to two carbonyl compounds. 15.30 Only the hydroxyl groups on C-1 and C-4 can be involved, since only these two can lead to a five- membered cyclic ether. Any other combination of hydroxyl groups would lead to a strained three-membered or four- membered ring and is unfavorable under conditions of acid catalysis. 15.31 Hydroxylation of alkenes with osmium tetraoxide is a syn addition. A racemic mixture of the 2R,3S and 2S,3R stereoisomers is formed from cis-2-pentene. trans-2-Pentene gives a racemic mixture of the 2R,3R and 2S,3S stereoisomers. OsO 4 , (CH 3 ) 3 COOH (CH 3 ) 3 COH, HO H11002 H H 3 C CH 2 CH 3 H trans-2-Pentene OH OH H 3 C CH 2 CH 3 H H 2R,3R-2,3-Pentanediol H 3 C H CH 2 CH 3 H HO OH 2S,3S-2,3-Pentanediol H11001 OsO 4 , (CH 3 ) 3 COOH (CH 3 ) 3 COH, HO H11002 H H CH 2 CH 3 CH 3 cis-2-Pentene OH OH CH 3 CH 2 CH 3 H H 2S,3R-2,3-Pentanediol CH 3 H CH 2 CH 3 H HO OH 2R,3S-2,3-Pentanediol H11001 1,2,4-Butanetriol HOCH 2 CHCH 2 CH 2 OH OH H H11001 heat H11001 H 2 O 3-Hydroxyoxolane (C 4 H 8 O 2 ) HO O HIO 4 H 3 C CH 2 OHHO H11001 O HCH O H 3 C (74%) 1. LiAlH 4 2. H 2 O H 3 C COCH 3 O CH 3 CO O H11001H11001CH 3 CH 2 OH CH 3 OH H 3 C CH 2 OHHO (96%) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 15.32 (a) The task of converting a ketone to an alkene requires first the reduction of the ketone to an alcohol and then dehydration. In practice the two-step transformation has been carried out in 54% yield by treating the ketone with sodium borohydride and then heating the resulting alcohol with p-toluenesulfonic acid. Of course, sodium borohydride may be replaced by other suitable reducing agents, and p-toluenesulfonic acid is not the only acid that could be used in the dehydration step. (b) This problem and the next one illustrate the value of reasoning backward. The desired prod- uct, cyclohexanol, can be prepared cleanly from cyclohexanone. Once cyclohexanone is recognized to be a key intermediate, the synthetic pathway becomes apparent—what is needed is a method to convert the indicated starting material to cyclohexa- none. The reagent ideally suited to this task is periodic acid. The synthetic sequence to be fol- lowed is therefore (c) No direct method allows a second hydroxyl group to be introduced at C-2 of 1-phenylcyclo- hexanol in a single step. We recognize the product as a vicinal diol and recall that such com- pounds are available by hydroxylation of alkenes. This tells us that we must first dehydrate the tertiary alcohol, then hydroxylate the resulting alkene. The syn stereoselectivity of the hydroxylation step ensures that the product will have its hydroxyl groups cis, as the problem requires. H 2 SO 4 heat (CH 3 ) 3 COOH (CH 3 ) 3 COH OsO 4 , HO H11002 C 6 H 5 OH 1-Phenylcyclohexanol C 6 H 5 1-Phenylcyclohexene C 6 H 5 OH OH 1-Phenyl-cis-1,2- cyclohexanediol C 6 H 5 OH OH C 6 H 5 OH C 6 H 5 HIO 4 NaBH 4 CH 3 OH CH 2 OH OH 1-(Hydroxymethyl)- cyclohexanol O Cyclohexanone OH Cyclohexanol OH CH 2 OH OH O NaBH 4 CH 3 OH O OH H H11001 heat 386 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 387 15.33 Because the target molecule is an eight-carbon secondary alcohol and the problem restricts our choices of starting materials to alcohols of five carbons or fewer, we are led to consider building up the carbon chain by a Grignard reaction. The disconnection shown leads to a three-carbon aldehyde and a five-carbon Grignard reagent. Starting with the corresponding alcohols, the following synthetic scheme seems reasonable. First, propanal is prepared. After converting 2-pentanol to its bromo derivative, a solution of the Grignard reagent is prepared. Reaction of the Grignard reagent with the aldehyde yields the desired 4-methyl-3-heptanol. 15.34 Our target molecule is void of functionality and so requires us to focus attention on the carbon skele- ton. Notice that it can be considered to arise from three ethyl groups. Considering the problem retrosynthetically, we can see that a key intermediate having the carbon skeleton of the desired product is 3-methyl-3-pentanol. This becomes apparent from the fact that alkanes may be prepared from alkenes, which in turn are available from alcohols. The desired alcohol may be prepared from reaction of an acetate ester with a Grignard reagent, ethylmagnesium bromide. CH 3 CORCH 3 CH 2 CHCH 2 CH 3 CH 3 CH 2 CCH 2 CH 3 OCH 3 OH H11001 2CH 3 CH 2 MgBr CH 3 CH 3 CH 2 CH 2 CH 3 CH CH 3 3-Methylpentane 1. diethyl ether 2. H 3 O H11001 1-Methylbutylmagnesium bromide Propanal CH 3 CH 2 CH O CH 3 CHCH 2 CH 2 CH 3 MgBr 4-Methyl-3-heptanol CH 3 CHCH 2 CH 2 CH 3 HOCHCH 2 CH 3 H11001 CH 3 CHCH 2 CH 2 CH 3 OH CH 3 CHCH 2 CH 2 CH 3 Br 2-Pentanol 2-Bromopentane 1-Methylbutylmagnesium bromide CH 3 CHCH 2 CH 2 CH 3 MgBr PBr 3 Mg diethyl ether CH 3 CH 2 CH 2 OH 1-Propanol Propanal PCC or PDC CH 2 Cl 2 CH 3 CH 2 CH O H11001CH 3 CH 2 CH O CHCH 2 CH 2 CH 3 H11002 CH 3 CHCH 2 CH 2 CH 3 CH 3 OH CH 3 CH 2 CH 4-Methyl-3-heptanol Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The carbon skeleton can be assembled in one step by the reaction of ethylmagnesium bromide and ethyl acetate. The resulting tertiary alcohol is converted to the desired hydrocarbon by acid-catalyzed dehydration and catalytic hydrogenation of the resulting mixture of alkenes. Because the problem requires that ethanol be the ultimate starting material, we need to show the preparation of the ethylmagnesium bromide and ethyl acetate used in constructing the carbon skeleton. 15.35 (a) Retrosynthetically, we can see that the cis carbon–carbon double bond is available by hydro- genation of the corresponding alkyne over the Lindlar catalyst. CH 3 CH 2 CH CHCH 2 CH 2 OH CH 3 CH 2 C CCH 2 CH 2 OH CH 3 COH O Acetic acid CH 3 COCH 2 CH 3 O Ethyl acetate H + CH 3 CH 2 OHH11001 CH 3 CH 2 OH CH 3 COH K 2 Cr 2 O 7 H 2 SO 4 , H 2 O, heat O Ethanol Acetic acid Mg diethyl ether CH 3 CH 2 Br Ethyl bromideEthanol CH 3 CH 2 OH Ethylmagnesium bromide CH 3 CH 2 MgBr PBr 3 CH 3 CCH 2 CH 3 CH 3 CH(CH 2 CH 3 ) 2 3-Methylpentane 3-Methyl-3-pentanol 3-Methyl-2-pentene CH 3 C CH 2 CH 3 CH 2 CH 3 H11001 OH H H11001 CHCH 3 C(CH 2 CH 3 ) 2 H 2 C 2-Ethyl-1-butene H11001trans)(cis H 2 , Ni 1. diethyl ether 2. H 3 O H11001 CH 3 CCH 2 CH 3 Ethylmagnesium bromide Ethyl acetate 3-Methyl-3-pentanol 2CH 3 CH 2 MgBr CH 3 COCH 2 CH 3 O CH 2 CH 3 H11001 OH 388 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 389 The @CH 2 CH 2 OH unit can be appended to an alkynide anion by reaction with ethylene oxide. The alkynide anion is derived from 1-butyne by alkylation of acetylene. This analysis suggests the following synthetic sequence: (b) The compound cited is the aldehyde derived by oxidation of the primary alcohol in part (a). Oxidize the alcohol with PDC or PCC in CH 2 Cl 2 . 15.36 Even though we are given the structure of the starting material, it is still better to reason backward from the target molecule rather than forward from the starting material. The desired product contains a cyano (@CN) group. The only method we have seen so far for introducing such a function into a molecule is by nucleophilic substitution. The last step in the syn- thesis must therefore be This step should work very well, since the substrate is a primary benzylic halide, cannot undergo elimination, and is very reactive in S N 2 reactions. The primary benzylic halide can be prepared from the corresponding alcohol by any of a number of methods. Suitable reagents include HBr, PBr 3 , or SOCl 2 . CH 2 OH OCH 3 CH 2 X OCH 3 CH 2 X CN H11002 OCH 3 H11001 X H11002 H11001 CH 2 CN OCH 3 PDC or PCC in CH 2 Cl 2CC H CH 3 CH 2 H CH 2 CH 2 OH cis-3-Hexen-1-ol cis-3-Hexenal CC H CH 3 CH 2 H CH 2 CH O 1. NaNH 2 , NH 3 1. NaNH 2 , NH 3 HC CH CH 3 CH 2 CCH 1-Butyne 3-Hexyn-1-ol CH 3 CH 2 C CCH 2 CH 2 OH Acetylene 2. CH 3 CH 2 Br CC H CH 3 CH 2 H CH 2 CH 2 OH cis-3-Hexen-1-ol H 2 Lindlar Pd 2. H 2 CCH 2 O CH 3 CH 2 C CCH 2 CH 2 OH H11001CH 3 CH 2 C H11002 C H 2 C CH 2 O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Now we only need to prepare the primary alcohol from the given starting aldehyde, which is ac- complished by reduction. Reduction can be achieved by catalytic hydrogenation, with lithium aluminum hydride, or with sodium borohydride. The actual sequence of reactions as carried out is as shown. Another three-step synthesis, which is reasonable but does not involve an alcohol as an interme- diate, is 15.37 (a) Addition of hydrogen chloride to cyclopentadiene takes place by way of the most stable car- bocation. In this case it is an allylic carbocation. HCl Cl H11002 H11001 H11001 (Allylic carbocation; more stable) Cl 3-Chlorocyclopentene (80–90%) (Compound A) (Not allylic; less stable) not Clemmensen or Wolff–Kishner reduction N-bromosuccinimide hH9263 CN H11002 CH OCH 3 O m-Methoxy- benzaldehyde CH 3 OCH 3 m-Methoxytoluene CH 2 Br OCH 3 m-Methoxybenzyl bromide CH 2 CN OCH 3 m-Methoxybenzyl cyanide H 2 , Pt ethanol (100% yield) HBr, benzene (98% yield) NaCN ethanol, water (87% yield) CH O OCH 3 m-Methoxy- benzaldehyde CH 2 OH OCH 3 m-Methoxybenzyl alcohol CH 2 Br OCH 3 m-Methoxybenzyl bromide CH 2 CN OCH 3 m-Methoxybenzyl cyanide CH OCH 3 O CH 2 OH OCH 3 390 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 391 Hydrolysis of 3-chlorocyclopentene gives the corresponding alcohol. Sodium bicarbonate in water is a weakly basic solvolysis medium. Oxidation of compound B (a secondary alcohol) gives the ketone 2-cyclopenten-1-one. (b) Thionyl chloride converts alcohols to alkyl chlorides. Ozonolysis cleaves the carbon–carbon double bond. Reduction of compound E yields the corresponding alcohol. (c) N-Bromosuccinimide is a reagent designed to accomplish benzylic bromination. NBS benzoyl peroxide, heat Br CH 3 1-Bromo-2-methylnaphthalene Br CH 2 Br 1-Bromo-2-(bromomethyl)naphthalene (compound G) HCCH 2 CH 2 CHCH 3 Cl O 4-Chloropentanal HOCH 2 CH 2 CH 2 CHCH 3 Cl 4-Chloro-1-pentanol (compound F) NaBH 4 H 2 C CHCH 2 CH 2 CHCH 3 Cl HCCH 2 CH 2 CHCH 3 H11001 HCH Cl O O 1. O 3 2. reductive workup Compound D 4-Chloropentanal (compound E) Formaldehyde H 2 C Cl CHCH 2 CH 2 CHCH 3 SOCl 2 pyridine H 2 C OH CHCH 2 CH 2 CHCH 3 5-Hexen-2-ol 5-Chloro-1-hexene (compound D) Na 2 Cr 2 O 7 H 2 SO 4 , H 2 O OH Compound B O 2-Cyclopenten-1-one (60–68%) (compound C) NaHCO 3 H 2 O Cl Compound A OH 2-Cyclopenten-1-ol (88%) (compound B) Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Hydrolysis of the benzylic bromide gives the corresponding benzylic alcohol. The bromine that is directly attached to the naphthalene ring does not react under these conditions. Oxidation of the primary alcohol with PCC gives the aldehyde. 15.38 The alcohol is tertiary and benzylic and yields a relatively stable carbocation. The alcohol is chiral, but the carbocation is not. Thus, irrespective of which enantiomer of 2-phenyl- 2-butanol is used, the same carbocation is formed. The carbocation reacts with ethanol to give an optically inactive mixture containing equal quantities of enantiomers (racemic). 15.39 The difference between the two ethers is that 1-O-benzylglycerol contains a vicinal diol function, but 2-O-benzylglycerol does not. Periodic acid will react with 1-O-benzylglycerol but not with 2-O- benzylglycerol. HOCH 2 CHCH 2 OH no reaction OCH 2 C 6 H 5 2-O-Benzylglycerol HIO 4 H11001C 6 H 5 CH 2 OCH 2 CHCH 2 OH OH 1-O-Benzylglycerol C 6 H 5 CH 2 OCH 2 CH O 2-Benzyloxyethanal HCH O Formaldehyde HIO 4 H H11001 H11001H11001 C CH 2 CH 3 CH 3 OCH 2 CH 3 2-Ethoxy-2-phenylbutane (50% R, 50% S) C H11001 CH 2 CH 3 CH 3 1-Methyl-1-phenylpropyl cation CH 3 CH 2 OH Ethanol C CH 2 CH 3 CH 3 OH C H11001 CH 2 CH 3 CH 3 H 2 SO 4 2-Phenyl-2-butanol 1-Methyl-1-phenylpropyl cation PCC CH 2 Cl 2 Br CH 2 OH (1-Bromo-2-naphthyl)methanol Br CH O 1-Bromonaphthalene-2-carboxaldehyde (compound I) H 2 O, CaCO 3 heat Br CH 2 Br 1-Bromo-2-(bromomethyl)naphthalene Br CH 2 OH (1-Bromo-2-naphthyl)methanol (compound H) 392 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 393 15.40 The formation of an alkanethiol by reaction of an alkyl halide or alkyl p-toluenesulfonate with thiourea occurs with inversion of configuration in the step in which the carbon–sulfur bond is formed. Thus, the formation of (R)-2-butanethiol requires (S)-sec-butyl p-toluenesulfonate, which then reacts with thiourea by an S N 2 pathway. The p-toluenesulfonate is formed from the corre- sponding alcohol by a reaction that does not involve any of the bonds to the stereogenic center. Therefore, begin with (S)-2-butanol. 15.41 (a) Cysteine contains an GSH group and is a thiol. Oxidation of thiols gives rise to disulfides. Biological oxidation of cysteine gives the disulfide cystine. (b) Oxidation of a thiol yields a series of acids, including a sulfinic acid and a sulfonic acid. Biological oxidation of cysteine can yield, in addition to the disulfide cystine, cysteine sulfinic acid and the sulfonic acid cysteic acid. 15.42 The ratio of carbon to hydrogen in the molecular formula is C n H 2nH110012 (C 8 H 18 O 2 ), and so the com- pound has no double bonds or rings. The compound cannot be a vicinal diol, because it does not react with periodic acid. The NMR spectrum is rather simple as all peaks are singlets. The 12-proton singlet at H9254 1.2 ppm must correspond to four equivalent methyl groups and the four-proton singlet at H9254 1.6 ppm to two equivalent methylene groups. No nonequivalent protons can be vicinal, because no splitting is ob- served. The two-proton singlet at H9254 2.0 ppm is due to the hydroxyl protons of the diol. oxidizeoxidize Cysteine HSCH 2 CHCO H11002 O H11001 NH 3 Cysteine sulfinic acid (C 3 H 7 NO 4 S) H11001 SCH 2 CHCO H11002 OO H11002 H11001 NH 3 HO Cysteic acid (C 3 H 7 NO 5 S) 2H11001 SCH 2 CHCO H11002 OO H11002 O H11002 H11001 NH 3 HO RSH Thiol Sulfinic acid Sulfonic acid RS H11001 OH O H11002 O H11002 RS 2H11001 OH O H11002 oxidize Cysteine 2HSCH 2 CHCO H11002 H11001 NH 3 O Cystine H11002 OCCHCH 2 S H11001 NH 3 O SCH 2 CHCO H11002 H11001 NH 3 O 2RSH Thiol Disulfide RSSR oxidize H CH 3 CH 3 CH 2 COH (S)-2-Butanol p-toluenesulfonyl chloride retention of configuration H CH 3 CH 3 CH 2 COTs (S)-sec-Butyl p-toluenesulfonate H CH 3 CH 2 CH 3 CHS (R)-2-Butanethiol 1. (H 2 N) 2 C S inversion of configuration 2. NaOH Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The compound is 2,5-dimethyl-2,5-hexanediol. 15.43 The molecular formula of compound A (C 8 H 10 O) corresponds to an index of hydrogen deficiency of 4. The 4 hydrogen signal at H9254 7.2 ppm in the 1 H NMR spectrum suggests these unsaturations are due to a disubstituted benzene ring. That the ring is para-substituted is supported by the symmetry of the signal; it is a pair of doublets, not a quartet. The broad signal (1H) at H9254 2.1 ppm undergoes rapid exchange with D 2 O, indicating it is the pro- ton of the hydroxyl group of an alcohol. As the remaining signals are singlets, with areas of 2H and 3H, respectively, compound A can be identified as 4-methylbenzyl alcohol. 15.44 (a) This compound has only two different types of carbons. One type of carbon comes at low field and is most likely a carbon bonded to oxygen and three other equivalent carbons. The spec- trum leads to the conclusion that this compound is tert-butyl alcohol. (b) Four different types of carbons occur in this compound. The only C 4 H 10 O isomers that have four nonequivalent carbons are CH 3 CH 2 CH 2 CH 2 OH, , and CH 3 OCH 2 CH 2 CH 3 . The lowest field signal, the one at 69.2 ppm from the carbon that bears the oxygen substituent, is a methine (CH). The compound is therefore 2-butanol. (c) This compound has two equivalent CH 3 groups, as indicated by the signal at 18.9 ppm. Its lowest field carbon is a CH 2 , and so the group @CH 2 O must be present. The compound is 2- methyl-1-propanol. 15.45 The compound has only three carbons, none of which is a CH 3 group. Two of the carbon signals arise from CH 2 groups; the other corresponds to a CH group. The only structure consistent with the observed data is that of 3-chloro-1,2-propanediol. HOCH 2 CH 2 Cl OH CH H 3 CCH 2 OH CH 3 CH 30.8 ppm 69.4 ppm18.9 ppm CH 3 CHCH 2 CH 3 OH CH 3 CHCH 2 CH 3 OH H 3 COH CH 3 CH 3 C 31.2 ppm 68.9 ppm CH 2 OHH 3 C H9254 4.7 ppm H9254 2.1 ppm H9254 7.2 ppm H9254 2.4 ppm CH 3 CCH 2 CH 2 CCH 3 CH 3 CH 3 OH OH 394 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 395 The structure cannot be correct. It would exhibit only two peaks in its 13 C NMR spectrum, because the two terminal carbons are equivalent to each other. 15.46 The observation of a peak at m/z 31 in the mass spectrum of the compound suggests the presence of a primary alcohol. This fragment is most likely . On the basis of this fact and the appearance of four different carbons in the 13 C NMR spectrum, the compound is 2-ethyl-1-butanol. 15.47–15.49 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. For each of the following reactions give the structure of the missing reactant or reagent. (a) (b) (c) (d) (e) A-2. For the following reactions of 2-phenylethanol, C 6 H 5 CH 2 CH 2 OH, give the correct reagent or product(s) omitted from the equation. (a) (b) (c) (d) C 6 H 5 CH 2 CH 2 OH C 6 H 5 CH 2 CO 2 H ? C 6 H 5 CH 2 CH 2 OH (2 mol) H 2 O H H11001 heat H11001 ? C 6 H 5 CH 2 CH 2 OH CH 3 CO 2 CH 2 CH 2 ? C 6 H 5 CH 2 CH 2 OH PCC CH 2 Cl 2 ? C 6 H 5 CH 2 SHC 6 H 5 CH 2 Br 1. ? 2. NaOH CH 3 CH 3 OH OH ? C 6 H 5 CH 2 CHCH 2 OH CH 3 C 6 H 5 CH 2 CCH 2 CH 3 ? 1. diethyl ether 2. H 3 O H11001 ? 2CH 3 CH 2 MgBr CH 3 CH 2 OHH11001H11001C 6 H 5 C(CH 2 CH 3 ) 2 OH 1. LiAlH 4 2. H 2 O ? HOH CH 3 CH 2 CH 3 CH 2 CH 2 OHCH 44 ppm 65 ppm 23 ppm 11 ppm H 2 COH H11001 HOCH 2 CHCH 2 OH Cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-3. Write the structure of the major organic product formed in the reaction of 2-propanol with each of the following reagents: (a) Sodium amide (NaNH 2 ) (b) Potassium dichromate (K 2 Cr 2 O 7 ) in aqueous sulfuric acid, heat (c) PDC in dichloromethane (d) Acetic acid in the presence of dissolved hydrogen chloride (e) in the presence of pyridine ( f ) in the presence of pyridine (g) in the presence of pyridine A-4. Outline two synthetic schemes for the preparation of 3-methyl-1-butanol using different Grignard reagents. A-5. Give the structure of the reactant, reagent, or product omitted from each of the following. Show stereochemistry where important. (a) (b) (c) A-6. Give the reagents necessary to carry out each of the following transformations: (a) Conversion of benzyl alcohol (C 6 H 5 CH 2 OH) to benzaldehyde (b) Conversion of benzyl alcohol to benzoic acid (C 6 H 5 CO 2 H) (c) Conversion of to (d) Conversion of cyclohexene to cis-1,2-cyclohexanediol A-7. Provide structures for compounds A to C in the following reaction scheme: K 2 Cr 2 O 7 H H11001 , H 2 O CH 3 OH, H H11001 1. LiAlH 4 2. H 2 O H H11001 , heat A(C 5 H 12 O 2 ) B(C 5 H 8 O 3 ) C(C 6 H 10 O 3 ) A H11001 CH 3 OH H 3 C O CHCH 2 CH 2 CH 2 OHH 2 C CHCH 2 CH 2 CO 2 HH 2 C (C 6 H 5 CH O) OsO 4 , (CH 3 ) 3 COOH (CH 3 ) 3 COH, HO H11002 ? 2,3-butanediol (chiral diastereomer) H H11001 heat ? (a diol) CH 3 O OH OH H CH 3 HIO 4 ? CH 3 COCCH 3 O O CClCH 3 CH 2 O SO 2 ClH 3 C (CH 3 COH) O 396 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 397 A-8. Using any necessary organic or inorganic reagents, outline a scheme for each of the follow- ing conversions. PART B B-1. Ethanethiol (CH 3 CH 2 SH) is a gas at room temperature, but ethanol is a liquid. The reason for this is (a) The C@S@H bonds in ethanethiol are linear. (b) The C@O@H bonds in ethanol are linear. (c) Ethanol has a lower molecular weight. (d) Ethanethiol has a higher boiling point. (e) Ethanethiol is less polar. B-2. Which of the following would yield a secondary alcohol after the indicated reaction, followed by hydrolysis if necessary? (a) LiAlH 4 H11001 a ketone (b)CH 3 CH 2 MgBr H11001 an aldehyde (c) 2-Butene H11001 aqueous H 2 SO 4 (d) All of these B-3. What is the major product of the following reaction? (a)(c) (b)(d) B-4. Which of the esters shown, after reduction with LiAlH 4 and aqueous workup, will yield two molecules of only a single alcohol? (a)CH 3 CH 2 CO 2 CH 2 CH 3 (b)C 6 H 5 CO 2 C 6 H 5 (c)C 6 H 5 CO 2 CH 2 C 6 H 5 (d) None of these CO 2 CH 3 OH CH 2 OH O CO 2 H OH CH 2 OH OH NaBH 4 CH 3 OH ? CO 2 H O C 6 H 5 CH 3 (c)C 6 H 5 CH 2 CH 2 CO 2 CH 2 CH 3 ? (b) ? CH O CCH 2 CH 3 O CHCH 3 (CH 3 ) 2 C (CH 3 ) 2 CHCCH 3 (a) ? O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-5. For the following reaction, select the statement that best describes the situation. (a) The alcohol is oxidized to an acid, and the Cr(VI) is reduced. (b) The alcohol is oxidized to an aldehyde, and the Cr(VI) is reduced. (c) The alcohol is reduced to an aldehyde, and the Cr(III) is oxidized. (d) The alcohol is oxidized to a ketone, and the Cr(VI) is reduced. B-6. What is the product from the following esterification? (a)(c) (b d) B-7. The following substance acts as a coenzyme in which of the following biological reactions? (a) Alcohol oxidation (c) Aldehyde reduction (b) Ketone reduction (d) None of these B-8. Which of the following alcohols gives the best yield of dialkyl ether on being heated with a trace of sulfuric acid? (a) 1-Pentanol (c) Cyclopentanol (b) 2-Pentanol (d) 2-Methyl-2-butanol B-9. What is the major organic product of the following sequence of reactions? (a)(c) (CH 3 ) 2 CHCH 2 CH 2 OH (b d) (CH 3 ) 2 CHCH 2 CH 2 CH 2 OH B-10. What is the product of the following reaction? CC H H 3 C H CH 3 CH 3 CH 3 HO OHH H 3 CH 3 CH 3 H HHO OH 2 CH 3 CH 3 H OH OH H 1 OsO 4 (cat), (CH 3 ) 3 COOH (CH 3 ) 3 COH, HO H11002 (CH 3 ) 2 CHCH 2 CHCH 3 OH (CH 3 )CHCHCH 2 CH 3 OH PBr 3 Mg H 3 O H11001 ?(CH 3 ) 2 CHCH 2 OH CH 2 H 2 C O (R H11005 adenine dinucleotide) O CNH 2 NH11001 R CH 3 CH 2 COCH 2 C 6 H 5 18 O 18 OCH 2 CH 3 C 6 H 5 CH 2 C O 18 OCH 2 CH 3 C 6 H 5 CH 2 C 18 O C 6 H 5 CH 2 COCH 2 CH 3 18 O CH 3 CH 2 C 6 H 5 CH 2 CO 2 H H11001 18 OH H H11001 heat ? RCH 2 OH PCC [C 5 H 5 NH H11001 ClCrO 3 H11002 ]H11001 398 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALCOHOLS, DIOLS, AND THIOLS 399 (a) Only 1 (d) A 1:1 mixture of 2 and 3. (b) Only 2 (e) A 1:1:1 mixture of 1, 2, and 3. (c) Only 3 B-11. Which reaction is the best method for preparing (R)-2-butanol? (a) (b) (c) (d) (e) None of these would be effective. B-12. An organic compound B is formed by the reaction of ethylmagnesium iodide (CH 3 CH 2 MgI) with a substance A, followed by treatment with dilute aqueous acid. Compound B does not react with PCC or PDC in dichloromethane. Which of the following is a possible candidate for A? (a)(d) (b e) None of these (c) B-13. Which alcohol of molecular formula C 5 H 12 O has the fewest signals in its 13 C NMR spectrum? (a) 1-Pentanol (d) 3-Methyl-2-butanol (b) 2-Pentanol (e) 2,2-Dimethyl-1-propanol (c) 2-Methyl-2-butanol B-14. Which of the following reagents would carry out the following transformation? (D H11005 2 H, the mass-2 isotope of hydrogen) (a) NaBD 4 in CH 3 OH (b) NaBD 4 in CH 3 OD (c) LiAlH 4 , then D 2 O (d) LiAlD 4 , then D 2 O (e) NaBH 4 in CH 3 OD CCH 3 O CCH 3 OH D ? H 2 CCH 2 O H 2 CO CH 3 CH 2 CCH 3 O CH 3 CH O 1. CH 3 CH 2 Li, diethyl ether 2. H 3 O H11001 CH 3 CH O 1. CH 3 MgBr, diethyl ether 2. H 3 O H11001 O CH 3 CH 2 CH 1. LiAlH 4 , diethyl ether 2. H 2 O O OCCH 3 C H H 3 C CH 3 CH 2 1. LiAlH 4 , diethyl ether 2. H 2 O O CH 3 CH 2 CCH 3 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-15. Which sequence of steps describes the best synthesis of 2-methyl-3-pentanone? (a) 1. 1-Propanol H11001 (CH 3 ) 2 CHMgBr, diethyl ether 2. H 3 O H11001 3. PDC, CH 2 Cl 2 (b) 1. 1-Propanol H11001 Na 2 Cr 2 O 7 , H 2 SO 4 , H 2 O, heat 2. SOCl 2 3. (CH 3 ) 2 CHCl, AlCl 3 (c) 1. 1-Propanol H11001 PCC, CH 2 Cl 2 2. (CH 3 ) 2 CHLi, diethyl ether 3. H 3 O H11001 4. Na 2 Cr 2 O 7 , H 2 SO 4 , H 2 O, heat (d) 1. 2-Propanol H11001 Na 2 Cr 2 O 7 , H 2 SO 4 , H 2 O, heat 2. CH 3 CH 2 CH 2 Li, diethyl ether 3. H 3 O H11001 4. PCC, CH 2 Cl 2 O 2-Methyl-3-pentanone 400 ALCOHOLS, DIOLS, AND THIOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website