CHAPTER 17 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP SOLUTIONS TO TEXT PROBLEMS 17.1 (b) The longest continuous chain in glutaraldehyde has five carbons and terminates in aldehyde functions at both ends. Pentanedial is an acceptable IUPAC name for this compound. (c) The three-carbon parent chain has a double bond between C-2 and C-3 and a phenyl sub- stituent at C-3. (d) Vanillin can be named as a derivative of benzaldehyde. Remember to cite the remaining sub- stituents in alphabetical order. HO CH O CH 3 O 4-Hydroxy-3-methoxybenzaldehyde (vanillin) 3-Phenyl-2-propenal (cinnamaldehyde) C 6 H 5 CH CHCH O 2 1 3 Pentanedial (glutaraldehyde) HCCH 2 CH 2 CH 2 CH O O 1 23 4 5 426 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 427 17.2 (b) First write the structure from the name given. Ethyl isopropyl ketone has an ethyl group and an isopropyl group bonded to a carbonyl group. Ethyl isopropyl ketone may be alternatively named 2-methyl-3-pentanone. Its longest contin- uous chain has five carbons. The carbonyl carbon is C-3 irrespective of the direction in which the chain is numbered, and so we choose the direction that gives the lower number to the po- sition that bears the methyl group. (c) Methyl 2,2-dimethylpropyl ketone has a methyl group and a 2,2-dimethylpropyl group bonded to a carbonyl group. The longest continuous chain has five carbons, and the carbonyl carbon is C-2. Thus, methyl 2,2-dimethylpropyl ketone may also be named 4,4-dimethyl-2-pentanone. (d) The structure corresponding to allyl methyl ketone is Because the carbonyl group is given the lowest possible number in the chain, the substitutive name is 4-penten-2-one not 1-penten-4-one. 17.3 No. Lithium aluminum hydride is the only reagent we have discussed that is capable of reducing carboxylic acids (Section 15.3). 17.4 The target molecule, 2-butanone, contains four carbon atoms. The problem states that all of the car- bons originate in acetic acid, which has two carbon atoms. This suggests the following disconnections: The necessary aldehyde (acetaldehyde) is prepared from acetic acid by reduction followed by oxidation in an anhydrous medium. Ethylmagnesium bromide may be obtained from acetic acid by the following sequence: CH 3 CH 2 OH Ethanol (Prepared as previously) CH 3 CH 2 Br Ethyl bromide CH 3 CH 2 MgBr Ethylmagnesium bromide HBr or PBr 3 Mg diethyl ether CH 3 CO 2 H Acetic acid CH 3 CH 2 OH Ethanol CH 3 CH O Acetaldehyde 1. LiAlH 4 2. H 2 O PDC CH 2 Cl 2 CH 3 CHCH 2 CH 3 OH CH 3 CCH 2 CH 3 O O CH 3 CH H11001 CH 2 CH 3 H11002 2-Butanone CH 3 CCH 2 CH CH 2 O CH 3 CCH 2 CCH 3 CH 3 OCH 3 CH 3 CH 2 CCHCH 3 O CH 3 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 428 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP The preparation of 2-butanone is completed as follows: 17.5 Chloral is trichloroethanal, . Chloral hydrate is the addition product of chloral and water. 17.6 Methacrylonitrile is formed by the dehydration of acetone cyanohydrin, and thus has the structure shown. 17.7 The overall reaction is HCl is a strong acid and, when dissolved in ethanol, transfers a proton to ethanol to give ethylox- onium ion. Thus, we can represent the acid catalyst as the conjugate acid of ethanol. The first three steps correspond to acid-catalyzed addition of ethanol to the carbonyl group to yield a hemiacetal. Step 1: Step 2: C 6 H 5 CH H11001 O H11001 C 6 H 5 CH OH CH 2 CH 3 H O H11001 H CH 2 CH 3 H O H11001 CH 2 CH 3 H OC 6 H 5 CH OH H11001 H11001 C 6 H 5 CH CH 2 CH 3 H HH11001 O O Ethanol 2CH 3 CH 2 OHH11001 Benzaldehyde diethyl acetal C 6 H 5 (OCH 2 CH 3 ) 2 Water H 2 OH11001 Benzaldehyde C 6 H 5 CH O HCl MethacrylonitrileAcetone cyanohydrin CH 3 CCH 3 OH CN H H11001 , heat (H11002H 2 O) CH 3 CCH 2 CN Chloral hydrate Cl 3 CCH OH OH CCl 3 CH O H11001 CH 3 CH 2 MgBr Ethylmagnesium bromide Acetaldehyde CH 3 CH O 2-Butanone CH 3 CCH 2 CH 3 O 2-Butanol CH 3 CHCH 2 CH 3 OH 1. diethyl ether 2. H 3 O H11001 K 2 Cr 2 O 7 H 2 SO 4 , H 2 O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Step 3: Formation of the hemiacetal is followed by loss of water to give a carbocation. Step 4: Step 5: The next two steps describe the capture of the carbocation by ethanol to give the acetal: Step 6: Step 7: 17.8 (b) 1,3-Propanediol forms acetals that contain a six-membered 1,3-dioxane ring. H H11001 H11001 C 6 H 5 H 2-Phenyl-1,3-dioxane HOCH 2 CH 2 CH 2 OH 1,3-Propanediol C 6 H 5 CH O Benzaldehyde O O H11001 OH CH 2 CH 3 C 6 H 5 CH OCH 2 CH 3 OCH 2 CH 3 Acetal C 6 H 5 CH OCH 2 CH 3 O CH 3 CH 2 H H11001 H11001 O CH 2 CH 3 H H H11001 C 6 H 5 CH OCH 2 CH 3 O CH 3 CH 2 H H11001 H11001C 6 H 5 CH OCH 2 CH 3 O CH 2 CH 3 H H11001 C 6 H 5 CH H11001 OHHOCH 2 CH 3 H11001 H11001 C 6 H 5 CH HH O OCH 2 CH 3 H11001C 6 H 5 C HOH O H CH 2 CH 3 OCH 2 CH 3 H11001 H O H11001 H H CH 2 CH 3 H11001C 6 H 5 C HO H OCH 2 CH 3 H11001C 6 H 5 CH OH O CH 2 CH 3 O H11001 H CH 2 CH 3 H H11001 H11001 Hemiacetal C 6 H 5 CH OH H O H CH 2 CH 3 OCH 2 CH 3 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 429 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) The cyclic acetal derived from isobutyl methyl ketone and ethylene glycol bears an isobutyl group and a methyl group at C-2 of a 1,3-dioxolane ring. (d) Because the starting diol is 2,2-dimethyl-1,3-propanediol, the cyclic acetal is six-membered and bears two methyl substituents at C-5 in addition to isobutyl and methyl groups at C-2. 17.9 The overall reaction is The mechanism of acetal hydrolysis is the reverse of acetal formation. The first four steps convert the acetal to the hemiacetal. Step 1: Step 2: Step 3: H11001C 6 H 5 CH OCH 2 CH 3 H11001 H H O C 6 H 5 CH OCH 2 CH 3 O HH H11001 H11001 O H CH 2 CH 3 C 6 H 5 CH OCH 2 CH 3 H11001 C 6 H 5 CH OCH 2 CH 3 O H11001 HCH 3 CH 2 H11001H11001C 6 H 5 CH O H CH 2 CH 3 OCH 2 CH 3 OCH 2 CH 3 H H11001 O H CH 2 CH 3 C 6 H 5 CH OCH 2 CH 3 O H11001 HCH 3 CH 2 H11001H11001C 6 H 5 CH(OCH 2 CH 3 ) 2 Benzaldehyde diethyl acetal H 2 O Water C 6 H 5 CH O Benzaldehyde 2CH 3 CH 2 OH Ethanol HCl H11001 H H11001 (CH 3 ) 2 CHCH 2 CCH 3 O Isobutyl methyl ketone HOCH 2 CCH 2 OH CH 3 CH 3 2,2-Dimethyl-1,3- propanediol 2-Isobutyl-2,5,5-trimethyl- 1,3-dioxane CH 3 (CH 3 ) 2 CHCH 2 CH 3 H 3 C OO Ethylene glycol HOCH 2 CH 2 OHH11001 H H11001 Isobutyl methyl ketone (CH 3 ) 2 CHCH 2 CCH 3 O 2-Isobutyl-2-methyl-1,3-dioxolane (CH 3 ) 2 CHCH 2 O O CH 3 430 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Step 4: Step 5: Step 6: Step 7: 17.10 The conversion requires reduction; however, the conditions necessary (LiAlH 4 ) would also reduce the ketone carbonyl. The ketone functionality is therefore protected as the cyclic acetal. Reduction of the carboxylic acid may now be carried out. Hydrolysis to remove the protecting group completes the synthesis. CH 2 OH CH 2 OHCH 3 C O H 2 O, HCl 4-Acetylbenzyl alcohol O O C H 3 C COH O CH 2 OH 1. LiAlH 4 2. H 2 O O O C H 3 C O O C H 3 C 4-Acetylbenzoic acid COH O CH 3 C O HOCH 2 CH 2 OH p-toluenesulfonic acid, benzene COH O O O C H 3 C O H CH 2 CH 3 H11001 O H CH 2 CH 3 H11001C 6 H 5 CH O H11001 H11001 C 6 H 5 CH O H H O H CH 2 CH 3 H11001 H11001 O H CH 2 CH 3 H11001C 6 H 5 CH O C 6 H 5 CH OH H O H CH 2 CH 3 OH H CH 2 CH 3 H11001 H11001 O H CH 2 CH 3 H11001OCH 2 CH 3 H11001C 6 H 5 CH HO C 6 H 5 CH OH O H CH 2 CH 3 OH H CH 2 CH 3 H11001 H11001OCH 2 CH 3 C 6 H 5 C O H11001 H HH H11001C 6 H 5 C OCH 2 CH 3 H HO Hemiacetal ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 431 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 17.11 (b) Nucleophilic addition of butylamine to benzaldehyde gives the carbinolamine. Dehydration of the carbinolamine produces the imine. (c) Cyclohexanone and tert-butylamine react according to the equation (d) 17.12 (b) Pyrrolidine, a secondary amine, adds to 3-pentanone to give a carbinolamine. Dehydration produces the enamine. 3-Pyrrolidino-2-penteneCarbinolamine intermediate CH 3 CH 2 CCH 2 CH 3 OH H11001 H 2 O N CH 3 CH CCH 2 CH 3 N H11001 3-Pentanone CH 3 CH 2 CCH 2 CH 3 O Carbinolamine intermediate CH 3 CH 2 CCH 2 CH 3 OH Pyrrolidine H N N H11001 CyclohexylamineAcetophenone Carbinolamine intermediate C 6 H 5 CCH 3 OH NH NH 2 C 6 H 5 CCH 3 O N-(1-Phenylethylidene)- cyclohexylamine C 6 H 5 CCH 3 N H11002H 2 O H11002H 2 O H11001 O Cyclohexanone HO NC(CH 3 ) 3 H Carbinolamine intermediate (CH 3 ) 3 CNH 2 tert-Butylamine NC(CH 3 ) 3 N-Cyclohexylidene- tert-butylamine N-Benzylidenebutylamine NCH 2 CH 2 CH 2 CH 3 CH OH H H11002H 2 O CH NCH 2 CH 2 CH 2 CH 3 Carbinolamine intermediate CH 3 CH 2 CH 2 CH 2 NH 2 CH O Benzaldehyde Butylamine H11001 CH NCH 2 CH 2 CH 2 CH 3 OH H 432 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) 17.13 (b) Here we see an example of the Wittig reaction applied to diene synthesis by use of an ylide containing a carbon–carbon double bond. (c) Methylene transfer from methylenetriphenylphosphorane is one of the most commonly used Wittig reactions. 17.14 A second resonance structure can be written for a phosphorus ylide with a double bond between phosphorus and carbon. As a third-row element, phosphorus can have more than 8 electrons in its valence shell. 17.15 (b) Two Wittig reaction routes lead to 1-pentene. One is represented retrosynthetically by the disconnection The other route is 1-Pentene CH 3 CH 2 CH 2 CH CH 2 H11001 Formaldehyde O HCH Butylidenetriphenylphosphorane CH 3 CH 2 CH 2 CH H11001 H11002 P(C 6 H 5 ) 3 H11001 1-Pentene CH 3 CH 2 CH 2 CH CH 2 CH 3 CH 2 CH 2 CH O Butanal Methylenetriphenyl- phosphorane H11001 H11002 CH 2 (C 6 H 5 ) 3 P Methylenetriphenylphosphorane (C 6 H 5 ) 3 PCH 2 CH 2 (C 6 H 5 ) 3 P H11001 H11002 CCH 3 O Cyclohexyl methyl ketone CCH 3 CH 2 2-Cyclohexylpropene (66%) H11001 H11001 Triphenylphosphine oxide O H11002 (C 6 H 5 ) 3 P H11001 Methylenetriphenyl- phosphorane (C 6 H 5 ) 3 P H11001 CH 2 H11002 H11001 Allylidenetriphenylphosphorane 1,3-Heptadiene (52%) Triphenylphosphine oxide Butanal CH 3 CH 2 CH 2 CH CH 3 CH 2 CH 2 CH O (C 6 H 5 ) 3 PCH 2 CH 2 CHCHCHCH H11001 (C 6 H 5 ) 3 P O H11002 H11001H11001 H11001 Carbinolamine intermediate 1-Piperidino-1- phenylethene PiperidineAcetophenone C 6 H 5 CCH 3 O H C 6 H 5 CCH 3 OH H11002H 2 O C 6 H 5 C CH 2 N NN ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 433 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 17.16 Ylides are prepared by the reaction of an alkyl halide with triphenylphosphine, followed by treat- ment with strong base. 2-Bromobutane is the alkyl halide needed in this case. 17.17 The overall reaction is In the first step, the peroxy acid adds to the carbonyl group of the ketone to form a peroxy monoester of a gem-diol. The intermediate then undergoes rearrangement. Alkyl group migration occurs at the same time as cleavage of the O@O bond of the peroxy ester. In general, the more substituted group migrates. 17.18 The formation of a carboxylic acid from Baeyer–Villiger oxidation of an aldehyde requires hydrogen migration. C 6 H 5 COOH O O CH O 2 N m-Nitrobenzaldehyde O COH O 2 N m-Nitrobenzoic acid OH CH O OCC 6 H 5 O O 2 N H 3 C O C O O HOCC 6 H 5 H11001 C OCC 6 H 5 O OH O H 3 C O C 6 H 5 COOHH11001 O CCH 3 Peroxy monoester OH OOCC 6 H 5 O CCH 3 Peroxybenzoic acid O C 6 H 5 COOHH11001 Cyclohexyl methyl ketone O CCH 3 Benzoic acid O C 6 H 5 COHH11001 Cyclohexyl acetate O OCCH 3 H11001Br H11002 CH 3 CHCH 2 CH 3 H11001 (C 6 H 5 ) 3 P (1-Methylpropyl)triphenyl- phosphonium bromide Sodiomethyl methyl sulfoxide O NaCH 2 SCH 3 Ylide CH 3 H11001 H11002 CCH 2 CH 3 (C 6 H 5 ) 3 P H11001 Br H11002 CH 3 CHCH 2 CH 3 H11001 (C 6 H 5 ) 3 P (1-Methylpropyl)triphenyl- phosphonium bromide CH 3 CHCH 2 CH 3 Br 2-Bromobutane (C 6 H 5 ) 3 P Triphenyl- phosphine 434 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 17.19 (a) First consider all the isomeric aldehydes of molecular formula C 5 H 10 O. There are three isomeric ketones: (b) Reduction of an aldehyde to a primary alcohol does not introduce a stereogenic center into the molecule. The only aldehydes that yield chiral alcohols on reduction are therefore those that already contain a stereogenic center. Among the ketones, 2-pentanone and 3-methyl-butanone are reduced to chiral alcohols. NaBH 4 CH 3 OH O 3-Methyl-2-butanone OH 3-Methyl-2-butanol (chiral but racemic) NaBH 4 CH 3 OH O 3-Pentanone OH 3-Pentanol (achiral) NaBH 4 CH 3 OH O 2-Pentanone OH 2-Pentanol (chiral but racemic) (R)-2-Methylbutanal (R)-2-Methyl-1-butanol NaBH 4 CH 3 OH H O H OH H H (S)-2-Methylbutanal (S)-2-Methyl-1-butanol NaBH 4 CH 3 OH O OH H H 3-Pentanone2-Pentanone 3-Methyl-2-butanone O O O H H (S)-2-Methylbutanal O (R)-2-Methylbutanal O 2,2-Dimethylpropanal H O H H O H Pentanal O H 3-Methylbutanal ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 435 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) All the aldehydes yield chiral alcohols on reaction with methylmagnesium iodide. Thus, A stereogenic center is introduced in each case. None of the ketones yield chiral alcohols. 17.20 (a) Chloral is the trichloro derivative of ethanal (acetaldehyde). (b) Pivaldehyde has two methyl groups attached to C-2 of propanal. (c) Acrolein has a double bond between C-2 and C-3 of a three-carbon aldehyde. O H 2 C CHCH 2-Propenal (acrolein) OCH 3 CH 3 C CH 3 CH 2,2-Dimethylpropanal (pivaldehyde) O CH 3 CH 2 CH Propanal O CH 3 CH OCl Cl C Cl CH Trichloroethanal (chloral) Ethanal O 3-Methyl-2-butanone 1. CH 3 MgI 2. H 3 O H11001 2,3-Dimethyl-2-butanol (achiral) OH O 3-Pentanone 1. CH 3 MgI 2. H 3 O H11001 3-Methyl-3-pentanol (achiral) OH O 2-Pentanone 2-Methyl-2-pentanol (achiral) 1. CH 3 MgI 2. H 3 O H11001 OH H OH C 4 H 9 CCH 3 O C 4 H 9 CH 1. CH 3 MgI 2. H 3 O H11001 436 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Crotonaldehyde has a trans double bond between C-2 and C-3 of a four-carbon aldehyde. (e) Citral has two double bonds: one between C-2 and C-3 and the other between C-6 and C-7. The one at C-2 has the E configuration. There are methyl substituents at C-3 and C-7. ( f ) Diacetone alcohol is (g) The parent ketone is 2-cyclohexenone. Carvone has an isopropenyl group at C-5 and a methyl group at C-2. (h) Biacetyl is 2,3-butanedione. It has a four-carbon chain that incorporates ketone carbonyls as C-2 and C-3. 2,3-Butanedione (biacetyl) CH 3 CCCH 3 OO 5-Isopropenyl-2-methyl-2- cyclohexenone (carvone) O CH 3 CH 3 CH 2 C 2-Cyclohexenone O 6 5 2 1 3 4 O OH H11013 O CH 3 CCH 2 C(CH 3 ) 2 OH 4-Hydroxy-4-methyl- 2-pentanone H CH O 8 7 6 5 4 2 13 (E)-3,7-Dimethyl-2,6-octadienal (citral) CC H H 3 C CH H O (E)-2-Butenal (crotonaldehyde) ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 437 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 17.21 (a) Lithium aluminum hydride reduces aldehydes to primary alcohols. (b) Sodium borohydride reduces aldehydes to primary alcohols. (c) Aldehydes can be reduced to primary alcohols by catalytic hydrogenation. (d) Aldehydes react with Grignard reagents to form secondary alcohols. (e) Sodium acetylide adds to the carbonyl group of propanal to give an acetylenic alcohol. ( f ) Alkyl- or aryllithium reagents react with aldehydes in much the same way that Grignard reagents do. (g) Aldehydes are converted to acetals on reaction with alcohols in the presence of an acid catalyst. O CH 3 CH 2 CH Propanal H11001 2CH 3 OH Methanol CH 3 CH 2 CH(OCH 3 ) 2 Propanal dimethyl acetal HCl O CH 3 CH 2 CH Propanal OH CH 3 CH 2 CHC 6 H 5 1-Phenyl-1-propanol 1. C 6 H 5 Li, diethyl ether 2. H 3 O H11001 CH 3 CH 2 CH O 2. H 3 O H11001 CNa,1. HC liquid ammonia CH 3 CH 2 CHC CH OH Propanal 1-Pentyn-3-ol O CH 3 CH 2 CH Propanal OH CH 3 CH 2 CHCH 3 2-Butanol 1. CH 3 MgI, diethyl ether 2. H 3 O H11001 O CH 3 CH 2 CH Propanal CH 3 CH 2 CH 2 OH 1-Propanol H 2 Ni O CH 3 CH 2 CH Propanal CH 3 CH 2 CH 2 OH 1-Propanol NaBH 4 CH 3 OH O CH 3 CH 2 CH Propanal CH 3 CH 2 CH 2 OH 1-Propanol 1. LiAlH 4 2. H 2 O 438 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (h) Cyclic acetal formation occurs when aldehydes react with ethylene glycol. (i) Aldehydes react with primary amines to yield imines. ( j) Secondary amines combine with aldehydes to yield enamines. (k) Oximes are formed on reaction of hydroxylamine with aldehydes. (l) Hydrazine reacts with aldehydes to form hydrazones. (m) Hydrazone formation is the first step in the Wolff–Kishner reduction (Section 12.8). (n) The reaction of an aldehyde with p-nitrophenylhydrazine is analogous to that with hydrazine. H 2 OH11001 Propanal Propanal p-nitrophenylhydrazone CH 3 CH 2 CH NNH NO 2 CH 3 CH 2 CH O H11001 p-Nitrophenylhydrazine O 2 N NHNH 2 CH 3 CH 2 CH NNH 2 Propanal hydrazone N 2 H11001CH 3 CH 2 CH 3 Propane NaOH triethylene glycol, heat CH 3 CH 2 CH O Propanal CH 3 CH 2 CH NNH 2 Propanal hydrazone H 2 NNH 2 CH 3 CH 2 CH O Propanal CH 3 CH 2 CH NOH Propanal oxime H 2 NOH H11001 (CH 3 ) 2 NH Dimethylamine 1-(Dimethylamino)propene CH 3 CH CH N(CH 3 ) 2 O CH 3 CH 2 CH Propanal p-toluenesulfonic acid benzene H11002H 2 O H11001 C 6 H 5 NH 2 Aniline N-Propylideneaniline CH 3 CH 2 CH NC 6 H 5 O CH 3 CH 2 CH Propanal H11001 p-toluenesulfonic acid benzene HOCH 2 CH 2 OH Ethylene glycol O CH 3 CH 2 CH Propanal 2-Ethyl-1,3-dioxolane CH 2 CH 3 H OO ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 439 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (o) Semicarbazide converts aldehydes to the corresponding semicarbazone. (p) Phosphorus ylides convert aldehydes to alkenes by a Wittig reaction. (q) Acidification of solutions of sodium cyanide generates HCN, which reacts with aldehydes to form cyanohydrins. (r) Chromic acid oxidizes aldehydes to carboxylic acids. 17.22 (a) Lithium aluminum hydride reduces ketones to secondary alcohols. (b) Sodium borohydride converts ketones to secondary alcohols. (c) Catalytic hydrogenation of ketones yields secondary alcohols. O Cyclopentanone OHH Cyclopentanol H 2 Ni O Cyclopentanone OHH Cyclopentanol NaBH 4 CH 3 OH 1. LiAlH 4 2. H 2 O O Cyclopentanone OHH Cyclopentanol Propanoic acid CH 3 CH 2 CO 2 H Propanal CH 3 CH 2 CH O H 2 CrO 4 Propanal cyanohydrinHydrogen cyanide HCN Propanal CH 3 CH 2 CH O CH 3 CH 2 CHCN OH H11001 Propanal CH 3 CH 2 CH O 2-Pentene CH 3 CH 2 CH CHCH 3 H11001 Ethylidenetriphenyl- phosphorane (C 6 H 5 ) 3 P CHCH 3 H11001 H11002 H11001 Triphenylphosphine oxide (C 6 H 5 ) 3 PO H11001 H11002 H 2 O Propanal CH 3 CH 2 CH O Semicarbazide H 2 NNHCNH 2 O Propanal semicarbazone CH 3 CH 2 CH NNHCNH 2 O H11001 H11001 440 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Grignard reagents react with ketones to form tertiary alcohols. (e) Addition of sodium acetylide to cyclopentanone yields a tertiary acetylenic alcohol. ( f ) Phenyllithium adds to the carbonyl group of cyclopentanone to yield 1-phenylcyclopentanol. (g) The equilibrium constant for acetal formation from ketones is generally unfavorable. (h) Cyclic acetal formation is favored even for ketones. (i) Ketones react with primary amines to form imines. ( j) Dimethylamine reacts with cyclopentanone to yield an enamine. H11001 O Cyclopentanone (CH 3 ) 2 NH Dimethylamine 1-(Dimethylamino)- cyclopentene N(CH 3 ) 2 p-toluenesulfonic acid benzene H11001 O Cyclopentanone NC 6 H 5 N-Cyclopentylideneaniline C 6 H 5 NH 2 Aniline H11001 O Cyclopentanone HOCH 2 CH 2 OH Ethylene glycol 1,4-Dioxaspiro[4.4]nonane OO p-toluenesulfonic acid benzene H11001 K < 1 O Cyclopentanone 2CH 3 OH Methanol OCH 3 CH 3 O Cyclopentanone dimethyl acetal HCl O Cyclopentanone OHC 6 H 5 1-Phenylcyclopentanol 1. C 6 H 5 Li, diethyl ether 2. H 3 O H11001 O Cyclopentanone 1. HC liquid ammonia 2. H 3 O H11001 CNa, 1-Ethynylcyclopentanol OHCHC O Cyclopentanone OHH 3 C 1-Methylcyclopentanol 1. CH 3 MgI, diethyl ether 2. H 3 O H11001 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 441 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (k) An oxime is formed when cyclopentanone is treated with hydroxylamine. (l ) Hydrazine reacts with cyclopentanone to form a hydrazone. (m) Heating a hydrazone in base with a high-boiling alcohol as solvent converts it to an alkane. (n)Ap-nitrophenylhydrazone is formed. (o) Cyclopentanone is converted to a semicarbazone on reaction with semicarbazide. (p) A Wittig reaction takes place, forming ethylidenecyclopentane. H11001 Triphenylphosphine oxide (C 6 H 5 ) 3 PO H11001 H11002 H11001 Ethylidenetriphenyl- phosphorane (C 6 H 5 ) 3 P CHCH 3 H11001 H11002 Ethylidenecyclo- pentane CHCH 3 Cyclopentanone O H11001H11001H 2 O Cyclopentanone H 2 NNHCNH 2 Semicarbazide O NNHCNH 2 Cyclopentanone semicarbazone O O H11001 NNH NO 2 Cyclopentanone p-nitrophenylhydrazone O 2 N NHNH 2 p-NitrophenylhydrazineCyclopentanone O Cyclopentane H11001 N 2 NNH 2 Cyclopentanone hydrazone NaOH triethylene glycol, heat O Cyclopentanone NNH 2 Cyclopentanone hydrazone H 2 NNH 2 O Cyclopentanone NOH Cyclopentanone oxime H 2 NOH 442 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (q) Cyanohydrin formation takes place. (r) Cyclopentanone is not oxidized readily with chromic acid. 17.23 (a) The first step in analyzing this problem is to write the structure of the starting ketone in stereo- chemical detail. Reduction of the ketone introduces a new stereogenic center, which may have either the R or the S configuration; the configuration of the original stereogenic center is unaffected. In prac- tice the 2R,3S diastereomer is observed to form in greater amounts than the 2S,3S (ratio 2.5:1 for LiAlH 4 reduction). (b) Reduction of the ketone can yield either cis- or trans-4-tert-butylcyclohexanol. It has been observed that the major product obtained on reduction with either lithium alu- minum hydride or sodium borohydride is the trans alcohol (trans/cis H333609:1). (c) The two reduction products are the exo and endo alcohols. The major product is observed to be the endo alcohol (endo/exo 9:1) for reduction with NaBH 4 or LiAlH 4 . The stereoselectivity observed in this reaction is due to decreased steric hindrance to attack of the hydride reagent from the exo face of the molecule, giving rise to the endo alcohol. H OH endo-Bicyclo[2.2.1]- heptan-2-ol OH H exo-Bicyclo[2.2.1]- heptan-2-ol O Bicyclo[2.2.1]- heptan-2-one H11001 H11001 4-tert-Butylcyclo- hexanone H (CH 3 ) 3 C O trans-4-tert- Butylcyclohexanol H (CH 3 ) 3 C OH H cis-4-tert- Butylcyclohexanol H (CH 3 ) 3 C H OH H11001 metal hydride reduction (S)-3-Phenyl-2- butanone O H C 6 H 5 (2R,3S)-3-Phenyl- 2-butanol HO H H C 6 H 5 (2S,3S)-3-Phenyl- 2-butanol H OH H C 6 H 5 O Cyclopentanone Cyclopentanone cyanohydrin CNHO NaCN H 2 SO 4 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 443 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) The hydroxyl group may be on the same side as the double bond or on the opposite side. The anti alcohol is observed to be formed in greater amounts (85:15) on reduction of the ke- tone with LiAlH 4 . Steric factors governing attack of the hydride reagent again explain the major product observed. 17.24 (a) Aldehydes undergo nucleophilic addition faster than ketones. Steric crowding in the rate- determining step of the ketone reaction raises the energy of the transition state, giving rise to a slower rate of reaction. Thus benzaldehyde is reduced by sodium borohydride more rapidly than is acetophenone. The measured relative rates are (b) The presence of an electronegative substituent on the H9251-carbon atom causes a dramatic in- crease in K hydr . Trichloroethanal (chloral) is almost completely converted to its geminal diol (chloral hydrate) in aqueous solution. Electron-withdrawing groups such as Cl 3 C destabilize carbonyl groups to which they are at- tached and make the energy change favoring the products of nucleophilic addition more fa- vorable. (c) Recall that the equilibrium constants for nucleophilic addition to carbonyl groups are gov- erned by a combination of electronic effects and steric effects. Electronically there is little difference between acetone and 3,3-dimethyl-2-butanone, but sterically there is a significant difference. The cyanohydrin products are more crowded than the starting ketones, and so the K rel H11005 H33360 20,000 Cl 3 CCH O O CH 3 CH H11001 H 2 O Cl 3 CCH OH OH 2,2,2-Trichloro-1,1-ethanediol (chloral hydrate) Cl 3 CCH O Trichloroethanal (chloral) k rel H11005 H11005 440 C 6 H 5 CH O O C 6 H 5 CCH 3 HO H syn-Bicyclo[2.2.1]- hept-2-en-7-ol HOH anti-Bicyclo[2.2.1]- hept-2-en-7-ol H11001 Bicyclo[2.2.1]- hept-2-en-7-one O 444 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website bulkier the alkyl groups that are attached to the carbonyl, the more strained and less stable will be the cyanohydrin. (d) Steric effects influence the rate of nucleophilic addition to these two ketones. Carbon is on its way from tricoordinate to tetracoordinate at the transition state, and alkyl groups are forced closer together than they are in the ketone. The transition state is of lower energy when R is smaller. Acetone (for which R is methyl) is reduced faster than 3,3-dimethyl-2-butanone (where R is tert-butyl). (e) In this problem we examine the rate of hydrolysis of acetals to the corresponding ketone or aldehyde. The rate-determining step is carbocation formation. Hybridization at carbon changes from sp 3 to sp 2 ; crowding at this carbon is relieved as the carbocation is formed. The more crowded acetal (R H11005 CH 3 ) forms a carbocation faster than the less crowded one (R H11005 H). Another factor of even greater importance is the extent of C OCH 2 CH 3 CH 3 CH 2 OHH11001 R R H11001 C R OCH 2 CH 3 H OCH 2 CH 3 R H11001 k rel H11005 H11005 12 CH 3 CCH 3 O O CH 3 CC(CH 3 ) 3 NaBH 4 Transition state R H H 3 C BH 3 CO H9254H11002 H9254H11002 R H H 3 C C OBH 3 H11002 sp 3 CO H 3 C R sp 2 K rel 40H11005H11005 CH 3 CCH 3 O O CH 3 CC(CH 3 ) 3 H11001CO Ketone HCN Hydrogen cyanide CH 3 CCN OH R Cyanohydrin [less strained for R H11005 CH 3 than for R H11005 C(CH 3 ) 3 ] H 3 C R ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 445 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website stabilization of the carbocation intermediate; the more stable carbocation (R H11005 CH 3 ) is formed faster than the less stable one (R H11005 H). 17.25 (a) The reaction as written is the reverse of cyanohydrin formation, and the principles that govern equilibria in nucleophilic addition to carbonyl groups apply in reverse order to the dissocia- tion of cyanohydrins to aldehydes and ketones. Cyanohydrins of ketones dissociate more at equilibrium than do cyanohydrins of aldehydes. More strain due to crowding is relieved when a ketone cyanohydrin dissociates and a more stabilized carbonyl group is formed. The equi- librium constant K diss is larger for than it is for (b) Cyanohydrins of ketones have a more favorable equilibrium constant for dissociation than do cyanohydrins of aldehydes. Crowding is relieved to a greater extent when a ketone cyanohydrin dissociates and a more stable carbonyl group is formed. The measured dissociation constants are 17.26 (a) The reaction of an aldehyde with 1,3-propanediol in the presence of p-toluenesulfonic acid forms a cyclic acetal. H11001 CH 3 O CH 3 O Br CH 3 OCH O 2-Bromo-3,4,5- trimethoxybenzaldehyde O O CH 3 O CH 3 O Br CH 3 O 2-(2H11032-Bromo-3H11032,4H11032,5H11032- trimethoxyphenyl)-1,3-dioxane (81%) HOCH 2 CH 2 CH 2 OH 1,3-Propanediol p-toluenesulfonic acid benzene, heat C 6 H 5 CCN OH CH 3 Acetophenone cyanohydrin H11001 HCN K H11005 1.3 Acetophenone C 6 H 5 CCH 3 O C 6 H 5 CHCN OH Benzaldehyde cyanohydrin H11001 HCN K H11005 4.7 H11003 10 H110023 C 6 H 5 CH O Benzaldehyde CH 3 CH 2 CHCN OH Propanal cyanohydrin H11001 HCN Hydrogen cyanide CH 3 CH 2 CH O Propanal K diss CH 3 CCH 3 OH CN Acetone cyanohydrin H11001 HCN Hydrogen cyanide CH 3 CCH 3 O Acetone K diss k rel H11005H11005 1.8 H11003 10 7 (CH 3 ) 2 C(OCH 2 CH 3 ) 2 CH 2 (OCH 2 CH 3 ) 2 446 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) The reagent CH 3 ONH 2 is called O-methylhydroxylamine, and it reacts with aldehydes in a manner similar to hydroxylamine. (c) Propanal reacts with 1,1-dimethylhydrazine to yield the corresponding hydrazone. (d) Acid-catalyzed hydrolysis of the acetal gives the aldehyde in 87% yield. (e) Hydrogen cyanide adds to carbonyl groups to form cyanohydrins. ( f ) The reagent is a secondary amine known as morpholine. Secondary amines react with ke- tones to give enamines. (g) Migration of the alkyl group in a Baeyer–Villiger oxidation occurs with retention of configuration. H11001 Peroxybenzoic acid O C 6 H 5 COOH (R)-3-Methyl-3-phenyl-2- pentanone O H 3 C CCH 3 CH 3 CH 2 (R)-1-Methyl-1-phenylpropyl acetate O H 3 C OCCH 3 CH 3 CH 2 HN OH11001 Acetophenone Morpholine Carbinolamine intermediate C 6 H 5 CCH 3 O H H11001 H H11001 (H11002H 2 O) C 6 H 5 CCH 3 OH O N 1-Morpholinostyrene (57–64%) C 6 H 5 CCH 2 O N NaCN HCl C 6 H 5 CCN OH CH 3 Acetophenone cyanohydrin C 6 H 5 CCH 3 O Acetophenone H 3 C CHCH 2 CH 2 CH 3 O O H 2 O, HCl heat H 3 C CHCH 2 CH 2 CH CH 3 O 4-(p-Methylphenyl)pentanal H11001 (CH 3 ) 2 NNH 2 1,1-Dimethylhydrazine NN(CH 3 ) 2 CH 3 CH 2 CH Propanal dimethylhydrazone O CH 3 CH 2 CH Propanal H11001 OCH 3 HO CH O 4-Hydroxy-2- methoxybenzaldehyde CH 3 ONH 2 O-Methyl- hydroxylamine OCH 3 HO CH NOCH 3 4-Hydroxy-2-methoxybenzaldehyde O-methyloxime ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 447 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 17.27 Wolff–Kishner reduction converts a carbonyl group (C?O) to a methylene group (CH 2 ). Treatment of the alkene with m-chloroperoxybenzoic acid produces an epoxide, compound B. Epoxides undergo reduction with lithium aluminum hydride to form alcohols (Section 16.12). Chromic acid oxidizes the alcohol to a ketone. 17.28 Hydration of formaldehyde by H 2 17 O produces a gem-diol in which the labeled and unlabeled hydroxyl groups are equivalent. When this gem-diol reverts to formaldehyde, loss of either of the hydroxyl groups is equally likely and leads to eventual replacement of the mass-16 isotope of oxygen by 17 O. This reaction has been monitored by 17 O NMR spectroscopy; 17 O gives an NMR signal, but 16 O does not. 17.29 First write out the chemical equation for the reaction that takes place. Vicinal diols (1,2-diols) react with aldehydes to give cyclic acetals. H11001 H H11001 C 6 H 5 CH O Benzaldehyde HOCH 2 CH(CH 2 ) 5 CH 3 OH 1,2-Octanediol (CH 2 ) 5 CH 3 OO C 6 H 5 4-Hexyl-2-phenyl- 1,3-dioxolane H 2 17 OH11001 H 2 16 OH11001 16 OC H H 17 OC H H H 17 O H 16 O CH 2 H 2 CrO 4 HO Bicyclo[4.3.0]nonan-3-ol Bicyclo[4.3.0]nonan-3-one (compound D, 75%) O 1. LiAlH 4 2. H 2 O O 3,4-Epoxybicyclo[4.3.0]nonane HO Bicyclo[4.3.0]nonan-3-ol (compound C, 90%) Bicyclo[4.3.0]non-3-ene 3,4-Epoxybicyclo[4.3.0]nonane (compound B, 92%) O COOH Cl O N 2 H 4 , KOH HOCH 2 CH 2 OH 130H11034C O Bicyclo[4.3.0]non- 3-en-8-one Bicyclo[4.3.0]non-3-ene (compound A, 90%) 448 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Notice that the phenyl and hexyl substituents may be either cis or trans to each other. The two prod- ucts are the cis and trans stereoisomers. 17.30 Cyclic hemiacetals are formed by intramolecular nucleophilic addition of a hydroxyl group to a carbonyl. The ring oxygen is derived from the hydroxyl group; the carbonyl oxygen becomes the hydroxyl oxygen of the hemiacetal. (a) This compound is the cyclic hemiacetal of 5-hydroxypentanal. Indeed, 5-hydroxypentanal seems to exist entirely as the cyclic hemiacetal. Its infrared spec- trum lacks absorption in the carbonyl region. (b) The carbon connected to two oxygens is the one that is derived from the carbonyl group. Using retrosynthetic symbolism, disconnect the ring oxygen from this carbon. The next two compounds are cyclic acetals. The original carbonyl group is identifiable as the one that bears two oxygen substituents, which originate as hydroxyl oxygens of a diol. (c) O O CH 3 CH 2 CH 3 H11013CH 3 CH 3 CH 2 OH OH O CH 3 CH 2 CHCHCH 2 CH 2 CH 2 CCH 3 OH OH O Brevicomin 6,7-Dihydroxy- 2-nonanone OOH O H HC O H11013 HCCH 2 CH 2 CHCH OH CH 2 CHCH O 4-Hydroxy-5,7-octadienal HOCH 2 CH 2 CH 2 CH 2 CH H11013 O H 2 C H 2 C O H CO CH 2 CH 2 H O OH C R O H O O C R OH Cyclic hemiacetal OO (CH 2 ) 5 CH 3 C 6 H 5 H H cis-4-Hexyl-2-phenyl- 1,3-dioxolane OO (CH 2 ) 5 CH 3 C 6 H 5 H H trans-4-Hexyl-2-phenyl- 1,3-dioxolane ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 449 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) 17.31 (a) The Z stereoisomer of CH 3 CH?NCH 3 has its higher ranked substituents on the same side of the double bond, The lone pair of nitrogen is lower in rank than any other substituent. (b) Higher ranked groups are on opposite sides of the carbon–nitrogen double bond in the E oxime of acetaldehyde. (c)(Z)-2-Butanone hydrazone is (d)(E)-Acetophenone semicarbazone is 17.32 Cyclopentanone reacts with peroxybenzoic acid to form a peroxy monoester. The alkyl group that migrates is the ring itself, leading to formation of a six-membered lactone. H11001H11001 O O 5-Pentanolide C 6 H 5 COOH O Peroxybenzoic acid Cyclopentanone C 6 H 5 COH O Benzoic acid O O OCC 6 H 5 HO O CN H 3 C NHCNH 2 Higher Higher O CN H 3 C NH 2 Higher CH 3 CH 2 Higher (E)-Acetaldehyde oxime CN H H 3 C OH Higher Higher Higher CN H CH 3 H 3 C Higher Lower (Z )-N-Ethylidenemethylamine O O CH 2 CH 3 OH HOCH 2 Talaromycin A CH 2 CH 3 HOCH 2 OH OH OH O C 2,8-Di(hydroxymethyl)-1,3- dihydroxy-5-decanone 450 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 17.33 (a) The bacterial enzyme cyclohexanone monooxygenase was described in Section 17.16 as able to catalyze a biological Baeyer–Villiger reaction. Compound A is 4-methylcyclohexanone. (b) The product of Baeyer–Villiger oxidation of 4-methylcyclohexanone with peroxyacetic acid would be the racemic cyclic ester (lactone), not the single enantiomer shown in part (a) from the enzyme-catalyzed oxidation. 17.34 (a) Nucleophilic ring opening of the epoxide occurs by attack of methoxide at the less hindered carbon. The anion formed in this step loses a chloride ion to form the carbon–oxygen double bond of the product. (b) Nucleophilic addition of methoxide ion to the aldehyde carbonyl generates an oxyanion, which can close to an epoxide by an intramolecular nucleophilic substitution reaction. The epoxide formed in this process then undergoes nucleophilic ring opening on attack by a second methoxide ion. 17.35 Amygdalin is a derivative of the cyanohydrin formed from benzaldehyde; thus the structure (without stereochemistry) is C 6 H 5 CH CN OR R H11005 H, benzaldehyde cyanohydrin CHOCH 3 (CH 3 ) 3 CCH OCH 3 O H11002 (CH 3 ) 3 CCHCHOCH 3 OCH 3 OH H11001(CH 3 ) 3 CCH CHOCH 3 O OCH 3 H11002 CH 3 OH H11001(CH 3 ) 3 CCH CHOCH 3 O Cl H11002 H11001(CH 3 ) 3 CCHCH Cl OCH 3 H11002 O CHOCH 3 (CH 3 ) 3 CCH Cl O H11002 CH 2 OCH 3 (CH 3 ) 3 CC O Cl H11002 (CH 3 ) 3 CCCH 2 OCH 3 Cl H11002 H11001 O CH 2 OCH 3 (CH 3 ) 3 CC O Cl H11002 H11001C CH 2 Cl (CH 3 ) 3 C O OCH 3 H11002 4-Methylcyclohexanone (Compound A) O CH 3 O 2 , cyclohexanone monoxygenase, and coenzymes Compound B O O H 3 C ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 451 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The order of decreasing sequence rule precedence is HO H11022 CN H11022 C 6 H 5 H11022 H. The groups are arranged in a clockwise orientation in order of decreasing precedence in the R enantiomer. 17.36 (a) The target molecule is the diethyl acetal of acetaldehyde (ethanal). Acetaldehyde may be prepared by oxidation of ethanol. Reaction with ethanol in the presence of hydrogen chloride yields the desired acetal. (b) In this case the target molecule is a cyclic acetal of acetaldehyde. Acetaldehyde has been prepared in part (a). Recalling that vicinal diols are available from the hydroxylation of alkenes, 1,2-ethanediol may be prepared by the sequence Hydrolysis of ethylene oxide is also reasonable. HOCH 2 CH 2 OH CH 3 COOH Ethylene Ethylene oxide 1,2-Ethanediol H 2 CCH 2 H 2 C CH 2 H 2 O HO H11002 O O CH 3 CH 2 OH Ethanol HOCH 2 CH 2 OH 1,2-Ethanediol H 2 CCH 2 Ethylene H 2 SO 4 heat OsO 4 , (CH 3 ) 3 COOH (CH 3 ) 3 COH, HO H11002 2-Methyl-1,3-dioxolane HOCH 2 CH 2 OHandCH 3 CH O OO H CH 3 CH 3 CH(OCH 2 CH 3 ) 2 Acetaldehyde diethyl acetal H11001CH 3 CH O Acetaldehyde 2CH 3 CH 2 OH Ethanol HCl CH 3 CH 2 OH Ethanol CH 3 CH O Acetaldehyde PCC CH 2 Cl 2 CH 3 CH O and CH 3 CH 2 OHCH 3 CH(OCH 2 CH 3 ) 2 Acetaldehyde diethyl acetal (R)-Benzaldehyde cyanohydrin C 6 H 5 H OH CN C 452 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Reaction of acetaldehyde with 1,2-ethanediol yields the cyclic acetal. (c) The target molecule is, in this case, the cyclic acetal of 1,2-ethanediol and formaldehyde. The preparation of 1,2-ethanediol was described in part (b). One method of preparing formaldehyde is by ozonolysis of ethylene. Another method is periodate cleavage of 1,2-ethanediol. Cyclic acetal formation is then carried out in the usual way. (d) Acetylenic alcohols are best prepared from carbonyl compounds and acetylide anions. Acetaldehyde is available as in part (a). Alkynes such as acetylene are available from the corresponding alkene by bromination followed by double dehydrobromination. Using ethyl- ene, prepared in part (b), the sequence becomes H 2 CCH 2 Ethylene Acetylene BrCH 2 CH 2 Br 1,2-Dibromoethane Br 2 NaNH 2 NH 3 HC CH H11001CH 3 CH O 3-Butyn-2-ol OH CH 3 CHC CH H11002 CCH H11001 H H11001 OO 1,3-Dioxolane HOCH 2 CH 2 OH 1,2-Ethanediol HCH O Formaldehyde 2HCH O Formaldehyde HOCH 2 CH 2 OH 1,2-Ethanediol HIO 4 2HCH O Formaldehyde H 2 CCH 2 Ethylene CH 3 CH 2 OH Ethanol H 2 SO 4 heat 1. O 3 2. H 2 O, Zn OO 1,3-Dioxolane HOCH 2 CH 2 OHandHCH O 2-Methyl-1,3-dioxolane1,2-Ethanediol HOCH 2 CH 2 OHH11001 Acetaldehyde CH 3 CH O OO H CH 3 H H11001 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 453 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Then (e) The target aldehyde may be prepared from the corresponding alcohol. The best route to this alcohol is through reaction of an acetylide ion with ethylene oxide. Oxidation with PCC or PDC is appropriate for the final step. ( f ) The target molecule has four carbon atoms, suggesting a route involving reaction of an ethyl Grignard reagent with ethylene oxide. Ethylmagnesium bromide is prepared in the usual way. Reaction of the Grignard reagent with ethylene oxide, prepared in part (b), completes the synthesis. CH 3 CH 2 CH 2 CH 2 OHCH 3 CH 2 MgBr H 2 C CH 2 H11001 2. H 3 O H11001 1. diethyl ether Ethylmagnesium bromide Ethylene oxide 1-Butanol O CH 3 CH 2 OH CH 3 CH 2 Br CH 3 CH 2 MgBr HBr or PBr 3 Mg diethyl ether Ethanol Bromoethane Ethylmagnesium bromide CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 H 2 C CH 2 H11001 H11002 O 3-Butyn-1-ol 3-Butynal CCH 2 CH 2 OHHC PCC or PDC CH 2 Cl 2 CCH 2 CHHC O H11001 1. diethyl ether 2. H 3 O H11001 H 2 CCH 2 O Ethylene oxide [prepared in part (b)] HC CCH 2 CH 2 OH 3-Butyn-1-ol HC CNa Sodium acetylide [prepared in part (d)] 3-Butynal O HCCH 2 CCH 3-Butyn-1-ol HOCH 2 CH 2 CCH NaNH 2 Acetylene HC CH Sodium acetylide HC CNa 3-Butyn-2-ol HC OH CCHCH 3 1. CH 3 CH O 2. H 3 O H11001 454 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 17.37 (a) Friedel–Crafts acylation of benzene with benzoyl chloride is a direct route to benzophenone. (b) On analyzing the overall transformation retrosynthetically, we see that the target molecule may be prepared by a Grignard synthesis followed by oxidation of the alcohol formed. In the desired synthesis, benzyl alcohol must first be oxidized to benzaldehyde. Reaction of benzaldehyde with the Grignard reagent of bromobenzene followed by oxidation of the resulting secondary alcohol gives benzophenone. (c) Hydrolysis of bromodiphenylmethane yields the corresponding alcohol, which can be oxi- dized to benzophenone as in part (b). (d) The starting material is the dimethyl acetal of benzophenone. All that is required is acid- catalyzed hydrolysis. 2H 2 O H H11001 Dimethoxydiphenyl- methane C 6 H 5 CC 6 H 5 OCH 3 OCH 3 C 6 H 5 CC 6 H 5 O BenzophenoneWater Methanol H11001 2CH 3 OHH11001 C 6 H 5 CHC 6 H 5 H 2 O Bromodiphenylmethane Diphenylmethanol C 6 H 5 CHC 6 H 5 OHBr oxidize C 6 H 5 CC 6 H 5 O Benzophenone C 6 H 5 MgBrC 6 H 5 CH O 1. diethyl ether 2. H 3 O H11001 Benzaldehyde Phenylmagnesium bromide Diphenylmethanol H11001 C 6 H 5 CHC 6 H 5 OH PDC CH 2 Cl 2 C 6 H 5 CC 6 H 5 O Benzophenone C 6 H 5 CH 2 OH C 6 H 5 CH O PDC CH 2 Cl 2 Benzyl alcohol Benzaldehyde C 6 H 5 MgBrH11001C 6 H 5 CHC 6 H 5 OH C 6 H 5 CC 6 H 5 O C 6 H 5 CH O Benzoyl chloride Benzene Benzophenone CCl O H11001 AlCl 3 C O ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 455 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) Oxidative cleavage of the alkene yields benzophenone. Ozonolysis may be used. 17.38 The two alcohols given as starting materials contain all the carbon atoms of the desired product. What is needed is to attach the two groups together so that the two primary alcohol carbons become doubly bonded to each other. This can be accomplished by using a Wittig reaction as the key step. Alternatively, allyl alcohol could be oxidized to CH 2 ?CHCHO for subsequent reaction with the ylide derived from CH 3 (CH 2 ) 8 CH?CHCH 2 CH?CHCH 2 CH 2 OH via its bromide and triphenyl- phosphonium salt. 17.39 The expected course of the reaction would be hydrolysis of the acetal to the corresponding aldehyde. H11001 H 2 O HCl C 6 H 5 CHCH(OCH 3 ) 2 OH Compound A (mandelaldehyde dimethyl acetal) C 6 H 5 CHCH OH O Mandelaldehyde 2CH 3 OH Methanol O CHCH 2 CHCHCH 2 CHCH 3 (CH 2 ) 8 CH 3,6-Hexadecadienal Allylidenetriphenylphosphorane 1,3,6,9-Nonadecatetraene CH 3 (CH 2 ) 8 CH CHCH 2 CH CHCH 2 CH CHCH CH 2 H11001 CH 2 (C 6 H 5 ) 3 P H11001 H11002 CHCH Allyltriphenylphosphonium bromide Allylidenetriphenylphosphorane (C 6 H 5 ) 3 P H 2 C CHCH 2 BrH 2 C CHCH 2 OH Allyl alcohol Allyl bromide PBr 3 CH 3 CH 2 CH 2 CH 2 Li, THF (C 6 H 5 ) 3 PCH 2 CH CH 2 Br H11002 H11001 (C 6 H 5 ) 3 P CH 2 CHCH H11002 H11001 CHCH 2 CHCH 3 (CH 2 ) 8 CH CHCH 2 CH 2 OH 3,6-Hexadecadien-1-ol 3,6-Hexadecadienal PCC CH 2 Cl 2 CHCH 2 CHCH 3 (CH 2 ) 8 CH CHCH 2 CH O CH 2 CHCHCHCH 2 CHCH 3 (CH 2 ) 8 CH CHCH 2 CH 3,6-Hexadecadien-1-ol Allyl alcohol CH 3 (CH 2 ) 8 CH CHCH 2 CH CHCH 2 CH 2 OH and HOCH 2 CH CH 2 1. O 3 2. H 2 O, Zn 1,1,2,2-Tetraphenylethene 2(C 6 H 5 ) 2 C O Benzophenone (C 6 H 5 ) 2 C C(C 6 H 5 ) 2 456 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The molecular formula of the observed product (compound B, C 16 H 16 O 4 ) is exactly twice that of mandelaldehyde. This suggests that it might be a dimer of mandelaldehyde resulting from hemiac- etal formation between the hydroxyl group of one mandelaldehyde molecule and the carbonyl group of another. Because compound B lacks carbonyl absorption in its infrared spectrum, the cyclic structure is indicated. 17.40 (a) Recalling that alkanes may be prepared by hydrogenation of the appropriate alkene, a synthe- sis of the desired product becomes apparent. What is needed is to convert @C?O into @C?CH 2 ; a Wittig reaction is appropriate. The two-step procedure that was followed used a Wittig reaction to form the carbon–carbon bond, then catalytic hydrogenation of the resulting alkene. (b) In putting together the carbon skeleton of the target molecule, a methyl group has to be added to the original carbonyl carbon. The logical way to do this is by way of a Grignard reagent. H11001 1. diethyl ether 2. H 3 O H11001 O C 6 H 5 C Cyclopentyl phenyl ketone CH 3 OH C 6 H 5 C 1-Cyclopentyl-1- phenylethanol CH 3 MgI Methylmagnesium iodide C 6 H 5 H 3 C C CH 3 OH C 6 H 5 C H11001 O C 6 H 5 C CH 3 H11002 THF H 2 , Pt acetic acid H11001O 5,5-Dimethylcyclononanone CH 2 5,5-Dimethyl-1- methylenecyclononane (59%) CH 3 1,1,5-Trimethylcyclononane (73%) (C 6 H 5 ) 3 PCH 2 H11001 H11002 H 3 C H 3 C H 3 C H 3 C H 3 C H 3 C ? O H 3 C H 3 C 5,5-Dimethylcyclononanone H 3 C CH 3 H H 3 C 1,1,5-Trimethylcyclononane C 6 H 5 CH CH O OH O HO CHCHC 6 H 5 C 6 H 5 CH C O H HO CHC 6 H 5 HC O OH OO C 6 H 5 C 6 H 5 OH HO Compound B ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 457 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Acid-catalyzed dehydration yields the more highly substituted alkene, the desired product, in accordance with the Zaitsev rule. (c) Analyzing the transformation retrosynthetically, keeping in mind the starting materials stated in the problem, we see that the carbon skeleton may be constructed in a straightforward manner. Proceeding with the synthesis in the forward direction, reaction between the Grignard reagent of o-bromotoluene and 5-hexenal produces most of the desired carbon skeleton. Oxidation of the resulting alcohol to the ketone followed by a Wittig reaction leads to the final product. Acid-catalyzed dehydration of the corresponding tertiary alcohol would not be suitable, be- cause the major elimination product would have the more highly substituted double bond. H H11001 heat 2-(o-Methylphenyl)-6-hepten-2-ol 6-(o-Methylphenyl)-1,5-heptadiene C(CH 2 ) 3 CH CH 3 OH CH 3 CH 2 C CH 3 CH 3 CHCH 2 CH 2 CH CH 2 PCC CH 2 Cl 2 1-(o-Methylphenyl)-5-hexen-1-ol 1-(o-Methylphenyl)-5-hexen-1-one CH(CH 2 ) 3 CH OH CH 3 CH 2 C(CH 2 ) 3 CH O CH 3 CH 2 2-(o-Methylphenyl)-1,6-heptadiene C(CH 2 ) 3 CH CH 2 CH 3 CH 2 (C 6 H 5 ) 3 PCH 2 H11001 H11002 1. diethyl ether 2. H 3 O H11001 o-Methylphenylmagnesium bromide 5-Hexenal 1-(o-Methylphenyl)-5-hexen-1-ol MgBr CH 3 O H 2 C CHCH 2 CH 2 CH 2 CH CH(CH 2 ) 3 CH OH CH 3 CH 2 H11001 CH 3 CH(CH 2 ) 3 CH OH CH 2 CH 3 MgBr HC(CH 2 ) 3 CH O CH 2 H11001 CH 3 C(CH 2 ) 3 CH CH 2 CH 2 CH 3 C(CH 2 ) 3 CH (C 6 H 5 ) 3 P O CH 2 H11001 H11001 CH 2 H11002 C 6 H 5 C OH CH 3 C 6 H 5 H 3 C 1-Cyclopentyl-1-phenylethanol (1-Phenylethylidene)cyclopentane H 3 PO 4 heat 458 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Remember that terminal acetylenes can serve as sources of methyl ketones by hydration. This gives us a clue as to how to proceed, since the acetylenic ketone may be prepared from the starting acetylenic alcohol. The first synthetic step is oxidation of the primary alcohol to the aldehyde and construction of the carbon skeleton by a Grignard reaction. Oxidation of the secondary alcohol to a ketone and hydration of the terminal triple bond com- plete the synthesis. (e) The desired product is a benzylic ether. To prepare it, the aldehyde must first be reduced to the corresponding primary alcohol. Sodium borohydride was used in the preparation described in the literature, but lithium aluminum hydride or catalytic hydrogenation would also be possi- ble. Once the alcohol is prepared, it can be converted to its alkoxide ion and this alkoxide ion treated with methyl iodide. Alternatively, the alcohol could be treated with hydrogen bromide or with phosphorus tribro- mide to give the benzylic bromide and the bromide then allowed to react with sodium methoxide. 17.41 Step 1 of the synthesis is formation of a cyclic acetal protecting group; the necessary reagents are ethylene glycol (HOCH 2 CH 2 OH) and p-toluenesulfonic acid, with heating in benzene. In step 2 the ester function is reduced to a primary alcohol. Lithium aluminum hydride (LiAlH 4 ) is the reagent of choice. Oxidation with PCC in CH 2 Cl 2 converts the primary alcohol to an aldehyde in step 3. Cl CH 3 CH 2 Br Cl CH 3 CH 2 OH Cl CH 3 CH 2 OCH 3 HBr or PBr 3 NaOCH 3 CH 3 OH Cl H 3 C CH O Cl CH 3 CH 2 OH Cl CH 3 CH 2 OCH 3 NaBH 4 H 2 O 1. NaH 2. CH 3 I 1-Undecyn-5-ol HC CCH 2 CH 2 CH(CH 2 ) 5 CH 3 OH 2,5-Undecanedione CH 3 CCH 2 CH 2 C(CH 2 ) 5 CH 3 O O 1-Undecyn-5-one O HC CCH 2 CH 2 C(CH 2 ) 5 CH 3 PDC CH 2 Cl 2 H 2 O, H 2 SO 4 HgSO 4 4-Pentyn-1-ol HC CCH 2 CH 2 CH 2 OH 1-Undecyn-5-ol HC CCH 2 CH 2 CH(CH 2 ) 5 CH 3 OH 4-Pentynal O HC CCH 2 CH 2 CH PDC CH 2 Cl 2 1. CH 3 (CH 2 ) 5 MgBr 2. H 3 O H11001 HC CCH 2 CH 2 C(CH 2 ) 5 CH 3 O HC CCH 2 CH 2 CH O HC CCH 2 CH 2 CH 2 OHH11001 CH 3 (CH 2 ) 4 CH 2 H11002 CH 3 CCH 2 CH 2 C(CH 2 ) 5 CH 3 O O HC CCH 2 CH 2 C(CH 2 ) 5 CH 3 O ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 459 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Wolff–Kishner reduction (N 2 H 4 , KOH, ethylene glycol, heat) converts the aldehyde group to a methyl group in step 4. The synthesis is completed in step 5 by hydrolysis (H 3 O H11001 ) of the acetal- protecting group. 17.42 We need to assess the extent of resonance donation to the carbonyl group by the H9266 electrons of the aromatic rings. Such resonance for benzaldehyde may be written as Electron-releasing groups such as methoxy at positions ortho and para to the aldehyde function increase the “single-bond character” of the aldehyde by stabilizing the dipolar resonance forms and increasing their contribution to the overall electron distribution in the molecule. Electron- withdrawing groups such as nitro decrease this single-bond character. The aldehyde with the low- est carbonyl stretching frequency is 2,4,6-trimethoxybenzaldehyde; the one with the highest is 2,4,6-trinitrobenzaldehyde. The measured values are 17.43 The signal in the 1 H NMR spectrum at H9254 9.7 ppm tells us that the compound is an aldehyde rather than a ketone. The 2H signal at H9254 2.4 ppm indicates that the group adjacent to the carbonyl is a CH 2 group. The remaining signals support the assignment of the compound as butanal. CH 3 CH 2 CH 2 CH O Doublet of triplets at 2.4 ppm Triplet at 1 ppm Sextet at 1.7 ppm Signal at 9.7 ppm CHO CH 3 O OCH 3 OCH 3 CHO O 2 NNO 2 NO 2 CHO 2,4,6-Trimethoxybenzaldehyde (1665 cm H110021 ) 2,4,6-Trinitrobenzaldehyde (1715 cm H110021 ) Benzaldehyde (1700 cm H110021 ) CH O CH O H11002 H11001 CH O H11002 H11001 CH O H11002 H11001 1 45 3 O O O CH 3 CH 3 CH 2 OHCOCH 3 O O COCH 3 O 2 O O O O O O O CH 460 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 17.44 A carbonyl group is evident from the strong infrared absorption at 1710 cm H110021 . Since all the 1 H NMR signals are singlets, there are no nonequivalent hydrogens in a vicinal or “three-bond” relationship. The three-proton signal at H9254 2.1 ppm, and the 2-proton signal at H9254 2.3 ppm can be understood as arising from a unit. The intense 9-proton singlet at H9254 1.0 ppm is due to the three equiva- lent methyl groups of a (CH 3 ) 3 C unit. The compound is 4,4-dimethyl-2-pentanone. 17.45 The molecular formula of compounds A and B (C 6 H 10 O 2 ) indicates an index of hydrogen deficiency of 2. Because we are told the compounds are diketones, the two carbonyl groups account for all the unsaturations. The 1 H NMR spectrum of compound A has only two peaks, both singlets, at H9254 2.2 and 2.8 ppm. Their intensity ratio (6:4) is consistent with two equivalent methyl groups and two equivalent meth- ylene groups. The chemical shifts are appropriate for The simplicity of the spectrum can be understood if we are dealing with a symmetric diketone. The correct structure is Compound B is an isomer of compound A. The triplet–quartet pattern in the 1 H NMR spectrum is consistent with an ethyl group and, because the triplet is equivalent to 6 protons and the quartet to 4, it is likely that two equivalent ethyl groups are present. The two ethyl groups account for four car- bons, and because the problem stipulates that the molecule is a diketone, all the carbons are ac- counted for. The only C 6 H 10 O 2 diketone with two equivalent ethyl groups is 3,4-hexanedione. 17.46 From its molecular formula (C 11 H 14 O), the compound has a total of five double bonds and rings. The presence of signals in the region H9254 7 to 8 ppm suggests an aromatic ring is present, accounting for four of the elements of unsaturation. The presence of a strong peak at 1700 cm H110021 in the infrared spectrum indicates the presence of a carbonyl group, accounting for the remaining element of 3,4-Hexanedione (compound B) CH 3 CH 2 C CCH 2 CH 3 OO 1.3 ppm triplet 2.8 ppm quartet CH 3 CCH 2 CH 2 CCH 3 O O Equivalent methylene groups do not split each other. 2,5-Hexanedione (compound A) andCH 3 C O CH 2 C O CH 3 CCH 2 C(CH 3 ) 3 O 2.1 ppm singlet 1.0 ppm singlet 2.3 ppm singlet 4,4-Dimethyl-2-pentanone CH 2 CCH 3 O ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 461 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website unsaturation. The highest field peak in the NMR spectrum is a 3-proton triplet, corresponding to the methyl group of a CH 3 CH 2 unit. The 2-proton signal at H9254 3.0 ppm corresponds to a CH 2 unit adja- cent to the carbonyl group and, because it is a triplet, suggests the grouping CH 2 CH 2 C?O. The compound is butyl phenyl ketone (1-phenyl-1-pentanone). 17.47 With a molecular formula of C 7 H 14 O, the compound has an index of hydrogen deficiency of 1. We are told that it is a ketone, so it has no rings or double bonds other than the one belonging to its C?O group. The peak at 211 ppm in the 13 C NMR spectrum corresponds to the carbonyl carbon. Only three other signals occur in the spectrum, and so there are only three types of carbons other than the carbonyl carbon. This suggests that the compound is the symmetrical ketone 4-heptanone. 17.48 Compounds A and B are isomers and have an index of hydrogen deficiency of 5. Signals in the region 125–140 ppm in their 13 C NMR spectra suggest an aromatic ring, and a peak at 200 ppm indicates a carbonyl group. An aromatic ring contributes one ring and three double bonds, and a carbonyl group contributes one double bond, and so the index of hydrogen deficiency of 5 is sat- isfied by a benzene ring and a carbonyl group. The carbonyl group is attached directly to the ben- zene ring, as evidenced by the presence of a peak at m/z 105 in the mass spectra of compounds A and B. Each 13 C NMR spectrum shows four aromatic signals, and so the rings are monosubstituted. Compound A has three unique carbons in addition to C 6 H 5 C?O and so must be 1-phenyl-1- butanone. Compound B has only two additional signals and so must be 2-methyl-1-phenyl-1- propanone. 17.49–17.50 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. Compound A Compound B CCH 2 CH 2 CH 3 O CCH(CH 3 ) 2 O CO H11001 mH11408z 105 4-Heptanone (all chemical shifts in ppm) CH 3 CH 2 CH 2 CCH 2 CH 2 CH 3 O 141745211451714 Butyl phenyl ketone CCH 2 CH 2 CH 2 CH 3 O 3.0 ppm triplet 1.0 ppm triplet Multiplets at 1.4 and 1.7 ppm 462 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website SELF-TEST PART A A-1. Give the correct IUPAC name for each of the following: A-2. Write the structural formulas for (a)(E)-3-Hexen-2-one (b) 3-Cyclopropyl-2,4-pentanedione (c) 3-Ethyl-4-phenylpentanal A-3. For each of the following reactions supply the structure of the missing reactant, reagent, or product: CH 3 CH 2 CH 2 CH O ?(h) Na 2 Cr 2 O 7 H H11001 , H 2 O C 6 H 5 C N(CH 3 ) 2 CHCH 3 (g) H11001 ?? CH 3 CH 2 CH 2 CH 2CH 3 CH 2 OH( f ) H11001 ? O HCl O CH 3 C 6 H 5 NHNH 2 ?(e) H11001 (d)OH11001 ? CHCH 2 CH 3 ?(c) (two products) H 2 O, H H11001 O O (CH 3 ) 2 CHCH C 6 H 5 CH C 6 H 5 CH NOH(b) H11001 ? O O HCN ?(a) H11001 CN H11002 O (d)(CH 3 ) 3 CCCH 2 CH(CH 3 ) 2 (b) O Br (c) CH 3 O CH 3 CH 2 CHCHCH 2 CH(a) CH 3 O CH 3 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 463 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-4. Write the structures of the products, compounds A through E, of the reaction steps shown. A-5. Give the reagents necessary to convert cyclohexanone into each of the following com- pounds. More than one step may be necessary. A-6. (a) What two organic compounds react together (in the presence of an acid catalyst) to give the compound shown, plus a molecule of water? (b) Draw the structure of the open-chain form of the following cyclic acetal: A-7. Outline reaction schemes to carry out each of the following interconversions, using any necessary organic or inorganic reagents. (b)to O OH CH 3 O HO CH 3 CH 3 (CH 3 ) 2 Cto CHCH 3 (CH 3 ) 2 C(a)O O O O H 3 C CH 3 H 3 C CH 3 CH 3 CH 3 O O (d) O O CH 2 (b) (c) O O CH 3 (a) C 6 H 5 CH 2 CHCH 3 OH D(b) PCC CH 2 Cl 2 D H11001 CH 3 COOH O E H11001 CH 3 COH O (C 6 H 5 ) 3 P H11001 (CH 3 ) 2 CHCH 2 Br A(a) A H11001 CH 3 CH 2 CH 2 CH 2 Li B H11001 C 4 H 10 B H11001 benzaldehyde C H11001 (C 6 H 5 ) 3 P O H11002 H11001 464 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-8. Write a stepwise mechanism for the formation of CH 3 CH(OCH 3 ) 2 from acetaldehyde and methanol under conditions of acid catalysis. A-9. Suggest a structure for an unknown compound, C 9 H 10 O, that exhibits a strong infrared ab- sorption at 1710 cm H110021 and has a 1 H NMR spectrum that consists of three singlets at H9254 2.1 ppm (3H), 3.7 ppm (2H), and 7.2 ppm (5H). PART B B-1. Which of the compounds shown is (are) correctly named as pentane derivatives, either as pentanals or pentanones? (a) 1 only (b) 2 only (c) 3 only (d) 1 and 3 (e) None of them B-2. The compound shown is best classified as a(an) (a) Carbinolamine (d) Imine (b) Enamine (e) Oxime (c) Hydrazone B-3. When a nucleophile encounters a ketone, the site of attack is (a) The carbon atom of the carbonyl (b) The oxygen atom of the carbonyl (c) Both the carbon and oxygen atoms, with equal probability (d) No attack occurs—ketones do not react with nucleophiles. B-4. What reagent and/or reaction conditions would you choose to bring about the following con- version? (a) 1. LiAlH 4 ,2.H 2 O(c)H 2 O, H 2 SO 4 , heat (b)H 2 O, NaOH, heat (d) PCC, CH 2 Cl 2 B-5. Rank the following in order of increasing value of the equilibrium constant for hydration, K hyd (smallest value first). (a)1H11021 2 H11021 3(b)3H11021 1 H11021 2(c)2H11021 1 H11021 3(d)2H11021 3 H11021 1 O 1 2 O 3 O (CH 3 ) 3 CCC(CH 3 ) 3 O O CH O HOCH 2 CH 2 OHH11001 (CH 3 ) 3 CCH 2 CH?NCH 3 3 O H 2 O H O 1 (c) O CO 2 H O CH 2 Br ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 465 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-6. The structure would be best classified as a(n) (a) Acetal (c) Hydrate (b) Hemiacetal (d) Cyanohydrin B-7. Which of the following pairs of reactants is most effective in forming an enamine? (d) None of these forms an enamine. B-8. Which of the following species is an ylide? (a)(C 6 H 5 ) 3 P H11001 CH 2 CH 3 Br H11002 (d) None of these B-9. Which pair of the following compounds could serve as the reagents X and Y in the follow- ing reaction sequence? XY XY (a)1 5 (d)2 5 (b)1 4 (e)3 4 (c)2 4 (CH 3 ) 2 CHCH 2 Br 1 (CH 3 ) 2 CHBr 23 CH 3 CH 2 CHBr CH 3 4 (CH 3 ) 2 CHCH O 5 CH 3 CCH 3 O X (C 6 H 5 ) 3 P Y CH 3 CH 2 CH 2 CH 2 Li (CH 3 ) 2 CHCH CH 3 CH 2 CH 3 C (C 6 H 5 ) 3 PCHCH 3 (b) H11001 H11002 (C 6 H 5 ) 3 P(c) O CHCH 3 CH 2 (CH 3 ) 3 CNH 2 O (c) H11001 (CH 3 ) 3 CCH O (b) H11001 (CH 3 ) 2 NH CH 3 CH 2 CH(a) H11001 [(CH 3 ) 2 CH] 2 NH O O OH H 466 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-10. The final product of the following sequence of reactions is. B-11. Which of the following sets of reagents, used in the order shown, would successfully ac- complish the conversion shown? B-12. Which of the following species is the conjugate acid of the hemiacetal formed by reaction of benzaldehyde with methanol containing a trace of acid? C 6 H 5 CH(c) H11001 OCH 3 H OH C 6 H 5 CH OH (e) H11001 C 6 H 5 CH OCH 3 (b) H11001 C 6 H 5 CH(d) OH 2 OH H11001 C 6 H 5 CH(a) OCH 3 OH (a) CH 3 CH 2 CH 2 MgBr; H 3 O H11001 ; PCC, CH 2 Cl 2 (b) CH 3 CH 2 CH 2 MgBr; H 3 O H11001 ; H 2 SO 4 , heat; PCC, CH 2 Cl 2 CHCH 2 CH 3 ; B 2 H 6 ; H 2 O 2 , HO H11002 (c) (C 6 H 5 ) 3 P H11001 H11002 H11001 H11002 CHCH 2 CH 3 ; H 2 SO 4 , H 2 O(d) (C 6 H 5 ) 3 P CCH 3 O CHCHCH 2 CH 3 CH 3 OH ? (CH 3 O) 2 CHCH 2 CH 2 CH 2 CH(e) O HCCH 2 CH 2 CH 2 CH(d) O O HCCH 2 CH 2 CH 2 CH 2 OH(c) O CH 3 CCH 2 CH 2 CH 2 CH 2 OH(b) O CH 3 OCCH 2 CH 2 CH 2 CH 2 OH(a) O (CH 3 O) 2 CHCH 2 CH 2 CH 2 Br ? Mg H 2 C O H 3 O H11001 heat ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 467 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-13. Which sequence represents the best synthesis of hexanal? CH 3 CH 2 CH 2 CH 2 CH 2 CH?O Hexanal (a)1.CH 3 CH 2 CH 2 CH 2 Br H11001 NaC>CH (c)1. 2. H 2 O, H 2 SO 4 , HgSO 4 2. CH 3 MgBr, diethyl ether 3. H 3 O H11001 4. PCC, CH 2 Cl 2 (b)1. (d)1. CH 3 CH 2 CH 2 CH 2 MgBr H11001 H 2 C CH 2 O CH 3 CH 2 CH 2 CH 2 CCH 3 O CH 3 CH 2 CH 2 CH CH 3 COOHCH 2 H11001 O 468 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 2. 2. H 3 O H11001 3. LiAlH 4 3. PCC, CH 2 Cl 2 4. H 2 O 5. PCC, CH 2 Cl 2 B-14. The amino ketone shown undergoes a spontaneous cyclization on standing. What is the product of this intramolecular reaction? (a)(d) (b)(e) (c) NH 2 CH 3 N CH 3 CH 3 NH 2 CH 3 N CH 3 CH 3 CH 3 O CCH 3 CH 3 CH 2 CHNH 2 O CH 3 COOH O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-15. Which of the following compounds would have a 1 H NMR spectrum consisting of three sin- glets? B-16. Which of the following compounds would have the fewest number of signals in its 13 C NMR spectrum? CH 3 CHCCH 2 CH 3 (d) O CH 3 CH 3 CH 2 CH 2 CCH 2 CH 2 CH 3 (b) O CH 3 CHCCHCH 3 (c) O CH 3 CH 3 CH 3 CCH 2 CCH 3 (a) CH 3 O CH 3 CH 2 CH 2 CH(d) O CH 3 CH 2 CH 2 CCH 2 CH 2 CH 3 (b) O HCCH 2 CH 2 CH 2 CH(c) O O CH 3 CCH 2 CCH 3 (a) CH 3 O CH 3 ALDEHYDES AND KETONES: NUCLEOPHILIC ADDITION TO THE CARBONYL GROUP 469 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website