676 CHAPTER 24 PHENOLS SOLUTIONS TO TEXT PROBLEMS 24.1 (b) A benzyl group (C 6 H 5 CH 2 G) is ortho to the phenolic hydroxyl group in o-benzylphenol. (c) Naphthalene is numbered as shown. 3-Nitro-1-naphthol has a hydroxyl group at C-1 and a nitro group at C-3. (d) Resorcinol is 1,3-benzenediol. 4-Chlororesorcinol is therefore OH Cl OH OH NO 2 3-Nitro-1-naphtholNaphthalene 54 8 1 3 2 6 7 OH CH 2 C 6 H 5 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 24.2 Intramolecular hydrogen bonding between the hydroxyl group and the ester carbonyl can occur when these groups are ortho to each other. Intramolecular hydrogen bonds form at the expense of intermolecular ones, and intramolecularly hydrogen-bonded phenols have lower boiling points than isomers in which only intermolecular hydrogen-bonding is possible. 24.3 (b) A cyano group withdraws electrons from the ring by resonance. A p-cyano substituent is con- jugated directly with the negatively charged oxygen and stabilizes the anion more than does an m-cyano substituent. p-Cyanophenol is slightly more acidic than m-cyanophenol, the K a values being 1.0 H11003 10 H110028 and 2.8 H11003 10 H110029 , respectively. (c) The electron-withdrawing inductive effect of the fluorine substituent will be more pronounced at the ortho position than at the para. o-Fluorophenol (K a H11005 1.9 H11003 10 H110029 ) is a stronger acid than p-fluorophenol (K a H11005 1.3 H11003 10 H1100210 ). 24.4 The text points out that the reaction proceeds by the addition–elimination mechanism of nucleophilic aromatic substitution. Under the strongly basic conditions of the reaction, p-toluenesulfonic acid is first converted to its anion. Nucleophilic addition of hydroxide ion gives a cyclohexadienyl anion intermediate. Loss of sulfite ion (SO 3 2H11002 ) gives p-cresol. H11001 p-Cresol H 3 COHSO 3 2H11002 Cyclohexadienyl anion H 3 C SO 3 H11002 OH H11002 H11001 OH H11002 Hydroxidep-Toluenesulfonate ion H 3 CSO 3 H11002 Cyclohexadienyl anion H 3 C SO 3 H11002 OH H11002 H11001H11001H 3 CSO O O H p-Toluenesulfonic acid H 3 CSO H11002 O O p-Toluenesulfonate ion OH H11002 Hydroxide ion HOH Water OCN H11002 O CN H11002 OCH 3 O O C H Methyl salicylate PHENOLS 677 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 678 PHENOLS It is also possible that the elimination stage of the reaction proceeds as follows: 24.5 The text states that the hydrolysis of chlorobenzene in base follows an elimination–addition mechanism. 24.6 (b) The reaction is Friedel–Crafts alkylation. Proton transfer from sulfuric acid to 2-methyl- propene gives tert-butyl cation. Because the position para to the hydroxyl substituent already bears a bromine, the tert-butyl cation attacks the ring at the position ortho to the hydroxyl. (c) Acidification of sodium nitrite produces nitrous acid, which nitrosates the strongly activated aromatic ring of phenols. CH(CH 3 ) 2 OH H 3 C NaNO 2 HCl, H 2 O CH(CH 3 ) 2 OH N O H 3 C 2-Isopropyl-5-methylphenol 2-Isopropyl-5-methyl-4-nitrosophenol (isolated yield, 87%) H 2 SO 4 (CH 3 ) 2 C OH CH 3 CH 2 Br (CH 3 ) 3 C OH CH 3 Br H11001 4-Bromo-2- methylphenol 4-Bromo-2-tert-butyl- 6-methylphenol (isolated yield, 70%) 2-Methylpropene H11002 OH OH Phenol OH H11002 H11001 Benzyne H 2 O OH H11002 H 2 OH11001H11001Cl H11002 H11001 H Cl Chlorobenzene Benzyne H11001H11001H 2 O Cyclohexadienyl anion intermediate H 3 C SO 3 H11002 OH H H11002 H 3 C SO 3 H11002 OH H H SO 3 2H11002 H 2 OH11001H11001 OH H11002 H 3 C H H O OH H11002 p-Methylphenoxide ion H 3 C O H11002 HO H11002 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d ) Friedel–Crafts acylation occurs ortho to the hydroxyl group. 24.7 (b) The hydroxyl group of 2-naphthol is converted to the corresponding acetate ester. (c) Benzoyl chloride acylates the hydroxyl group of phenol. 24.8 Epoxides are sensitive to nucleophilic ring-opening reactions. Phenoxide ion attacks the less hin- dered carbon to yield 1-phenoxy-2-propanol. 24.9 The aryl halide must be one that is reactive toward nucleophilic aromatic substitution by the addition–elimination mechanism. p-Fluoronitrobenzene is far more reactive than fluorobenzene. The reaction shown yields p-nitrophenyl phenyl ether in 92% yield. 24.10 Substituted allyl aryl ethers undergo a Claisen rearrangement similar to the reaction described in text Section 24.13 for allyl phenyl ether. 2-Butenyl phenyl ether rearranges on heating to give o-(1- methyl-2-propenyl)phenol. O H OH o-(1-Methyl-2-propenyl)- phenol rearrangement enolization 2-Butenyl phenyl ether O H11001OK Potassium phenoxide NO 2 F p-Fluoronitrobenzene ONO 2 p-Nitrophenyl phenyl ether 150H11034C OCH 2 CHCH 3 OH 1-Phenoxy-2-propanol O H11002 Phenoxide ion HO H11002 , H 2 O 1,2-Epoxypropane H 2 C O CHCH 3 H11001H11001 Phenyl benzoate OC O Phenol OH Benzoyl chloride O CCl Hydrogen chloride HCl H11001H11001 NaOH 2-Naphthyl acetate OCCH 3 O 2-Naphthol OH Sodium acetate CH 3 CONa O Acetic anhydride CH 3 COCCH 3 O O AlCl 3 OH CH 3 O CH 3 CH 2 CClH11001 p-Cresol Propanoyl chloride CCH 2 CH 3 OH CH 3 O 1-(2-Hydroxy-5-methylphenyl)- 1-propanone (isolated yield, 87%) PHENOLS 679 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 24.11 (a) The parent compound is benzaldehyde. Vanillin bears a methoxy group (CH 3 O) at C-3 and a hydroxyl group (HO) at C-4. (b, c) Thymol and carvacrol differ with respect to the position of the hydroxyl group. (d) An allyl substituent is GCH 2 CH?CH 2 . (e) Benzoic acid is C 6 H 5 CO 2 H. Gallic acid bears three hydroxyl groups, located at C-3, C-4, and C-5. ( f ) Benzyl alcohol is C 6 H 5 CH 2 OH. Salicyl alcohol bears a hydroxyl group at the ortho position. Salicyl alcohol (o-hydroxybenzyl alcohol) CH 2 OH OH Gallic acid (3,4,5-trihydroxybenzoic acid) CO 2 H OHHO OH 1 2 3 4 5 6 OH OCH 3 CH 2 CH CH 2 1 2 3 4 5 6 Eugenol (4-allyl-2-methoxyphenol) HO CH 3 CH(CH 3 ) 2 3 2 1 6 5 4 Thymol (2-isopropyl-5-methylphenol) HO CH 3 CH(CH 3 ) 2 3 2 1 6 5 4 Carvacrol (5-isopropyl-2-methylphenol) OCH 3 C HO OH 3 2 1 6 5 4 Vanillin (4-hydroxy-3-methoxybenzaldehyde) 680 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 24.12 (a) The compound is named as a derivative of phenol. The substituents (ethyl and nitro) are cited in alphabetical order with numbers assigned in the direction that gives the lowest number at the first point of difference. (b) An isomer of the compound in part (a) is 4-ethyl-3-nitrophenol. (c) The parent compound is phenol. It bears, in alphabetical order, a benzyl group at C-4 and a chlorine at C-2. (d) This compound is named as a derivative of anisole, C 6 H 5 OCH 3 . Because multiplicative pre- fixes (di, tri-, etc.) are not considered when alphabetizing substituents, isopropyl precedes dimethyl. (e) The compound is an aryl ester of trichloroacetic acid. The aryl group is 2,5-dichlorophenyl. 2,5-Dichlorophenyl trichloroacetate Cl Cl 1 2 3 4 5 6 OCCCl 3 O 4-Isopropyl-2,6- dimethylanisole OCH 3 CH 3 H 3 C CH(CH 3 ) 2 1 2 3 4 5 6 Cl CH 2 HO 1 2 3 4 56 4-Benzyl-2-chlorophenol OH NO 2 CH 2 CH 3 1 2 3 4 5 6 4-Ethyl-3-nitrophenol 3-Ethyl-4-nitrophenol OH CH 2 CH 3 NO 2 1 2 3 4 5 6 PHENOLS 681 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 24.13 (a) The reaction is an acid–base reaction. Phenol is the acid; sodium hydroxide is the base. (b) Sodium phenoxide reacts with ethyl bromide to yield ethyl phenyl ether in a Williamson reaction. Phenoxide ion acts as a nucleophile. (c) p-Toluenesulfonate esters behave much like alkyl halides in nucleophilic substitution reac- tions. Phenoxide ion displaces p-toluenesulfonate from the primary carbon. (d ) Carboxylic acid anhydrides react with phenoxide anions to yield aryl esters. (e) Acyl chlorides convert phenols to aryl esters. ( f ) Phenols react as nucleophiles toward epoxides. CH 3 OH H11001 m-Cresol Ethylene oxide H 2 C CH 2 O 2-(3-Methylphenoxy)ethanol H 3 C OCH 2 CH 2 OH OH OC HCl CH 3 CH 3 CCl H11001H11001 O O o-Cresol Benzoyl chloride 2-Methylphenyl benzoate Hydrogen chloride C 6 H 5 ONa C 6 H 5 OCCH 3 CH 3 COCCH 3 CH 3 CONaH11001H11001 Sodium phenoxide Acetic anhydride Phenyl acetate Sodium acetate O O O O CH 3 CH 2 CH 2 CH 2 OSC 6 H 5 ONa C 6 H 5 OCH 2 CH 2 CH 2 CH 3 NaOSH11001H11001 Sodium phenoxide Butyl p-toluenesulfonate Sodium p-toluenesulfonateButyl phenyl ether O O O O CH 3 CH 3 CH 3 CH 2 BrC 6 H 5 ONa C 6 H 5 OCH 2 CH 3 NaBrH11001H11001 Sodium phenoxide Sodium bromide Ethyl bromide Ethyl phenyl ether OH H11001H11001NaOH ONa H 2 O Phenol (stronger acid) Water (weaker acid) Sodium hydroxide (stronger base) Sodium phenoxide (weaker base) 682 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The reaction as written conforms to the requirements of the problem that a balanced equation be written. Of course, the reaction will be much faster if catalyzed by acid or base, but the cat- alysts do not enter into the equation representing the overall process. (g) Bromination of the aromatic ring of 2,6-dichlorophenol occurs para to the hydroxy group. The more activating group (GOH) determines the orientation of the product. (h) In aqueous solution bromination occurs at all the open positions that are ortho and para to the hydroxyl group. (i) Hydrogen bromide cleaves ethers to give an alkyl halide and a phenol. 24.14 (a) Strongly electron-withdrawing groups, particularly those such as GNO 2 , increase the acidity of phenols by resonance stabilization of the resulting phenoxide anion. Electron-releasing substituents such as GCH 3 exert a very small acid-weakening effect. OH CH 3 H 3 C CH 3 2,4,6-Trimethylphenol, less acidic (K a H11005 1.3 H11003 10 H1100211 , pK a H11005 10.9) OH NO 2 O 2 N NO 2 2,4,6-Trinitrophenol, more acidic (K a H11005 3.8 H11003 10 H110021 , pK a H11005 0.4) H11001 Hydrogen bromide HBr Isopropyl phenyl ether OCH(CH 3 ) 2 H11001 Isopropyl bromide (CH 3 ) 2 CHBr Phenol OH heat OH CH 3 CH 3 Br Br 2HBr2Br 2 OH H11001H11001 Brominep-Cresol 2,6-Dibromo-4- methylphenol Hydrogen bromide H 2 O OH Br HBrBr 2 Cl Cl OH Cl Cl H11001H11001 2,6-Dichlorophenol Bromine 4-Bromo-2,6- dichlorophenol Hydrogen bromide PHENOLS 683 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Picric acid (2,4,6-trinitrophenol) is a stronger acid by far than 2,4,6-trimethylphenol. All three nitro groups participate in resonance stabilization of the picrate anion. (b) Stabilization of a phenoxide anion is most effective when electron-withdrawing groups are present at the ortho and para positions, because it is these carbons that bear most of the nega- tive charge in phenoxide anion. 2,6-Dichlorophenol is therefore expected to be (and is) a stronger acid than 3,5-dichloro- phenol. (c) The same principle is at work here as in part (b). A nitro group para to the phenol oxygen is directly conjugated to it and stabilizes the anion better than one at the meta position. OH NO 2 4-Nitrophenol, stronger acid (K a H11005 1.0 H11003 10 H110028 , pK a H11005 7.2) OH NO 2 3-Nitrophenol, weaker acid (K a H11005 4.1 H11003 10 H110029 , pK a H11005 8.4) OH ClCl 3,5-Dichlorophenol, less acidic (K a H11005 6.5 H11003 10 H110029 , pK a H11005 8.2) OH Cl Cl 2,6-Dichlorophenol, more acidic (K a H11005 1.6 H11003 10 H110027 , pK a H11005 6.8) O H11002 H11002 O H11002 O H11002 O N H11001 N H11001 O O O H11002H11002 O O H11002 O N H11001 O H11002 N H11001 N H11001 O O O H11002H11002 O O H11002 O N H11001 H11002 O N H11001 N H11001 O O O H11002H11002 O O H11002 N H11001 O O H11002 H11002 N H11001 N H11001 O O O H11002H11002 O O H11002 N H11001 O O 684 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) A cyano group is strongly electron-withdrawing, and so 4-cyanophenol is a stronger acid than phenol. There is resonance stabilization of the 4-cyanophenoxide anion. (e) The 5-nitro group in 2,5-dinitrophenol is meta to the hydroxyl group and so does not stabilize the resulting anion as much as does an ortho or a para nitro group. 24.15 (a) The rate-determining step of ester hydrolysis in basic solution is formation of the tetrahedral intermediate. Because this intermediate is negatively charged, there will be a small effect favoring its for- mation when the aryl group bears an electron-withdrawing substituent. Furthermore, this in- termediate can either return to starting materials or proceed to products. The proportion of the tetrahedral intermediate that goes on to products increases as the leav- ing group ArO H11002 becomes less basic. This is strongly affected by substituents; electron- withdrawing groups stabilize ArO H11002 . The prediction is that m-nitrophenyl acetate undergoes hydrolysis in basic solution faster than phenol. Indeed, this is observed to be the case; m-nitrophenyl acetate reacts some ten times faster than does phenyl acetate at 25°C. HO H11002 H11001OCCH 3 O 2 N O m-Nitrophenyl acetate (more reactive) H11001 CH 3 COH O O H11002 O 2 N m-Nitrophenoxide anion (a better leaving group than phenoxide because it is less basic) H11001 CH 3 COHArO H11002 O H11001 CH 3 CO H11002 ArO H11002 O CCH 3 ArO O OH H11002 HO H11002 ArOCCH 3 O H11002 OH H11001ArOCCH 3 O HO H11002 slow OH NO 2 O 2 N 2,6-Dinitrophenol, more acidic (K a H11005 2.0 H11003 10 H110024 , pK a H11005 3.7) OH NO 2 O 2 N 2,5-Dinitrophenol, less acidic (K a H11005 6.0 H11003 10 H110026 , pK a H11005 5.2) OCN H11002 O CN H11002 OH CN 4-Cyanophenol, more acidic (K a H11005 1.1 H11003 10 H110028 , pK a H11005 8.0) OH Phenol, less acidic (K a H11005 1 H11003 10 H1100210 , pK a H11005 10) PHENOLS 685 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) The same principle applies here as in part (a). p-Nitrophenyl acetate reacts faster than m- nitrophenyl acetate (by about 45%) largely because p-nitrophenoxide is less basic and thus a better leaving group than m-nitrophenoxide. Resonance in p-nitrophenoxide is particularly effective because the p-nitro group is directly conjugated to the oxyanion; direct conjugation of these groups is absent in m-nitrophenoxide. (c) The reaction of ethyl bromide with a phenol is an S N 2 reaction in which the oxygen of the phe- nol is the nucleophile. The reaction is much faster with sodium phenoxide than with phenol, because an anion is more nucleophilic than a corresponding neutral molecule. Faster reaction: Slower reaction: (d) The answer here also depends on the nucleophilicity of the attacking species, which is a phe- noxide anion in both reactions. The more nucleophilic anion is phenoxide ion, because it is more basic than p-nitrophenoxide. Rate measurements reveal that sodium phenoxide reacts 17 times faster with ethylene oxide (in ethanol at 70°C) than does its p-nitro derivative. (e) This reaction is electrophilic aromatic substitution. Because a hydroxy substituent is more activating than an acetate group, phenol undergoes bromination faster than does phenyl acetate. Resonance involving ester group reduces tendency of oxygen to donate electrons to ring. O CCH 3 O O CCH 3 O H11002 H11001 More basic; better nucleophile H11002 O Better delocalization of negative charge makes this less basic and less nucleophilic. O 2 N H11002 O ArO H11002 ArOCH 2 CH 2 O H11002 CH 2 H 2 C O H11001 H11001ArOCH 2 CH 3 Br H11002 BrCH 2 CH 3 HH ArO ArO H11002 H11001ArOCH 2 CH 3 Br H11002 BrCH 2 CH 3 O H11002 O H11002 O H11001 N H11002 O H11001 H11002 ON O 686 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 24.16 Nucleophilic aromatic substitution by the elimination–addition mechanism is impossible, owing to the absence of any protons that might be abstracted from the substrate. The addition–elimination pathway is available, however. This pathway is favorable because the cyclohexadienyl anion intermediate formed in the rate- determining step is stabilized by the electron-withdrawing inductive effect of its fluorine substituents. 24.17 (a) Allyl bromide is a reactive alkylating agent and converts the free hydroxyl group of the aryl compound (a natural product known as guaiacol) to its corresponding allyl ether. (b) Sodium phenoxide acts as a nucleophile in this reaction and is converted to an ether. (c) Orientation in nitration is governed by the most activating substituent, in this case the hydroxyl group. (d) Allyl aryl ethers undergo a Claisen rearrangement on heating. Heating p-acetamidophenyl allyl ether gave an 83% yield of 4-acetamido-2-allylphenol. heat p-Acetamidophenyl allyl ether OCH 2 CH CH 2 CH 3 CNH O 4-Acetamido-2-allylphenol OHCH 3 CNH O CH 2 CH CH 2 OCH 3 CH O HO Vanillin OCH 3 CH O O 2 N HO 4-Hydroxy-3-methoxy-5- nitrobenzaldehyde (83%) HNO 3 acetic acid, heat H11001 Sodium phenoxide ONa 3-Chloro-1,2- propanediol ClCH 2 CHCH 2 OH OH 3-Phenoxy-1,2- propanediol (61–63%) OCH 2 CHCH 2 OH OH H11001 Guaiacol OCH 3 OH K 2 CO 3 acetone Allyl bromide H 2 C CHCH 2 Br 2-Allyloxyanisole (80–90%) OCH 3 OCH 2 CH CH 2 HO H11002 H11001 Hexafluorobenzene F F F FF F slow fast F F F F OH F F H11002 Pentafluorophenol F H11002 F F F OHF F H11001 PHENOLS 687 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) The hydroxyl group, as the most activating substituent, controls the orientation of elec- trophilic aromatic substitution. Bromination takes place ortho to the hydroxyl group. ( f ) Oxidation of hydroquinone derivatives (p-dihydroxybenzenes) with Cr(VI) reagents is a method for preparing quinones. (g) Aryl esters undergo a reaction known as the Fries rearrangement on being treated with aluminum chloride, which converts them to acyl phenols. Acylation takes place para to the hydroxyl in this case. (h) Nucleophilic aromatic substitution takes place to yield a diaryl ether. The nucleophile is the phenoxide ion derived from 2,6-dimethylphenol. H11001 OH CH 3 H 3 C 2,6-Dimethylphenol ONO 2 CH 3 CH 3 2,6-Dimethylphenyl p-nitrophenyl ether (82%) NO 2 Cl p-Chloro- nitrobenzene NaOH AlCl 3 5-Isopropyl-2- methylphenyl acetate CH(CH 3 ) 2 CH 3 OCCH 3 O 4-Hydroxy-2-isopropyl- 5-methylacetophenone (90%) CH(CH 3 ) 2 CH 3 OH CH 3 C O OH OH Cl 2-Chloro-1,4- benzenediol O O Cl 2-Chloro-1,4- benzoquinone (88%) K 2 Cr 2 O 7 H 2 SO 4 Br 2 H11001 NO 2 OH OCH 2 CH 3 2-Ethoxy-4- nitrophenol NO 2 OH OCH 2 CH 3 Br 2-Bromo-6-ethoxy-4- nitrophenol (65%) acetic acid 688 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (i) Chlorination with excess chlorine occurs at all available positions that are ortho and para to the hydroxyl group. ( j ) Amines react with esters to give amides. In the case of a phenyl ester, phenol is the leaving group. (k) Aryl diazonium salts attack electron-rich aromatic rings, such as those of phenols, to give the products of electrophilic aromatic substitution. 24.18 In the first step p-nitrophenol is alkylated on its phenolic oxygen with ethyl bromide. Reduction of the nitro group gives the corresponding arylamine. Treatment of p-ethoxyaniline with acetic anhydride gives phenacetin. H11001 p-Ethoxyaniline OCH 2 CH 3 H 2 N Acetic anhydride CH 3 COCCH 3 O O p-Ethoxyacetanilide (phenacetin) OCH 2 CH 3 CH 3 CNH O OCH 2 CH 3 O 2 N Ethyl p-nitrophenyl ether p-Ethoxyaniline OCH 2 CH 3 H 2 N 1. Fe, HCl 2. HO H11002 HO H11002 H11001 Ethyl bromide CH 3 CH 2 Br p-Nitrophenol OHO 2 N Ethyl p-nitrophenyl ether OCH 2 CH 3 O 2 N H11001 Cl OH Cl Cl 2,4,5-Trichlorophenol Benzenediazonium chloride H11001 C 6 H 5 NNCl H11002 2-Benzeneazo-3,4,6-trichlorophenol (80%) Cl OH Cl ClC 6 H 5 NN H11001 NH 2 CH 3 o-Methylaniline Phenyl salicylate C O O OH heat H11001 Phenol OH N-(o-Methylphenyl)salicylamide (isolated yield, 73–77%) H 3 C C NH O OH H11001 Cl Cl OH 2,5-Dichlorophenol 2Cl 2 Chlorine H11001 2HCl Hydrogen chloride Cl Cl Cl Cl OH 2,3,4,6-Tetrachlorophenol (isolated yield, 100%) acetic acid PHENOLS 689 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 24.19 The three parts of this problem make up the series of steps by which o-bromophenol is prepared. (a) Because direct bromination of phenol yields both o-bromophenol and p-bromophenol, it is essential that the para position be blocked prior to the bromination step. In practice, what is done is to disulfonate phenol, which blocks the para and one of the ortho positions. (b) Bromination then can be accomplished cleanly at the open position ortho to the hydroxyl group. (c) After bromination the sulfonic acid groups are removed by acid-catalyzed hydrolysis. 24.20 Nitration of 3,5-dimethylphenol gives a mixture of the 2-nitro and 4-nitro derivatives. H11001 OH H 3 CCH 3 3,5-Dimethylphenol OH H 3 CCH 3 NO 2 3,5-Dimethyl-4- nitrophenol OH H 3 CCH 3 NO 2 3,5-Dimethyl-2- nitrophenol HNO 3 H 2 O H 2 OH11001 H H11001 heat OH Br o-Bromophenol (compound C) OH SO 3 HBr SO 3 H Compound B Br 2 H11001 1. HO H11002 2. H H11001 OH SO 3 H SO 3 H Compound A OH SO 3 HBr SO 3 H 5-Bromo-4-hydroxy-1,3- benzenedisulfonic acid (compound B) H11001 OH Phenol 2H 2 SO 4 SO 3 H OH SO 3 H 4-Hydroxy-1,3- benzenedisulfonic acid (compound A) heat 690 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The more volatile compound (compound A), isolated by steam distillation, is the 2-nitro derivative. Intramolecular hydrogen bonding is possible between the nitro group and the hydroxyl group. The 4-nitro derivative participates in intermolecular hydrogen bonds and has a much higher boiling point; it is compound B. 24.21 The relationship between the target molecule and the starting materials tells us that two processes are required, formation of a diaryl ether linkage and nitration of an aromatic ring. The proper order of carrying out these two separate processes is what needs to be considered. The critical step is ether formation, a step that is feasible for the reactants shown: The reason this reaction is suitable is that it involves nucleophilic aromatic substitution by the addition–elimination mechanism on a p-nitro-substituted aryl halide. Indeed, this reaction has been carried out and gives an 80–82% yield. A reasonable synthesis would therefore begin with the preparation of p-chloronitrobenzene. Separation of the p-nitro-substituted aryl halide and reaction with phenoxide ion complete the synthesis. The following alternative route is less satisfactory: Diphenyl ether O 2-Nitrophenyl phenyl ether O O 2 N 4-Nitrophenyl phenyl ether ONO 2 HNO 3 H 2 SO 4 H11001 H11001 Chlorobenzene Diphenyl ether OCl Phenol OH base Cl Chlorobenzene o-Chloronitrobenzene Cl NO 2 HNO 3 H 2 SO 4 p-Chloronitrobenzene Cl NO 2 H11001 H11001 p-Chloronitrobenzene Cl NO 2 4-Nitrophenyl phenyl ether O NO 2 Phenol OH KOH heat C 6 H 5 OH H11001C 6 H 5 ONO 2 Cl NO 2 C 6 H 5 Cl H 3 CCH 3 H OO H11002 O N H11001 Intramolecular hydrogen bonding in 3,5-dimethyl-2-nitrophenol PHENOLS 691 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The difficulty with this route concerns the preparation of diphenyl ether. Direct reaction of phenox- ide ion with chlorobenzene is very slow and requires high temperatures because chlorobenzene is a poor substrate for nucleophilic substitution. A third route is also unsatisfactory because it, too, requires nucleophilic substitution on chlorobenzene. 24.22 The overall transformation that needs to be effected is A reasonable place to begin is with the attachment of the side chain. The aldehyde function allows for chain extension by a Wittig reaction. Hydrogenation of the double bond and hydrogen halide cleavage of the ether functions complete the synthesis. CH 3 O CH 2 CH 2 (CH 2 ) 12 CH 3 OCH 3 CH 3 O CH CH(CH 2 ) 12 CH 3 OCH 3 H 2 Pt HBr heat HO CH 2 CH 2 (CH 2 ) 12 CH 3 OH 3-Pentadecylcatechol CH 3 O CH O OCH 3 2,3-Dimethoxy- benzaldehyde H11001 CH 3 O CH CH(CH 2 ) 12 CH 3 OCH 3 H11001 H11002 CH 3 (CH 2 ) 12 CH P(C 6 H 5 ) 3 CH 2 CH 2 RAr CH CHRAr H11001CH O Ar CHR(C 6 H 5 ) 3 P H11001 H11002 CH 3 O CH O OCH 3 2,3-Dimethoxybenzaldehyde HO (CH 2 ) 14 CH 3 OH 3-Pentadecylcatechol H11001 4-Nitrophenyl phenyl ether ONO 2 base Chlorobenzene Cl p-Nitrophenol OH NO 2 OH Phenol o-Nitrophenol OH NO 2 HNO 3 p-Nitrophenol OH NO 2 H11001 692 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Other synthetic routes are of course possible. One of the earliest approaches used a Grignard reaction to attach the side chain. The resulting secondary alcohol can then be dehydrated to the same alkene intermediate prepared in the preceding synthetic scheme. Again, hydrogenation of the double bond and ether cleavage leads to the desired 3-pentadecylcatechol. 24.23 Recall that the Claisen rearrangement converts an aryl allyl ether to an ortho-substituted allyl phe- nol. The presence of an allyl substituent in the product ortho to an aryl ether thus suggests the fol- lowing retrosynthesis: As reported in the literature synthesis, the starting phenol may be converted to the corresponding allyl ether by reaction with allyl bromide in the presence of base. This step was accomplished in 80% yield. Heating the allyl ether yields the o-allyl phenol. The synthesis is completed by methylation of the phenolic oxygen and saponification of the acetate ester. The final three steps of the synthesis proceeded in an 82% overall yield. OH CH 3 O H 3 C CH 2 CH CH 2 OCCH 3 O OCH 3 CH 3 O H 3 C CH 2 CH CH 2 OCCH 3 O OCH 3 CH 3 O H 3 C CH 2 CH CH 2 OH K 2 CO 3 CH 3 I 1. KOH, CH 3 OH 2. H H11001 OH CH 3 O H 3 C OCCH 3 O OCH 2 CH OCCH 3 O CH 3 O H 3 C CH 2 OH OCCH 3 O CH 3 O H 3 C CH 2 CH CH 2 200H11034C 3 hK 2 CO 3 H 2 C CHCH 2 Br OCH 3 OH CH 3 O H 3 C CH 2 CH CH 2 OCH 2 CH OCCH 3 O CH 3 O H 3 C CH 2 CH 3 O CHCH 2 (CH 2 ) 12 CH 3 OH OCH 3 CH 3 O CH CH(CH 2 ) 12 CH 3 OCH 3 H H11001 CH 3 O CH O OCH 3 2,3-Dimethoxy- benzaldehyde CH 3 O CHCH 2 (CH 2 ) 12 CH 3 OH OCH 3 H11001 CH 3 (CH 2 ) 12 CH 2 MgBr 1. diethyl ether 2. H 3 O H11001 PHENOLS 693 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 24.24 The driving force for this reaction is the stabilization that results from formation of the aromatic ring. A reasonable series of steps begins with protonation of the carbonyl oxygen. 24.25 Bromination of p-hydroxybenzoic acid takes place in the normal fashion at both positions ortho to the hydroxy group. A third bromination step, this time at the para position, leads to the intermediate shown. Aromatization of this intermediate occurs by decarboxylation. H11001H11001CO 2 H H11001 2,4,6-Tribromophenol OH Br Br Br OH Br C O O H H11001 H11001 Br 2 H11001 Br H11002 OH BrBr CO 2 H OH BrBr Br CO 2 H H11001 Br 2 OH BrBr CO 2 H 3,5-Dibromo-4- hydroxybenzoic acid OH CO 2 H p-Hydroxybenzoic acid H11002H H11001 Protonated ketone can rearrange by alkyl migration. OH H H H H H11001 OH HH H Aromatization of this intermediate occurs by loss of a proton. OH H H H H H11001 H H11001 O H H H H H11001 OH H H H H OH H H H H H11001 Resonance forms of protonated ketone 694 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 24.26 Electrophilic attack of bromine on 2,4,6-tribromophenol leads to a cationic intermediate. Loss of the hydroxyl proton from this intermediate generates the observed product. 24.27 A good way to approach this problem is to assume that bromine attacks the aromatic ring of the phenol in the usual way, that is, para to the hydroxyl group. This cation cannot yield the product of electrophilic aromatic substitution by loss of a proton from the ring but can lose a proton from oxygen to give a cyclohexadienone derivative. This cyclohexadienone is the compound C 18 H 29 BrO, and the peaks at 1655 and 1630 cm H110021 in the infrared are consistent with C?O and C?C stretching vibrations. The compound’s symmetry is consistent with the observed 1 H NMR spectrum; two equivalent tert-butyl groups at C-2 and C-6 appear as an 18-proton singlet at H9254 1.3 ppm, the other tert-butyl group is a 9-proton singlet at H9254 1.2 ppm, and the 2 equivalent vinyl protons of the ring appear as a singlet at H9254 6.9 ppm. 24.28 Because the starting material is an acetal and the reaction conditions lead to hydrolysis with the production of 1,2-ethanediol, a reasonable reaction course is O O O Compound A O O Compound B HOCH 2 CH 2 OH 1,2-Ethanediol H 2 O, H H11001 H11001 4-Bromo-2,4,6-tri-tert- butyl-2,5-cyclohexadienone Br C(CH 3 ) 3 O (CH 3 ) 3 C C(CH 3 ) 3 H11002H H11001 Br C(CH 3 ) 3 O (CH 3 ) 3 C C(CH 3 ) 3 H11001 H 2,4,6-Tri-tert-butylphenol OH (CH 3 ) 3 C C(CH 3 ) 3 C(CH 3 ) 3 Br C(CH 3 ) 3 OH (CH 3 ) 3 C C(CH 3 ) 3 H11001 Br 2 O H Br Br Br Br H11001 2,4,4,6-Tetrabromo- cyclohexadienone Br Br O Br Br H11001 Br 2 H11001 Br H11002 OH Br Br Br Br H11001 2,4,6-Tribromophenol OH Br Br Br PHENOLS 695 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Indeed, dione B satisfies the spectroscopic criteria. Carbonyl bands are seen in the infrared spec- trum, and compound B has two sets of protons to be seen in its 1 H NMR spectrum. The two vinyl protons are equivalent and appear at low field, H9254 6.7 ppm; the 4 methylene protons are equivalent to each other and are seen at H9254 2.9 ppm. Compound B is the doubly ketonic tautomeric form of hydroquinone, compound C, to which it isomerizes on standing in water. 24.29 A reasonable first step is protonation of the hydroxyl oxygen. The weak oxygen–oxygen bond can now be cleaved, with loss of water as the leaving group. This intermediate bears a positively charged oxygen with only six electrons in its valence shell. Like a carbocation, such a species is highly electrophilic. The electrophilic oxygen attacks the H9266 system of the neighboring aromatic ring to give an unstable intermediate. Ring opening of this intermediate is assisted by one of the lone pairs of oxygen and restores the aromaticity of the ring. The cation formed by ring opening is captured by a water molecule to yield the hemiacetal product. C(CH 3 ) 2 H11001 O O HH OC(CH 3 ) 2 H H11001 OH H11001 C(CH 3 ) 2 H11001 O CCH 3 CH 3 H11001 O CCH 3 CH 3 O H11001 CCH 3 CH 3 H11001 O H 2 OH11001CCH 3 CH 3 O H11001 H11001 CCH 3 CH 3 OOH 2 H H11001 H11001 H11001 CCH 3 CH 3 OOH 2 CCH 3 CH 3 OOH Cumene hydroperoxide O O Compound B Compound C (hydroquinone) HO OH 696 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 24.30 (a) The molecular formula of the compound (C 9 H 12 O) tells us that it has a total of four double bonds and rings (index of hydrogen deficiency H11005 4). The prominent peak in the infrared spec- trum is the hydroxyl absorption of an alcohol or a phenol at 3300 cm H110021 . Peaks in the H9254 110–160 ppm region of the 13 C NMR spectrum suggest an aromatic ring, which accounts for six of the nine carbon atoms and all its double bonds and rings. The presence of four peaks in this region, two of which are C and two CH, indicates a para- disubstituted aromatic derivative. That the remaining three carbons are sp 3 -hybridized is indicated by the upfield absorptions at H9254 15, 26, and 38 ppm. None of these carbons has a chemical shift below H9254 40 ppm, and so none of them can be bonded to the hydroxyl group. Thus the hydroxyl group must be bonded to the aromatic ring. The compound is 4-propylphenol. (b) Once again the molecular formula (C 9 H 11 BrO) indicates a total of four double bonds and rings. The four peaks in the H9254 110 –160 ppm region of the spectrum, three of which represent CH, suggest a monosubstituted aromatic ring. The remaining atoms to be accounted for are O and Br. Because all the unsaturations are accounted for by the benzene ring and the infrared spectrum lacks any hydroxyl absorption, the oxygen atom must be part of an ether function. The three CH 2 groups indicated by the ab- sorptions at H9254 32, 35, and 66 ppm in the 13 C NMR spectrum allow the compound to be identi- fied as 3-bromopropyl phenyl ether. SELF-TEST PART A A-1. Which is the stronger acid, m-hydroxybenzaldehyde or p-hydroxybenzaldehyde? Explain your answer, using resonance structures. A-2. The cresols are methyl-substituted phenols. Predict the major products to be obtained from the reactions of o-, m-, and p-cresol with dilute nitric acid. A-3. Give the structure of the product from the reaction of p-cresol with propanoyl chloride, , in the presence of AlCl 3 . What product is obtained in the absence of AlCl 3 ? A-4. Provide the structure of the reactant, reagent, or product omitted from each of the following: (a) (b) OCH 2 CH(CH 3 ) 2 ? (two compounds) CH 3 OCH(CH 3 ) 2 ? (two products) HBr CH 3 CH 2 CCl O OCH 2 CH 2 CH 2 Br 3-Bromopropyl phenyl ether HO CH 2 CH 2 CH 3 4-Propylphenol PHENOLS 697 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) (d) A-5. Provide the structures of compounds A and B in the following sequence of reactions: A-6. Prepare p-tert-butylphenol from tert-butylbenzene using any necessary organic or inorganic reagents. PART B B-1. Rank the following in order of decreasing acid strength (most acidic → least acidic): (a)2H11022 4 H11022 1 H11022 3(c)1H11022 3 H11022 4 H11022 2 (b)3H11022 1 H11022 2 H11022 4(d )3H11022 1 H11022 4 H11022 2 B-2. Rank the following compounds in order of increasing acidity (weakest acid first). (a)2H11021 3 H11021 1(c)3H11021 1 H11021 2(e)1H11021 2 H11021 3 (b)3H11021 2 H11021 d )2H11021 1 H11021 3 OH OH OHO 2 N NO 2 NO 2 Cl Cl Cl CH 3 CH 3 H 3 C 12 OH NO 2 NO 2 OH OH OH 1234 A B (C 11 H 14 O) K 2 CO 3 OH CH 3 CH 2 CH CHCH 2 Br heat HBr heat O ? (C 8 H 9 BrO) CO 2 H O H11002 Na H11001 1. ? 2. H H11001 OH 698 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-3. Which of the following phenols has the largest pK a value (i.e., is least acidic)? (a) (c) (b)(d ) B-4. Which of the following reactions is a more effective method for preparing phenyl propyl ether? I: II: (a) Reaction I is more effective. (b) Reaction II is more effective. (c) Both reactions I and II are effective. (d) Neither reaction I nor reaction II is effective. B-5. What reactant gives the product shown on heating with aluminum chloride? (a)(c) (b d ) B-6. What are the products of the following reaction? (a)(d ) (b e) (c) OH BrCH 2 CH 2 BrBr H11001 Br BrCH 2 CH 2 BrH11001OH BrCH 2 CH 2 BrH11001 Br BrCH 2 CH 2 OHH11001OCH 2 CH 2 BrBr OCH 2 CH 2 OH excess HBr heat OH O C CH 3 H 3 C COH O O H 3 C CO O H 3 C OC O H 3 C C O OH C 6 H 5 BrCH 3 CH 2 CH 2 ONa H11001 C 6 H 5 ONa CH 3 CH 2 CH 2 BrH11001 OHCNOHH 3 C OHO 2 NOHCl PHENOLS 699 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-7. Which of the following sets of reagents, used in the order shown, would enable preparation of p-chlorophenol from p-chloronitrobenzene? (a) 1. Fe, HCl; 2. NaOH; 3. NaNO 2 , H 2 SO 4 ; 4. H 3 PO 2 (b) 1. Fe, HCl; 2. NaOH; 3. NaNO 2 , H 2 SO 4 ; 4. H 2 O, heat (c) 1. Fe, HCl; 2. NaOH; 3. NaNO 2 , H 2 SO 4 ; 4. ethanol (d) 1. NaOH, heat; 2. HCl B-8. What is the product obtained by heating the following allylic ether of phenol? (a)(c) (b)(d ) HO CHCH CH 2 C 6 H 5 OC 6 H 5 CH 2 CH CH 2 OH CHCH CH 2 C 6 H 5 OH CH 2 CH CHC 6 H 5 OCH 2 CH 200H11034C CHC 6 H 5 ? 700 PHENOLS Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website