CHAPTER 23 ARYL HALIDES SOLUTIONS TO TEXT PROBLEMS 23.1 There are four isomers of C 7 H 7 Cl that contain a benzene ring, namely, o, m, and p-chlorotoluene and benzyl chloride. Of this group only benzyl chloride is not an aryl halide; its halogen is not attached to the aromatic ring but to an sp 3 -hybridized carbon. Benzyl chloride has the weakest carbon–halogen bond, its measured carbon–chlorine bond dissociation energy being only 293 kJ/mol (70 kcal/mol). Homolytic cleavage of this bond produces a resonance-stabilized benzyl radical. 23.2 (b) The negatively charged sulfur in is a good nucleophile, which displaces chloride from 1-chloro-2,4-dinitrobenzene. Cl NO 2 NO 2 1-Chloro-2,4- dinitrobenzene SCH 2 C 6 H 5 NO 2 Cl H11002 H11001 NO 2 Benzyl 2,4- dinitrophenyl sulfide C 6 H 5 CH 2 SNa H11001H11002 C 6 H 5 CH 2 S H11002 Na H11001 Benzyl chloride Benzyl radical CH 2 H11001 Chlorine atom ClCH 2 Cl Cl CH 3 p-Chlorotoluene CH 2 Cl Benzyl chloride Cl CH 3 m-Chlorotoluene CH 3 Cl o-Chlorotoluene 656 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ARYL HALIDES 657 (c) The nitrogen in ammonia has an unshared electron pair and is nucleophilic; it displaces chloride from 1-chloro-2,4-dinitrobenzene. (d) As with ammonia, methylamine is nucleophilic and displaces chloride. 23.3 The most stable resonance structure for the cyclohexadienyl anion formed by reaction of methox- ide ion with o-fluoronitrobenzene involves the nitro group and has the negative charge on oxygen. 23.4 The positions that are activated toward nucleophilic attack are those that are ortho and para to the nitro group. Among the carbons that bear a bromine leaving group in 1,2,3-tribromo-5-nitrobenzene, only C-2 satisfies this requirement. 23.5 Nucleophilic addition occurs in the rate-determining step at one of the six equivalent carbons of hexafluorobenzene to give the cyclohexadienyl anion intermediate. H11001 Hexafluorobenzene FF F FF F Methoxide ion OCH 3 H11002 Cyclohexadienyl anion intermediate F F F F F F OCH 3 H11002 1,2,3-Tribromo- 5-nitrobenzene Br BrBr NO 2 1,3-Dibromo-2-ethoxy- 5-nitrobenzene OCH 2 CH 3 BrBr NO 2 NaOCH 2 CH 3 CH 3 O F N H11001 O H11002 H11002 O Cl NO 2 NO 2 1-Chloro-2,4- dinitrobenzene NHCH 3 NO 2 NO 2 N-Methyl-2,4- dinitroaniline CH 3 NH 2 Cl NO 2 NO 2 1-Chloro-2,4- dinitrobenzene NH 2 NO 2 NO 2 2,4-Dinitroaniline NH 3 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 658 ARYL HALIDES Elimination of fluoride ion from the cyclohexadienyl anion intermediate restores the aromaticity of the ring and completes the reaction. 23.6 4-Chloropyridine is more reactive toward nucleophiles than 3-chloropyridine because the anionic intermediate formed by reaction of 4-chloropyridine has its charge on nitrogen. Because nitrogen is more electronegative than carbon, the intermediate is more stable. 23.7 The aryl halide is incapable of elimination and so cannot form the benzyne intermediate necessary for substitution by the elimination–addition pathway. 23.8 The aryne intermediate from p-iodotoluene can undergo addition of hydroxide ion at the position meta to the methyl group or para to it. The two isomeric phenols are m- and p-methylphenol. CH 3 OH p-Methylphenol H11001 CH 3 OH m-Methylphenol CH 3 I p-Iodotoluene CH 3 NaOH, H 2 O (elimination phase) NaOH, H 2 O (addition phase) CH 3 Br CH 3 2-Bromo-1,3- dimethylbenzene (No protons ortho to bromine; elimination is impossible.) H11001 3-Chloropyridine N Cl Y H11002 Anionic intermediate (less stable) N Cl Y H11002 H11001 4-Chloropyridine N Cl Y H11002 Anionic intermediate (more stable) N ClY H11002 H11001 2,3,4,5,6-Pentafluoroanisole FF F OCH 3 F F Fluoride ion F H11002 Cyclohexadienyl anion intermediate F F F F F F OCH 3 H11002 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 23.9 The “triple bond” of benzyne adds to the diene system of furan. 23.10 (a)(b) (c)(d) (e)(f ) (g) (h) (i)(j) 23.11 (a) Chlorine is a weakly deactivating, ortho, para-directing substituent. Chlorobenzene Cl AlCl 3 H11001H11001 o-Chloroacetophenone Cl CCH 3 O Acetyl chloride CH 3 CCl O p-Chloroacetophenone Cl C O CH 3 9-Fluorophenanthrene F 3 2 110 9 8 7 6 5 4 1,8-Dichloronaphthalene ClCl 1 2 3 45 6 7 8 2-Chloronaphthalene Cl 1 2 3 45 6 7 8 CH 2 Cl Br p-Bromobenzyl chloride ClCHCH 3 1-Chloro-1-phenylethane (Note: This compound is not an aryl halide.) Br O 2 N Cl 2-Bromo-1-chloro-4- nitrobenzene 4,4H11032-Diiodobiphenyl II 11H11032 22H1103233H11032 44H11032 55H1103266H11032 CH CH 2 F p-Fluorostyrene OCH 3 BrBr 2,6-Dibromoanisole CH 3 Cl m-Chlorotoluene F Br Mg, THF heat O O ARYL HALIDES 659 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) Bromobenzene reacts with magnesium to give a Grignard reagent. (c) Protonation of the Grignard reagent in part (b) converts it to benzene. (d) Aryl halides react with lithium in much the same way that alkyl halides do, to form organo- lithium reagents. (e) With a base as strong as sodium amide, nucleophilic aromatic substitution by the elimination–addition mechanism takes place. ( f ) The benzyne intermediate from p-bromotoluene gives a mixture of m- and p-methylaniline. (g) Nucleophilic aromatic substitution of bromide by ammonia occurs by the addition–elimination mechanism. (h) The bromine attached to the benzylic carbon is far more reactive than the one on the ring and is the one replaced by the nucleophile. NaCN p-Bromobenzyl bromide Br CH 2 Br p-Bromobenzyl cyanide Br CH 2 CN NH 3 1-Bromo-4- nitrobenzene Br NO 2 p-Nitroaniline NO 2 NH 2 NaNH 2 NH 3 NaNH 2 NH 3 p-Bromotoluene Br CH 3 4-Methylbenzyne CH 3 m-Methylaniline NH 2 CH 3 p-Methylaniline NH 2 CH 3 H11001 NaNH 2 NH 3 NaNH 2 NH 3 BenzyneBromobenzene Br Aniline NH 2 diethyl ether H11001 Iodobenzene C 6 H 5 I Phenyllithium C 6 H 5 Li Lithium 2Li H11001 Lithium iodide LiI H 2 O HCl Benzene C 6 H 6 Phenylmagnesium bromide C 6 H 5 MgBr diethyl ether H11001 Bromobenzene C 6 H 5 Br Phenylmagnesium bromide C 6 H 5 MgBrMg 660 ARYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (i) The aromatic ring of N, N-dimethylaniline is very reactive and is attacked by p-chlorobenzene- diazonium ion. ( j) Hexafluorobenzene undergoes substitution of one of its fluorines on reaction with nucleo- philes such as sodium hydrogen sulfide. 23.12 (a) Since the tert-butoxy group replaces fluoride at the position occupied by the leaving group, substitution likely occurs by the addition–elimination mechanism. (b) In nucleophilic aromatic substitution reactions that proceed by the addition–elimination mechanism, aryl fluorides react faster than aryl bromides. Because the aryl bromide is more reactive in this case, it must be reacting by a different mechanism, which is most likely elimination–addition. 23.13 (a) Two benzyne intermediates are equally likely to be formed. Reaction with amide ion can occur in two different directions with each benzyne, giving three possible products. They are formed in a 1:2:1 ratio. Asterisk (*) refers to 14 C. H11001 NaNH 2 NH 3 Cl * * * ::1 NH 2 * 2 NH 2 * NH 2 1Ratio: * Bromobenzene Br tert-Butyl phenyl ether OC(CH 3 ) 3 Benzyne KOC(CH 3 ) 3 DMSO KOC(CH 3 ) 3 DMSO tert-Butoxide ion (CH 3 ) 3 CO H11002 H11001 o-Fluorotoluene CH 3 F H11002F H11002 tert-Butyl o-methylphenyl ether CH 3 OC(CH 3 ) 3 CH 3 F OC(CH 3 ) 3 H11002 Hexafluorobenzene Sodium hydrogen sulfide NaSHH11001 F F F FF F 2,3,4,5,6-Pentafluoro- benzenethiol SH F F FF F N,N-Dimethylaniline (CH 3 ) 2 N H11001 4-(4H11032-Chlorophenylazo)-N,N-dimethylaniline (CH 3 ) 2 N ClN N p-Chlorobenzenediazonium ion ClN N H11001 ARYL HALIDES 661 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) Only one benzyne intermediate is possible, leading to two products in a 1:1 ratio. D refers to 2 H (deuterium). 23.14 (a) o-Chloronitrobenzene is more reactive than chlorobenzene, because the cyclohexadienyl anion intermediate is stabilized by the nitro group. Comparing the rate constants for the two aryl halides in this reaction reveals that o-chloro- nitrobenzene is more than 20 billion times more reactive at 50°C. (b) The cyclohexadienyl anion intermediate is more stable, and is formed faster, when the electron-withdrawing nitro group is ortho to chlorine. o-Chloronitrobenzene reacts faster than m-chloronitrobenzene. The measured difference is a factor of approximately 40,000 at 50°C. (c) 4-Chloro-3-nitroacetophenone is more reactive, because the ring bears two powerful electron- withdrawing groups in positions where they can stabilize the cyclohexadienyl anion intermediate. (d) Nitro groups activate aryl halides toward nucleophilic aromatic substitution best when they are ortho or para to the leaving group. is more reactive than 2-Fluoro-1,3- dinitrobenzene F O 2 N NO 2 1-Fluoro-3,5- dinitrobenzene O 2 NNO 2 F CH 3 OCl C N H11001 CH 3 O H11002 H11002H11002 CH 3 OCl N O H11001 H11002 C CH 3 CH 3 OCl C N CH 3 O H11002 O O O O O O H11001 H11002 CH 3 OCl N H11001 O H11002 O H11002 O CH 3 OCl N H11001 O H11002 D H11001 Cl D D NH 2 D NH 2 D NaNH 2 NH 3 D same as H20874H20874 1Ratio: 1: 662 ARYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) The aryl halide with nitro groups ortho and para to the bromide leaving group is more reactive than the aryl halide with only one nitro group. 23.15 (a) The nucleophile is the lithium salt of pyrrolidine, which reacts with bromobenzene by an elimination–addition mechanism. (b) The nucleophile in this case is piperidine. The substrate, 1-bromo-2,4-dinitrobenzene, is very reactive in nucleophilic aromatic substitution by the addition–elimination mechanism. (c) Of the two bromine atoms, one is ortho and the other meta to the nitro group. Nitro groups activate positions ortho and para to themselves toward nucleophilic aromatic substitution, and so it will be the bromine ortho to the nitro group that is displaced. 23.16 Because isomeric products are formed by reaction of 1- and 2-bromonaphthalene with piperidine at elevated temperatures, it is reasonable to conclude that these reactions do not involve a common 1,4-Dibromo-2- nitrobenzene Br Br NO 2 H11001 Piperidine N H N-(4-Bromo-2-nitrophenyl)- piperidine Br NO 2 N 1-Bromo-2,4- dinitrobenzene Br NO 2 NO 2 H11001 Piperidine N H N-(2,4-Dinitrophenyl)- piperidine NO 2 NO 2 N Bromobenzene Br LiBrH11001H11001 Lithium pyrrolidide LiN N-Phenylpyrrolidine (observed yield, 84%) N is more reactive than 1,4-Dibromo-2- nitrobenzene NO 2 Br Br 1-Bromo-2,4- dinitrobenzene Br NO 2 NO 2 ARYL HALIDES 663 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website intermediate and hence follow an addition–elimination pathway. Piperidine acts as a nucleophile and substitutes for bromine on the same carbon atom from which bromine is lost. When the strong base sodium piperidide is used, reaction occurs by the elimination–addition pathway via a “naphthalyne” intermediate. Only one mode of elimination is possible from 1-bromo- naphthalene. This intermediate can yield both A and B in the addition stage. Two modes of elimination are possible from 2-bromonaphthalene: Both naphthalyne intermediates are probably formed from 2-bromonaphthalene because there is no reason to expect elimination to occur only in one direction. Br elimination stage H11001 Compounds A and B addition stage Compound B only addition stage Compound A N Compound B N sodium piperidide piperidine H11001 Br H11001H11001H11001NaBr N H N Na H11001 H11002 H11001 Piperidine N H 2-Bromonaphthalene Br Compound B N 1-Bromonaphthalene Br H11001 Piperidine N H Compound A N 664 ARYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 23.17 Reaction of a nitro-substituted aryl halide with a good nucleophile leads to nucleophilic aromatic substitution. Methoxide will displace fluoride from the ring, preferentially at the positions ortho and para to the nitro group. 23.18 (a) This reaction is nucleophilic aromatic substitution by the addition–elimination mechanism. The nucleophile, , displaces chloride directly from the aromatic ring. The product in this case was isolated in 57% yield. (b) The nucleophile, hydrazine, will react with 1-chloro-2,4-dinitrobenzene by an addition– elimination mechanism as shown. The nitrogen atoms of hydrazine each has an unshared electron pair and hydrazine is fairly nucleophilic. The product, 2,4-dinitrophenylhydrazine, is formed in quantitative yield. (c) The problem requires you to track the starting material through two transformations. The first of these is nitration of m-dichlorobenzene, an electrophilic aromatic substitution reaction. Because the final product of the sequence has four nitrogen atoms (C 6 H 6 N 4 O 4 ), 2,4-dichloro- 1-nitrobenzene is an unlikely starting material for the second transformation. Stepwise Cl NO 2 Cl 2,4-Dichloro-1- nitrobenzene ClCl m-Dichlorobenzene HNO 3 H 2 SO 4 H11001 Cl NO 2 NO 2 1-Chloro-2,4- dinitrobenzene H 2 NNH 2 Hydrazine 2,4-Dinitrophenyl- hydrazine H 2 N NO 2 NO 2 NH H11002H H11001 H11002Cl H11002 Cl NO 2 N H11001 H11002 OO H11002 H 2 NNH 2 H11001 C 6 H 5 CH 2 S H11002 H11001 C 6 H 5 CH 2 SK Cl NO 2 CH 3 4-Chloro-3- nitrotoluene SCH 2 C 6 H 5 NO 2 CH 3 4-(Benzylthio)-3- nitrotoluene NaOCH 3 CH 3 OH H11001 NO 2 F F F F F 1,2,3,4,5-Pentafluoro- 6-nitrobenzene NO 2 OCH 3 F F F F 2,3,4,5-Tetrafluoro- 6-nitroanisole NO 2 F F OCH 3 F F 2,3,5,6-Tetrafluoro- 4-nitroanisole ARYL HALIDES 665 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website nucleophilic aromatic substitution of both chlorines is possible but leads to a compound with the wrong molecular formula (C 6 H 7 N 3 O 2 ). To obtain a final product with the correct molecular formula, the original nitration reaction must lead not to a mononitro but to a dinitro derivative. This is reasonable in view of the fact that this reaction is carried out at elevated temperature (120°C). This two-step sequence has been carried out with product yields of 70–71% in the first step and 88–95% in the second step. (d) This problem also involves two transformations, nitration and nucleophilic aromatic substitu- tion. Nitration will take place ortho to chlorine (meta to trifluoromethyl). (e) The primary alkyl halide is more reactive toward nucleophilic substitution than the aryl halide. A phosphonium salt forms by an S N 2 process. ( f ) N-Bromosuccinimide (NBS) is a reagent used to substitute benzylic and allylic hydrogens with bromine. The benzylic bromide undergoes S N 2 substitution with the nucleophile, methanethiolate. As in part (e), the alkyl halide is more reactive toward substitution than the aryl halide. NBS benzoyl peroxide, CCl 4 , heat NaSCH 3 OCH 3 H 3 C Br 2-Bromo-5- methoxytoluene OCH 3 CH 2 Br Br 2-Bromo-5-methoxy- benzyl bromide OCH 3 CH 3 SCH 2 Br 2-Bromo-5-methoxybenzyl methyl sulfide H11001ICH 2 Br p-Iodobenzyl bromide ICH 2 P(C 6 H 5 ) 3 Br H11002 (p-Iodobenzyl)triphenyl- phosphonium bromide (86%) (C 6 H 5 ) 3 P Triphenyl phosphine H11001 HNO 3 H 2 SO 4 NaOCH 3 CH 3 OH 1-Chloro-4- (trifluoromethyl)- benzene CF 3 Cl 1-Chloro-2-nitro-4- (trifluoromethyl)- benzene NO 2 CF 3 Cl 2-Nitro-4- (trifluoromethyl)- anisole NO 2 CF 3 OCH 3 Cl NO 2 O 2 N Cl NH 2 NO 2 O 2 N H 2 N 1,5-Diamino-2,4-dinitrobenzene (C 6 H 6 N 4 O 4 ) ClCl m-Dichlorobenzene NH 3 ethylene glycol, 140H11034C HNO 3 , H 2 SO 4 120H11034C Cl NO 2 Cl 2,4-Dichloro-1- nitrobenzene NH 2 NO 2 H 2 N 2,4-Diamino-1- nitrobenzene NH 3 666 ARYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 23.19 The reaction of p-bromotoluene with aqueous sodium hydroxide at elevated temperature proceeds by way of a benzyne intermediate. The same benzyne intermediate is formed when p-chlorotoluene is the reactant, and so the product ratio must be identical regardless of whether the leaving group is bromide or chloride. 23.20 Dinitration of p-chloro(trifluoromethyl)benzene will take place at the ring positions ortho to the chlorine. Compound A is 2-chloro-5-(trifluoromethyl)-1,3-dinitrobenzene. Trifluralin is formed by nucleophilic aromatic substitution of chlorine by dipropylamine. Trifluralin is N, N-dipropyl-4- (trifluoromethyl)-2,6-dinitroaniline. 23.21 p-Chlorobenzenethiolate reacts with p-nitrobenzyl chloride by an S N 2 process to give compound A. Reduction of the nitro group yields the aniline derivative, compound B. Chlorbenside is then formed by a Sandmeyer reaction in which the diazonium ion is replaced by chlorine. CH 2 SO 2 N Cl Compound A 1. Fe, HCl 2. NaOH CH 2 SH 2 N Cl Compound B CH 2 SCl Cl Chlorbenside 1. NaNO 2 , HCl 2. CuCl H11001CH 2 ClO 2 N p-Nitrobenzyl chloride CH 2 SO 2 N Cl p-Chlorophenyl p-nitrobenzyl sulfide (compound A) ClNaS Sodium p-chlorobenzenethiolate CF 3 Cl NO 2 O 2 N 2-Chloro-5-(trifluoromethyl)- 1,3-dinitrobenzene (compound A) CF 3 N(CH 2 CH 2 CH 3 ) 2 NO 2 O 2 N N,N-dipropyl-4-(trifluoromethyl)- 2,6-dinitroaniline (trifluralin) CF 3 Cl p-Chloro- (trifluoromethyl)- benzene HNO 3 H 2 SO 4 , heat (CH 3 CH 2 CH 2 ) 2 NH CH 3 Br H11001 CH 3 OH m-Methylphenol CH 3 OH p-Methylphenol NaOH, H 2 O 300H11034C CH 3 ARYL HALIDES 667 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 23.22 p-Chloro(trifluoromethyl)benzene undergoes nucleophilic substitution by the alkoxide anion to give compound A. Prozac (Fluoxetine hydrochloride) differs from compound A in having an GNHCH 3 group in place of GN(CH 3 ) 2 . 23.23 Benzyne is formed by loss of nitrogen and carbon dioxide. 23.24 o-Bromofluorobenzene yields benzyne on reaction with magnesium (see text Section 23.9). Triptycene is the Diels–Alder cycloaddition product from the reaction of benzyne with anthracene (compound A). Although anthracene is aromatic, it is able to undergo cycloaddition at the center ring with a dienophile because the adduct retains the stabilization energy of two benzene rings. 23.25 (a) Ethoxide ion adds to the aromatic ring to give a cyclohexadienyl anion. H11001 Na H11001 H11002 H11002 OCH 3 OCH 2 CH 3 NO 2 NO 2 N O O H11001 Meisenheimer complex NaOCH 2 CH 3 Sodium ethoxide OCH 3 NO 2 NO 2 N O O H11001 H11002 2,4,6-Trinitroanisole H11001 Mg, THF heat F Br o-Bromofluoro- benzene Anthracene (compound A) Triptycene H11001H11001 NN C O O H11002 H11001 Benzenediazonium-2- carboxylate Benzyne NN Nitrogen OOC Carbon dioxide OCHCH 2 CH 2 NHCH 3 F 3 C Prozac OCHCH 2 CH 2 N(CH 3 ) 2 F 3 C Compound A H11001 ClF 3 C OCHCH 2 CH 2 N(CH 3 ) 2 F 3 C 3-(p-(Trifluoromethyl)phenoxy)- N,N-dimethyl-3-phenyl-1-propanamine (Compound A) CHCH 2 CH 2 N(CH 3 ) 2 ONa H11001H11002 668 ARYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) The same Meisenheimer complex results when ethyl 2,4,6-trinitrophenyl ether reacts with sodium methoxide. 23.26 Methoxide ion may add to 2,4,6-trinitroanisole either at the ring carbon that bears the methoxyl group or at an unsubstituted ring carbon. The two Meisenheimer complexes are the sodium salts of the anions shown. It was observed that compound A was the more stable of the two. Compound B was present immediately after adding sodium methoxide to 2,4,6-trinitroanisole but underwent relatively rapid isomerization to compound A. 23.27 (a) The first reaction that occurs is an acid–base reaction between diethyl malonate and sodium amide. A second equivalent of sodium amide converts bromobenzene to benzyne. The anion of diethyl malonate adds to benzyne. This anion then abstracts a proton from ammonia to give the observed product. H11001 H11001CH(COOCH 2 CH 3 ) 3 H11002 Ammonia NH 2 H Diethyl 2-phenylmalonate CH(COOCH 2 CH 3 ) 2 H Amide anion NH 2 H11002 H11002 CH(COOCH 2 CH 3 ) 2 CH(COOCH 2 CH 3 ) 2 H11001 H11002 Benzyne Anion of diethyl malonate H11001H11001NaNH 2 Sodium amide NH 3 Ammonia H11001 NaBr Sodium bromide Bromobenzene Br Benzyne H11001H11001CH 2 (COOCH 2 CH 3 ) 2 Diethyl malonate NaNH 2 Sodium amide CH(COOCH 2 CH 3 ) 2 Na H11001 H11002 Diethyl sodiomalonate NH 3 Ammonia NaOCH 3 CH 3 OH H11001 NO 2 NO 2 O 2 N OCH 3 2,4,6-Trinitroanisole NO 2 N H11001 O H11002H11002 O O 2 N OCH 3 CH 3 O A NO 2 NO 2 H11001 N OCH 3 H OCH 3 O H11002 H11002 O B Na H11001 H11002 H11002 OCH 3 OCH 2 CH 3 NO 2 NO 2 N O O H11001 Meisenheimer complex H11001 NaOCH 3 Sodium methoxide OCH 2 CH 3 NO 2 NO 2 N O O H11001 H11002 Ethyl 2,4,6-trinitrophenyl ether ARYL HALIDES 669 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) The ester is deprotonated by the strong base sodium amide, after which the ester enolate under- goes an elimination reaction to form a benzyne intermediate. Cyclization to the final product occurs by intramolecular attack of the ester enolate on the reactive triple bond of the aryne. (c) In the presence of very strong bases, aryl halides undergo nucleophilic aromatic substitution by an elimination–addition mechanism. The structure of the product indicates that a nitrogen of the side chain acts as a nucleophile in the addition step. (d) On treatment with base, intramolecular nucleophilic aromatic substitution leads to the observed product. FF F F F OCH 2 CH 2 OH F F F F O OFF F F F OCH 2 CH 2 O H11002 O O F F F F F H11002 K 2 CO 3 heat NaNH 2 ether NaNH 2 , NH 3 NH 3 NCH 2 CH 2 NHCH 3 Cl CH 3 NCH 2 CH 2 NCH 3 Cl CH 3 H11002 CH 3 CH 3 N H11002 N CH 3 CH 3 N N N CH 3 CH 2 CH 2 N CH 3 H11002 NaNH 2 NH 3 NaNH 2 CH 2 CH 2 CH 2 CH 2 COCH 2 CH 3 Cl O Ethyl 5-(2-chlorophenyl)pentanoate Ester enolate CH 2 CH 2 CH 2 CHCOCH 2 CH 3 Cl O H11002 COOCH 2 CH 3 CH 2 CH 2 CH 2 CH COOCH 2 CH 3 H11002 Aryne intermediate COOCH 2 CH 3 Ethyl 1,2,3,4-tetrahydronaphthalene- 1-carboxylate H11002 670 ARYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 23.28 Polychlorinated biphenyls (PCBs) are derived from biphenyl as the base structure. It is numbered as shown. (a) There are three monochloro derivatives of biphenyl: (b) The two chlorine substituents may be in the same ring (six isomers): The two chlorine substituents may be in different rings (six isomers): There are therefore a total of 12 isomeric dichlorobiphenyls. 3,4H11032-Dichlorobiphenyl Cl Cl Cl Cl 4,4H11032-Dichlorobiphenyl Cl Cl 3,3H11032-Dichlorobiphenyl Cl Cl 2,4H11032-Dichlorobiphenyl Cl Cl 2,3H11032-Dichlorobiphenyl Cl Cl 2,2H11032-Dichlorobiphenyl Cl Cl 2,3-Dichlorobiphenyl Cl Cl 2,4-Dichlorobiphenyl Cl Cl 2,5-Dichlorobiphenyl ClCl 2,6-Dichlorobiphenyl Cl Cl 3,4-Dichlorobiphenyl ClCl 3,5-Dichlorobiphenyl Cl 2-Chlorobiphenyl (o-chlorobiphenyl) Cl 3-Chlorobiphenyl (m-chlorobiphenyl) Cl 4-Chlorobiphenyl (p-chlorobiphenyl) 1H11032 2H11032 23H11032 3 4H11032 4 5H11032 6H11032 65 1 ARYL HALIDES 671 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) The number of octachlorobiphenyls will be equal to the number of dichlorobiphenyls (12). In both cases we are dealing with a situation in which eight of the ten substituents of the biphenyl system are the same and considering how the remaining two may be arranged. In the dichloro- biphenyls described in part (b), eight substituents are hydrogen and two are chlorine; in the octachlorobiphenyls, eight substituents are chlorine and two are hydrogen. (d) The number of nonachloro isomers (nine chlorines, one hydrogen) must equal the number of monochloro isomers (one chlorine, nine hydrogens). There are therefore three nonachloro de- rivatives of biphenyl. 23.29 The principal isotopes of chlorine are 35 Cl and 37 Cl. A cluster of five peaks indicates that dichlorodiphenyldichloroethane (DDE) contains four chlorines. m/z for C 14 H 8 Cl 4 316 35 Cl 35 Cl 35 Cl 35 Cl 318 35 Cl 35 Cl 35 Cl 37 Cl 320 35 Cl 35 Cl 37 Cl 37 Cl 322 35 Cl 37 Cl 37 Cl 37 Cl 324 37 Cl 37 Cl 37 Cl 37 Cl The peak at mH11408z 316 therefore corresponds to a compound C 14 H 8 Cl 4 in which all four chlorines are 35 Cl. The respective molecular formulas indicate that DDE is the dehydrochlorination product of dichlorodiphenyltrichloroethane (DDT). The structure of DDT was given in the statement of the problem. This permits the structure of DDE to be assigned. SELF-TEST PART A A-1. Give the product(s) obtained from each of the following reactions: (a) ? (two products) Cl CF 3 KNH 2 NH 3 DDE (only reasonable dehydrochlorination product of DDT) ClCl C C ClCl H11002HCl C 14 H 9 Cl 5 DDT C 14 H 8 Cl 4 DDE 672 ARYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) (c) A-2. Draw the structure of the intermediate formed in each reaction of problem A-1. A-3. Suggest synthetic schemes by which chlorobenzene may be converted into (a) 2,4-Dinitroanisole (1-methoxy-2,4-dinitrobenzene) (b) p-Isopropylaniline A-4. Write a mechanism using resonance structures to show how a nitro group directs ortho, para in nucleophilic aromatic substitution. A-5. What is the cycloaddition product of the following reaction? What is the structure of the short-lived intermediate formed in this reaction? PART B B-1. The reaction most likely occurs by which of the following mechanisms? (a) Addition–elimination (b) Elimination–addition (c) Both (a) and (b) (d) Neither of these B-2. Rank the following in order of decreasing rate of reaction with ethoxide ion (CH 3 CH 2 O H11002 ) in a nucleophilic aromatic substitution reaction: (a)3H11022 4 H11022 1 H11022 2(c)3H11022 4 H11022 2 H11022 1 (b)2H11022 1 H11022 4 H11022 3(d)4H11022 3 H11022 2 H11022 1 NO 2 NO 2 Br 4 NO 2 NO 2 Br 3 NO 2 Br 2 NO 2 Br 1 NO 2 O 2 N Cl NO 2 O 2 N OCH 3 CH 3 O H11002 H11001 C 11 H 10 Br F Mg, THF heat ? (two products) I C(CH 3 ) 3 NaNH 2 NH 3 ? (monosubstitution) NO 2 Cl Cl CH 3 O H11002 CH 3 OH ARYL HALIDES 673 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-3. The reaction most likely involves which of the following aromatic substitution mechanisms? (a) Addition–elimination (b) Electrophilic substitution (c) Elimination–addition (d) Both (a) and (c) B-4. Identify the principal organic product of the following reaction: (a)(d) (b)(e) (c) B-5. Which of the following compounds gives a single benzyne intermediate on reaction with sodium amide? (a) 1 only (b) 1 and 3 (c) 3 only (d) 1 and 2 Cl CH 2 CH 3 Cl CH 2 CH 3 Cl CH 2 CH 3 312 O 2 N F Br SCH 3 O 2 N F SCH 3 O 2 N F Br CH 3 S O 2 N Br SCH 3 CH 3 S F Br NaSCH 3 O 2 N F Br H11001 BrO 2 N CH 3 H11001 NH 2 O 2 N CH 3 NH 2 O 2 N CH 3 KNH 2 NH 3 674 ARYL HALIDES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-6. Which one of the following compounds can be efficiently prepared by a procedure in which nucleophilic aromatic substitution is the last step? (a)(b)(c)(d ) B-7. Which one of the following undergoes nucleophilic aromatic substitution at the fastest rate? (a)(b)(c)(d )(e) B-8. What combination of reactants will give the species shown as a reactive intermediate? (a) 1-Bromo-4-nitrobenzene and NaOH (b) 4-Nitrophenol and HBr (c) 4-Nitrophenol, Br 2 , and FeBr 3 (d) Bromobenzene and HONO 2 (e) Nitrobenzene, Br 2 , and water HO Br N H11001 O H11002 O H11002 N(CH 3 ) 2 OCH 3 Cl Cl Cl ClNO 2 Cl CH 3 Cl NO 2 OCH 3 CH 3 SO 3 H C(CH 3 ) 3 OCH 2 CH CH 2 Br CH 3 H 3 C NH 2 ARYL HALIDES 675 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website