Hornback_Ch10_348-403 12/16/04 12:50 PM Page 352 352 CHAPTER 10 I SYNTHETIC USES OF SUBSTITUTION AND ELIMINATION REACTIONS provide better control of the reaction and a higher yield is a common strategy in organic synthesis. PROBLEM 10.3 On the basis of the bond cleavage shown for this reaction in Figure 10.1, predict the ste- reochemistry of the product. Explain. O X OCCH3 W KOH CH3CH2 C± H H2O H3C PROBLEM 10.4 Show the products of these reactions: Cl – CH3CO2 NaOH a) DMSO H2O Br – CH CO KOH b) 3 2 DMF H2O CH3 10.3 Preparation of Ethers Ethers can be prepared by using an alcohol or its conjugate base, an alkoxide ion, as the nucleophile. A general equation for the reaction with alkoxide ion is . – – R–O. ϩ R'–L R–O–R' ϩ .L . When an alkoxide ion is used as the nucleophile, the reaction is called a Williamson ether synthesis. Because the basicity of an alkoxide ion is comparable to that of hy- droxide ion, much of the discussion about the use of hydroxide as a nucleophile also ap- plies here. Thus, alkoxide ions react by the SN2 mechanism and are subject to the usual SN2 limitations. They give good yields with primary alkyl halides and sulfonate esters but are usually not used with secondary and tertiary substrates because elimination re- actions predominate. The alkoxide ion nucleophile is often prepared from the alcohol by reaction with sodium metal, as shown in the following equation for the formation of ethoxide ion from ethanol: . –. + 2 CH3CH2O–H ϩ 2 Na 2 CH3CH2O Na ϩ H–H . Because phenols are stronger acids than alcohols, nucleophilic phenoxide ions can be prepared by reacting the phenol with bases such as hydroxide ion or carbonate ion. Hornback_Ch10_348-403 12/16/04 12:50 PM Page 353 10.3 I PREPARATION OF ETHERS 353 . – + . OH . O. K ϩ K2CO3 ϩ KHCO3 Several examples of the Williamson ether synthesis are given in the following equations: 1) Na, hexanol OH O (92%) 2) CH3CH2–I In equations like this, the reagents over the arrow are added in a sequence of separate « Important steps, not all at once. Thus, in step 1, sodium metal is added to excess hexanol, which Convention is both a reactant and the solvent for the reaction. Only after the reaction of the sodium and the alcohol is complete and the conjugate base of the alcohol has formed is the reagent shown in step 2 added. In the second step, the alkoxide ion acts as a nucleophile, replacing the leaving group of iodoethane to form the ether. – – OH 1) Na O CH2CO2 – (80%) 2) Cl–CH2CO2 – OTs CH –O OCH 3 3 (65%) OH OCH2CO2H Cl Cl 1) K2CO3, H2O 2) ClC H2CO2H Cl Cl 2,4-Dichlorophenoxyacetic acid (2,4-D, an important herbicide) OCH2CH2N(CH3)2 W CH 1) Na (CH3)2NCH2CH2OH Br W CH Diphenhydramine 2) Benadryl (an antihistamine) Hornback_Ch10_348-403 12/16/04 12:50 PM Page 354 354 CHAPTER 10 I SYNTHETIC USES OF SUBSTITUTION AND ELIMINATION REACTIONS PROBLEM 10.5 Show the products of these reactions: Br W – EtOH a) CH3CH2CH2 ϩ CH3CH2O OH OH Cl W 1) Na NaOH CH3CH2CH2CH2 b) c) 2) CH 3I EtOH CH3 BioLink PROBLEM 10.6 Diphenhydramine can also be synthesized by heating bromodiphenylmethane with the amino alcohol shown here. Offer a reason why the oxygen, rather than the nitrogen, of this compound acts as the nucleophile. What factor favors the N? What factor favors the O? Which factor is winning in this case? H3C– NCH2CH2OH – H3C An unsymmetrical ether can usually be prepared by two different Williamson ether syntheses. For example, the preparation of ethyl isopropyl ether could be accomplished by the reaction of ethoxide ion (nucleophile) with isopropyl bromide (electrophile) or by the reaction of isopropoxide ion (nucleophile) with ethyl bromide (electrophile), as shown in Figure 10.2. Which of these routes is better? Because alkoxide ions are strong Figure 10.2 W CH2CH3 . – Br O . O. TWO POSSIBLE SYNTHESES . – W a W b W OF ETHYL ISOPROPYL ETHER . CH3CH2O. ϩ CH3CHCH3 CH3CHCH3 CH3CH2–Br ϩ CH3CHCH3 . Ethoxide ion 2-Bromopropane Ethyl isopropyl ether Bromoethane Isopropoxide ion a This reaction has the bromine attached to a b In contrast, the bromine is attached to a secondary carbon. With a strong base like primary carbon in this reaction. Much less ethoxide ion, the major reaction is elimination elimination occurs, and the yield of the desired (E2) rather than substitution (SN2), resulting in ether is higher here than in the other reaction. a poor yield of the desired ether. This is a better method for the synthesis of ethyl isopropyl ether. Hornback_Ch10_348-403 12/16/04 12:50 PM Page 355 10.3 I PREPARATION OF ETHERS 355 bases, an unacceptable amount of elimination occurs if the leaving group is attached to a secondary carbon. Therefore, the route using the primary halide (ethyl bromide) will give a higher yield of the substitution product. PROBLEM 10.7 Explain which route would provide a better synthesis of these ethers: CH3 CH3 CH3 – W W W – a) CH3O ϩ CH3CCl CH3COCH3 CH3I ϩ CH3CO W W W CH3 CH3 CH3 CH3 – W CH2O CH2OCHCH2CH3 CH2Br – Br O W W b) ϩ CH3CHCH2CH3 ϩ CH3CHCH2CH3 PRACTICE PROBLEM 10.1 Show a method for synthesizing this ether from an alcohol and an alkyl halide: OCH2CH3 Solution To minimize competing elimination by the E2 mechanism, treat the conjugate base of the secondary alcohol with the primary alkyl halide: OH OCH2CH3 1) Na 2) CH3CH2Br PROBLEM 10.8 Suggest a synthesis of these ethers starting with an alcohol and an alkyl halide: OCH2CH2CH3 O a) CH3OCH2CH2CH2CH3 b) c) Ethers can also be prepared by using alcohols as the nucleophiles: R–O–H ϩ R'–L –£ R–O–R' ϩ HL If the leaving group is bonded to a secondary or tertiary carbon, the reaction usually fol- lows the SN1 mechanism and is the preferred method in order to avoid problems with Hornback_Ch10_348-403 12/16/04 12:50 PM Page 356 356 CHAPTER 10 I SYNTHETIC USES OF SUBSTITUTION AND ELIMINATION REACTIONS elimination. An alcohol must also be used as the nucleophile when the reaction is run under acidic conditions because alkoxide ions cannot exist in acid. Examples are pro- vided by the following equations. In the first example, in which ethanol is the solvent, the reaction is an ethanolysis. CH3 CH3 W EtOH W Ph±C±Cl ϩ CH3CH2OH Ph±C±OCH2CH3 ϩ HCl (87%) W W CH3 CH3 OH H2SO4 O 2 ϩ H2O CH3(CH2)4OH 1-Pentanol Dipentyl ether PROBLEM 10.9 Show the products of these reactions: Cl CH3 W CH3OH H2SO4 a) b) CH3COH W CH3CH2OH CH3 H2SO4 c) Ph2CHOH ϩ HOCH2CH2Cl PROBLEM 10.10 Show all the steps in the mechanism for the reaction of 1-pentanol with sulfuric acid to form dipentyl ether. Finally, it is worth noting that the formation of cyclic ethers by intramolecular nu- cleophilic substitutions is quite favorable if the resulting ring is three, five, or six mem- bered, as shown in the following reactions: O–H ϩ NaOH O ϩ NaCl H2O Cl ϩ H2O (73%) OH OH O H3PO4 ϩ H2O (97%) Hornback_Ch10_348-403 12/16/04 12:50 PM Page 357 10.4 I PREPARATION OF ESTERS 357 PROBLEM 10.11 Show the steps in the mechanism for the reaction of trans-2-chlorocyclohexanol with sodium hydroxide shown in the previous equation. Explain why cis-2-chlorocyclohexanol does not give a similar reaction. PROBLEM 10.12 Show the product, including stereochemistry, for this reaction: HO Br ± ± NaOH H C±C H H2O H3C CH3 PROBLEM 10.13 Because of the acidic conditions, this reaction proceeds by an SN1 mechanism. Which hydroxy group acts as the leaving group in the reaction? Show all the steps in the mech- anism for this reaction: OH OH O H3PO4 10.4 Preparation of Esters Esters can be prepared by employing carboxylate salts as nucleophiles, as shown in the following equation: O O X . – X . – R–C–O . ϩ R'–L R–C–O–R' ϩ . L . Because carboxylate salts are only weakly basic, elimination is not a problem when the leaving group is attached to a primary or secondary carbon. Several examples are pro- vided in the following equations: CH3 CH3 O I O CH3 . X . –. W acetone X W – H3C CO . ϩ CH3CH(CH2)5CH3 H3C CO–CH(CH2)5CH3 ϩ I . (100%) CH3 CH3 O . X . O . Cl OCCH3 X . –. CH3CO2H – ϩ CH3CO . ϩ Cl (80%) . O O X . X . –. CH3CO2H – CH3CO . ϩ ClCH2 NO2 CH3CO–CH2 NO2 ϩ Cl (80%) . Hornback_Ch10_348-403 12/16/04 12:50 PM Page 358 358 CHAPTER 10 I SYNTHETIC USES OF SUBSTITUTION AND ELIMINATION REACTIONS PROBLEM 10.14 Show the products of these reactions: Br – DMSO a) ϩ CH3CO2 H3C Cl – acetone b) ϩ CH3CH2CO2 – CO2 DMF c) CH3CH2CH2CH2Br ϩ 10.5 Preparation of Alkyl Halides The preparation of alkyl halides by substitution reactions usually starts from alcohols be- cause alcohols are widely available. Hydroxide ion is a poor leaving group, so the OH must first be converted into a better leaving group, either by protonation in acid or by conversion to a sulfonate or similar ester (see Section 8.9), as illustrated in the following equations: H . .– . – W . X. H A . R–O–H R–O–H R–X ϩ H2O . + . .– . X. R'SO2Cl . – R–O–H R–O–SO2R' R–X ϩ R'SO3 . Protonation of the alcohol can be accomplished by using the halogen acids, HCl, HBr, and HI, which also provide the nucleophile for the reaction.