FOUR REACTIONS OF ALCOHOLS 1 4 REACTIONS OF ALCOHOLS 1. 2. 3. 4. SUBSTITUTION WITH HALIDES TO FORM ALKYL HALIDES ELIMINATION OF WATER (DEHYDRATION TO ALKENES) ACID/BASE REACTIONS WITH STRONG BASES/ACIDS OXIDATION TO ALDEHYDES, KETONES, OR CARBOXYLIC ACIDS. Alcohols have several polar atoms, several sites of reactivity. Alcohols can undergo substitution, elimination and acid/base reactions. An alcohol, like HOH, can act as a Nu:- or base as well as an E+ or acid. + d H3C E + d - .. O .. - Nu: & base + d H E+ & acid 2 1. SUBSTITUTION WITH HALIDES 1. SUBSTITUTION WITH HALIDES TO FORM ALKYL HALIDES Hydrohalic acids (HCl, HBr, HI) convert alcohols to alkyl halides SN1 or SN2 R OH + H alcohol X R X + H OH alkyl halide Even in concentrated form, these acids contain a lot of H2O H2O is a protic solvent. It stabilizes C+’s and weakens (solvates) Nu:’s alcohol (substrate) HI HBr HCl Me SN2 1 unhindered SN2 2 SN1 3 SN1 At room temperature, the SN2 reaction is slow. 3° alcohols react rapidly 2° alcohols react moderately 1° and Me°alcohols are unreactive 3 1. SUBSTITUTION WITH HALIDES Mechanism: H C C C C .. ..OH .. Cl .. : C conc. (37%) + H O .. H C C fast SN1 rxn. C H2O C C C .. Cl .. : conc. (37%) isopropyl moderate alcohol SN1 rxn. (2º alcohol) C C C H2O + H O .. H C C+ 2º C+ C .. Cl : .. + H2O C t-butyl chloride C C C C 3º C+ H .. ..OH C+ C t-butyl alcohol (3º alcohol) C .. : Cl : .. .. .. C : Cl : C C .. Cl : .. + H2O isopropyl chloride The hydroxyl (-OH) group is a poor leaving group unless protonated. 4 1. SUBSTITUTION WITH HALIDES H C C .. ..OH .. Cl .. : H2O 1º C+ conc. (37%) C C needs heat ethyl alcohol slow SN2 rxn. + H O .. H C C+ not formed too unstable (1º alcohol) C .. .. : Cl : - C .. Cl : .. + H2O ethyl chloride In these mechanisms, even though H-Cl is shown as the acidic species, it is understood that H2O is present (37% HCl contains 63% H2O). H-Cl is largely present as hydronium ion (H3O+ Cl-) and H3O+ protonates the alcohol in the first step. Name the mechanism & show the product of the following reaction conc. aq. HBr CH2 CH CH2 allyl alcohol 2-propen-1-ol CH2 OH SN1 CH CH2 Br + H2O allyl bromide 3-bromo-1-propene 5 1. SUBSTITUTION WITH HALIDES 1° and 2°alcohol reactivity can be improved by reacting with a polar aprotic reagent/solvent, such as phosphorus tribromide (PBr3) or thionyl chloride (SOCl2). These are fuming, corrosive liquids (nasty) but they do the job well. 1° and 2° alcohols react quickly via SN2 without heating. (1º alcohol) R OH + E+ Cl O Cl S Fast SN2 H - O R + Cl + O S Cl R Cl + SO2 + HCl E (1º alcohol) R OH + E+ Br P Br Br - Br R + H + O Fast SN2 PBr2 R Br + HOPBr2 E Note that SOCl2 produces alkyl chlorides & PBr3 produces alkyl bromides. 6 1. SUBSTITUTION WITH HALIDES The difference in the rate of reaction of alcohols with conc. HCl is used as a qualitative test (the Lucas Test) to determine the degree of substitution of an alcohol. Lucas reagent is conc. HCl, saturated with ZnCl2 salt. The Zn+2 ion coordinates (bonds) with the alcohol oxygen even better than H+ and speeds up the rate at which the C+ can form. About ½ mL of alcohol and 3 mL Lucas reagent are mixed in a test tube. 3°, allyl and benzyl alcohols react instantly (via SN1) to produce an insoluble alkyl halide, which appears initially as a hazy emulsion that separates into two liquid layers on standing. 2° alcohols also react quickly (SN1) in Lucas reagent (R-Cl separates). 1° alcohols do not react (no R-Cl layer forms). 2° and 3° alcohols can be differentiated by repeating the test in conc. HCl (no ZnCl2 present). 3° alcohols react quickly but the solution remains clear in 2° alcohols. 7 2. DEHYDRATION OF ALCOHOLS TO ALKENES 2. DEHYDRATION TO ALKENES WITH H2SO4 OR H3PO4 AND HEAT H OH C C alcohol H+HSO4dehydration E1 C C + H2O alkene H2SO4 and H3PO4 acids are dehydrating; they react strongly with water and will remove H2O from other chemicals. H2SO4 is sometimes used as a desiccant. When heated with alcohols, elimination (E1) occurs. The more stable, more substituted alkene (Zaitsev product) is the major product. With only gently heating (50°C) 20% H2SO4 dehydrates 3° alcohols. 2°alcohols require 60% H2SO4 @ 100°C. 1°alcohols need 95% H2SO4 @ 170°C (!), however, much of the product is lost. The 1° alcohol is badly charred. 8 2. DEHYDRATION OF ALCOHOLS TO ALKENES Mechanism: Fast E1 rxn. 20% + H HSO4- CH3 .. O .. H 1-methylcyclohexanol (3º alcohol) .. ..OH C isopropyl alcohol (2º alcohol) THF 50ºC C CH3 CH3 + H H H2O 3º C+ C 60% 100ºC C C + H O H moderate .. OH .. ethyl alcohol (1º alcohol) C H E1 C + H2O H HSO4- 1-methylcyclohexene HSO4- C C C H2O + + H2SO4 2º C+ propene E1 95% 170ºC slow E1 rxn. C+ H2O E1 rxn. H+ HSO4C E1 + H O .. H H+ HSO4- C C CH3 C H C Unstable 1 C+ requires high temp. + H O .. H C H H2O C+ 1º C+ C C + H2O + H2SO4 HSO4ethylene 9 2. DEHYDRATION OF ALCOHOLS TO ALKENES Note that even 1° alcohols dehydrate via the E1 mechanism. The high temperature (170°C) allows the formation of the unstable 1°C+. With weak nucleophiles, no reaction will occur unless a C+ forms. HSO4- (or H2PO4-) are very weak, non basic Nu:-’s. They have no propensity to remove H+ from a b-carbon via the E2 mechanism. Recall that E1 and SN1 occur in competition. High temperature favors elimination over substitution. Draw and name the major products of the following reactions. CH3 OH H2SO4 CH 3 H 1-methylcyclopentene Zaitsev product (major product) 2-methylcyclopentanol 10 2. DEHYDRATION OF ALCOHOLS TO ALKENES Recall that in unimolecular reactions, C+ rearrangements can occur. With that in mind, draw the major (Zaitsev) product from the dehydration of neohexyl alcohol. CH3 H3C C CH2 CH2 OH H3C CH3 CH3 CH3 H2SO4 C CH2 + CH2 H3C + C CH CH3 CH CH3 CH3 CH3 neohexyl alcohol 2,3-dimethyl-2-butene H3C H2O + Zaitsev product (major product) H3C CH3 CH3 C H3C C CH3 C + CH3 11 2. DEHYDRATION OF ALCOHOLS TO ALKENES 1° and 2° alcohols are charred by the severe conditions of acid dehydration and the E1 mechanism. There is a way to safely dehydrate these alcohols with good yields even at room temperature. The reagents needed however are unpleasant and corrosive. Phosphorus oxychloride (POCl3) in pyridine solvent (C5H5N:) will dehydrate 1°, 2° and 3° alcohols to alkenes via an E2 mechanism. pKb = 5.3 .. OH .. Cl + O P E+ Cl H :N H .. P O E2 H O + Cl pyridine is a moderately strong base & very bulky Cl Cl good leaving group cyclopentene Recall that strong, bulky bases like t-butoxide cause E2 even w. 1°substrates. 12 2. DEHYDRATION OF ALCOHOLS TO ALKENES Pyridine, like t-butoxide causes E2 reactions even w. 1°substrates. Pyridine is very bulky because of its large p-system of overlapping p-orbitals. H2O :N - - HSO4- alkyl halide (substrate) good Nu strong base e.g., bromide e.g., ethoxide strong bulky base e.g., t-butoxide Br C2H5O (CH3)3CO Me SN2 SN2 SN2 no reaction 1° SN2 SN2 E2 (SN2) no reaction 2 SN2 E2 E2 SN1, E1 3 SN1 E2 E2 SN1, E1 - - good Nu - good Nu nonbasic - very poor Nu nonbasic e.g., acetic acid CH3COOH HSO4- and H2O are nonbasic, weak nucleophiles (like CH3COOH). They will not react unless a C+ forms. This occurs rapidly with 3° alcohols, moderately with 2° alcohols and only at high temp. with 1° alcohols. 13 PROBLEMS Write equations showing how the following transformations can be carried out. More than one step may be necessary. Mechanism are not required but show all reagents used. OH CH2 CH2 OH CH H2O E2 POCl3, CH CH2 CH3 Markovnikov H2SO4 product H2O N Br 1° OH CH2 CH3 Markovnikov product KOH SN2 H2SO4 H2O OH POCl3, CH2 1° CH2 N E2 14 3. ACID/BASE REACTIONS OF ALCOHOLS 3. ACID & BASE REACTIONS WITH STRONG BASES & ACIDS Very strong bases abstract (remove) the weakly acidic H+ (pKa = 16). C .. O .. + H alcohol (as acid) B: .. O ..: C strong base + BH+ alkoxide pKb = -2 pKa = 16 Very strong acids protonate the weakly basic –OH group (pKb = 16). C .. O .. H H alcohol (as base) + H A C + O .. H + A - strong acid protonated alcohol pKb = 16 pKa = -2 The acid/base behavior of alcohols is much the same as that of H2O. 15 PROBLEMS Draw products, show mechanisms and calculate extent of reaction (%). CH3 CH2 .. O .. Na+HCO3H pKa = 16 ethanol CH3 CH2 .. O .. H pKb = 7.6 sodium ethoxide CH3 pKeq = pKa + pKb -14 CH3 CH3 CH2 + CH2 Na :NH2- CH3 pKb = -21 + H2CO3 .. O: .. Na+ + H2O pKeq = 16 -1.74 -14 = +.3 Extent of reaction < 50% .. H Na+ pKeq = 16 + 7.6 -14 = 9.4 Extent of reaction = 0% Na+ OHpKb = -1.74 .. O .. CH2 .. O: .. CH2 .. O: .. Na+ + pKeq = 16 -21 -14 = -19 Extent of reaction = 100% :NH3 16 3. REACTION OF ALCOHOLS WITH A BASE Active metals such as Na, Li, K, Ca, etc. are very strong bases. They deprotonate alcohols liberating H2 gas Draw products and show mechanisms for the following. CH3CH2O + H ethanol Na pKb = ca. -30 . H3C CH H3C O H + CH3 / 2 H2 Extent of reaction = 100% H3C CH H3C O- K+ + 1 /2 H2 potassium isopropoxide CH3 H + CH3 CH3CHCH2O 1 + sodium ethoxide pKeq = 16 -30 -14 = -28 K isopropyl alcohol CH3CHCH2O CH3CH2O- Na+ H . . Ca (CH3CHCH2O- )2 Ca+ + isobutyl alcohol H2 calcium diisobutoxide 17 3. REACTION OF ALCOHOLS WITH A BASE Show the reaction of neopentyl alcohol with aluminum metal. CH3 3 CH3CCH2OH + .. Al . CH3 neopentyl alcohol CH3 (CH3CCH2O- )3 Al+3 + 3 / 2 H2 CH3 aluminum trineopentoxide The evolution of H2 gas from alcohols in the presence Na metal is a quick qualitative test for alcohols. The reaction is vigorous but not explosive as it is in H2O. Aldehydes (pKa = 17) and ketones (pKa = 19) release only trace amounts of H2 gas when mixed with Na metal. [The aldehydes and ketones tested (and the test tube) must be free of water, i.e., dry! or a false positive result will be seen.] 18 ACIDITY OF ALCOHOLS Alcohols typically have a pKa of ca. 16, e.g., CH3CH2OH and CH3OH. Alcohol (CH3)3COH pKa CH3CH2OH HOH CH3OH 16 15.74 15.5 CF3CH2OH (CF3)3COH 12 5.4 18 Tert-butyl alcohol is less acidic because its conjugate base (alkoxide) is bulky and not easily solvated by water. Electron withdrawing groups such as –F make alcohols more acidic. They stabilize the alkoxide (by induction). A weaker conjugate base necessitates a stronger acid. Note the increased acidity of 2,2,2-trifluoroethanol and especially nonafluoro – t-butyl alcohol. Unlike resonance effects, the inductive effect is only significant over a short range. 19 PROBLEMS Write equations showing how the following transformations can be carried out. More than one step may be necessary. Mechanism are not required but show all reagents used. OH O CH3 CH3 Na CH2CH3 or CH3CH2Cl Na+NH2- SN2 - + O Na CH3 CH2 CH2 OH Gilman Reagent HBr fast SN1 CH2 Br 1 2 LI + Cu 2 H3O+ 20 4. OXIDATION OF ALCOHOLS 4. OXIDATION OF ALCOHOLS ( TO CARBONYLS) H :O : oxidation :O : C C reduction alcohol carbonyl compound Alcohols are oxidized to carbonyl compounds: aldehydes, ketones and carboxylic acids. In turn, these carbonyl compounds can be reduced to alcohols. The most common oxidants are oxides of nonmetals that are in a high oxidation state, e.g.,HNO3, KMnO4, Na2Cr2O7 in H2SO4, and NaOCl in HAc. Calculate the oxidation state of: (1) + N + (-2)3 = 0 N = +5 N in HNO3 Mn + (-2)4 = -1 Mn = +7 Mn in MnO4(1)2 + Cr2 + (-2)7 = 0 Cr = 12/2 = +6 Cr in Na2Cr2O7 (-2) + Cl = -1 Cl = +1 21 Cl in OCl- 4. OXIDATION OF ALCOHOLS Oxidants are usually classified into one of 3 groups (strengths). 1. Mild oxidants include Collins reagent (CrO3 in pyridine), and PCC (pyridinium chlorochromate, (C5H5NCrO3Cl in dichloromethane solvent.) These oxidants are dissolved in cold anhydrous solvents. (no water– this weakens the oxidant.) 2. Moderate-to-strong oxidants are cold to warm, aqueous solutions of HNO3, acidic or basic KMnO4, NaOCl in aq. HAc, Jones reagent (CrO3 in aq. H2SO4 + acetone). 3. Severe oxidants are hot, aqueous solutions of HNO3, acidic or basic aq. KMnO4, or Na2Cr2O7 in H2SO4 KMnO4 and Cr+6 reagents are often used for qualitative tests. When reacting, these oxidizing agents are reduced and show distinct color changes. List the color changes that occur. neutral or OH+3 Cr+6 MnO2 brown Cr orange green or blue precipitate MnO4- purple + H Mn+2 colorless 22 4. OXIDATION OF ALCOHOLS 1 alcohols are easily oxidized. They have 2 -hydrogens. Mild oxidants remove 1 hydrogen yielding an aldehyde. Moderate-to-strong (or severe) oxidants remove both hydrogens yielding a carboxylic acid. mild oxidation (oxidants in non aq. solvents) OH R PCC (in CH2Cl2) Collins reagent (CrO3 in pyridine) CH2 O R C H aldehyde O 1º alcohol moderate or strong oxidation (oxidants in aq. solvents) Jones reagent (CrO3 in aq. H2SO4 & acetone) KMnO4, HNO3, etc. R C OH carboxylic acid 23 4. OXIDATION OF ALCOHOLS 2 alcohols have only 1 -hydrogen. Mild or moderate-to-strong alcohols remove the only -hydrogen yielding a ketone. Under severe oxidation conditions, the ketone product is further oxidized. In the study of alkynes, we learned that a ketone is always in equilibrium with a minute amount of an enol, its ‘tautomer’. In our study of alkenes we learned that hot KMnO4, hot H2CrO4, or hot conc. HNO3 oxidize enols, cleaving the C-to-C p-bond and yielding carboxylic acids or ketones (which will again be cleaved until only carboxylic acids remain). O mild to moderate oxidation OH R CH R 2º alcohol (in non aq. solvents or any cold aq. solvent) PCC (in CH2Cl2) Collins reagent (CrO3 in pyridine) Jones reagent (CrO3 in aq. H2SO4 & acetone) cold acidic or basic KMnO4, HNO3, etc. severe oxidation (hot aq. KMnO4 or HNO3) R R O H R ketone R ketone O C C R C enol R O 2 R C OH carboxylic acid 24 4. OXIDATION OF ALCOHOLS 3 alcohols are not oxidized by any of the above oxidants except hot KMnO4 or hot conc. HNO3 which dehydrate them to alkenes and subsequently cleave the alkenes to carboxylic acids. Recall that 3 alcohols are easiest to dehydrate to alkenes. no rxn. any anhydrous or cold aq. oxidant OH R C R R 3º alcohol O severe oxidation (hot aq. KMnO4 or HNO3) R dehydration CH R alkene 2 R C OH carboxylic acid The mechanism of oxidations are either not fully understood or are very complex. The student will need to memorize the oxidation products of alcohols with the various classes of oxidants. 25 PROBLEMS Draw the products of the following oxidations. CH2OH Jones reagent 1-hexanol OH cold Na2CrO4 2-hexanol in H2SO4 CH2OH 1-hexanol PCC or Collins reagent hexanoic acid COH O O 2-hexanone hexanal CH O 26 PROBLEMS Draw the products of the following oxidations. H3C cold KMnO4 aqueous H3C OH O 3-methylcyclohexanol H3C hot concentrated HNO3 O H3C C Cleavage occurs on either side of the -carbon H3C O OH C OH OH OH C + C OH O O 27 IODOFORM REACTION (OXIDATION) The Iodoform reaction: is a qualitative oxidation test used to identify the structure of a 1° or 2° alcohol. The alcohol is treated with a solution of iodine and a solution of sodium hydroxide. NaOH + I2 NaOI (sodium hypoiodite) NaOI is analogous to sodium hypochlorite (NaOCl, the so-called ‘liquid chlorine’ used as laundry bleach,e.g., Javex, and used as swimming pool disinfectant). NaOI is a mild oxidant. It oxidizes 1° alcohols to aldehydes and 2° alcohols to ketones; but will not oxidize 3° alcohols. If the 1° or 2° alcohol had a methyl group (-CH3) attached to the -carbon, the three hydrogens of the –CH3 group are substituted by iodine (-CH3 -CI3) and the product cleaves producing iodoform (CHI3), a pale yellow precipitate. CH3CH2OH 1° alcohol NaOI O 3 NaOI CI3-C-H CH3-C-H aldehyde O -CH3 -CI3 Na+ OHcleaves O + -O-C-H CHI3 + Na iodoform 28 IODOFORM REACTION (OXIDATION) Ethanol is the only 1° alcohol that gives a positive iodoform test because it is the only 1° that has a –CH3 group attached directly to the -carbon. By contrast, many 2° alcohols have –CH3 group attached directly to their -carbon and this can be identified by the iodoform test. Write equations for the iodoform reaction with 2-butanol. CH3CH2CHCH3 + OH CH3CH2CCH3 + OXIDATION CH3CH2CCH3 + NaI + H2O NaOI IODINATION 3 NaOI O O CH3CH2CCI3 + O O CLEAVAGE CH3CH2CCI3 + NaOH 3 NaOH CHI3 + + Na-O-C-CH2-CH3 O 29 PROBLEMS Using the Lucas test, oxidation with Jones reagent, and the iodoform test, show how a student could distinguish between 2-pentanol and 3-pentanol. 2-pentanol OH Lucas test ZnCl2, con. HCl Jones reagent CrO3 in H2SO4 Iodoform NaOI 3-pentanol 2-chloropentane Cl Cl 3-chloropentane 2-pentanone O OH O O 3-pentanone O- Na+ no iodoform precipitate + CHI3 30 PROBLEMS Write equations showing how the following transformations can be carried out. More than one step may be necessary. Mechanism are not required but show all reagents used. OH OH CH3 Cr+6, H+ 1 CH3MgBr O CH2 CH2 POCl3, OH CH2OH KMnO4 H2O N or CH + 2 H3O CH2 1 LiAlH4 + 2 H3O O C OH hot KMnO4 31 SOLUBILITY AND BOILING POINTS OF ALCOHOLS Alcohols are the first organics we have studied that are polar. Alkanes, alkenes, alkynes and arenes are all hydrocarbons with non polar bonds. Even alkyl halides are only weakly polar. These compounds all exhibit characteristically low boiling points are are insoluble in polar solvent like water. HC and RX n-pentane carbon tetrachloride n-butyl bromide H2O Solubility (g/100 mL, 25 C) 0.05 0.08 0.06 Describe the trend in solubility of alcohols re: MW and branching. Explain the unusually high solubility of a polyol such as 1,4-butanediol. Alcohols/phenol H2O Solubility (g/100 mL, 20 C) n-butyl alcohol sec-butyl alcohol isobutyl alcohol tert-butyl alcohol n-pentyl alcohol n-hexyl alcohol 1-heptanol phenol 1-octanol 1,4-butanediol 9 12 10 miscible 2.7 0.6 0.2 6.7 0.05 miscible Explain why cyclohexanol boils at 160 °C and cyclohexene at 83 °C. 32 PHENOLS Phenols have the hydroxyl (-OH) functional group of an alcohol directly bonded to an aromatic ring; a fully conjugated ring; a ring with a complete pattern of alternating single and double bonds. alcohols OH cyclohexyl alcohol CH3 phenols OH phenol OH CH3CHCH2OH isobuty alcohol b-naphthol Despite their similar appearance, the chemistry of phenols is different than that of alcohols. 33 PHENOLS Phenol is manufactured in two steps from benzene. Benzene undergoes Electrophilic Aromatic Substitution (EAS) with H2SO4 and SO3 (a mixture called ‘oleum’) producing benzenesulfonic acid. The sulfonic acid group is replaced by the –OH group by reacting with NaOH at high temperature (300°C). Write equations showing the preparation of phenol from benzene. SO3, H2SO4 40 ºC O S 1 NaOH 300°C OH + 2 H3O O OH phenol benzenesulfonic acid The hydroxyl group donates electron density to the aromatic ring by resonance, making phenols reactive toward EAS. 34 REACTIONS OF PHENOLS Phenols have the same hydroxyl (-OH) functional group as alcohols but, while alcohols and H2O are essentially neutral (pKa = 16), phenols are weak acids (pKa = 10). pKeq = 9.9 -1.74 -14 = -5.8 Extent -of reaction = 100% .. O .. .. O: .. H pKa = 9.9 phenol .. O .. Na+ HCO3- pKb = 7.6 + H2O sodium- phenoxide .. O: .. H + pKb = 4.1 Na+ OH- + Na+ Na+ + H2CO3 pKeq = 9.9 + 7.6 -14 = 3.5 Extent of reaction = 0% Phenols react 100% with aqueous NaOH but not with NaHCO3 (pKb = 7.6). This difference is often used differentiate phenols from other stronger organic acids, such as carboxylic acids (pKa = 5). Carboxylic acids react 100% with both NaOH and NaHCO3. 35 REACTIONS OF PHENOLS Phenols are 106 times more acidic than alcohols (pKa 10 vs. pKa 16) because the conjugate base, phenoxide ion (pKb = 4), is 106 times more stable than the alkoxide (pKb = -2). In the alkoxide, the negative charge resides on only one atom, the oxygen. In the phenoxide, the negative charge is ‘delocalized’ (spread out) over 4 atoms by resonance throughout the fully conjugated p system of the aromatic ring. .. O ..: .. .. O .. : .. O .. .. .. O .. Sodium phenoxide has 4 different resonance structures. 36 REACTIONS OF PHENOLS Phenols become stronger acids when electron withdrawing groups are attached to the phenyl ring, e.g., -Br, -NO2, -CF3, etc. Phenols become weaker acids when electron donating groups are attached to the phenyl ring, e.g., -CH3, -OCH3, -NH2, etc. Draw these structures p-aminophenol phenol p-nitrophenol 2,4,6-trinitrophenol pKa 10.5 9.9 7.2 0.60 .. HO .. .. HO .. .. NH2 NO2 O2N .. HO .. O2N NO2 37 REACTIONS OF PHENOLS Recall from our study of electrophilic aromatic substitution (EAS) reactions that electron donor groups make the ring a better Nu:(electron donor); thus accelerating EAS reactions of aromatics. These same ring activators (electron donors) will reduce the acidity of phenols by making conjugate base (phenoxide) less stable. Ring deactivators are electron withdrawing groups. They increase the acidity of phenols by making phenoxides more stable. o- and pdirecting NH2 OCH3 o- and pdirecting F CH3 m-directing Br O O CH C OH O NH C CH3 activators NO2 Reactivity Reactivity OH SO3H H Cl deactivators I O O COCH3 C CH3 C N + N R3 deactivators 38 PROBLEM Rank the following phenols in order of acidity, where 1 is most acidic. Then name their conjugate bases. m-cresol m-nitrophenol benzoic acid p-chlorophenol phenol OH HO HO OH CH3 C 4 5 OH 3 1 O NO2 Cl 2 .. .. : .. O CH3 .. O: .. C .. O .. : : O: .. :O .. Cl NO2 O m-cresoxide phenoxide benzoate p-chlorophenoxide m-nitrophenoxide 39 PROBLEM Write equations showing how the following transformation can be carried out. More than one step may be necessary. Mechanism are not required but show all reagents used. an ether phenyl n-propyl ether O CH2CH2CH3 CH3CH2CH2I H2SO4 SN2 SO3 300° C SO3H 1 NaOH + 2 H3O benzenesulfonic acid OH NaOH O- Na+ acid/base sodium phenoxide 40 Do the practice problems in your purchased notes. 41 - - - alkyl halide (substrate) good Nu strong base e.g., bromide e.g., ethoxide strong bulky base e.g., t-butoxide Br C2H5O (CH3)3CO Me SN2 SN2 SN2 no reaction 1° SN2 SN2 E2 (SN2) no reaction 2 SN2 E2 E2 SN1, E1 3 SN1 E2 E2 SN1, E1 - - good Nu - good Nu nonbasic - very poor Nu nonbasic e.g., acetic acid CH3COOH 42