Alcohols, Ethers and Phenols

ALCOHOLS, ETHERS AND PHENOLS E M RAOALCOHOLS Physical properties: Solubility: Alcohols are quite soluble in water as they can form hydrogen bond with water. But alcohols having equal to or more than 6 carbons per OH group are insoluble in water as they have predominating hydrophobic (hydrocarbon) chain. Boiling point: Alcohols have higher boiling points than hydrocarbons of similar molecular weight due to hydrogen bonding. For a given molecular formula, more the branching lesser is the boiling point. 0 0 3 2 2 2 3 3 2 0 0 3 2 3 3 3 CH CH CH CH OH 118C CH CH(CH )CH OH 108C CH CH CH(OH)CH 100 C (CH ) COH 83C More the number of OH groups more is the boiling point. 0 0 3 2 2 2 CH CH OH 97 C HOCH CH OH 198 C Methods of preparation: 1. Hydration of alkenes: Water can be added to alkenes in three different ways. Some alkenes give three different isomers in the three methods. CH 3 -CH-CH=CH 2 CH 3 dil. H 2 SO 4 CH 3 CH 3 -C-CH 2 -CH 3 OH Acid catalysed hydration: CH 3 -CH-CH=CH 2 CH 3 1.Hg(OAc) 2 /H 2 O 2. NaBH 4 CH 3 CH 3 -CH-CH-CH 3 OH Oxymercuration-demercuration: CH 3 -CH-CH=CH 2 CH 3 1. B 2 H 6 ;THF 2. H 2 O 2 ;OH - CH 3 -CH-CH 2 CH 2 OH CH 3 Hydroboration-oxidation Acid catalyzed hydration: Mechanism: CH 3 -CH=CH 2 H + slow CH 3 -CH-CH 3 + H 2 O CH 3 -CH-CH 3 OH 2 + CH 3 -CH-CH 3 OH -H + Acids that have weakly nucleophilic conjugate bases are used in this reaction e.g. H2SO4, H3PO4 etc. Being weakly nucleophilic, HSO4 - and H2PO4 - offer little competition to H2O. In the case of H2SO4, small amount of ROSO3H may be formed. As soon as formed, it undergoes hydrolysis to form ROH because HSO4 - is a very good leaving group. Reactivity of alkenes towards acid catalyzed hydration: As formation carbocation is the rate limiting step, the alkene which forms the more stable carbocation will be more reactive. ALKENE Relative rate of reaction with H 3 O + CH2=CH2 1 CH3-CH=CH2 6 1.6 10 × (CH3)2C=CH2 11 2.5 10 × ALCOHOLS, ETHERS AND PHENOLS E M RAO Examples: H + OH H + OH OH 1. 2. and its enantiomer and its enantiomer + Oxymercuration-demercuration: Mechanism: (Q) CH 3 -CH-CH=CH 2 CH 3 Hg(OAc) 2 Hg(OAc) H 2 O CH 3 CH 2 -CH-CH-CH 2 -Hg(OAc) OH 2 + CH 3 CH 2 -CH-CH-CH 2 - Hg(OAc) OH CH 2 -CH-CH-CH 2 - Hg(OAc) OH NaBH 4 CH 3 CH 2 -CH-CH-CH OH 1. 2. H 2 O as a nucleophile attacks the carbon that can form more stable carbocation (as in the case of cyclic halonium ion). (cyclic mercurinium ion) C H CH 3 H 3 C + Hg(OAc) C H CH 3 H 3 C + CH 3 Cyclic mercurinium ion is similar to cyclic halonium ion. Like in the case of halogenations, there is no fully fledged carbocation in this reaction too. The net addition of H and OH is syn and anti both (although oxymercuration is anti addition, demercuration step is a mix of both syn and anti). If H2O is replaced with any other polar solvent, the product is obtained accordingly. Examples: 1.Hg(OAc) 2 /H 2 O 2. NaBH 4 OH 1.Hg(OAc) 2 /MeOH 2. NaBH 4 OCH 3 OH 2. NaBH 4 O 1.Hg(OAc) 2 1. 2. 3. Hydroboration-oxidation: Diborane in THF exists as the following complex. O BF 3 + THF can be replaced by diglyme, CH3OCH2CH2OCH2CH2OCH3. Mechanism: ALCOHOLS, ETHERS AND PHENOLS E M RAO CH 2 CH 3 CH H BH 2 CH 3 CH 2 -CH 2 BH 2 Both H and BH 2 are added from the same side i.e.syn addition 1. Same reaction repeats until two more alkenes molecules are added. Finally we get tri- alkyl boride B CH 2 CH 2 CH 3 H 3 CH 2 CH 2 C CH 2 CH 2 CH 3 Tri-alkyl boride Boron attacks less hindered carbon because of the crowding in trialkyl boride. Trialkyl boride undergoes oxidation in the second step. 2. H 2 O 2 +OH - H-O-O - +H 2 O 3. (CH 3 CH 2 CH 2 ) 3 B +H-O-O - (CH 3 CH 2 CH 2 ) 3 B O-OH B CH 2 CH 2 CH 3 H 3 CH 2 CH 2 C CH 2 CH 2 CH 3 O OH 4. (CH 3 CH 2 CH 2 ) 2 B(OCH 2 CH 2 CH 3 ) Step 4 involves migration of alkyl group from boron to oxygen. As is the case with all migrations, here also it is simultaneous. Steps 3 and 4 repeat two more times until trialkyl borate is formed. Trialkyl borate undergoes hydrolysis to give 3 moles of alcohol and H 3 BO 3 . 5. (CH 3 CH 2 CH 2 O) 3 B +OH - 3 CH 3 CH 2 CH 2 OH +H 3 BO 3 The net addition of H and OH is syn. Examples: 1.B 2 H 6 ;THF 2.H 2 O 2 , OH - OH 1. 1.B 2 H 6 ;THF 2.H 2 O 2 , OH - OH 2. + and its enantiomer EXERCISE I: 1. OH 1.Hg(OAc) 2 2. NaBH 4 2. 1.B 2 H 6 ;THF 2.H 2 O 2 , OH - 3. H 3 O + 4. 1.B 2 D 6 ;THF 2.H 2 O 2 , OH - 5. Et 1.Hg(OAc) 2 / PhOH 2. NaBH 4 6. O CH 3 OH H + ALCOHOLS, ETHERS AND PHENOLS E M RAO 2. From the nucleophilic substitution reactions of alkyl halides: CH 3 CH 2 Br NaOH aq. acetone CH 3 CH 2 OH (CH 3 ) 3 CBr H 2 O Acetone (CH 3 ) 3 COH S N 2 S N 1 3. From epoxides: Epoxides are three membered cyclic ethers. They are highly reactive due to ring strain. Nucleophiles attack epoxides and thereby convert them into alcohols. O 1.LiAlH 4 2.H + CH 3 CH 2 OH O 1.LiAlH 4 2.H + CH 3 CHCH 3 OH O 2.H + CH 3 CH 2 CH 2 OH O CH 3 CHCH 2 CH 3 OH 1.CH 3 MgBr/dry ether 2.H + 1.CH 3 MgBr/dry ether Mechanism: O CH 3 CH 2 OH H - CH 3 CH 2 O - H + When epoxides undergo attack directly (without first being protonated) by a nucleophile, the attack occurs at less hindered carbon. But a protonated epoxide undergoes attack by a nucleophile at more substituted carbon. We will see the details about this later in epoxides. Epoxides react with LiAlH4, NaBH4, RLi, RMgX and even with R2CuLi. O 1.(CH 3 ) 2 CuLi 2.H 2 O OH 2. By reduction of carbonyl compounds (aldehydes, ketones, carboxylic acids and their derivatives): Alcohols can be prepared by nucleophilic addition reactions of aldehydes and ketones. They can also be prepared by the nucleophilic acyl substitution reactions of carboxylic acids and their derivatives. Let us see the mechanism of both the reactions first and then have a look at different reagents. O (H)R R Nu - O - Nu R R(H) OH Nu R R(H) H + rds Nucleophilic addition reactions of aldehydes/ketones: When the nucleophile is H - (LiAlH4 or NaBH4) or R - (RLi/RMgX), the product will be an alcohol. α,β- unsaturated aldehydes and ketones behave differently toward nucleophile because the nucleophile may attack the C-C double bond in addition to carbonyl carbon. The reactions of α,β- unsaturated aldehydes and ketones will be discussed in aldehydes and ketones chapter. O L R Nu - O - Nu R L rds Nucleophilic acyl substitution reactions of acid chlorides, anhydrides and esters: O Nu R L - Cl / OR / OCOR ALCOHOLS, ETHERS AND PHENOLS E M RAO The above mechanism excludes carboxylic acids and amides because the mechanism in their case is slightly different due to the presence of acidic hydrogen. The mechanism is discussed later. Different reagents used for reduction of carbonyl compounds 1. LiAlH4 in dry ether followed by acidification . LiAlH4 reduces aldehydes, ketones, carboxylic acids, amides, anhydrides, esters and acid halides. Alcohols are the products from all the above reductions except from amides. Amides gove amines with LiAlH4. LiAlH4, being a strong hydride donor, can only be used in a non-polar solvents like ethers. 1.LiAlH 4 2.H + O H R R-CH 2 -OH 1.LiAlH 4 2.H + O R R R-CHR-OH 1.LiAlH 4 R-C-Cl O R-C-H O The aldehyde further gets reduced and ultimately gives alcohol RCH 2 OH. 1.LiAlH 4 O R-C-H +R'O - O The aldehyde further gets reduced and the ester ultimately gives two alcohols RCH 2 OH and R'OH. R-C-OR' R-C-H O The aldehyde further gets reduced and ultimately gives alcohol RCH 2 OH. O R-C-OH H - O R-C-O - H - O R-C-O - Al +2 H 1.LiAlH 4 2.H + O R-C-NH 2 R-C-H NH R-CH 2 -NH 2 Mechanism: O R-C-NH 2 H - O R-C-NH - H - O R-C-NH - Al +2 H imine H - NH - R-C-H H H + R-CH 2 -NH 2 1.LiAlH 4 R-C-H +R'-C-O - O Both the carboxylate ion (R'COO - )and the aldehyde (RCHO) further get reduced and the anhydride ultimately gives two alcohols RCH 2 OH and R'CH 2 OH. R-C-O-C-R' O O O 2. NaBH4 in C2H5OH NaBH4 reduces aldehydes, ketones and acid halides only. It does not reduce esters, anhydrides, amides and carboxylic acids. And the mechanism is similar to that of LiAlH4. NaBH4, being a very weak hydride donor (due to very high covalent character of B-H bond), can directly be used in polar solvents. Organometallic compounds: RLi, RMgX and R2CuLi are the commonly used organometallic compounds. Out of these, RLi is the strongest R - donor and R2CuLi the least. Observe the following table. ALCOHOLS, ETHERS AND PHENOLS E M RAO C-M bond Difference in electronegativity Percent ionic character * C-Li 2.5-1 =1.5 60 C-Mg 2.5-1.2 =1.3 52 C-Zn 2.5-1.6 =0.9 36 C-Cu 2.5-1.9 =0.6 24 * C M C E -E Percentioniccharacter= ×100 E 3. RLi in dry ether followed by acidification . Organo lithium compounds can reduce aldehydes, ketones, esters, anhydrides, acid chlorides into alcohols. They convert carboxylic acids into ketones only. They do not react with amides (at the most acid-base reaction may happen). Mechanism of conversion of acid into ketone is given below. OLi R-C-OLi R H + O R-C-OLi +RH O R-C-OH R - R - OH R-C-OH R O R-C-R 4. RMgX in dry ether followed by acidification . Grignard reagents convert aldehydes, ketones, acid halides and esters into alcohols. They do not react with carboxylic acids and amides (at the most acid- base reaction may happen in both the cases). 5. R2CuLi/ R2Cd Gilman’s reagent (R2CuLi) and R2Cd can only reduce acid chlorides. They do not react with any other carbonyl compounds including aldehydes and ketones. or R 2 Cd O R-C-Cl R 2 CuLi O R-C-R 6. H2/Ni or Pt Catalytic hydrogenation converts aldehydes and ketones into alcohols. It converts esters into alcohols only under drastic conditions. H2/Ni does not react with carboxylic acids, amides, acid chlorides and anhydrides. H2 in the presence of Lindlar’s catalyst (Pd-BaSO4) converts acid chlorides into aldehydes. O H R R-CH 2 -OH O R R R-CHR-OH H 2 /Ni H 2 /Ni O R'O R R-CH 2 -OH +R'OH H 2 /Ni O Cl R O H R H 2 Pd-BaSO 4 200 0 C/10 atm 7.B2H6/THF followed by hydrolysis Diborane converts aldehydes, ketones and carboxylic acids into alcohols. The mechanism of diborane reduction is similar to that of hydroboration oxidation of alkenes. ALCOHOLS, ETHERS AND PHENOLS E M RAO O H R R-CH 2 -OH O R R R-CHR-OH 1. B 2 H 6 /THF O HO R R-CH 2 -OH 2. NaOH 1. B 2 H 6 /THF 2. NaOH 1. B 2 H 6 /THF 2. NaOH EXERCISE-II 1. Show how do you prepare n-butyl alcohol using C2H5MgBr? 2. 2.H + 1.LiAlH 4 /THF O A +B 3. 1.CH 3 MgBr/THF O O A 2.H + 4. 1.CH 3 MgBr/THF O O A 2.H + 5. 2.H + 1.LiAlH 4 /THF O-C-CH 3 H 3 CO-C O O A +B +C 6. 2.H + 1.LiAlH 4 /THF O O C-OEt A +B Chemical reactions of alcohols 1. Conversion of alcohols into alkyl halides Alcohols have poor leaving group i.e. OH - . As result, alcohols do not undergo direct nucleophilic substitution reactions under normal conditions. But they can be made to undergo nucleophilic substitution reactions by converting OH - into a better leaving group. Reagents like HCl, PCl3, PCl5 and SOCl2 can convert alcohols into alkyl halides, ROH +NaCl No reaction ROH NaCl H 2 SO 4 RCl +H 2 O ROH +HCl RCl +H 2 O ROH +NaBr No reaction ROH NaBr H 2 SO 4 RBr +H 2 O ROH +HBr RBr +H 2 O ROH +NaI No reaction ROH NaI H 2 SO 4 No reaction ROH +HI RI +H 2 O ROH NaI H 3 PO 4 RI +H 2 O Note: Protonation makes OH - into a better leaving group i.e. H2O. As a result H + is necessary in all the above reactions. H2SO4, being a mild oxidizing agent, oxidizes I - to I2. As a result, ROH cannot be converted into RI with NaI/H2SO4. ALCOHOLS, ETHERS AND PHENOLS E M RAO I. HX. Observe the following examples. MECHANISM R-OH H + R-OH 2 + X - RX +H 2 O S N 2 mechanism if the R is primary. R + +H 2 O R + +X - RX S N 1 if the R is secondary or tertiary. Some important points: 1. Reactivity of alcohols toward HX – 3 0 >2 0 >1 0 . 2. Reactivity of different halides – HF <HCl <HBr <HI 2 0 or 1 0 alcohols react with HCl to give very low yield. In order to increase the yield, a Lewis acid (generally ZnCl2) is added. HCl +ZnCl2 mixture is called as Lucas reagent. ZnCl2 complexes with the alcohol making OH - a better leaving group. R-OH ZnCl 2 + - II. PCl3 / PBr3 / P +I2 Phosphorous trihalides convert alcohols into alkyl halides. Observe the following mechanism. R-OH + P Cl Cl Cl P Cl Cl R-O H + Cl - R-Cl + P Cl OH Cl Second step is S N 2 mechanism when R is primary or secondary. Phosphorous halides are not used for 3 0 alcohols. P(OH)Cl 2 can convert two more moles of alcohols into RCl. 3 ROH +PCl 3 3 RCl + P HO OH OH P H OH O OH III. PCl5: PCl5 also converts alcohols into alkyl halides and the mechanism is similar to that of PCl3. But one mole of PCl5 converts only one mole of alcohol into alkyl halide. ROH +PCl 5 RCl +POCl 3 +HCl IV. SOCl2 (SNi mechanism) SOCl2 converts alcohol into alkyl halide. The mechanism of this reaction depends on presence or absence of a bulky base such as pyridine. ALCOHOLS, ETHERS AND PHENOLS E M RAO OH H R R 1 + O S Cl Cl O H R R 1 S O Cl +HCl Substitution Nuceophilic internal, where nucleophile is part of the leaving group. Such a front side attack leads to retention of configuration. In the absence of pyridine: In the presence of pyridine: OH H R R 1 + O S Cl Cl O H R R 1 S O Cl + N H + Cl - A normal SN 2 reaction which leads to inversion of configuration. Cl +SO 2 H R R 1 R +SO 2 H Cl R 1 SOCl2 readily reacts with a solvent like water giving SO2 and HCl. As a result, a solvent like ether is taken in the above reaction. HCl gets evaporated in such a solvent and the internal attack becomes major. A bulky base like pyridine converts HCl into a salt, pyridinium chloride. And as a result, chloride ion is retained in the solution. Yield of this reaction is very high due to the formation of a gas SO2. V. Alkyl tosylates: Another method is converting alcohol into tosylate by treating it with tosyl chloride (TsCl). H 3 C S O O Cl Toluene sulfonyl chloride (TsCl) ROH +TsCl pyridine ROTs +HCl Mechanism: H 3 C S O O Cl ROH H 3 C S O O OR H + N H 3 C S O O OR ROTs As TsO - is a very good leaving group, ROTs undergoes nucleophilic substitution reactions very easily. Another advantage of ROTs is that it is insoluble in polar media. When the reaction medium is polar, ROTs gets precipitated and the precipitate can be treated with different nucleophiles as illustrated below. ALCOHOLS, ETHERS AND PHENOLS E M RAO ROTs Cl - RCl+TsO - ROTs I - RI+TsO - ROTs CN - RCN+TsO - ROTs N 3 - RN 3 +TsO - ROTs CH 3 COO - R-O-C-CH 3 +TsO - O ROTs CH 3 O - ROCH 3 +TsO - Illustration:1 OH HBr PBr 3 A B (i) OH HBr C PBr 3 D (ii) OH HCl E PCl 3 F G (iii) Write the structures of all the unknowns emphasising on stereochemistry whereever appropriate. SOCl 2 Br Br Br Br Cl Cl Cl A B C D E F G Answers: + 2. Fischer esterification and ester hydrolysis: Alcohols react with carboxylic acids in acidic medium to form esters. The reaction is reversible and as a result, under suitable conditions, ester can be hydrolysed back to carboxylic acid and alcohol. Although ester can’t be formed in basic medium, it can be hydrolysed in basic medium which is called saponification. Mechanism in acidic medium: ALCOHOLS, ETHERS AND PHENOLS E M RAO R-C-OH +R'ÓH O H + R-C-OR' +H 2 O O R-C-OH O H + R-C-OH OH + R'OH rds R-C-OH OH R'OH + R-C-OH OH R'O OH R'O + R-C-OR' OH + R-C-OR' +H + O Esterification: Rate =k R-C-OH OH + =k k eq [RCOOH] [H + ] [R'OH] =k' [RCOOH] where k' =k k eq [H + ] [R'OH] R-C-OH 2 During esterification, alcohol is taken as a solvent to drive the equilibrium in the forward direction. As solvent is generally taken in excess, its concentration hardly changes. The role of acid in esterification is catalyst. Its concentration also hardly changes. As a result, acid catalysed esterification is a pseudo-first order reaction. R-C-OR' +H 2 O O H + R-C-OH +R'OH O R-C-OR' O H + R-C-OR' OH + rds R-C-OR' OH HOH + R-C-OR' OH OH R-C-OH-R' OH OH + R-C-OH OH + R-C-OH +H + O Ester Hydrolysis (A AC 2): Rate =k R-C-OR' OH + =k k eq [RCOOR'] [H + ] [H 2 O] =k' [RCOOR'] where k' =k k eq [H + ] [H 2 O] HOH +R'OH During ester hydrolysis, water is taken as a solvent to drive the equilibrium in the forward direction. As solvent is generally taken in excess, its concentration hardly changes. The role of acid in ester hydrolysis is catalyst. Its concentration also hardly changes. As a result, acid catalysed ester hydrolysis is a pseudo-first order reaction. Mechanism in basic medium: R-C-OH O R'OH NaOH R-C-O - O R'OH or even R'O - No reaction Partial positive charge on the carbonyl carbon in carboxylate ion is too small to react even with alkoxide ion. As a result, one can't prepare ester in basic medium. But ester can be hydrolysed in basic medium which is called as saponification. Mechanism of ester hydrolysis in basic medium (BAC2): ALCOHOLS, ETHERS AND PHENOLS E M RAO R-C-OH +R'Ó - O R-C-OR' O R-C-OR' O - OH OH - acid base reaction which is highly favourable R-C-O - +R'ÓH O first step is rds. Rate =k [RCOOR'] [NaOH] Intramolecular ester formation: CH 3 -CH-C-OH OHO H + A O O O O CH 3 -CH-CH 2 -C-OH OH O o-hydroxy ester |-hydroxy ester CH 3 -CH=CH-C-OH O CH 3 -CH-CH 2 -CH 2 -C-OH OH O H + A H + A ¸-hydroxy ester O O Intra molecular esterification is difficult because of the ring strain. Dehydration is difficult because the carbocation at o position is unstable. Intra molecular esterification is difficult because of the ring strain. Dehydration is favorable because of the stability of the alkene. Intra molecular esterification is favourable. a lactone 3. Oxidation of alcohols: R-CH 2 -OH Oxidation state of carbon =-1 [O] R-C-H O Oxidation state of carbon =+1 [O] R-C-OH O Oxidation state of carbon =+2 DIFFERENT REAGENTS USED IN THE OXIDATION OF ALCOHOLS: A. Chromic acid Chromic acid is prepared by dissolving either chromium (VI) oxide or potassium dichromate in aqueous sulfuric acid. 2 4 3 2 2 4 2 4 2 2 2 7 2 2 7 2 4 H SO CrO + H O H CrO Chromium Chromicacid (VI)oxide H SO H O K Cr O H Cr O 2H CrO Chromicacid ÷÷÷÷÷ ÷÷÷÷÷ ÷÷÷÷ Chromic acid converts primary alcohols into carboxylic acids, secondary alcohols into ketones. It does not react with tertiary alcohols. Thus, the prerequisite for the oxidation of an alcohol is at least one H on the carbon bearing the OH group. A solution of chromic acid Mechanism of chromic acid oxidation: ALCOHOLS, ETHERS AND PHENOLS E M RAO OH H + HO-Cr-OH O O fast and reversible H O O O-Cr-OH +H 2 O H O O O-Cr-OH H 2 O slow and rate determining O +H 3 O + +HCrO 3 - Chromium (IV) then participates in further oxidations by a similar mechanism and eventually is transformed to Cr (III) which is green in color. An aldehyde is oxidized further to carboxylic acid through its hydrate form. O H 2 O +H 3 O + +HCrO 3 - R-C-H C OH OH H R An aldehyde hydrate O O HO-Cr-OH C OH H R O O-Cr-OH O RCOOH We will study about aldehyde hydrates in the aldehyde and ketones chapter. B. KMnO4: KMnO4 is similar to chromic acid in every respect. It oxidises primary alcohols to carboxylic acids, secondary alcohols to ketones and does not react with tertiary alcohol. But when KMnO4 is used in acidic medium at high temperature, the tertiary alcohol may first convert into alkene and the alkene is further oxidized. O C CH 3 OH CH 3 H 3 C H + /A KMnO 4 A CH 3 -C-CH 3 +CO 2 C. PCC, PDC and CrO3/Pyridine/Cold (Collin’s reagent): N H + CrO 3 Cl - Pyridinium chloro chromate N H + 2 Cr 2 O 7 Pyridinium dichromate These three reagents convert primary alcohols into aldehydes, secondary alcohols into ketones and they do not react with tertiary alcohols. PCC oxidations are generally carried out in aprotic solvents such as CH2Cl2. PCC has no effect on carbon-carbon double bonds OH PCC CH 2 Cl 2 H O PCC does not oxidize aldehydes further because the PCC is not used in water but rather in an organic solvent, CH2Cl2. Without water, the aldehyde can’t be in equilibrium with its hydrate. Recall that only the hydrate of the aldehyde is susceptible to further oxidation not the aldehyde itself. ALCOHOLS, ETHERS AND PHENOLS E M RAO C. Cu/350 0 C Copper at high temperature oxidises primary alcohols into aldehydes, secondary alcohols into ketones. Tertiary alcohols get dehydrated by copper at 350 0 C. CH 2 H 3 C H 3 C CH 3 CH 2 OH Cu/350 0 C CH 3 -C-H O CH 3 CHOHCH 3 Cu/350 0 C CH 3 -C-CH 3 O C CH 3 CH 3 OH H 3 C Cu/350 0 C D. MnO2 MnO2 oxidises only allyl and benzyl alcohols. If they are primary, aldehydes are the final products. If they are secondary, ketones are the final products. OH MnO 2 H O CH 2 OH CHO MnO 2 Exercise-III: 1. O 1.CH 3 MgBr 2. H 2 O A (Mixture of two compounds) A H + A B (Mixture of two compounds) Describe A and B. 2. C O OC 2 H 5 C 2 H 5 O Diethylcarbonate 1.ex CH 3 MgBr 2. H 2 O A +B If both A and B are alcohols, give their structures. 3. A (C 4 H 8 O 2 ) 1.ex CH 3 MgBr 2. H 2 O B If A is an ester and B is the only alcohol produced in the reaction, find out the structure 4. Starting with butane, synthesize the following two compounds . (A) CH3CH2CH2CH2D (B) CH3CHDCH2CH3 5. ALCOHOLS, ETHERS AND PHENOLS E M RAO O CH 3 O HO O 1.ex CH 3 MgBr 2. H 2 O OH HO O What is the number of moles of CH 3 MgBr required for the above conversion? 6. Using ethyl bromide as the starting compound, synthesize the following alcohols in not more than three steps. (A) 2-Butanol (B) 1-Propanol (C) 1-Butanol 7. Benzyl methyl ketone 1.LiAlH 4 2.NH 4 Cl 3.PCl 3 4.KO-t-Bu A B 1.H 2 /Ni 2. H + /A 8. 3-Ethyl-3-pentanol 1. i-PrMgBr 2. Br 9. 1-Hexen-3-ol 1.NaH 2. S O O OMe MeO 10. Ph O O 1.PhMgCl 2. H + 1.LiAlH 4 2.H + 11. 2-Butanol 1.HBr 2. LDA 3.BH 3 /THF 4.H 2 O 2 /OH - 12. 1.LiAlH 4 2.H + O O OMe O Ph O 1.NaBH 4 2.H + H 2 /Pt 13. ALCOHOLS, ETHERS AND PHENOLS E M RAO 1.LiAlH 4 2.H + 1.NaBH 4 2.H + O O 1.CH 3 MgBr 2.H + 14. 1.LiAlH 4 2.H + 1.NaBH 4 2.H + H N O 1.H 2 /Pt 15. 1.ex CH 3 MgBr 2. H 2 O O OH OH O OH OH OH O What is the number of moles of Grignard reagent used in the above reaction? 16. CO 2 Et O 1.LiAlH 4 2.H 2 O A +B If A and B are isomers, what is the relation between them? POLYHYDROXY COMPOUNDS Methods of preparation 1. Reaction of Baeyer’s reagent with alkenes: Alkenes give vicinal diols with cold alkaline KMnO4 (Baeyer’s reagent). Both hydroxyl groups are added from the same side (syn addition). Examples: ALCOHOLS, ETHERS AND PHENOLS E M RAO CH 3 H H 3 C H cold KMnO 4 CH 3 CH 3 H OH HO H CH 3 CH 3 HO H H OH + cold KMnO 4 OH OH + OH OH enantiomers enantiomers cold KMnO 4 OH OH 1. 2. 3. 2. Reaction of alkenes with OsO4/H2O; NaHSO3 Alkenes with OsO4/H2O; NaHSO3 give vicinal diols. Similar to the previous case, here also the hydroxyl groups are added from the same side (syn addition). 3. Reaction of alkenes first with peroxy acetic acid followed by treatment with H + or OH - . Alkenes react with peroxy carboxylic acids to give epoxides which then react with H + or OH - to give vicinal diols. Overall addition of hydroxyl groups is anti. Examples CH 3 H H 3 C H 1. RCOOOH CH 3 CH 3 H OH H OH 1. 2. H + OH OH + OH OH enantiomers OH OH 2. 3. 1. RCOOOH 2. H + 1. RCOOOH 2. H + OH OH + enantiomers Chemical Reactions 1. Pinacol-pinacolone rearrangement When vicinal diols are treated with H + , aldehydes or ketones are obtained. ALCOHOLS, ETHERS AND PHENOLS E M RAO H 3 C-C-C-CH 3 OOH H C CH 3 H 3 H + H 3 C-C-C-CH 3 O H C CH 3 H 3 + H 3 C-C-C-CH 3 OCH 3 H CH 3 + + H 3 C-C-C-CH 3 OCH 3 H CH 3 H 3 C-C-C-CH 3 O CH 3 CH 3 Pinacol Pinacolone Methyl shift Formation of the first carbocation is the rate determining step. It is surprising, at first look, that a 3 0 carbocation is getting rearranged. But see the relative stability of the carbocations: CH 3 -O-CH 2 + > Ph-CH 2 + Ph 2 CH >(CH 3 ) 3 C > + + It is often confusing in this reaction as to which group migrates. Whether it depends upon the migratory aptitude or the stability of the carbocation that is obtained after the migration. It is found that in some examples the first factor (migratory aptitude) predominates, in other examples, the second factor (carbocation stability) predominates. Migratory aptitude H >Ph >R(3 0 ) >R(2 0 ) >R(1 0 ) >CH3 OCH 3 NO 2 OCH 3 > > Phenyl ring migration is an example of intramolecular electrophilic substitution. Presence of an electron donating group at para position makes benzene ring a better migrating group. Presence of electron withdrawing group does the opposite. But presence of any group at the ortho group makes migration of benzene ring difficult because of steric hindrance. Stereochemistry of rearrangements: The configuration of the migrating group is retained during migration because the migrating group never becomes completely free. The configuration of the migratory terminus (the carbon atom to which migration takes place) is found to be predominantly inverted in some cases and completely inverted in other cases. The configuration is found to be completely inverted in the case of amino alcohols and cyclic diols. This suggests the formation of ion pair in the rate determining step. Ph HO Ph H H 3 C Ph H 3 C H Ph HO Ph NH 2 H H 3 C NaNO 2 HCl N 2 + O Ph Examples: ALCOHOLS, ETHERS AND PHENOLS E M RAO OH CH 3 CH 3 OH H + O CH 3 CH 3 2. C C CH 3 CH 3 Br CH 3 OH H 3 C AgNO 3 C C CH 3 CH 3 CH 3 O H 3 C C C CH 3 CH 3 NH 2 CH 3 OH H 3 C NaNO 2 C C CH 3 CH 3 CH 3 O H 3 C HCl 3. 4. C C CH 3 Ph OH Ph OH Ph H + C C Ph Ph Ph O H 3 C 1. cold KMnO 4 H + CHO 5. cold KMnO 4 H + O 6. O + OH HO H + A H O 7. ALCOHOLS, ETHERS AND PHENOLS E M RAO Retropinacol rearrangement O O H + OH O + H 2 O OH O OH tautomerism OH OH tautomerism O O OH Major O H + OH + OH + OH 1. 2. 2. Reaction of vicinal diols with HIO 4 (periodic oxidation) Vicinal diols are oxidized by HIO4. And in the process HIO4 is reduced to HIO3. If AgNO3 is added to after the reaction, a white precipitate of AgIO3 is obtained. C C OH OH HIO 4 C C O O I O O O _ C C O O + +IO 3 - Examples OH OH OH OH OH OH OH OH HIO 4 HIO 4 H O O H H O O H 1. 2. = = OH OH HIO 4 No reaction In chair form, dihedral angle between two equatorial bonds is 60 0 . Dihedral angle between one equatorial form and one axial form is also 60 0 . As a result, there is no much difference between the reactivity of cis-1,2-cyclohexanol and its trans isomer. Trans isomer is slightly less ALCOHOLS, ETHERS AND PHENOLS E M RAO reactive because some of it exists in another conformation in which the two hydroxyl groups are anti to each other. OH OH OH OH OH HIO 4 H O O H 3 (H 3 C) 3 C OH OH 4. (H 3 C) 3 C OH (H 3 C) 3 C (H 3 C) 3 C HIO 4 no reaction C(CH 3 ) 3 = = C C O OH HIO 4 C C O O HO + 5. C C O O HIO 4 C C O O OH OH + 6. C C C O OH HIO 4 OH C C C O O O O + + 7. CH 2 C C OH HIO 4 OH 8 No reaction Now it is easily understood that an alternate reagent for O3; Zn/H + can be cold alkaline KMnO4 followed by HIO4. Cold KMnO4 followed by HIO4 is called as Limaeux reagent. Exercise-IV C C CH 3 Ph OH Ph OH Ph H + /A 1. OH CH 3 CH 3 OH 2. H + /A C C CH 3 CH 3 Br CH 3 OH H 3 C AgNO 3 3. C C CH 3 CH 3 NH 2 CH 3 OH H 3 C NaNO 2 HCl 4. cold KMnO 4 5. H + /A cold KMnO 4 6. H + /A O OH H + 7. O O H 3 O + 8. ALCOHOLS, ETHERS AND PHENOLS E M RAO O H 3 O + 9. H + /A OH HO 10. Exercise-V 1. When one mole of each of the following compounds is treated with HIO4, what will the products be, and how many moles of HIO4 will be consumed? (A) CH3CHOHCH2OH (B) CH3CHOHCHO (C) CH2OHCHOHCH2OCH3 (D) CH2OHCH(OCH3)CH2OH (E) cis-1,2-Cyclopentanediol (F) CH2OH(CHOH)3CHO (G) CH2OH(CHOH)3CH2OH 2. Assign the structures to each of the following compounds. 4 3 3 4 2 4 4 2 4 4 4 4 2 4 A+1mol HIO CH COCH +HCHO B+1mol HIO OHC(CH ) CHO C+1mol HIO HOOC(CH ) CHO D+1mol HIO 2HOOC-CHO E+3mol HIO 2HCOOH+2HCHO F+3mol HIO 2HCOOH+HCHO+CO G+5mol HIO 5HCOOH+HCHO ÷÷÷ ÷÷÷ ÷÷÷ ÷÷÷ ÷÷÷ ÷÷÷ ÷÷÷ 3. OH OH HIO 4 1.LiAlH 4 2.H 2 O A B A is pure while B is a mixture. How many isomers does B contain? 4. HO OH ex.Cu/A H I O 4 P C C C H 2 C l 2 H + / A e x . C r O 3 / H + E A+B F C D 1. Exercise-VI Convert the following. CH 2 OH OH OH 2. ALCOHOLS, ETHERS AND PHENOLS E M RAO C 6 H 5 OH C 6 H 5 OH OH +its enantiomer C 6 H 5 OH OH +its enantiomer A B 3. O H 3 C OH O 4. Br CH 3 CHO 5. H H O O Without using ozone. 6. H O O EXERCISE-VII Write the structures of all the unknowns. O O O 1.AlCl 3 A Zn-Hg HCl 2.H 2 O B PCl 3 1.AlCl 3 2.H 2 O Zn-Hg HCl Se A C D E F + 1.LAH 2.H 2 O G PPA H J I K N P L M O 1 . L A H 2 .H 2O N a B H4 / C2 H5 O H C 6 H 6 /PPA C 6 H 6 /PPA Zn-Hg HCl C 6 H 6 /PPA NaBH 4 /C 2 H 5 OH C 6 H 6 /PPA Se/A Se/A NaBH 4 /C 2 H 5 OH Note: PPA stands for polyphosphoric acid, which is a source of H + . ETHERS Physical properties ALCOHOLS, ETHERS AND PHENOLS E M RAO Ethers and alcohols have comparable solubility in water as both of them can form hydrogen bonding with water. But boiling points of ethers are low when compared to those of alcohols as ethers can’t form hydrogen bonds among themselves. CH3CH2OCH2CH3 Diethyl ether CH3CH2CH2CH2OH 1-Butanol Boiling point: 35°C 117°C Solubility in water: 7.5 g/100 mL 9 g/100 mL Methods of preparation 1.Williamson’s synthesis: R-L +RO - R-O-R +L - L - Cl, Br, I, OSO 3 CH 3 , OTs, N 2 + etc. As RO - is a strong base, best yields are obtained when R in R-L is CH3 or 1 0 . With 2 0 and 3 0 substrates, RO - gives predominantly elimination products. Examples: CH 3 CH 2 I CH 3 ONa CH 3 OH CH 3 CH 2 OCH 3 +NaI 1. CH 3 -O-S-O-CH 3 CH 3 ONa CH 3 OH CH 3 OCH 3 + 2. O O CH 3 -O-S-O-Na O O 3. CH 3 CH 2 OH +CH 2 -N _ + diazomethane CH 3 CH 2 O - +CH 3 -N N N + CH 3 CH 2 OCH 3 +N 2 4. HO-CH 2 -CH 2 -CH 2 -CH 2 Br NaOH O +NaBr acid-base reaction 2. Alkoxymercuration-demercuration: This method is a slight variation of oxymercuration-demercuration. In oxymercuration- demercuration, water is taken as solvent in the first step. In alkoxymercuration-demercuration, alcohol is taken as solvent. Everything else is same. In alkoxymercuration-demercuration, there is no formation of fully fledged carbocation and as a result there are no rearrangements. But the addition is according to Markonikov’s rule. Examples: 1. (CH 3 ) 2 C=CH 2 1. Hg(OAc) 2 /CH 3 OH 2. NaBH 4 (CH 3 ) 2 C-CH 3 OCH 3 2. HO-CH 2 -CH 2 -CH 2 -CH=CH 2 1. Hg(OAc) 2 2. NaBH 4 O CH 3 3. Intermolecular dehydration of alcohols: When alcohols are heated at temperatures lower than those required for intra-molecular dehydration, they give ethers. This method best works for methyl and 1 0 alcohols. 2 0 and 3 0 alcohols give alkenes rather. ALCOHOLS, ETHERS AND PHENOLS E M RAO CH 3 CH 2 OH H + 17O 0 C CH 2 =CH 2 CH 3 CH 2 OH H + 14O 0 C CH 3 CH 2 OCH 2 CH 3 Elimination Substitution Chemical reactions: Ethers do not contain acidic hydrogen. As a result, they do not react with Na, NaOH or any other base. As such they do not undergo nucleophilic substitution reactions because RO - is not a good leaving group. But they do undergo nucleophilic substitution reactions under acidic conditions. 1.Ether cleavage reactions: R-O-R dil.H + R-O-R H + R-O-R H + H 2 O ROH +ROH R + +ROH 2 ROH S N 1 S N 2 1. With dil. acids A R-O-R R-O-R H + R-O-R H + Br - RBr +ROH R + +ROH S N 1 S N 2 conc. HBr Br - RBr +ROH 2. With conc. acids A Whether it is SN1 or SN2, depends on the nature of the two alkyl groups. It is experimentally found that when one of the alkyl groups is 3 0 , benzylic or allylic (or any other carbon chain that can form stable carbocation), the mechanism follows S N 1 pathway. Otherwise it is S N 2. Examples: 1. CH 3 CH 2 OCH 3 dil. acid conc. HBr CH 3 CH 2 OH +CH 3 OH CH 3 CH 2 OH +CH 3 Br ex. HBr CH 3 CH 2 Br +CH 3 Br 2. CH 3 CHOCH 3 dil. acid conc. HBr CH 3 CH(OH)CH 3 +CH 3 OH CH 3 CH(OH)CH 3 +CH 3 Br ex. HBr CH 3 CH(Br)CH 3 +CH 3 Br CH 3 S N 2 S N 2 ALCOHOLS, ETHERS AND PHENOLS E M RAO 3. (CH 3 ) 3 C-O-CH 3 conc. HI ex. HI (CH 3 ) 3 C-I +CH 3 OH (CH 3 ) 3 C-I +CH 3 I S N 1 4. O conc. HI ex. HI No reaction No reaction 5. O NO 2 conc.HI OH I NO 2 + ex. HI OH I NO 2 + ArS N O ex. HI conc. HI No reaction No reaction 6. 7. O CH=CH 2 dil. H + O CH-CH 3 OH a hemiacetal OH +CH 3 CHO In this example, first water gets added to the double bond as it is highly reactive. The resultant hemiacetal, being unstable, decomposes. O CH-CH 3 OH H + 8. CH 3 -OCH 2 -CH 2 -OCH 3 ex. HI 2CH 3 I +CH 3 CH 2 I In this example, the two carbon chain is first converted into I-CH 2 -CH 2 -I which, being unstable, decomposes into ethene. Ethene further reacts with HI. 2.Claisen rearrangement: When allyl phenyl ethers or allyl vinyl ethers are heated, they undergo rearrangement. Examples: ALCOHOLS, ETHERS AND PHENOLS E M RAO O A O O OH 1. O O A O OH O 2. O A O O Allyl-vinyl ether 3. EPOXIDES Epoxides are three membered cyclic ethers. O O Ethylene oxide (or) oxirane (or) epoxy ethane Propylene oxide (or) 2-methyloxirane (or) 1,2-epoxypropane Methods of preparation: 1.Reaction of alkenes with peroxy acids: Alkenes, when treated with an organic peroxy acids, yield epoxides. CH 3 H H H 3 C RCO 3 H CH 3 H H H 3 C C O H O O R O CH 3 H 3 C + RCOOH CH 2 Cl 2 ALCOHOLS, ETHERS AND PHENOLS E M RAO All the bond fissions and bond formations are simultaneous. As a result, there is no scope for bond rotation in alkene. Consequently, the reaction is stereospecific. Cis alkenes give cis epoxides and trans alkenes give trans epoxides exclusively. 2. Reactions of halohydrins with bulky bases: Halohydrins, when treated with bulky bases, yield epoxides. Bulky bases are used to avoid direct replacement of halogen. C C Br HO pyridine O pyridine O H 3 C H H CH 3 CH 3 CH 3 H OH H Br (S,R) (S,S) Chemical reactions: Unlike open chain ethers, epoxides are highly reactive toward nucleophiles. This is because of the ring strain. O Nu - - O-CH 2 -CH 2 -Nu H + HO-CH 2 -CH 2 -Nu In the case of unsymmetrical epoxides, two factors may influence the preferred site of nucleophilic attack. 1. Magnitude of partial positive charge 2. Steric hindrance. Which of the two predominates depends on the conditions. Nucleophile attacks epoxide under two conditions: in acidic medium and in basic medium. In acidic medium (SN1 like), epoxide is first protonated. Nucleophile attacks in the second step. Under these conditions, it is invariably the first factor (magnitude of partial positive charge) which predominates. The nucleophile preferably attacks the carbon which can stabilize the carbocation better. The important point here is that there is no formation of fully fledged carbocation and as a result there are no rearrangements. O H + O H + CH 3 OH CH 3 -CH-CH 2 OH OCH 3 In basic medium (SN2 like), the nucleophile directly attacks the epoxide and the protonation occurs in the second step. Under these conditions, it is invariably the second factor (steric hindrance) which predominates. The nucleophile preferably attacks the less hindered carbon. O H + CH 3 O - CH 3 -CH-CH-OCH 3 CH 3 -CH-CH-OCH 3 O - OH Examples: ALCOHOLS, ETHERS AND PHENOLS E M RAO O 1. CN - 2. H 3 O + CH 3 -CH-CH-CN OH O OH - CH 3 -CH-CH-OH OH 18 18 O H + CH 3 -CH-CH-OH OH 18 18 H 2 O O 1.AlCl 3 CH 3 -CH-CH-OH CH 3 2.CH 3 MgBr 3.H 2 O O CH 3 -CH-CH-CH 3 OH 1.CH 3 MgBr 2.H 2 O 1. 2. 3. 4. 5. H 2 O PHENOLS Compounds in which a hydroxyl group is bonded to an aromatic ring are called phenols. The chemical behavior of phenols is different in some respects from that of the alcohols. A corresponding difference in reactivity was observed in comparing aryl halides, such as bromobenzene, with alkyl halides, such as butyl bromide and tert-butyl chloride. Thus, nucleophilic substitution and elimination reactions were common for alkyl halides, but rare with aryl halides. This distinction carries over when comparing alcohols and phenols, so for all practical purposes substitution and/or elimination of the phenolic hydroxyl group does not occur. PHYSICAL PROPERTIES Phenol is not appreciably soluble in water (8.2 g/ 100 mL of water). But it is quite soluble in aqueous NaOH because of the soluble salt it forms with NaOH. At the same time it is not soluble in aqueous NaHCO3. OH NaOH ONa +H 2 O OH NO 2 NaHCO 3 No reaction OH NaHCO 3 No reaction ALCOHOLS, ETHERS AND PHENOLS E M RAO OH NO 2 NaHCO 3 NO 2 ONa NO 2 NO 2 +H 2 O +CO 2 OH NO 2 NaHCO 3 NO 2 ONa NO 2 NO 2 +H 2 O +CO 2 O 2 N O 2 N METHODS OF PREPARATION 1.Dow’s process: Cl NaOH / 350 0 C High pressure ONa + O + ONa Minor products Major product Dow’s process occurs through benzyne mechanism. The major product is obtained when OH - attacks the benzyne. The minor products are obtained when phenoxide ion attacks the benzyne. 2. Alkali fusion of benzene sulfonic acid SO 3 H NaOH / 350 0 C High pressure ONa Major product Mechanism of this reaction is similar to that of Dow’s process i.e, benzyne mechanism. 3. Preparation from benzene diazonium chloride When benzene diazonium chloride is warmed with water, phenol is produced. NO 2 NH 2 N 2 Cl OH H 2 Ni NaNO 2 HCl H 2 O warming 4. Preparation of phenol from p-nitrochlorobenzene p-Nitrochlorobenzene is first treated with NaOH. Chlorine is replaced with –OH in a nucleophilic aromatic substitution reaction (addition-elimination). Later the unwanted –NO2 group can be removed. ALCOHOLS, ETHERS AND PHENOLS E M RAO Cl ONa ONa OH NaOH H 2 Ni NaNO 2 NO 2 NO 2 NH 2 N 2 Cl OH HCl H 3 PO 2 5. Preparation of phenol from cumene hydroperoxide When cumene hydroperoxide is treated with H3O + , phenol and acetone are obtained. CH 2 =CHCH 3 H 3 PO 4 C CH 3 H 3 C O-OH cumene hydroperoxide H + OH +CH 3 COCH 3 cumene O 2 Mechanism Ph-C-O-OH CH 3 CH 3 H + Ph-C-O-OH 2 CH 3 CH 3 + CH 3 CH 3 C-OPh + CH 3 CH 3 C-OPh H 2 O + CH 3 CH 3 C-OPh HO Hemiacetal CH 3 CH 3 HO H + C OPh + H CH 3 COCH 3 +PhOH H 2 O rds Migration and removal of water occurs simultaneously in the rate determining step. Hemiacetals are unstable in acidic medium and decompose to aldehydes or ketones (hemiacetals and acetals are discussed in detail in aldehydes and ketones). Migratory aptitude: H >Ph >R(3 0 ) >R(2 0 ) >R(1 0 ) >CH3 OCH 3 NO 2 OCH 3 > > Phenyl ring migration is an example of intramolecular electrophilic substitution (ipso attack). Presence of an electron donating group at para position makes benzene ring a better migrating group. Presence of electron withdrawing group does the opposite. But presence of any group at the ortho group makes migration of benzene ring difficult because of steric hindrance. Examples: p-CH 3 -C 6 H 4 -C CH 3 Ph OOH H + p-Cresol +acetophenone 1. 2. Ph-CH 2 -OOH H + Benzaldehyde ALCOHOLS, ETHERS AND PHENOLS E M RAO OOH H + O H OH H H O O O as a result of hydride migration as a result of alkyl migration as a result of alkenyl migration 3. 6. Fries rearrangement O-C-R O OH C-R + OH C-R O O AlCl 3 Major at low temperature Major at high temperature The mechanism of Fries rearrangement is similar to that of Friedel-Craft’s acylation. It is often possible to select conditions so that either ortho or para is the major product. High temperature favors ortho isomer and low temperature favors para product. There is no clarity as to whether the reaction is intermolecular or intramolecular (whether there is a formation of fully fledged acylium ion or not). CHEMICAL REACTIONS OF PHENOL 1. Phenol as an acid: Phenol +Na PhONa +H 2 Phenol +NaOH PhONa +H 2 O ALCOHOLS, ETHERS AND PHENOLS E M RAO 2. Reactions with RX, acid chlorides (RCOCl), and anhydrides (RCOOCOR): OH NaOH O CH 2 CH 2 -Cl _ OCH 2 CH 3 OH OCCH 3 CH 3 C-Cl O O OH OCCH 3 CH 3 C-O-CCH 3 O O O pyridine +CH 3 COOH +HCl +NaCl Pyridine (in the case of anhydride) helps converting phenol into phenoxide ion, which is a better nucleophile. Pyridine is not generally required in the case of acid chloride as it is highly reactive. 3. Reaction of phenol with Br2/H2O OH Br 2 H 2 O OH Br Br Br OH Br 2 CS 2 OH Br Mechanism OH OH + _ Br-Br OH + Br tautomerism OH Br Reactions proceeds until all ortho and para positions are substituted. 2,4,6-Tribromophenol is not soluble in water and is obtained as a white precipitate. In CS2, only monobromination occurs because of the following reasons: 1. Phenol remains undissociated in CS2. Phenoxide ion is far more reactive than phenol. 2. Heterolytic fission of Br-Br bond is not favorable in non polar media. 3. Introduction of each bromine makes the benzene ring less reactive. Other compounds, which react with bromine water similar to phenol, are anisole and aniline. ALCOHOLS, ETHERS AND PHENOLS E M RAO NH 2 NH 2 Br Br Br Br 2 H 2 O OCH 3 OCH 3 Br Br Br Br 2 H 2 O 4. Reimer-Tiemann Reaction OH 1.CHCl 3 / KOH OH CHO Major 2.H + Mechanism CHCl 3 +KOH CCl 2 +KCl +H 2 O Dichlorocarbene is an electrophile as its central atom has only sextet of electrons. But it is a weak electrophile due to back bonding and as a result it attacks only highl y activated rings, eg: phenol, pyrrole, furan etc. O CCl 2 O CCl 2 tautomerism O - CHCl 2 _ _ OH CHO hydrolysis Salicylaldehyde The reaction is thermodynamically controlled and as a result, more stable product is the major product. Some important points regarding Reimer-Tiemann reaction. 1.CHCl3/KOH gives abnormal products with some aromatic compounds. See the example of pyrrole. Pyrrole, when treated with CHCl3/KOH, gives 3-chrolopyridine as the major product. In the first step dichlorocarbene undergoes cycloaddition to pyrrole and then the ring expands. Although there is a loss of aromaticity in the first step, it is regained ultimately. N N H CHCl 3 KOH H Cl Cl N Cl H + N Cl N H CHCl 3 / KOH N H CHO Minor product Normal Reimer-Tiemann reaction followed by H + 2. Aniline, which is otherwise similar to phenol in its properties, does not undergo Reimer- Tiemann reaction. It gives a different product with CHCl3/KOH. We will discuss this in the chapter amines. NH 2 CHCl 3 KOH NC Isocyanide Carbylamine reaction 3. Phenol with CCl4/KOH gives salycilic acid. ALCOHOLS, ETHERS AND PHENOLS E M RAO CHCl 3 / KOH followed by H + OH OH Salicylic acid COOH 4. Furan reacts with CHCl3/KOH and gives the expected product. O CHCl 3 KOH O CHO Major product 5. Tropolone (a highly activated aromatic ring) too undergoes Reimer-Tiemann reaction. CHCl 3 / KOH followed by H + O HO O HO OHC Tropolone 5.Kolbe-Schmitt Reaction (Carboxylation of phenoxide ion) Carefully observe the following example. OH 1. NaOH 2.CO 2 OH COOH Major product 3.H 3 O + RMgX 1.CO 2 2.H 3 O + RCOOH O C O R _ RCO O _ Carboxylation generalised: O O COO tautomerism O - COO _ OH COOH hydrolysis Salicylic acid O C O _ _ The Kolbe-Schmitt reaction is equilibrium controlled and as a result the major product is the more stable isomer i.e, salicylic acid. Salicylic acid is used in the preparation of acetyl salicylic acid which is an ingredient of aspirin, an analgesic. OH COOH OCCH 3 COOH O CH 3 CCl O or CH 3 COCCH 3 O O acetyl salicylic acid Summary of the reactions of phenol: When reactions of phenol are carefully observed, one thing becomes obvious. That is, in some reactions oxygen is the attacking site and in other reactions it is carbon. ALCOHOLS, ETHERS AND PHENOLS E M RAO OH Br 2 / H 2 O Reimer Teimann Reaction RX / RCOCl / anhydride Kolbe's reactio Carbon attacks Carbon attacks Carbon attacks Oxygen attacks Why phenol or phenoxide ion behaves differently? This is generally explained on the basis of Hard and Soft acid-base concept which is beyond the scope of J EE syllabi. Practical organic chemistry of phenols: Phenols can be identified by the following reagents: 1. Na: Phenols, like any other compounds having acidic hydrogen, liberate H2 with Na. Other compounds which liberate H2 with Na are; terminal alkynes, alcohols, carboxylic acids, sulfonic acids etc. 2. Br2/H2O: Phenols decolorize bromine water and give white precipitate. Other compounds which also decolorize bromine water are alkenes, alkynes, aniline, anisole, and some aldoses (we will study about aldoses in Biomolecules). 3. FeCl3: Phenol/substituted phenols give colored (ranging from orange to red) with FeCl3. In fact all enols (phenol is an enol) give colored compounds with FeCl3. See the following examples. O O O O Both the above compounds give colored compounds with FeCl3, as their enols are in appreciable amounts. Acetone, acetaldehyde etc do not give colored compounds with FeCl3, as enols are in negligible amounts. PhOH +FeCl 3 Fe(OPh) 3 a colored compound Exercise-VIII 1. Show how each of the following ethers is prepared by Williamson’s synthesis? If there is any ether which can’t be prepared by Williamson’s synthesis, propose an alternate method. (A) O (B) O (C) OMe (D) O (E) OEt 2. Draw the structural formulas of the major products obtained when each of the following reactions. (A) O 1 mol HI A (B) OMe 1 mol HI A (C) O 1 mol HI A (D) OEt 1 mol HI A (E) OCH 3 ex HI A (F) O 1 mol HI A ALCOHOLS, ETHERS AND PHENOLS E M RAO (G) O 2 N O ex HI A (H) ex HI A O O (I) O 1 eq HI PCC CH 2 Cl 2 (J) 1.PhCO 3 H 2.PhOH,H + 3. Write the major products for the following reaction. 1. OH 1. NaOH 2. Br 3.A 2. + 2 4 14 12 H /H O HIO A(C H O) B 2mol Benzaldehyde ÷÷÷÷÷÷ ÷÷÷÷÷ A is optically active while B is not. Find the structural formulas of A and B. 4. Write all the unknown reagents in the following sequence of reactions. Br Br OH OH O OH OMe OH C CH OMe O H H O O O 1 2 3 4 5 6 7 8 9 10 11 12 Exercise-IX 1.Complete the following equations. ALCOHOLS, ETHERS AND PHENOLS E M RAO (A) phenol +Br2 0 2 5 C,CS ÷÷÷÷ (B) phenol +conc. H2SO4 0 25 C ÷÷÷÷ (C) phenol +conc. H2SO4 0 100 C ÷÷÷÷÷ (D) p-cresol +p-toluenesulfonyl chloride - OH ÷÷÷÷ (E) phenol +phthalic anhydride 3 AlCl ÷÷÷÷÷ (F) p-cresol +Br2 2 H O ÷÷÷÷ (G) phenol +C6H5COCl pyridine ÷÷÷÷÷÷ (H) phenol +(C6H5CO)2O pyridine ÷÷÷÷÷÷ (I) phenol +NaOH ÷÷÷ (J ) product of (I) +CH3OSO2OCH3 ÷÷÷ (K) product of (I) +CH3I ÷÷÷ 2. Desribe a simple chemical test for distinguishing following pairs of compounds a) 4-Chlorophenol and 4-chloro-1-phenylbenzene b) 4-methylphenol and 2,4,6-trinitrophenol c) 4-methylphenol and 4-methylbenzoic acid d) ethyl phenyl ether and 4-ethyl phenol e) phenyl vinyl ether and ethyl phenyl ether 3. Describe simple chemical tests that would serve to distinguish between; f) phenol and 0-xylene g) p-ethylphenol, p-methylanisole, and p-methylbenzyl alcohol h) 2,5-dimethylphenol, phenylbenzoate, m-toluic acid i) anisole and 0-toluidine j) acetylsalicylic acid, ethyl acetylsalicylate, ethyl salicylate, and salicylic acid k) m-dinitrobenzene, m-nitroaniline, m-nitrobenzoic acid, and m-nitrophenol 4. Describe simple chemical methods for the separation of the compounds of problem 3, parts a),c),d),and f), recovering each component in essentially pure form. Exercise-X Write the major products of the following reactions. 1. OH CH=CH 2 H + 2. O Br CH 3 ONa 3. O O O NaOH MeOH 4. OH OH H + 5. COOH HO O HOOC A H + 6. O OH O 1.NaBH 4 2. H + ALCOHOLS, ETHERS AND PHENOLS E M RAO EXERCISE-I 1. O 2. OH and its enantiomer 3. OH + 4. D OH 5. Et OPh 6. O OCH 3 EXERCISE-II 1. n-Butyl alcohol can be prepared by treating C2H5MgBr with ethylene oxide (epoxide) followed by hydrolysis. 2. OH OH & 3. OH OH 4. O OH 5. OH CH 2 OH +C 2 H 5 OH +CH 3 OH 6. HO CH 2 OH HO CH 2 OH + +C 2 H 5 OH Answers: Exercise-III ALCOHOLS, ETHERS AND PHENOLS E M RAO 1. OH HO & A (mixture of diastereomers) & B (mixture of enantiomers) 2. (CH3)3COH, 2C2H5OH 3. C O H O-CH CH 3 CH 3 4. (A) 1.Br2/hv 2.(CH3)3COK/(CH3)3COH 3.HBr/H2O2 4.Mg/THF 5.D2O (B) 1.Br2/hv 2.Mg/THF 3.D2O 5. 3 6. (A)1.Mg/THF 2.CH3CHO 3. H + (B)1.Mg/THF 2.HCHO 3. H + O 1.Mg/THF 2. 3.H + (C) 7. CH 2 -CH=CH 2 (A) CH=CH 2 -CH 2 (B) 8. 9. CH 2 =CH-CH-CH 2 -CH 2 -CH 3 OCH 3 10. Ph3COH +CH3CH2CH2CH2OH PhCH2OH +CH3CH2CH2CH2OH 11. CH3CH2CH2CH2OH 12. CH 2 OH OH HO HO OMe O O Ph O HO OMe O O Ph O 13. HO OH No reaction HO OH 14. H N No reaction No reaction 15. 3 16. HO CH 2 OH HO CH 2 OH Diastereomers Exercise-IV 1. C C CH 3 Ph Ph Ph O 2. O CH 3 CH 3 3. C C CH 3 CH 3 CH 3 O H 3 C 4. C C CH 3 CH 3 CH 3 O H 3 C ALCOHOLS, ETHERS AND PHENOLS E M RAO 5. CHO O & 6. O 7. O O 8. OH OH OH 9. OH 10. O Exercise-V 1. (A) CH3CHO, HCHO,1 (B) CH3CHO, HCOOH,1 (C) HCHO, OHCCH2OCH3,1 (D) No reaction (E) OHCCH2CH2CH2CH2CHO (F) HCHO,4HCOOH,4 (G) 2HCHO, 3HCOOH, 4 2. A (CH3)2C(OH)CH2OH B OH OH (Cis or trans) C O OH D HOOC(CHOH)2COOH E HOCH2(CHOH)2CH2OH F HOCH2COCH(OH)CHO G OHC(CHOH)4CH2OH 3. O O OH OH (A) (B) A pair of enantiomers and a meso compound 4. O CHO COOH OH CHO OH (A) (B) HCHO(C) (D) CHO (E) (F) Exercise-VI 1. 1.HIO4 2.H2/Ni 2. (A) 1.H + /A 2. KMnO4/OH - /cold (B) 1.H + /A 2.RCO3H/H + . 3. 1.CH3MgBr/THF 2. H + /A 3. KMnO4/OH - /cold 4.CrO3/H + . 4. 1.Mg/THF 2.H2O 3.NBS/CCl4 4.H2O 5.PCC/CH2Cl2 (or MnO2) 5. 1.Br2/hv 2.ROH/KOH/ A 3. KMnO4/OH - /cold 4.HIO4 6. 1. Br2/hv 2.NaOH 3. CrO3/H + 4. CH3MgBr/THF 5. H + 6. H + /A 7. O3; Zn/H + Exercise-VII ALCOHOLS, ETHERS AND PHENOLS E M RAO HO 2 C O (A) HO 2 C (B) ClOC (C) O (D) (E) (F) (G) OH HO 2 C OH (H) O (I) (J) OH OH (K) OH (L) (M) OH (N) (O) (P) Exercise-VIII 1. (A) Ethyl chloride +Sodium isopropoxide (B) Ethyl chloride +Sodium tertiarybutoxide (C) ONa CH 3 I + (D) 1.Hg(OAc) 2 /(CH 3 ) 3 COH 2.NaBH 4 (E) ALCOHOLS, ETHERS AND PHENOLS E M RAO ONa +EtBr 2. (A) (CH3)3CI +(CH3)2C=CH2 +CH3CH2OH (B) I + +MeOH (C) Cyclohexanol +Tertiarybutyl iodide + Isobutene (D) Cyclohexanol +Ethyl iodide (E) Phenol +Ethyl iodide (F) HO I HO + (G) I O 2 N + OH (H) 2 mol CH3CH2I (I) I CHO (J) OH OPh 3. 1. OH (Claisen rearrangement) 2. O Ph H H Ph Ph Ph H OH H OH (A) (B) 4. 1. C CH 3 OH H 3 C CH CH 3 OCH 3 Pair of enatiomers 2. C CH 3 H 3 CO H 3 C CH CH 3 OH Configuration is retained 3. C CH 3 OH C 2 H 5 CH 2 CH 3 4. C CH 3 OH C 2 H 5 CH 2 SH Configuration is retained 5. CH(OH)CH 3 Configuration is retained 6. C OH C-CH 3 +its enantiomer 5. 1. m-CPBA/CH2Cl2 2. CH÷CNa followed by H2O 3. Baeyer’s reagent 4. O3 followed by Zn/H + 5. HIO4 6. One eq TsCl followed by CH3ONa 7. KMnO4 8. H + /A (Pinacol-pinacolone rearrangement) 9. HBr 10. NBS/CCl4 ALCOHOLS, ETHERS AND PHENOLS E M RAO 11. CH3ONa/CH3OH/ A Exercise-IX 1. (A) Para-bromophenol (B) p-Hydroxysulphonic acid (C) o-Hydroxysulphonic acid (D) CH 3 TsO (E) O COOH HO (F) OH Br Br CH 3 (G) C OPh O (H) C OPh O +CH 3 COOH (I) Sodium phenoxide (J) Anisole (K) Anisole 2. a) Na (H2 gas is evolved in the first case) b) NaHCO3 (CO2 gas is evolved in the second case) c) NaHCO3 (CO2 gas is evolved in the second case) d) Na (H2 gas is evolved in the second case) e) Baeyer’s reagent or Br2/H2O (Both are decolorised by the first compound) 3. a) Na (H2 gas is evolved in the first case) or Br2/H2O (decolorised by the first compound) b) p-ethyl phenol decolorises Br2/H2O. p-methyl benzyl alcohol evolves H2 with Na. c) 2,5-Dimethyl phenol releases H2 gas with Na. m-Toluic acid evolves CO2 with NaHCO3. d) o-Toluidine gives offensive smell with CHCl3/KOH (Carbyl amine reaction) e) O-C-CH 3 COOH O Acetyl salicylic acid O-C-CH 3 COOC 2 H 5 O Ethyl acetyl salicylate OH COOC 2 H 5 Ethyl acetyl salicylate OH COOH Salicylic acid First add NaHCO3 to all beakers. 1 st and 4 th compounds liberate CO2 gas. Then add Br2/H2O to those two beakers. Only the fourth compound decolorises Br2/H2O. Add Na to the other two beakers. Only the third compound liberates H2. f) m-Nitroaniline gives offensive smell with CHCl3/KOH. m-Nitrobenzoic acid liberates CO2 gas with NaHCO3. m—Nitrophenol gives H2 gas with Na. 4. a) Aq. NaOH ALCOHOLS, ETHERS AND PHENOLS E M RAO c) Mixture aq.NaHCO 3 aqueous layer organic layer COONa CH 3 H + COOH CH 3 2,5-Dimethylphenol + Phenyl benzoate aq.NaOH aqueous layer organic layer ONa Me OH Me Me H + Me Phenyl benzoate d) Aq.HCl f) Mixture aq.NaHCO 3 aqueous layer organic layer COONa H + COOH m-Nitrophenol + m-Nitroaniline + m-Dinitrobenzene aq.NaOH aqueous layer organic layer ONa OH H + m-Nitroaniline + m-Dinitrobenzene NO 2 NO 2 NO 2 NO aq.HCl aqueous layer organic layer NH 3 + aq.NaOH NH 2 NO 2 NO 2 Exercise-X 1. O 2. O CH 2 OCH 3 3. O H 3 CO O OH 4. O 5. OH O O +CO 2 6. O O