30 Aryl Halides and Phenols

30 Aryl Halides and Phenols Objective Questions I (Only one correct option) OH 1 What will be the major product when m-cresol is reacted with propargyl bromide (HC ººC ¾ CH 2Br ) in presence of K 2CO3 in acetone? NH2 (a) (b) OH OH O (a) NH2 (b) CH3 CH3 OH 5. Which of the following compounds reacts with ethyl OH (c) (d) (c) magnesium bromide and also decolourises bromine water solution (d) OCH3 CN H3C H 3C CH2—CO2CH3 (a) 2. Increasing rate of S N1reaction in the following compounds is I OH I MeO (A) I (d) H3 C precipitate with AgNO3 ? I (D) Br Br (a) (b) (B) < (A) < (C) < (D) (d) (B) < (A) < (D) < (C) (b) (c) (d) N 3. p-hydroxybenzophenone upon reaction with bromine in carbon tetrachloride gives 7. The major product of the following reaction is Cl (i) HBr O O Br (ii) alc.KOH Br (b) O HO HO OH Br O (a) (c) (d) HO HO 4. The organic compound that gives following qualitative analysis is Cl Cl O Br (c) (b) O O 8. The major product of the following reaction is Inference Insoluble Decolourisation (d) OH OH (ii) NaOH solution Soluble (iii) Br2 /water Br Br H3CO (a) (A) < (B) < (C) < (D) (c) (A) < (B) < (D) < (C) Test (i) Dil. HCl O 6. Which of the following compounds will produce a (C) (a) CH2 CN (c) (B) CH (b) Br2 (excess) SO3H O Aryl Halides and Phenols OH O OH CH3 OH O Br (a) (b) Br (a) and CH3—OH (b) H3 C and Br SO3H OH OH Br Br (d) CH3 and H3C (c) 9. The major product of the following reaction is Br CH3 Br (i) aq. NaOH (ii) CH3I CH3 OH (b) CH3 (b) Br HO O (c) OH OCH3 (c) (ii) CrO3/H+ (iii) H2SO4/D O CH3 CH3 CH3 (d) HO (d) Br 14 The major product of the following reaction is CH3 (i) Br2 (ii) EtOH 10. The increasing order of the pK a values of the following compounds is Br OH OH OH OEt OH OEt (a) (b) NO2 NO2 B A (a) D < A < C < B (c) C < B < A < D C (b) B < C < A < D (d) B < C < D < A HO CH3 OEt OEt OMe D 11. The major product of the following reaction is O Br OEt (c) (d) 15. Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH3CO)2 O in the presence of catalytic amount of H2SO4 produces: AlCl3, D + O O OH OH H3C CH3 (a) O CH3 CH3 O (a) (b) (b) CO2H CO2H OH OH CH3 CH3 O (a) CH3 (a) H3 C (i) KOH (aq.) O OH and (d) 13. The major product of the following reaction is SO3H Br OH OH OH O Br Br (c) O C CH3 (c) CH3 Br (d) 12. The products formed in the reaction of cumene with O2 followed by treatment with dil. HCl are (c) O CO2H CH3 O OH (d) CO2H CH3 O O Aryl Halides and Phenols 16. Phenol reacts with methyl chloroformate in the presence of 19. OH NaOH to form product A. A reacts with Br 2 to form product B. A and B are respectively – OC2H5 + C2H5I Br OH (a) OCH3 anhy. C2H5OH OH and (a) C6H5OC2H5 (c) C6H5OC6H5 OCH3 O (b) C2H5OC2H5 (d) C6H5I 20. In the reaction of p-chlorotoluene with KNH2 in liq. NH3 , the O major product is O O (b) O and O O (a) o-toluidine (c) p-toluidine O 21. Phenol reacts with bromine in carbon disulphide at low temperature to give Br O O (c) O (a) m-bromophenol (c) p- bromophenol O and O OH (a) m-bromophenol (c) 2, 4-dibromophenol OH OCH3 and Br (b) o-and p-bromophenol (d) 2, 4, 6-tribromophenol Objective Questions II (One or more than one correct option) OCH3 O (b) o- and p-bromophenol (d) 2, 4, 6-tribromophenol 22. When phenol is treated with excess of bromine water, it gives O Br (d) (b) m-toluidine (d) p-chloroaniline O 17. For the identification of b-naphthol using dye test, it is necessary 23. The reactivity of compound Z with different halogens under appropriate conditions is given below to use (a) (b) (c) (d) dichloromethane solution of b-naphthol acidic solution of b-naphthol neutral solution of b -naphthol alkaline solution of b-naphthol OH 18. The major product of the following reaction is Z Br H3C (a) (b) (c) (d) Å PhS Na dimethyl formamide SPh H3C F (a) C(CH3)3 the steric effect on the halogen the steric effect of the tert-butyl group the electronic effect of the phenolic group the electronic effect of the tert-butyl group 24. The major product(s) of the following reaction is/are NO2 SPh X2 The observed pattern of electrophilic substitution can be explain by F H3C monohalo substituted derivative when X2 = I2 dihalo substituted derivative when X2 = Br2 trihalo substituted derivative when X2 = Cl 2 OH Aqueous Br2 (3.0 equivalents) F (b) OH SO3H NO2 OH Br Br NO2 Br Br SO3H Br H3C SPh H3C SPh (c) Br Br (P) (Q ) OH OH SPh Br (d) Br Br NO2 NO2 (a) P (b) Q Br Br Br SO3H (R) (S) (c) R (d) S Aryl Halides and Phenols 25. In the following reaction, the product (s) formed is/are OH CHCl3 OH– 28. When phenol is reacted with CHCl 3 and NaOH followed by acidification, salicylaldehyde is formed. Which of the following species are involved in the above mentioned reaction as intermediates ? ? CH3 OH H – CCl2 (a) O CHO OHC CHCl2 H 3C (Q) (c) (P) OH CHO CHCl2 CH3 (R) (S) (c) R (minor) H CHCl CHCl2 (d) 29. Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halide due to (a) the formation of less stable carbonium ion (b) resonance stabilisation (c) longer carbon halogen bond (d) sp2-hybridised carbon bonded to halogen Assertion and Reason Read the following questions and answer as per the direction given below : (a) Statement I is correct; Statement II is correct Statement II is the correct explanation of Statement I (b) Statement I is correct; Statement II is correct Statement II is not the correct explanation of Statement I (c) Statement I is correct; Statement II is incorrect (d) Statement I is incorrect; Statement II is correct NaOH(aq)/Br2 the intermediate(s) is/are s O O Br (a) (b) Br 30. Statement I Bromobenzene, upon reaction with Br2/Fe gives Br 1,4-dibromobenzene as the major product. Br s O s O (c) Statement II In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in directing the incoming electrophile. (d) Br 31. Statement I Phenol is more reactive than benzene towards electrophilic substitution reaction. Statement II In the case of phenol, the intermediate carbocation is more resonance stabilised. Br 27. The ether O—CH2 ,when treated 32. Statement I Benzonitrile is prepared by the reaction of chlorobenzene with potassium cyanide. with HI produces (a) O– (d) S (major) OH 26. In the reaction, (b) OH OH (a) P (major) (b) Q (minor) CHCl2 O CH3 H 3C OH O Statement II Cyanide (CN– ) is a strong nucleophile. CH2I (b) CH2OH 33. Statement I Aryl halides undergo nucleophilic substitution with ease. (c) I (d) OH Statement II The carbon halogen bond in aryl halides has partial double bond character. Aryl Halides and Phenols Passage Based Problems Subjective Questions Passage 1 Reimer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes as depicted below. OH ONa ½ OH ½ ½ CHO [I ] CHO aq. HCl 42. Carry out the following conversions. (i) Phenol to aspirin (ii) Benzoic acid to meta-fluorobenzoic acid in not more than three steps. 43. How would you synthesise 4-methoxyphenol from bromobenzene in not more than five steps? State clearly the reagents used in each step and show the structures of the intermediate compounds in your synthetic scheme. 44. What would be the major product in the following reaction? ½ ½ ½ CH3 CH3 CH3 (I) F (III) (II) NaOCH3 D 34. The structure of the intermediate I is ONa ONa CH2Cl NO2 CHCl2 (a) (b) 45. Explain briefly the formation of the products giving the structures of the intermediates. CH3 CH3 ONa ONa ¾ CH2OH —OMe NH3 ¾ CCl3 Br (d) (c) NaNH2 —OMe H 2N 46. Complete the following, giving the structures of the principal organic products CH3 CH3 35. The electrophile in this reaction is (a) · · CHCl + (b) CHCl 2 (c) · · · CCl 2 Me I + Cu heat ---- (d) CCl 3 36. Which one of the following reagents is used in the above reaction ? 47. How will you prepare m-bromoiodobenzene from benzene (in not more than 5-7 steps) ? (a) aq NaOH + CH3Cl (c) aq NaOH + CHCl 3 (b) aq NaOH + CH2Cl 2 (d) aq NaOH + CCl 4 Integer Type Questions 37. The number of resonance structures for N is OH 48. Explain the following in one or two sentences only : “Phenol is an acid, but it does not react with sodium bicarbonate.” 49. Complete the following with appropriate structures : OH OH NaOH N ? CHO Fill in the Blanks 38. Amongst the three isomers of nitrophenol, the one that is least soluble in water is ……… 39. Phenol is acidic due to resonance stabilisation of its conjugate base, namely …… 40. Formation of phenol from chlorobenzene is an example of ……… aromatic substitution. 41. The acidity of phenol is due to the ……… of its anion. 50. A compound of molecular formula C7 H8 O is insoluble in water and dilute sodium bicarbonate but dissolve in dilute NaOH solution and gives a characteristic colour with FeCl 3. On treatment with bromine water, it readily gives a precipitate of C7 H5OBr3. Write down the structure of the compound. 51. Give reason in one or two sentences for the following : “o-nitrophenol is steam volatile whereas p-nitrophenol is not.” Answers 1. (a) 2. (b) 3. (c) 4. (b) 25. (b,d) 26. (b,c) 27. (a, d) 28. (a,d) 5. (c) 6. (a) 7. (c) 8. (c) 29. (b,d) 30. (c) 31. (a) 32. (d) 9. (c) 10. (b) 11. (c) 12. (b) 33. (d) 34. (b) 35. (c) 36. (c) 38. ortho-nitrophenol 13. (a) 14. (d) 15. (a) 16. (c) 37. (9) 17. (d) 18. (a) 19. (a) 20. (b) 39. phenoxide ion 40. nucleophilic 41. phenoxide ion 21. (c) 22. (d) 23. (a,b,c) 24. (b) Hints & Solutions I 1 The major product when m-cresol reacts with propargyl bromide (HC ºº C ¾ CH2Br) in presence of K 2CO3 in acetone is given in the following reaction: O– OH r s –I rds CH3O Step 2 Step 1 1. K2CO3 Br O CH2 C –Brs CH CH2 C CH3 Þ Highly stable CH3¾O (D) Additional +R effect from p-position (DN) CH 3. p-hydroxy benzophenone upon reaction with bromine in carbon tetrachloride gives 3-bromo-4-hydroxy benzophenone. O O Major product Br2 In step 1 K 2CO3 act as a base and abstract H-atom from ¾ OH group. This leads to the formation of substituted phenoxide ion (highly stable). In step 2 substituted phenoxide ion on reaction with Br ¾ CH2 ¾ C ºº CH gives the required product. 2. More stable the carbocation intermediate, higher will be the rate of S N1 reaction. The reaction involving carbocation intermediate formation for the given compounds are as follows: I CH3 r –I s rds HO Br HO ¾ OH group attached on the benzene ring direct the incoming group at ortho and para-positions due to increase in electron density at o and p-positions. ¾ OH group also exhibit –I group that reduces the electron density to some extent at o and p-positions. But overall electron density increases at these positions of the ring due to resonance. Hence, attack of —Br occur at ortho position. Resonating structures are as follows: +OH OH Three positive hyperconjugation 2°-benzyl carbocation Þ Stable +OH s s (A ) +R (AN) +OH I MeO CH3 s –I rds MeO r s 4. Phenol (ArOH) is insoluble in dil. HCl and readily soluble in (BN) I CH3 r Þ More stable s –I rds (C ) H3 C Additional three hyperconjugation (C ¢) OH Þ Less stable –I effect from m-position (B ) H3 C CCl4 NaOH solution. It reacts with Br2/water to give 2, 4, 6-tribromophenol. It readily decolourises the yellow colour of Br2 water. Reactions involved are as follows : OH · Dil. HCl Phenol (Acidic in nature) no reaction Aryl Halides and Phenols – + OH ONa —Br (b) Stable carbocation (Aromatic, 6p system) NaOH · Phenol (Acidic in nature) —Br (c) Sodium phenoxide OH + Br (Unstable carbocation) OH (+R-effect) Br Br —Br (d) Br2 Water + Br N · N Unstable (Aryl carbocation) Phenol Br Br White ppts. (2, 4, 6-tribromophenol) So, only 5. Ethyl magnesium bromide is a Grignard reagent (GR), it Å constitutes C2H5– [ C2H5– M gBr in ether/aprotic medium] which can act as nucleophile as well as strong base. Bromine water (Br2/H2O, red) gets decolourised with phenol derivatives (option, c), anisole derivatives (option, b) etc., as C==C is present outside the ring (aliphatic, not aromatic). d+ d– CºN produces a precipitate of AgBr with AgNO3 solution. 7. In presence of HBr, reactant containing C==C undergoes electrophilic addition reaction and give substituted alkyl halide. On further reaction with alc. KOH, a ,b-elimination takes place that give corresponding diene. The diene undergoes enolisation to give stable product (phenol). Cl –I effect at a-position to C==C Et group can react d+ d– OCH3 Br H Br2/H2O (excess) H ipso-substitution takes place with the carbon bearing H H Br O Mg Br Et MgBr d+ d– Conjugated diene, stable O [Very stable, phenol] 8. In Br OH NºC H OH Br2/H2O (excess) [Et acts as a base] H attain aromaticity No reaction OH (c) Enolisation to OH Br (b) Acidic 'H' —Br Acidic b–H Carbocation formation in the O H electrophilic addition of HBr Alc. KOH –HCl –I-effect at b-position to C==C –HBr (a, b–elimination) Od– EtMgBr H O (Br2/H2O does not react) OCH3 Cl HBr CH2—C—OMe (a) (d) + Br + C2H6 SO3H ¾SO 3H group. After the attack of the electrophilic Br + in the rate determining step (rds) of the ArS E2 pathway desulphonation ( ¾SO 3 ) takes place with a faster rate. (+R) O—H d– (fast) –SO3 –H (Br ) O Br—Br SO3H d+ OH O—H (rds) Br2/H2O —Br O S O Br Br O H Et group can react but (Br2/H2O) does not react - OH Br Br 6. Only ionic halides (X ) give precipitate of AgX with AgNO3 solution. So, an organic bromide able to produce R Å (stable carbocation) and Br - in aqueous solution will give precipitate of AgBr with AgNO3. (i) 2 moles of Br2 (ii) Attack at ortho- 1 and ortho-2 by Br+ (iii) –2HBr Br A white ppt. of 2,4,6-tribromophenol (+ R-effect) (a) —Br (Bond energy also increases) + Br Unstable (Aryl carbocation) 9. Substituted phenols react with aq.NaOH to form sodium phenoxides which on reaction with CH3 I undergoes S N 2 reaction to give 2-methoxy-1-methyl benzene. Aryl Halides and Phenols CH3 CH3 OH CH3 s r O Na d+ CH3 NaOH (aq) Acid-base reaction –H2O 12 The given process is cumene process (Hock process) to synthesise phenol and acetone industrially. In Hock p rocess, Ph — group migrate and release H2O. OCH3 d– I SN2 reaction –NaI Å 2-methoxy-1methyl benzene 10. Acidic strength is inversely proportional to pK a value. The acidity of phenols is due to greater resonance stabilisation of phenoxide ion relative to phenol. Therefore, any substituent which stabilises the phenoxide ion more by dispersal of negative charge will tend to increase the acidity of phenol. Electron withdrawing groups ( ¾ NO2 ) increases the acidic strength of phenol whereas electron donating group ( ¾ OCH3 ) decreases the acidic strength of phenol. In case of ¾ NO2 group attached to phenol, the dispersal of negative charge is more pronounced at o- and p-position than at m-position. OH OH substitution in first step followed by oxidation and dehydration in last step. The most important fact is that, the Br group attached directly to aromatic ring will not undergo substitution in step 1. The road map of the given reaction is as follows: Br (A ) Br H OH O Br Isolated C=C Br2 /CCl4 O O Stabilises the complex by hyperconjugation O -AlCl3 CH3 O O O –H2O H * EtOH (Polar protic – medium as well SN1,–Br as nucleophile) 2°-benzyl carbocation (Stable) Å Br EtOH * –H 15. OH + CO2 + NaOH Followed by acidification X (CH3CO)2O conc. H2SO4 (Catalytic amount) HO H2 O Br * Å Racemised ± as well as inverted (+ and –) O—AlCl3 O O OEt * Br * Anti-addition (electrophilic) of Br2 with the isolated (non-aromatic) C==C –H /+H Friedel-Craft's acetylation reaction Br 14 The road map of the given reaction is : HO—AlCl3 H 3C O H2SO4/D –H2O (Intramolecular dehydration) electrophlic Br AlCl3 CrO3/H+ O it makes ring activating and p-directing It becomes ring activating and m-directing HO—AlCl3 OH It is aryl halide, remains unaffected under ordinary SN2 reaction conditions (D) The reaction follows Ar S E2 (Aromatic substitution pathway) as shown below :4 aq. KOH SN2 Br 11. It is an aromatic electrophilic substitution reaction (ArS E 2). H—O Hydrolysis 13. Key Idea The reaction involves hydrolysis or nucleophilic OMe (C) O + NO2 (B) H Ph-group migration H 2O < NO2 O OH OH < < H Cumene p-nitrophenol > o-nitrophenol and the correct order of the pK a values of give option is OH —O—O H+ O2 hn 5 atm Thus, order of acidic strength of nitrophenol is: OH + —O—O—H ? OH O The very first reaction in the above road map looks like Kolbe’s reaction which results to salicylic acid as O O H OH OH (i) CO2, NaOH (ii) Acidification COOH Salicylic acid (X) Aryl Halides and Phenols The salicylic acid with acetic anhydride [(CH3CO)2 O] in the presence of catalytic amount of conc. H2SO4 undergoes acylation to produce aspirin as 17. In basic (alkaline) solution naphthol exists as naphthoxide ion which is a strong o, p-directing group. O CH3 C OH + CH3 C KOH OCCH3 + CH3COOH Conc. H2SO4 COOH O OH O O O PLAN This problem can be solved by using the concept of synthesis of dye using electrophilic aromatic substitution reaction. III -d COOH -d O -d Acetyl salicylic acid (Aspirin) -d Aspirin is a non-narcotic analgesic (Pain killer). Thus, formation of dye can be shown as N 16. Given, OH OH + Methyl chloroformate [Ph—N NaOH Alkaline solution A Br2 Thus, (d) is the correct choice. B In the above road map, first reaction appears as acid base reaction followed by SN AE (Nucleophilic substitution through Addition and Elimination). Both the steps are shown below (i) Acid base reaction OH H3C Br SPh H3C F 18. F PhS– DMF O– + H2 O – OH N]Cl– N¾Ph OH NO2 NO2 S N 2 reaction bring about inversion of configuration. (ii) S N AE 19. C6H5OH + C2H5O– ¾® C6H5O– O O– O Cl O OCH3 C C O C2H5I ¾¾® C6H5— OC2H5 CH3 CH3 CH3 In the product of SN AE the attached group is ortho and para-directing due to following cross conjugation + KNH2 20. liq. NH3 (Benzyne mechanism) NH2 O d– O1 C 6 5 1 4 2 d– 3 O 2 CH3 Cl OH + Br2 d– 21. Cross conjugation due to which lone pair of oxygen 1 will be easily available to ring resulting to higher electron density at 2, 4, 6 position with respect to group. However from the stability point of view ortho positions are not preferred by substituents as group ¾ O ¾ C ¾ O ¾ CH3 is bulky. || O Hence, on further bromination of SN AE product para bromo derivative will be the preferred product i.e. O C O CH3 O O C O + Br2 Br O CH3 OH CS2 1000 temp. Br In carbon disulphide, no phenoxide ion exist, therefore only monobromination takes place. OH OH Br 22. Br + Br2–H2O Br precipitate + HBr Br2 + H2O ¾® HBrO + HBr It is a reversible reaction, but equilibrium is significantly shifted to left, also indicated as Br2 (aq). Aryl Halides and Phenols 23. PLAN This problem includes concept of effect of steric and electronic effect on reactivity of organic compounds. Os OH O H Steric effect of halogens are as follows Cl 2 < Br2 < I2 Electronic effect of phenolic group directs the approaching electrophile towards ortho and para positions. Tertiary butyl group has large size so it causes steric effect around aromatic nucleus. On the basis of above factors the products of the given reactions are as follows: OH CHCl2 s + CCl2 +H + –H + : CCl2 CH3 CH3 CH3 H OH I X2 = I2 CHO OH– C(CH3)3 C H or OH OH X2 = Br2 CH3 (S) Br C(CH3)3 C(CH3)3 Os OH X2 = Cl2 O Cl Cl C(CH3)3 O H 2O s CH3 Hence, orientation in electrophilic substitution reaction is decided by s CH3 CH3 CCl2 CH3 CHCl2 Q (minor) Thus, (b) and (d) are correct. (a) The steric effect of the halogen (b) The steric effect of the tert-butyl group (c) The electronic effect of the phenolic group 26. — OH in phenol is ortho/para directing group. O– So, (a), (b) and (c) are correct choices. O– PLAN ¾ OH group is activating group and is o- and p-directing. O s Br2 Also, ¾ SO3H is a better leaving group and is knocked out by Br - . OH OH Br O :CCl2 Cl Br2 water (3 equivalents) CH3 Major as stable due to intramolecular H-bonding. Br 24. O O Br Br Br O O Br2 water Br2 SO3H SO3H OH OH Br Br r Br – Br ortho-attack Br O– O– Br + SO3 + + S —O—H Br Br Br Br (major) para-attack 25. Br Br O O O– O Br 27. PLAN Phenolic compounds in alkaline solution react with chloroform (CHCl 3) at a temperature lower than that of CHCl 3 to form ortho-isomer as the major product (due to greater stability resulting from intramolecular hydrogen bonding). s –Cl – – CCl2 HO + H—CCl3 a H2O ¾ CCl3 dichloro carbene —O—CH2— + HI —OH + —CH2I Aryl Halides and Phenols O– 28. O O + H CCl – 2 CCl2 dichlorocarbene O (V) (VI) (IV) (III) O O– O O O O CHCl2 VII (VIII) IX All the above shown nine resonance structures are different. 29. X X 38. Ortho-nitrophenol : Due to intramolecular H-bonding. + O H O– – + N Above resonance makes X- a poor leaving group. Also, the carbon bearing X is sp2- hybridised. 30. Bromo group is deactivating due to dominance of inductive effect over resonance effect. However, orientation is determined by mesomeric effet of —Br. 31. 39. Phenoxide ion 40. Nucleophilic 41. Phenoxide ion : O– OH r H+ + O—H H E + + O + OH OH Intramolecular H-bonding in ortho-nitrophenol E+ Resonance stabilised conjugate base H E resonance stabilisation of intermediate carbocation O– OH OH COO– NaOH 42. (i) CO2 32. Statement I is incorrect, aryl halides do not undergo Kolbe’s reaction nucleophilic substitution reaction with ease. Cyanide ion (CN - ) is a strong nucleophile. H – CCl2 + CCl2 CH3 Aspirin ONa 34. COOH COOH CHCl2 COOH Sn/HCl HNO3 (ii) CH3 COOH Heat nucleophilic substitution with ease. Statement II is correct, resonance introduces partial double bond character to C—X bond. O OCOCH3 (CH3CO)2O 33. Statement I is incorrect, aryl halides do not usually undergo O– COOH H+ H2SO4 NO2 NH2 COOH CH3 Intermediate NaNO2/HBF4 35. Dichlorocarbene is the electrophile as shown above. 36. CHCl 3 + NaOH ¾® · · CCl 2 D (Dichlorocarbene) electrophile 37. O OH F meta-fluorobenzoic acid OCH3 Br Br O NaOH N (I) (II) 43. HNO3 CH3ONa H2SO4 Heat NO2 Zn/HCl NO2 Aryl Halides and Phenols OCH3 OCH3 NO2 OCH3 NH2 NaNO2/HCl H 2O Br2 SnCl2 NaNO2 0°C NaOH Fe HCl HCl/0°C Br + N2Cl– NH2 Br N2+Cl– OH I 4-methoxyphenol 44. KI(aq) OCH3 F CuI Br CH3ONa + NaF heat 48. Phenol is weaker acid than carbonic acid. Nucleophilic aromatic substitution occur which is assisted by electron withdrawing —NO2 group from para position. 45. OCH3 OCH3 Br – + NH2 NH–2 NH3 Benzyne —I 49. H2O Reimer-Tiemann reaction 50. The compound must contain a hydroxy group on the ring with – all three ortho/para positions vacant : OH OH Br more stable carbanion is formed Br + Br2—H2O CH3 CH3 2Cu Heat CH3— Br — —CH3 +2CuI Ullmann’s reaction NO2 51. Intramolecular H-bonding in ortho-nitrophenol lowers its boiling point. No such intramolecular H-bonding is possible with p-nitrophenol and rather it is associated together by intermolecular H-bonding which increases the boiling point. O + 47. CHO H+ + CHCl3 + NaOH OCH3 NH2 46. 2 Me— OH OH NO2 NO2 Br m-bromoiodo-benzene + Conc. HNO3/conc.H2SO4 N H O– O ortho-nitrophenol