26 Carboxylic Acids and Their Derivatives

26 Carboxylic Acids and Their Derivatives Topic 1 Carboxylic Acids Objective Questions I (Only one correct option) 1 Which dicarboxylic acid in presence of a dehydrating agent 5. The compound that undergoes decarboxylation most readily under mild condition is is least reactive to give an anhydride? COOH CH2COOH CH2 CO2H (a) CH2 COOH COOH O (a) (b) (b) COOH CH2 CH2 CO2H O COOH (c) (d) CH2 CH2 COOH C OH OH C 2. In the reaction, PCl 5 Alc. KOH CH3 COOH ¾¾® A ¾¾® B ¾¾¾® C The product C is (a) acetaldehyde (b) acetylene (c) ethylene (d) acetyl chloride 3. The compound that does not liberate CO2 , on treatment with aqueous sodium bicarbonate solution, is (a) benzoic acid (b) benzenesulphonic acid (c) salicylic acid (d) carbolic acid (Phenol) 4. An organic compound A upon reacting with NH3 gives B. On heating, B gives C. C in the presence of KOH reacts with Br2 to give CH3 CH2 NH2 . A is (a) CH3 COOH (b) CH3 CH2 CH2 COOH (c) CH3 ¾ CH ¾ COOH ½ CH3 (d) CH3 CH2 COOH CH2COOH O (c) O LiAlH 4 COOH COOH (d) 6. The carboxyl functional group ( ¾ COOH) is present in (a) (b) (c) (d) picric acid barbituric acid ascorbic acid aspirin 7. In the following reaction sequence, the correct structures of E , F and G are O O Ph OH * Heat [E] (* implies 13 C labelled carbon) O (a) E = Ph * CH3 F= Ph O (b) E = CH3 F= CH3 F= Ph Å G = CHI3 Å G = CHI3 Å G = CHI3 Å G = CH3I ONa ONa Ph O (d) E = * O * Ph O Ph O (c) E = [F] + [G] O * Ph I2 NaOH ONa * O * CH3 F= Ph ONa * Carboxylic Acids and Their Derivatives 8. When benzene sulphonic acid and p-nitrophenol are treated with NaHCO3 , the gases released respectively, are (a) SO2 , NO2 (b) SO2 , NO (c) SO2 , CO2 (d) CO2 , CO2 List–I P. + H2SO4 OH Me Q. 2. [Ag(NH 3 )2 ] OH Ph H 2N H Ph (b) SO2 Cl 2 (d) Cl 2 , H2 O OH + HNO2 Me 11. When propionic acid is treated with aqueous sodium bicarbonate, CO2 is liberated. The C of CO2 comes from (a) methyl group (b) carboxylic acid group (c) methylene group (d) bicarbonate group R. 3. Fehling solution Ph HO Ph Me OH + H2SO4 Me S. (b) H2 O2 (d) CH3 COOH 4. HCHO, NaOH Ph Br 12. Which of the following is basic ? (a) CH3 CH2 OH (c) HOCH2 CH2 OH Me Ph 10. Benzoyl chloride is prepared from benzoic acid by (a) Cl 2 , hn (c) SOCl 2 1. I2 , NaOH Ph HO 9. An enantiomerically pure acid is treated with racemic mixture of an alcohol having one chiral carbon.The ester formed will be (a) optically active mixture (b) pure enantiomer (c) meso compound (d) racemic mixture List–II H Ph OH + AgNO3 Me Matching Type Questions 5. NaOBr 13. The desired product X can be prepared by reacting the major product of the reactions in List-I with one or more appropriate reagents in List-II. (given, order of migratory aptitude : aryl > alkyl > hydrogen) O The correct option is (a) P ® 1; Q ® 2, 3; R ® 1, 4; S ® 2,4 (b) P ® 1, 5; Q ® 3,4; R ® 4, 5; S ® 3 (c) P ® 1, 5; Q ® 3,4; R ® 5; S ® 2,4 (d) P ® 1, 5; Q ® 2, 3; R ® 1,5; S ® 2,3 Ph OH Me Subjective Questions Ph 14. How will you bring about the following conversion? X “Ethanal to 2-hydroxy-3-butenoic acid.’’ Topic 2 Acid Derivatives Objective Questions I (Only one correct option) 2. The major product of the following reaction is O Cl 1. The major product of the following reaction is (i) AlCl3, heat O+ OH CH2OH H2SO4(cat.) O CHCl3 CO2Et O (ii) H2O O CO2H O (a) (a) (b) CO2Et Cl COOH (c) OEt O O O OH OH Cl (b) O Cl O (d) O (c) (d) O COOH O Cl 374 Carboxylic Acids and Their Derivatives 3. The increasing order of the reactivity of the following with 8. Different possible thermal decomposition pathways for LiAlH 4 is O (A ) (B) C2H5 NH2 C2H5 OCH3 O O (C) P –CO2 O (D) C2H5 (a) (b) (c) (d) peroxyesters are shown below. Match each pathway from Column t I with an appropriate structure from Column II and select the correct answer using the code given below the lists. O O Cl C2H5 O C2H5 O O R ( A ) < ( B ) < ( D ) < (C ) ( A ) < ( B ) < (C ) < ( D ) ( B ) < ( A ) < ( D ) < (C ) ( B ) < ( A ) < (C ) < ( D ) R¢ (Peroxyester) Q –CO2 R S 4. The major product obtained in the following reaction is R· + R¢O· R· + X· R· + R¢O· + Carbonyl compound RCO· 2+ R¢O· –CO2 RCO2· + R¢O· R· + X¢· + Carbonyl compound R· + R¢O· –CO2 CO2Et NaOEt/D Column I Column II O O O P. O O (b) (a) Pathway P 1. CO2Et O Pathway Q 2. O (d) (c) C6H5 O O CH3 O CO2Et R. CO2Et Pathway R 3. O C6H5CH2 CH3 O CH3 CH2C6H5 5. The decreasing order of ease of alkaline hydrolysis for the following esters is O COOC2H5 COOC2H5 Cl S. Pathway S 4. C6H5 II I COOC2H5 CH3O COOC2H5 O2N IV III (a) III > II > IV > I (c) II > III > I > IV (b) III > II > I > IV (d) IV > II > III > I 6. The compounds A and B in the following reaction are, respectively HCHO+HCl A AgCN B (a) A = Benzyl alcohol, B = Benzyl isocyanide (b) A = Benzyl alcohol, B = Benzyl cyanide (c) A = Benzyl chloride, B = Benzyl isocyanide (d) A = Benzyl chloride, B = Benzyl cyanide 7. The major product of following reaction is (b) RCONH2 (d) RCH2 NH2 O CH3 CH3 C6H5 10. Which of the following reactants on reaction with conc. NaOH followed by acidification gives the following lactone as the only product? O (i) AIH( i - Bu) 2 (ii) H 2O O Codes P Q R S (a) 1 3 4 2 (b) 2 4 3 1 (c) 4 1 2 3 (d) 3 2 1 4 9. A compound with molecular mass 180 is acylated with CH 3 COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is (a) 2 (b) 5 (c) 4 (d) 6 R ¾ C ºº N ¾¾¾¾¾¾¾®? (a) RCHO (c) RCOOH CH3 O O Q. CO2Et O C6H5CH2 O Carboxylic Acids and Their Derivatives COOCH3 COOH (a) (b) COOH CHO CHO COOH (c) (d) CHO 17. Identify the binary mixture(s) that can be separated into individual compounds, by differential extraction, as shown in the given scheme. NaOH (aq) Binary mixture containing Compound 1 and Compound 2 COOH 11. Benzamide on treatment with POCl 3 gives (a) aniline (c) chlorobenzene (b) benzonitrile (d) benzyl amine CH3MgBr (excess) (a) (c) CH3 H 3C OH (b) H5C2 C2H5 H5C2 OH (d) H3C C2H5 H5C2 OH H5C2 C2H5 H7C3 OH 13. The product of acid hydrolysis of P and Q can be distinguished by OCOCH P : H2C== CH3 3 H3C ; Q: (b) 2, 4-DNP (d) NaHSO 3 14. Hydrogenation of benzoyl chloride in the presence of Pd on BaSO4 gives (a) benzyl alcohol (c) benzoic acid (b) benzaldehyde (d) phenol 15. Acetamide is treated separately with the following reagents. Which one of these would give methyl amine ? (a) PCl 5 (b) NaOH + Br2 (c) Sodalime (d) Hot conc. H2 SO4 Objective Questions I (One or more than one correct options) 16. With reference to the scheme given, which of the given O O CrO3/H+ Compound 1 + (b) C6 H5 COOH and C6 H5 CH2 OH (c) C6 H5 CH2 OH and C6 H5 OH (d) C6 H5 CH2 OH and C6 H5 CH2 COOH 18. Reaction of RCONH2 with a mixture of Br2 and KOH gives R ¾ NH2 as the main product. The intermediates involved in this reaction are (a) RCONHBr (b) RNHBr (c) R ¾ N == C == O (d) RCONBr2 Read the following questions and answer as per the direction given below : (a) Statement I is true; Statement II is true; Statement II is a correct explanation of Statement I. (b) Statement I is true; Statement II is true; Statement II is not the correct explanation of Statement I. (c) Statement I is true; Statement II is false. (d) Statement I is false; Statement II is true. 19. Statement I p-hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid. Statement II o-hydroxybenzoic acid has intramolecular hydrogen bonding. 20. Statement I Acetic acid does not undergo haloform reaction. Statement II Acetic acid has no alpha hydrogen. Passage Based Questions statement(s) about T, U, V and W is/are correct? H 3C Compound 2 Compound 2 Assertion and Reason OCOCH3 (a) Lucas reagent (c) Fehling’s solution Compound 1 + (a) C6 H5 OH and C6 H5 COOH 12. Ethyl ester ¾¾¾® P, the product ‘P’ will be H 3C NaHCO3 (aq) T LiAlH4 (excess) (CH3CO)2O U V W (a) T is soluble in hot aqueous NaOH (b) U is optically active (c) Molecular formula of W is C10H18 O4 (d) V gives effervescence on treatment with aqueous NaHCO3 Passage 1 The reaction of compound P with CH3 MgBr (excess) in (C2 H5 )2 O followed by addition of H2 O givesQ. The compoundQ on treatment with H2 SO4 at 0° C gives R. The reaction of R with CH3 COCl in the presence of anhydrous AlCl 3 in CH2 Cl 2 followed by treatment with H2O produces compound S . [Et in compound P is ethyl group] (H3C)3C CO2Et P Q R S Carboxylic Acids and Their Derivatives O 21. The product S is COCH3 (a) (H3C)3C H3C (b) (H3C)3C CH3 (c) CH3 and O O (W) (V) COCH3 H3COC (d) (H3C)3C H3C HO3S (c) (H3C)3C O CH3 HOH2C (d) CH3 and CH2OH CH2OH (V) (W) COCH3 Passage 3 22. The reactions, Q to R and R to S , are (a) Aromatic sulfonation and Friedel-Crafts acylation (b) Friedel-Crafts alkylation and Friedel-Crafts acylation (c) Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation (d) Dehydration and Friedel-Crafts acylation RCONH2 is converted into RNH 2 by means of Hofmann’s bromamide degradation. O O Cl— Cl NH2 NHBr (i) (ii) O Passage 2 C P and Q are isomers of dicarboxylic acid C4 H4 O4 . Both decolourize Br2 / H2 O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4 . P as well as Q could produce one or more than one form S, T and U. COOH H H COOH OH OH H HO COOH S COOH T H OH COOH U 23. Compounds formed from P and Q are, respectively (a) Optically active S and optically active pair (T, U) (b) Optically inactive S and optically inactive pair (T, U) (c) Optically active pair (T, U) and optically active S (d) Optically inactive pair (T, U) and optically inactive S 24. In the following reaction sequences V and W are respectively H 2 / Ni Q ¾¾¾® V D +V AlCl3 (anhydrous) (i) Zn—Hg/HCl (ii) H3PO4 O (a) O and O (V) – + OM O N Cl H Cl NH2 (v) (vi) In this reaction, RCONHBr is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann’s degradation reaction is an intramolecular reaction. (2006, ´ = ) 25. How can the conversion of (i) to (ii) be brought about? (a) KBr (b) KBr + CH 3 ONa (c) KBr + KOH (d) Br 2 + KOH 26. Which is the rate determining step in Hofmann’s W bromamide degradation? (a) Formation of (i) (b) Formation of (ii) (c) Formation of (iii) (d) Formation of (iv) (1) and (2) undergoes Hofmann’s bromamide degradation? ( W) CH2OH 15 CONH2 CONH2 and CH2OH (V) – N—Br (iii) 27. What are the constituent amines formed when the mixture of O (b) Cl (iv) COOH OH HO H H O Cl N D ( W) (1) (2) Carboxylic Acids and Their Derivatives (a) D * 15 NH2, NH2, formaldehyde (H2 CO) as one of the products when compound (Z) is treated with HBr and subsequently ozonolysed. Mark the C* carbon in the entire scheme. NH2, D 15 NH2 (b) * Ba CO3 + H2 SO4 ¾® X 15 NH2, NH2 * ( C = C14 gas ) (i) Mg/ether LiAlH 4 Z CH2 == CH ¾ Br ¾¾¾¾® Y ¾¾¾® (ii) X (ii) H3 O+ D (c) 15 NH2, 15 (d) 36. Write the structures of the products A and B. NH2 NHD, O ½½ 18 H3O+ CH3 ¾ C ¾ O C2H5 ¾¾® A+ B NH2 37. Explain briefly the formation of the products giving the structures of the intermediates Fill in the Blank O O 28. Formic acid when heated with conc. H2 SO4 produces OC2H5 ………………… C OC2H5 C OC2H5 H 2C True/False 29. The boiling point of propanoic acid is less than that of n-butyl (i) OH– (ii) H+ COOH COOH 30. Hydrolysis of an ester in the presence of a dilute acid is known as saponification. 38. Write the structures of the products : (CH3CO) 2 O, heat CH3 CH2 NH2 ¾¾¾¾¾¾® Integer Type Question 31. The total number of carboxylic acid groups in the product P is O O O Br O alcohol, an alcohol of comparable molecular weight. O NaOEt (i) H3O+ , D P (ii) O3 (iii) H2O2 39. An ester A (C4H8O2), on treatment with excess methyl magnesium chloride followed by acidification, gives an alcohol B as the sole organic product. Alcohol B, on oxidation with NaOCl followed by acidification, gives acetic acid. Deduce the structures of A and B. Show the reactions involved. 40. Complete the following, giving the structures of the principal O organic products Me Subjective Questions CH2Cl DMF (A) C2H5ONa/EtOH C6H5CHO (B) + H3O /D Heat SOCl2 Identify A to D. A NO2 32. KCN + (COOEt)2 + EtONa (i) CH3NH2 conc. H2SO 4 (ii) (COOH)2 + (CH2OH)2 ¾¾¾¾® ¾® B (C) (D) 33. Compound A of molecular formula C9 H7 O2 Cl exists in keto form and predominantly in enolic form B. On oxidation with KMnO4 , A gives m-chlorobenzoic acid. Identify A and B. 34. (±) 2-phenylpropanoic acid on treatment with (+) 2-butanol gives (A) and (B). Deduce their structures and also establish stereochemical relation between them. 35. Identify X and Y in the following synthetic scheme and write their structures. Explain the formation of labelled H O+ 3 (iii) H3CCOCOC6H5 + NaOH ¾¾® C 41. A hydrocarbon A of the formula C8 H10 , on ozonolysis gives compound B (C4 H6 O2 ) only. The compound B can also be obtained from the alkyl bromide C (C3 H5 Br) upon treatment with magnesium in dry ether, followed by carbon dioxide and acidification. Identify A, B and C and also give equations for the reactions. 42. Complete the following sequence of reactions with appropriate structures Red-P A CH3 ¾ CH2 ¾ COOH ¾¾¾® Br 2 (i) Alc. KOH (excess) A ¾¾¾¾¾¾¾® B (ii) H+ Carboxylic Acids and Their Derivatives 43. Which of the following carboxylic acids undergoes decarboxylation easily ? Explain briefly. and 1.36 Å and both the carbon-oxygen bonds in sodium formate have the same value, i.e. 1.27Å.” 51. Write balanced equations for the following reaction : (i) C6H5COCH2COOH (ii) C6H5COCOOH (iii) C6H5CH(OH)COOH (iv) C6H5CH(NH2) COOH “Acetamide is reacted with bromine in the presence of potassium hydroxide.” 52. A liquid X, having a molecular formula C6 H12 O2 is hydrolysed with water in the presence of an acid to give a carboxylic acid Y and an alcohol Z. Oxidation of Z with chromic acid gives Y. What are the structures of X, Y and Z ? 44. Predict the major product in the following reaction : (i) CH MgBr (excess) 3 C6H5 ¾ CH2COCH3 ¾¾¾¾¾¾¾® (ii) H+ 45. In the following reactions, identify the compounds A, B, C and D. (i) PCl5 + SO2 ¾® A + B (ii) A + CH3COOH ¾® C + SO2 + HCl (iii) 2C + (CH3 )2 Cd ¾® 2D + CdCl2 (1994, 1M ´ 4 = 4M) 53. An ester A (C4H8O2) on treatment with excess of methyl magnesium chloride followed by acidification, gives an alcohol B as the sole organic product. Alcohol B, on oxidation with NaOCl followed by acidification, gives acetic acid. Deduce structures of A and B. Show the reactions involved. 54. Complete the following with appropriate structures : C H OH 46. Complete the following sequence of the reactions with appropriate structures (i) (ii) —SO3H —CONH2 2 5 (CH3CO)2 O ¾¾¾® CH3COOH + ? 55. Arrange the following in order of their increasing ease of Fuming hydrolysis : 1. NaOH (Fuse) 2. H+ H2SO4 CH3COOC2H5 , CH3COCl, (CH3CO)2 O, CH3CONH2 56. Give reasons in one or two sentences for the following : + H , H2O P2O5 “Formic acid is a stronger acid than acetic acid.” D 57. Write down the reactions involved in the preparation of the 47. In the following identify the compounds/reaction conditions represented by the alphabets A, B, and C : PCl “Propionic anhydride from propionaldehyde” [AgNO3 , NH4 OH, P2 O5 ] NH 5 3 A ¾¾® C6 H5 COOH ¾¾® P2 O5 following using the reagents indicated against in parenthesis. 58. Give reasons for the following in one or two sentences. H2 /Ni ‘‘Acetic acid can be halogenated in the presence of P and Cl 2 , but formic acid cannot be halogenated in the same way.’’ Why? B ¾¾® C6 H5 CN ¾¾® C 48. Arrange the following as stated: “Increasing order of acidic strength.” ClCH2COOH, CH3CH2COOH, ClCH2CH2COOH, 59. State with balanced equation, what happens when, “Acetic anhydride reacts with phenol in presence of a base.” (CH3 )2 CHCOOH, CH3COOH 60. Write the structural formula of main organic product formed 49. How will you bring about the following conversion? “Ethanoic acid to a mixture of methanoic acid and diphenyl ketone.” when ethyl acetate is treated with double the molar quantity of methyl magnesium bromide and the reaction mixture is poured into water. 61. Write the chemical equation to show what happens when, 50. Give reasons for : “Carbon-oxygen bond lengths in formic acid are 1.23 Å “Ethyl acetate is treated with sodium ethoxide in ethanol and the reaction mixture is acidified”. Answers Topic 1 5. (b) 6. (c) 7. (a) 8. (a) 1. (b) 2. (c) 3. (d) 4. (d) 9. (b) 10. (c) 11. (b) 12. (a) 5. (b) 6. (d) 7. (c) 8. (d) 13. (c) 14. (b) 15. (b) 16. (a,c,d) 9. (d) 10. (c) 11. (d) 12. (a) 17. (b,d) 18. (a,c) 19. (d) 20. (c) 13. (d) 21. (a) 22. (c) 23. (b) 24. (a) Topic 2 25. (d) 26. (d) 27. (b) 1. (d) 2. (d) 3. (a) 4. (b) 28. H2O and CO gas 31. (2) 29. False 30. False Hints & Solutions Topic 1 Carboxylic Acids NOTE l COOH 1. Heating of (CH2)4 COOH (adipic acid) with a dehydrating agent, l decarboxylates ( ¾CO2 ) to give a ketone (cyclopentanone), not an anhydride. COOH COOH D –CO2 –H2O Ordinary carboxylic acid require soda-lime catalyst for decarboxylation. Final step of decarboxylation in the above shown mechanism involve tautomerism, therefore, for decarboxylation of b-keto acid by above mechanism, the acid must contain an a-H]. 6. Structures of the various compounds are O OH Codes O NO2 O2N P ® 1, Q ® 3, R ® 4, S ® 2 Thus, (a) is the correct choice. HN NH O 2. This problem is based on successive reduction, chlorination and NO2 elimination reaction. To solve such problem, use the function of the given reagents. (i) LiAlH 4 causes reduction (ii) PCl 5 causes chlorination (iii) Alc. KOH causes elimination reaction picric acid (A) OH COOH HO LiAlH O barbituric acid (B) NO2 ¾4® CH3CH2OH CH3COOH ¾ ¾ ¾ O CH3 (A ) Alc.KOH - HCl ¾¾® CH3CH2Cl ¾¾¾® CH2 == CH2 (B ) 3. Na + + HCO3- PLAN NaHCO 3 s HCO-3 HCO-3 O is decomposed by acid releasing CO2 aspirin (D ) ascorbic acid (C ) (C ) Ethylene O O Heat * Ph—C—CH3 7. + H+ ¾® H2O + CO2 Ph * OH E If acid is stronger than HCO-3 then CO2 is released. Phenol is less acidic and thus does not liberate CO2 with NaHCO3. O ½½ NH3 4. CH3CH2 ¾ C ¾ OH ¾® F + NaHCO3 8. + H2O + CO2­ OH + NaHCO3 (C ) Br 2 , KOH ¾¾¾¾¾¾¾® CH3 ¾ CH2 ¾ NH2 Hoffmann’ s bromamide reaction R C H O D C –CO2 CH2 OH R C CH2 Tautomerism Six membered cyclic transitionstate of a b-keto acid R C + H2O + CO2­ O½ - H+ O R ¢ ¾¾® R ¾ C — OH ·· ½ OR ¢ Reaction occur at planar sp2 carbon giving racemic mixture of product. O O O H ½½ 9. R ¾ C ¾ OH + ONa O2N O 2N 5. It is a b-keto acid which undergo decarboxylation in very mild condtion, i.e. on simple heating. This occur through a six member cyclic transition state as O G SO3Na O ½½ D ¾® CH3 ¾ CH2 ¾ C ¾ NH2 (B ) I2 NaOH * Ph—COONa + CHI 3 SO3H (A ) CH3CH2COONH4 O OH HO PCl 5 CH3 10. C6H5COOH + SOCl 2 ¾® C6H5 ¾ COCl * 11. CH3 ¾ CH2 ¾ COOH + NaHCO3 ¾® CH3CH2COONa * + H2 O + CO2 Carboxylic Acids and Their Derivatives 12. Ethanol is the weakest acid among these, hence it is most basic. The reaction proceed as Ph 13. (d) HO For P, i.e. Ph Me HO OH + H2SO4 Ph Me Me Ph Ph The correct match is 1 i.e., I2 , NaOH and 5 i.e., NaOBr H aO /N The reactions proceed as Ph Ph Ph I2 O Pinacol + H2SO4 Pinacolone Me OH rearrangement Ph Me (Methyl shift) I2 Me Ph For Q, i.e. NH2 Ph Ph Ph H OH + HNO2 OH + AgNO3 Me Ph Deamination + HNO2 rearrangement H2N Ph Me OH (Methyl shift) h Fe Ph For R, i.e. HO Me H Ph Me g lin o ss Me Ph t lu Both oxidation Reactions [Ag(NH3)2]OH (Tollens's reagent) ‘X’ Ph The correct match is 1, 5 again. H ' OH Ph O Me OH H Rest procedure is same as seen for Q above i.e., via oxidation. OH ½ D NaOH 14. CH3CHO + HCHO ¾¾® H ¾ C — CH2 ¾ CHO ¾® ½ Aldol H OH ½ HCN CH2 == CH ¾ CHO ¾¾® CH2 == CH ¾ CH ¾ CN Cyanohydrin OH OH ½ ½ H+ CH2 == CH ¾ CH ¾ CN ¾® CH2 == CH ¾ CH ¾ COOH H2 O + H2SO4 dibromination rearrangement Ph O ion O Me + AgNO3 Me Br H Ph Me Ph OH The reaction proceed as OH The correct match is 2 i.e. [Ag(NH3)2]OH and 3 i.e., Fehling’s solution. The reactions proceed as Ph H The correct match is 2, 3. ‘ X’ Ph OH For S, i.e. Br O Ph NaOBr X NaOBr Ph Me Both haloform reactions O Ph Me Ph Me Both haloform reactions /N aO H HO Me O Pinacol + H2SO4 Pinacolone OH Me rearrangement Ph Me (Phenyl shift) Carboxylic Acids and Their Derivatives Topic 2 Acid Derivatives 1. Given reaction involves acidic hydrolysis of esters followed by the intramolecular cyclisation. The chemical equation is as follows: OH O ½½ C2H5 ¾ C ¾ Cl > C2H5 OH CH2OH H2SO4(cat.) CH2OH CHCl3 COEt C O O OH This leaving group ability (weak conjugate base) corresponds directly to the reactivity order. Hence, the correct reactivity order is: +C2H5OH O O O Most reactive C2H5 (D) (C) OH O O ½½ ½½ > C2H5 ¾ C ¾ OCH3 > CH3 C NH2 OH Least reactive (B) CH2 OH –H2O (A) O C OH O O 2. The major product of the given reaction is (d). This reaction proceed via Friedel-Craft acylation. Here, ¾ Cl group present on chlorobenzene is ortho and para-directing. It can be easily understood by resonating structures of chlorobenzene. r Cl Cl r Cl r Cl Note The -I effect of ¾ Cl and + m effects of O ½½ · · ¾ N H2¾ O ¾ C ¾ C2H5 and ¾ OCH3 groups while attached to ¾ C ¾ group to be reduced. Also add to the ½½ O group leaving ability. Cl 4. In presence of strong base, acidic H can easily be removed that s s s result in formation of anion. The resulting anion undergoes intramolecular nucleophilic addition which on hydrolysis followed by heating gives the required product. The given reaction proceed as follows : More acidic O O H H CO2Et + C O + AlCl3 O – O----Al + Cl3 O O (Nucleophilic addition) s H –H O cidic b more a CO2Et O H2 O s –OH CO2Et OH H2O COOH Cl O a, b-eliminatioin C C—O – + Cl —AlC l3 O 3. All the given compounds are acid derivatives, thus contain carbonyl group in them. LiAlH4 reduces these compounds through nucleophilic substitution via addition elimination (S NA E) reaction. The rate of reaction depends upon the following factors : (i) Size of alkyl group. (ii) Steric hinderance around the > C == O group. (iii) (+ ) ve charge on the C-atom of > C == O group. The alkyl groups are more or less same in the given compounds. Thus, the reactivity order of given compounds depends upon 2nd and 3rd factor written above. The cumulative effect of these two factors results to leaving group ability (LGA ) of the substituents in the following order : O ½½ Cl> O- ¾ C ¾ CH3 >O ¾ CH3 > NH2 Good leaving group s O O O Poor leaving group CO2Et s O O EtO Cl O s r EtONa –EtOH CO2Et –EtOH –OHs 5. Alkaline hydrolysis of an ester (carboxylic acid derivative) follows acyl S N2 mechanism. O s O O OH C + ROs C OR OH C + ROH Os Rate of S N 2 mechanism depends on the polarity of C O group of ¾COOR group. Electron withdrawing group (- R > - I ) increases the rate of S N2 reaction whereas electron donating group (+ R > + I ) decreases the rate of S N 2 reaction. Here, the nature of functional groups attached para to the benzene ring are: — NO2 > —Cl > — OCH3 (- R ) (- I ) (+ R ) So, the order of hydrolysis will be, III > II > I > IV (- R ) (- I ) (+ R ) Carboxylic Acids and Their Derivatives 6. The mechanism of the given reaction is as follows: SN2 it is electrophilic substitution Å y [CH2==OH]Cl , (Ar SE2): Haloformylation CH2 OH –HCl [CH2==O + H CH2Cl –H2O More covalent (Ag—CººN ) –AgCl HCl (SN1) Cl] CH2CN + Ag CN (A ) CH2N—C B (Major) (Benzyl isocyanide) (Minor) (Benzyl chloride) Thus, both benzyl cyanide and benzyl isocyanide are the products of reaction but benzyl isocyanide being the major product gives the correct option as (c). 7. Key Idea DIBAL-H is diisobutyl aluminium hydride, [(CH3 )2 CHCH2 ]2 AlH. It is a selective reducing agent. It reduces carboxylic acids, carboxylic acid derivatives and nitriles into aldehydes. It is an electrophilic reducing agent. The mechanism of the reaction is as follows: H Al (sp2) s Al (DIBAL-H) R R C C r N –NH3 –DIBAL-H N H Hydride donor (Reducing agent) N H 2O Al (An electrophilie) R CH O [Aldehyde] CH (An imine) (A nucleophile) [Nitrile] R So, R ¾ C H is the correct answer. ½½ O 8. PLAN T his problem can be solved by using the stability of radical obtained after frag mentation of peroxyester. Allylic radical are more stable than alkyl radical, so when there is a possibility of formation of allyl radical, it will undergo fragmentation through formation of allyl radical. i.e. fragmentation produces stable radical. On the basis of stability of radical, fragmentation can be done as Column I Column II Explanation O P. 1. O C6H5H2C O CH3 C6H5 — CH2+ CO2+ CH3O O O Q. 3. C6H5H2C CH3 CH3 O C6H5 — CH2+ CO2+ Ph—CH2 — C — CH3 O CH3 CH2C6H5 O R. 4. C6H5 O CH3 O CH3 O C6H5 — CO2+ CH3—C—CH3 C6H5 C6H5 O S. 2. C6H5 C6H5 —CO2+ CH3O O O CH3 Ph —CH2+ CH3—CO—CH3 C6H5+ CO2 –CO2 Ph + CH3—CO—Ph + CH3 + CO2 Carboxylic Acids and Their Derivatives O O ½½ ½½ 9. R ¾ NH2 + CH3 ¾ C ¾ Cl ¾¾® R ¾ NH ¾ C ¾ CH3 16. (a) T undergoes an ester hydrolysis in hot aqueous alkali as O (- HCl) Since, each ¾ COCH3 group displace one H atom in the O ½½ reaction of one mole of CH3 ¾ C ¾ Cl with one ¾ NH2 group, the molecular mass increases with 42 unit. Since the mass increases by (390 - 180) = 210, hence the 210 number of ¾ NH2 group is = 5. 42 CHO 10. COO – OH – CHO O H 3C Soluble in aqueous NaOH (b) LiAlH4 reduces ester to alcohol as “U ” No chiral carbon optically inactive. (c) U on treatment with excess of acetic anhydride forms a diester as (d) U on treatment with CrO3 |H+ undergo oxidation to diacid which gives effervescence with NaHCO3. O H+ H3C O O O ½½ POCl 3 11. C6H5 ¾ C ¾ NH2 ¾¾® C6H5 ¾ CN OMgBr O ½ ½½ CH3 MgBr 12. CH3 ¾ C ¾ OC2H5 ¾¾¾® CH3 ¾ C — CH3 excess ½ CH3 CH3 ½ H 2O ¾® CH3 ¾ C — OH ½ CH3 OH O ½ ½½ 13. P ¾¾® CH3 ¾ C == CH2 r CH3 ¾ C ¾ CH3 H+ NaOH ¾¾® CHI3 ¯ I2 O O ½½ ½ ½ R ¾ C ¾ NHBr + OH- ¾® R ¾ C ¾ N ¾ Br (B ) Yellow H 2O ¾® R ¾ N == C == O + Br (C ) 19. p-hydroxy benzoic acid has higher boiling point than o-hydroxy Fehling solution ¾¾¾¾¾® Cu 2 O ¯ benzoic acid because former prefers intermolecular H-bonding while the latter prefer intramolecular H-bonding. Red 14. ‘V’ (A ) Q ¾¾® CH3 ¾ CH== CH ¾ OH r CH3 CH2 OH CHO H2 Pd/BaSO4 CO2 must form salt with the given base so that the salt will be extracted in aqueous layer leaving other component in organic layer. (a) Both phenol and benzoic acid forms salt with NaOH, hence this mixture can’t be separated. (b) Benzoic acid forms salt with NaOH while benzyl alcohol does not, hence the mixture can be separated using NaOH. Also benzoic acid forms salt with NaHCO3 but benzyl alcohol does not, hence NaHCO3 can be used for separation. (c) Neither benzyl alcohol nor phenol forms salt with NaHCO3, mixture cannot be separated using NaHCO3. (d) C6H5CH2COOH forms salt with NaOH, C6H5CH2OH does not, hence mixture can be separated using NaOH. C6H5CH2COOH forms salt with NaHCO3. C6H5CH2OH does not, hence mixture can be separated using NaHCO3. O O ½½ ½½ 18. R ¾ C ¾ NH2 + OH + Br2 ¾® R ¾ C ¾ NHBr H+ COCl NaHCO3 OH COOH 17. For separation by differential extraction one of the component H+ H2 O CH3O C ONa HO T U + CrO3 CH2OH NaOH(aq) hot ‘‘Rosenmund reduction’’ O ½½ 15. CH3 ¾ C ¾ NH2 + Br2 + NaOH ¾® CH3NH2 “Hofmann’s bromamide reaction”. H HOOC— O C— —O H—O intermolecular H-bonding OH C O O H intramolecular H-bonding —OH Carboxylic Acids and Their Derivatives O ½½ 20. Compounds with CH3 ¾ C ¾ or CH3 ¾ CH(OH)–group gives haloform reaction but this reaction is given only by aldehydes, ketones and alcohols, so acetic acid does not give haloform reaction. However acetic acid has three a-H, therefore, statement I is true but statement II is false. COOH H—C H—C—OH H2O + O C—H dilute alkaline + OH—C—H HO—C—H KMnO4 Passage-1 COOH COOH H—C—OH COOH COOH COOH (fumaric acid) (T) d (+) (U) l (–) OH T and U (in 1 : 1 molar ratio) form optically inactive (racemic mixture) due to external compensation. (I) CH3MgBr(ex) OEt+ (ii) H2O Q O 24. H+/0°C PLAN Ni / H2 reduces (C == C) bond. Benzene undergoes Friedel-Crafts reaction Zn-Hg/HCl reduces carbonyl group (Clemmensen reduction) CH3COCl O AlCl3 CHCOOH R(F.C. Alkylation) O CH2COOH Ni/H2 CH2COOH HOOCHC S(F.C. Acylation) D CH2C O CH2C O 22. (a) succinic anhydride (V) 22. (c) 23. O O PLAN A lkenes decolourise Br2 water dil× KMnO4 c i s -isomer ¾¾¾¾® M e s o isomers by s y n addition CH2—C + dil× KMnO4 t r a n s -isomer ¾¾¾¾® d (+ ) and l(- ) isomers by s y n a d d it io n t h u s , r a c e m ic m ix t u r e . Formation of anhydride from dicarboxylic acid indicates c i s -isomer. P and Q are isomers of dicarboxylic acids. O CH2COOH O Å H3PO4 Zn–Hg/HCl Clemmensen HO Br water D anhydride P ¾® Thus, P is cis-isomer. O Br—CHCOOH D Br—CHCOOH CHCOOH COOH H 2O + O H—C—OH H—C dilute alkaline KMnO4 H—C—OH (P) (maleic acid) 25. Cl— CHCO O —C—NH2 + KOH + Br2 O (i) CHCO Cl— COOH H—C COOH O (W) O P and Q have ( C ==C ) bond Br2 Water CH2 CH2—C 2 decolourised P , Q ¾¾¾¾® CHCOOH C anhydrous AlCl3 COOH (S) Optically inactive due to internal compensation of rotation (meso-isomer) —C—NHBr an intermediate (ii) 26. Rearrangement of (iii) to (iv) is the rate determining step : O Slow Cl— Cl— —C – N—Br (iii) —N==C==O + Br– (iv) 27. The rate determining step of Hofmann’s bromamide reaction is unimolecular rearrangement of bromamide anion (iii) and no cross-products are formed when mixture of amides are taken. Carboxylic Acids and Their Derivatives 15 —CONH2 + D 33. Compound A of molecular formula C9 H7 O2Cl exist in keto and KOH Br2 —CONH2 15 (i) —NH2 + predominantly in enolic form B. Hence, A must be a carbonyl compound which contain a-H. Enolic forms of B predominates because of presence of intramolecular H-bonding. —NH2 Cl Cl [O] Cl tautomerism D D 28. HCOOH + H2SO4 ¾® H2O + CO ­ COOH conc. O O 29. Propanoic acid has higher boiling point than n-butanol because 30. Saponification is hydrolysis of ester in presence of dilute base rather in presence of dilute acid. 31. H CHO of more exhaustive H-bonding in former case. H O 34. The two stereoisomers of 2-phenyl propanoic acid in the racemic mixture are : PLAN R eactant is cyclic anhydride and chang es to dicarboxylic acid on hydrolysis. Also there is decarboxylation on heating if there is keto group w.r.t ¾ COOH group. Ozonolysis cleaves (C== C) bond and H2O2 oxidises ¾ CHO to ¾ COOH group. O O O O * COOH COO CH(CH3 )Et * ½ ½ CH3 C H(OH)Et H ¾ C — Ph ¾¾¾¾¾¾® H ¾ C — Ph ½ ½ CH3 CH3 (A ) * O (i) H3O+ O O COOCH(CH3 )Et COOH * ½ ½ CH3C H(OH)Et and Ph ¾ C — H ¾¾¾¾¾® Ph ¾ C — H ½ ½ CH3 CH3 OH OH O O Anhydride (B ) O O Here A and B are diastereomers. D decrboxylation O3 OHC O ½½ 35. H2C == CH ¾ Br ¾¾¾¾® H2C == CH ¾ C ¾ OH OHC (i) Mg/ether (ii) X (iii) H3O+ O O O Y LiAlH4 Z HOOC (P) O O 18 ½½ 18 ½½ H3O+ 36. CH3 ¾ C ¾ OC2H5 ¾¾® CH3 ¾ C ¾ OH + C2H5 ¾ OH O (A ) Thus, number of ¾COOH groups in P = 2. CH2CN KCN 32. C C (A) C6H5—CH—CH—C6H5 H3O+/D C6H5 C6H5 C==C HOOC CN H (C) CH3NH2 C6H5 CH3HN—C H (D ) NaOEt – CH COOC2H5 COOC2H5 OC2H5 OC2H5 Br O O CH3 COOC2H5 C6H5 C==C O OC2H5 (i) OH– (ii) H+ H5C2O—C—CH—CH—COOC2H5 (B) SOCl2 O 37. CH2 C6H5CHO OH (B ) O C2H5ONa/EtOH DMF * ¾¾® H2 C==CH ¾ CH2 OH H2O2 HOOC CH2Cl * 38. CH3CH2NH2 O ½½ ¾¾¾¾® CH3CH2NH ¾ C ¾ CH3 (CH3 CO) 2 O Heat COOH COOH Carboxylic Acids and Their Derivatives CH3 O ½ ½½ MeMgBr /H2 O 39. H ¾ C ¾ O ¾ CH — CH3 ¾¾¾¾¾® CH3 ¾ CH ¾ OH excess A ½ CH3 B NaOCl ¾¾¾® CH3COOH + CHCl 3 CH3 40. (i) NO2 45. (i) PCl 5 + SO2 ¾® POCl 3 + SOCl 2 B (COOEt)2 NO2 (A ) O ½½ (iii) 2CH3COCl + Cd(CH3 )2 ¾® 2CH3 ¾ C ¾ CH3 + CdCl 2 O SO3H CH2—C—COOEt SO3H NaOH H2SO4 fusion NO2 A COOH CH2 + (ii) COOH H+ CH2 O H+ P 2O 5 H2O O B CH3 O O ½ ½½ ½½ NaOH (iii) CH3 ¾ C ¾ C ¾ C6H5 ¾¾® C6H5 ¾ C ¾ COOH H2 O ½ OH O ½½ 47. C6H5COOH ¾¾® C6H5COCl ¾¾® C6H5 ¾ C ¾ NH2 PCl 5 NH3 A P2O5 B H2 /Ni ¾¾® C6 H5 ¾ CN ¾¾® C6 H5 CH2 NH2 C C 41. B is C3H5COOH and A is C3H5 ¾ C ºº C ¾ C3H5 48. Electron withdrawing inductive effect increases acid strength Also A on ozonolysis gives B indicates that there is no olefinic bond in C3H5-unit of A and it is cyclopropyl group. while electron donating inductive effect decreases acid strength. Þ H3C ¾ CH ¾ COOH < CH3CH2COOH < CH3COOH ½ CH3 + I-effect + I-effect and 42. COOH O (ii) HO OH SO3H CN CONH2 O OH fuming 46. (i) HO A (ii) SOCl 2 + CH3COOH ¾® CH3COCl + SO2 + HCl – CH2 EtO – OH O ½ ½½ (i) CH3 MgBr 44. C6H5CH2 ¾ C ¾ CH3 ¾¾¾¾® C H ¾ CH ¾ C — CH3 6 5 2 (ii) H+ ½ CH3 A= —C==C— B= —COOH + I-effect —Br C= < C H2 ¾ CH2COOH < C H2 ¾ COOH ½ ½ Cl Cl - I-effect - I-effect red-P CH3CH2COOH ¾¾¾® CH3 ¾ CH ¾ COOH Br 2 ½ A Br (HVZ reaction) (greater distance) (i) alc.KOH ¾¾¾¾® CH2 ==CH ¾ COOH (ii) H+ B C C6H5 H OH O – CO2 C6H5—C==CH2 enol O O ½½ 49. CH3 ¾ COOH ¾® CH3COCl ¾¾® C6H5 ¾ C ¾ CH3 PCl 5 C6 H6 AlCl 3 43. A b-keto acid undergo very fast decarboxylation O + I-effect O C6H5—C==CH3 keto OH C6H5 ½ ½ C6 H5 MgBr H+ ¾¾¾¾® C6H5 ¾ C — CH3 ¾® C6H5 ¾ C==CH2 H2 O D ½ C6H 5 O ½½ ¾¾¾® C6H5 ¾ C ¾ C6H5 + H ¾ COOH O3 /H2 O2 Carboxylic Acids and Their Derivatives Hydrolysis is an example of nucleophilic acyl substitution, hence the reactivity towards hydrolysis is : 50. Both formic acid and sodium formate exhibit the phenomenon of resonance as : O O½½ · · ½ + H ¾ C ¾ O ¾ H ¬® H ¾ C == O ¾ H ·· CH3CONH2 < CH3COOC2H5 < (CH3CO)2 O < CH3COCl ·· O O½½ ½ H ¾ C ¾ O- ¬® H ¾ C == O 56. O ½½ CH3 ¾ C ¾ OH < does not affect acidity. +I -effect decreases acid-strength In formic acid, the extent of delocalisation is less compared to sodium formate because of charge separation in the former case. Due to this reason, the bond length between carbon and sp3 oxygen in formic acid is slightly greater than the same between carbon and sp2 oxygen. In formate ion, there is no separation of charge and both the resonance structures are equivalent giving equal bond length of both carbon oxygen bonds. O ½½ H ¾ C ¾ OH NH4 OH 57. CH3CH2CHO + AgNO3 ¾¾® CH3CH2COOH P2O5 ¾¾® (CH3 CH2 CO)2 O D Red-P 58. CH3 ¾ COOH + Cl 2 ¾¾® CH 2 ¾ COOH (HVZ reaction) ½ Cl For this reaction to occur, presence of a a-H is essential requirement. Formic acid does not has any a-H, fails in HVZ reaction. NaOH 51. CH3CONH2 + Br2 ¾¾® CH3NH2 + Na 2CO3 Hoffman bromamide reaction 52. X is an ester and both its acid and alcohol fragments have same number of carbons. Hence, X is : 59. OH + (CH3CO)2O O + H CH3CH2COOCH2CH2CH3 ¾¾® CH3CH2COOH X H2 O Y + CH3CH2CH2OH Z [O] Z ¾¾® CH3CH2COOH OMgBr O ½ ½½ 60. CH3 ¾ C ¾ OC2H5 + 2CH3MgBr ¾® CH3 ¾ C ¾ CH3 ½ CH3 Y OH ½ ¾¾® CH3 ¾ C — CH3 ½ CH3 O CH3 ½½ ½ MeMgCl H+ /H2 O 53. H ¾ C ¾ O ¾ CH — CH3 ¾¾¾® ¾¾¾® A H 2O excess OH ½ CH3 ¾ C — CH3 ½ H B B + NaOCl ¾® CHCl 3 + CH3COOH O ½½ C2H5 OH 54. (CH3CO)2 O ¾¾¾® CH3COOH + CH3 ¾ C ¾ OC2H5 55. Among acid derivatives, the reactivity towards nucleophilic acyl substitution is in the order of : Amide < Ester < Anhydride < Acid chloride CH3 O OH– C H O- - 2 5 61. CH3COOC2H5 ¾¾® CH2COOC2H5 O ½½ CH3 ¾ C ¾ OC2 H2 O ½½ ¾¾¾¾¾¾® CH3 ¾ C ¾ CH2 ¾ COO2H5 Claisen condensation O ½½ ¾¾¾® C2H5OH + CH3 ¾ C ¾ CH2COOH H+ H2 O