25 Aldehydes and Ketones

25 Aldehydes and Ketones Objective Questions I (Only one correct option) 4. The major product of the following reaction is 1. The major product(s) obtained in the following reaction is/are O Cl (i) KOtBu (ii) O3/Me2S O (b) CHO and OHC ¾ CHO (a) OHC (b) OHC O O CHO (d) (c) CHO (c) OHC OtBu 5. An organic compound neither reacts with neutral ferric (d) OHC chloride solution nor with Fehling solution. It however, reacts with Grignard reagent and gives positive iodoform test. The compound is CHO 2. In the following reaction, HCl O Carbonyl compound + MeOH acetal Rate of the reaction is the highest for: - (a) Acetone as substrate and methanol in excess (b) Propanal as substrate and methanol in stoichiometric amount (c) Acetone as substrate and methanol in stoichiometric amount (d) Propanal as substrate and methanol in excess OH O OH O C2H5 O CH3 6. In the following reactions, products A and B are (b) H O O O CH3 CH3 O H3 C H3 C CH2 CH3 O A= HO [B] O CH3 (a) H CH3 H3O+ D [A ] (d) O C2H5 O H 3C (a) H CH3 (d) CH3 O NaOH D OHC CH3 (b) H (c) CH3 O CH3 (a) 3. The major product obtained in the following reaction is O (2) Conc. H2 SO4 /D O (a) Br (c) (1) t-BuOK CH3 CH3 ; B= CH3 Dil. NaOH [A ] Aldehydes and Ketones O CH3 CH3 ; B= CH3 A= (b) O O CH3 HO CH2 CH2 OH (a) O OH CH3 C H ; B= CH3 O (d) H2 C H A= H3C H 3C CH3 CH2 CH2 OH H (d) conversion will be H O HO CH3 CH3 7. The major product obtained in the following reaction is CHO H3C OH OH CH3 (a) [Ag(NH3 )2 ]+ OH- , H+ / CH3 OH, CH3 MgBr LiAlH4 (excess) (b) CH3 MgBr, H+ / CH3 OH, [Ag(NH3 )2 ]+ OH- CH3 (c) CH3 MgBr, [Ag(NH3 )2 ]+ OH- , H+ / CH3 OH NO2 (d) [Ag(NH3 )2 ]+ OH- , CH3 MgBr, H+ / CH3 OH OH OH (a) OCH3 OH C ; B= H 3C C 10. The correct sequence of reagents for the following O C CH2 (c) H3 C CH3 OH O O H CH2 (b) O HC (c) A= 3 H 3C OH OH 11. The major product of the following reaction sequence is O (b) CH3 NO2 OH CH3 (i) HCHO (Excess)/NaOH, Heat NH2 OH OH (ii) HCHO/H+ (Catalytic amount) OH O (c) (d) CH3 O O O OH (b) NH2 OH 8. The correct match between item ‘I’ and item ‘II’ is Item ‘I’ (Compound) HO Item ‘II’ (Reagent) O O (c) (A) Lysine (P) 1-naphthol (B) Furfural (Q) Ninhydrin (C) Benzyl alcohol (R) KMnO4 (D) Styrene (S) Ceric ammonium nitrate (d) OH 12. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis? CH3 C S Q S R ½ D P S R S (a) C ½ (c) OCH3 MeOH X ½ NaBH4 CH3 CH3 CH3 CH3 CH3 (d) ½ ½ O CH2 ½ CH3 (b) 9. The major product ‘X’ formed in the following reaction is O CH3 ½ B R P P P ½ Codes A (a) Q (b) R (c) Q (d) Q OH CH3 (a) NO2 OH O Aldehydes and Ketones 13. The major product in the following reaction is (2014 Adv.) 17. Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound E. Compound E on further treatment with aqueous KOH yields compound F. Compound F is O (i) CH3MgBr, dry ether, 0°C Cl (ii) Aqueous acid CH3 O (a) H 3C CH3 H2C O (c) —COOH (d) —CHO COOH NH2OH to form oxime (a) two different oximes are formed (b) three different oximes are formed (c) two oximes are optically active (d) all oximes are optically active CH2 CH3 (d) O (b) 18. The smallest ketone and its next homologue are reacted with CH3 (c) —CHO CH3 OH (b) (a) 19. Butan-2-one can be converted to propanoic acid by which of CH3 the following ? (a) NaOH, NaI/H+ (b) Fehling’s solution (c) NaOH, I2/H+ (d) Tollen’s reagent 14. The most suitable reagent for the conversion of R ¾ CH2 ¾ OH ® R ¾ CHO is (a) KMnO4 (b) K 2 Cr2 O7 (c) CrO3 (d) PCC (pyridinium chlorochromate) CHO 15. The major product H in the given reaction sequence is 95% H2 SO4 s CN G ¾¾¾¾® H CH3 ¾ CH2 ¾ CO ¾ CH3 ¾¾® 20. CH3COONa + (X ) MeO COOH (a) CH3 ¾ CH == C ¾ COOH ½ CH3 MeO (b) CH3 ¾ CH == C ¾ CN ½ CH3 What is X ? (a) CH3 COOH (c) (CH3 CO)2 O OH ½ (c) CH3 ¾ CH2 ¾ C ¾ COOH ½ CH3 21. The order of reactivity of phenyl magnesium bromide with the following compounds is O O ½½ ½½ (d) CH3 ¾ CH == C ¾ CO ¾ NH2 ½ CH3 H3 C transformation is OH Conc. aq. NaOH HO (a) 1 (c) 3 (b) 2 (d) 4 CH3 (I) 16. The number of aldol reaction(s) that occurs in the given CH3CHO + 4HCHO (b) BrCH2 COOH (d) HOC ¾ COOH OH OH H3 C O ½½ H Ph (II) Ph (III) (a) (II) > (III) > (I) (b) (I) > (III) > (II) (c) (II) > (I) > (III) (d) All of the above 22. A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives (a) benzyl alcohol and sodium formate (b) sodium benzoate and methyl alcohol (c) sodium benzoate and sodium formate (d) benzyl alcohol and methyl alcohol Aldehydes and Ketones 23. The appropriate reagent for the following transformation: O CH2CH3 CH3 HO (a) Zn (Hg), HCl (c) H2 /Ni Objective Questions II (One or more than one correct option) 33. Reagent(s) which can be used to bring about the following transformation is (are) HO O (b) NH2 NH2 , OH– (d) NaBH4 O C O O H O 24. Which of the following has the most acidic hydrogen? (b) 2, 4-hexanedione (d) 2, 3-hexanedione 25. The enol form of acetone, after treatment with D2 O, gives OH O COOH (a) 3-hexanone (c) 2, 5-hexanedione O C COOH (a) LiAlH 4 in ( C2 H5 )2 O (c) NaBH 4 in C2 H5 OH (b) BH 3 in THF (d) Raney Ni/H 2 in THF 34. The major product of the following reaction is O OD ½ (a) H3 C ¾ C ==CH2 OH ½ (c) H2 C==C ¾ CH2 D O ½½ (b) D3 C ¾ C ¾ CD3 OD ½ (d) D2 C==C ¾ CD3 26. Which of the following will react with water? (a) CHCl 3 (c) CCl 4 (b) 1-butanol (d) t-butyl alcohol 28. The compound that will not give iodoform on treatment with alkali and iodine is (a) acetone (c) diethyl ketone CH3 CH3 O O O CH3 (c) (d) CH3 35. After completion of the reactions (I and II), the organic compound (s) in the reaction mixtures is/are O (b) ethanol (d) isopropyl alcohol (a) trimethyl acetaldehyde (b) acetaldehyde (c) benzaldehyde (d) formaldehyde Br2 (1.0 mol) H3C Reaction I Br2 (1.0 mol) H3C CH3 CH3COOH (1.0 mol) O Q BrH2C CBr2 R O CH2Br S (a) (b) (c) (d) O CBr2 Br2C O (b) 3-pentanone (d) pentanal (b) Schiff ’s reagent (d) Fehling’s reagent CH2Br H3C P (b) CuO (d) Cu + Cu 2 O + CuO 32. The reagent with which both acetaldehyde and acetone react O H 3C 31. A compound that gives a positive iodoform test is easily is (a) Tollen’s reagent (c) Grignard’s reagent CH3 Aqueous/NaOH O Reaction II 30. When acetaldehyde is treated with Fehling’s solution, it (a) 1-pentanol (c) 2- pentanone O (b) (a) 29. The Cannizzaro’s reaction is not given by gives a precipitate of (a) Cu (c) Cu 2 O (ii) H+, Heat O CH3 (b) Cl 3 CCHO (d) ClCH2 CH2 Cl 27. Which of the following compounds is oxidised to prepare methyl ethyl ketone? (a) 2-propanol (c) 2-butanol (i) KOH, H2O H3C ONa T CHBr3 U reaction I : P and reaction II : P reaction I : U, acetone and reaction II : Q, acetone reaction I : T , U , acetone and reaction II : P reaction I : R, acetone and reaction II : S , acetone Aldehydes and Ketones 36. Tautomerism is exhibited by (a) —CH==CH—OH (c) Passage Based Questions (b) O== ==O (d) ==O == == ==O Passage 1 Two aliphatic aldehydes P and Q react in the presence of aqueous K 2 CO3 to give compound R, which upon treatment with HCN provides compound S . On acidification and heating, S gives the product shown below : O 37. A new carbon–carbon bond formation is possible in O O (a) Cannizzaro’s reaction (b) Friedel-Crafts’ reaction (c) Clemmensen’s reduction (d) Reimer-Tiemann reaction 44. The compounds P and Q respectively are CH3 38. Which of the following will undergo aldol condensation? CH (a) H3C (a) Acetaldehyde (b) Propanaldehyde (c) Benzaldehyde (d) Trideutero acetaldehyde H3C H C and H C O O CH3 39. Among the following compounds, which will react with acetone to give a product containing (a) C6 H5 NH2 (c) C6 H5 NHC6 H5 OH H 3C H3C O (b) H3C 40. Which of the following is an example of aldol condensation? H C and C H O C==N¾ ? (b) (CH3 )3 N (d) C6 H5 NHNH2 H CH O H3C CH2 CH C H (c) H 3C H C and CH3 O O Dil.NaOH (a) 2CH3 CHO ¾¾¾® CH3 CH(OH)CH2 CHO OH ½ Dil.NaOH (b) 2CH3 COCH3 ¾¾¾® H3 C ¾ C ¾ CH2 COCH3 ½ CH3 H3C precipitate with iodine and alkali ? (a) 2-hydroxy propane (b) Acetophenone (c) Methyl acetate (d) Acetamide (a) H3C O C C H3C H 2C (b) H OH O CH C CH H2C H 3C H C C H 3C H3C CH3 (c) H3C H CH OH CH3 O CH C (d) H3C CH CH OH H3C OH 46. The compound S is CH3 O 43. Base catalysed aldol condensation occurs with (a) propionaldehyde (b) benzaldehyde (c) 2-methyl propionaldehyde (d) 2, 2-dimethyl propionaldehyde O O O (b) Acetyl chloride (d) Benzaldehyde 42. Which of the following compounds will give a yellow H C 45. The compound R is 41. Which of the following compounds will react with ethanolic KCN? (a) Ethyl chloride (c) Chlorobenzene H H and CH3 Dil .NaOH (d) C6 H5 CHO + HCHO ¾¾¾® C6 H5 CH2 OH +HCOONa C (d) (c) 2HCHO ¾¾¾® CH3 OH + HCOONa Dil .NaOH CH2 CH (a) H 3C CH C CH H2C CN O H (b) H3C H3C C H2C C CN H H Aldehydes and Ketones CH3 CN CH CH (c) H3C CH H 2C CN (d) OH H3C H3C C H 2C OH 49. The structure of the product S, is O CH OH OH (a) (b) Me Passage 2 A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S. O3 /Zn -H2O 1. MeMgBr 2. , H2O O O Me (d) (c) Me Me OH– Passage 3 heat In the following sequence, product I , J and L are formed. K represents a reagent. P ¾¾¾¾¾® Q ¾¾¾¾¾® R ¾¾¾® S H+ 3. H2SO4 / heat 47. The structure of the carbonyl compound P, is Hex-3-ynal (i) NaBH4 (ii) PBr3 I (i) Mg/ether (ii) CO2 (iii) H3O+ CH3 (a) O CH3 CH3 C2H5 CH3 48. The structures of the products Q and R, respectively, are O (a) H3C CH3 CH3 H 3C H2 Pd/BaSO4 quinoline Br Br (c) H3C Br Br 51. The structures of compounds J and K , respectively, are (a) H3C COOH and SOCl2 O (b) H3C H CHO (b) H 3C CH3 CH3 H 3C COOH and SO2Cl2 (c) H3C COOH (d) H3C COOH O H CHO (c) Me Et O Me Me Me (a) H3C (b) H3C CH3 CHO (d) 52. The structure of product L is CHO Et Me Et L (b) H3C (d) H3C H COCH3 CH3 K O (a) H3C (d) O J 50. The structure of the product I is O (c) Cl H 3C (b) O O Me Me CHO (c) H3C CHO (d) H3C CHO and SOCl2 and CH3SO2Cl Aldehydes and Ketones 59. In the scheme given below, the total number of Match the Column intramolecular aldol condensation products formed fromY is 53. Match each of the compounds given in Column I with the reaction(s) that they can undergo, given in Column II. Column I 1. O3 2. Zn, H2O Column II A. Br p. Nucleophilic substitution HCl 60. (A), C6H12 ¾¾® (B), C6H13Cl + (C ), C6H13Cl B. q. Alcoholic KOH (B) ¾¾¾¾¾® (D), (an isomer of (A)) Elimination OH O zonolysis r. CHO (D) ¾¾¾¾® (E), (positive iodoform and negative Fehling’s solution test) Nucleophilic addition Ozonolysis (A) ¾¾¾¾® (F) + (G), (positive Tollen's test for both) Conc. NaOH (F) + (G) ¾¾¾¾® HCOONa + A primary alcohol OH D. s. Br Esterification with acetic anhydride Identify the compounds (A) to (D). 61. A compound C9H7O2Cl exists predominantly in enol form (A) and also in keto form (B). On oxidation with KMnO4 it gives m-chlorobenzoic acid as one of the products. Identify the compounds (A) and (B). NO2 t. Dehydrogenation 54. Match the compounds/ions in Column I with their 62. An alkene (A) C16 H16 on ozonolysis gives only one product properties/reactions in Column II. Column I (B) C8 H8 O. Compound (B) on reaction with NaOH / I2 yields sodium benzoate. Compound (B) reacts with KOH/NH2 NH2 yielding a hydrocarbon (C ) C8 H10 . Write the structures of compounds (B) and (C ). Based on this information two isomeric structures can be proposed for alkene (A). Write their structures and identify the isomer which on catalytic hydrogenation (H2 / Pd – C) gives a racemic mixture. Column II A. C6 H5 CHO p. gives precipitate with 2, 4-dinitrophenylhydrazine B. CH3 C ºº CH q. gives precipitate with AgNO3 – 2. heat Subjective Questions O C. 1. NaOH(aq) Y C. CN r. is a nucleophile D. I- s. is involved in cyanohydrin formation 63. Identify A, B and C, and give their structures. O CH3 CH3 Fill in the Blanks O 55. Fehling’s solution A consists of an aqueous solution of copper sulphate, while Fehling’s solution B consists of an alkaline solution …… (1987, 1M) 57. Benzaldehyde undergoes aldol condensation in an alkaline Ph base O 58. Consider NOTE (Stereoisomers are also reacted separately). C(C7H12O) 65. What would be the major product in the following reaction ? Integer Answer Type Questions all possible isomeric ketones including stereoisomers of MW = 100. All these isomers are independently reacted with NaBH4 . The total number of ketones that gives a racemic product(s) is/are H+ heat followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-acetyl cyclopentane D. The compound B on reaction with HBr gives compound E. Write the structures of A, B, C and E. Show, how D is formed from C. 56. The reaction of methyl magnesium iodide with acetone medium. A +B 64. An organic compound A, C6 H10 O, on reaction with CH3 MgBr True or False followed by hydrolysis gives secondary butanol. Br2 NaOH Br 66. (a) Compound A (C8 H8 O) on treatment with NH2OH. HCl given B and C. B and C rearrange to give D and E, respectively, on treatment with acid. B, C, D and E are all Aldehydes and Ketones isomers of molecular formula (C8 H9 NO). When D is boiled with alcoholic KOH, an oil F (C6H7 N) separates out. F reacts rapidly with CH3COCl to give back D. On the other hand, E on boiling with alkali followed by acidification gives a white solid G (C7 H6O2). Identify A-G. (b) Carry out the following transformation in not more than three steps. 1-butyne ¾® 2-pentanone 73 Complete the following reaction with appropriate structure : CH3 CH2 (i) KCN/H2SO4 C==O ¾¾¾¾¾® (ii) LiAlH4 H 74. Complete the following reaction with appropriate structure. NaOC 2 H5 in absolute C6H5CHO + CH3COOC2H5 ¾¾¾¾¾¾¾¾® A C2 H5 OH heat 67. Write the intermediate steps for each of the following 75. Write the structure of the major organic product expected reactions from the following reaction. H3O+ (i) C6H5CH(OH)C ºº CH ¾¾® C6H5CH == CHCHO MeO CHO + HCHO KOH – H (ii) O OH 76. Arrange the following in the increasing order of expected CH3 68. Complete the following reactions with appropriate structures of products/reagents. O CH3 COCH2 CHO, CH3 COCH3 , CH3 CHO, CH3 COCH2 COCH3 77. Give reason in one or two sentences : ½½ O enol content. “Iodoform is obtained by the reaction of acetone with hypoiodite but not with iodide”. CHC6H6 (i) LiAIH4 (ii) H +, heat [C] 78. A ketone A, which undergoes haloform reaction, gives condensation, gives benzaldehyde and two moles of B on ozonolysis. Compound B, on oxidation with silver ion gives oxalic acid. Identify the compounds A and B. compound B on reduction. B on heating with sulphuric acid gives compound C, which forms mono-ozonide D. D on hydrolysis in the presence of zinc dust gives only acetaldehyde. Identify A, B and C. Write down the reactions involved. 70. Acetophenone on reaction with hydroxylamine hydrochloride 79. Answer the followings with suitable equations wherever 69. An aldehyde A (C 11 H8 O), which does not undergo self aldol can produce two isomeric oximes. Write structures of the oxime. 71. Complete the following, giving the structures of the principal organic products, O necessary (i) suggest a reagent to distinguish acetaldehyde from acetone. (ii) what happens when excess chlorine is passed through boiling toluene in the presence of sunlight? 80. Complete the following with appropriate structures (i) + Ph3P CH2 ? A NaOH CH CH—CHO 81. How may the following transformation be carried out (in not (ii) ClCH2CH2CH2COPh + KOH + MeOH ¾® B more than six steps) “benzaldehyde to cyanobenzene”? O C6H5 (iii) H3C + NaOH 82. Give reason in one or two sentences for the following: H3O+ C “Hydrazones of aldehydes and ketones are not prepared in highly acidic medium”. O 72. Suggest appropriate structures for the missing compounds. (the number of carbon atoms remains the same throughout the reaction) CH3 Dil. KMnO4 A HIO4 B HO– 83. Write down product of the following reaction NaOH Propanal ¾¾¾® heat 84. Arrange the following in order of their increasing reactivity C towards HCN : CH3CHO, CH3COCH3 , HCHO, C2H5COCH3 CH3 Aldehydes and Ketones treated with bromine in presence of phosphorus yields a compound C which on hydrolysis gives a hydroxyl acid D. This acid can also be obtained from acetone by the reaction with hydrogen cyanide followed by hydrolysis. Identify the compounds A, B, C and D. 85. Write down the reactions involved in the preparation of the following using the reagents indicated against in parenthesis: “Acetoxime from acetaldehyde.” [K2Cr2O7 / H+ , Ca(OH)2 and NH2OH, HCl] 86. Show with balanced equation, what happens, when the 88. Outline the reaction sequence for the conversion of methanal to ethanol (the number of steps should not be more than three). following are mixed : “Chloral is heated with aqueous hydroxide” 87. An alkene A on ozonolysis yields acetone and an aldehyde. 89. Write the structural formula of the main organic product formed when methanal reacts with ammonia. The aldehyde is easily oxidised to an acid B. When B is Answers 1. 5. 9. 13. 17. 21. 25. 29. 33. (a) (d) (b) (d) (a) (c) (b) (b) (c) 2. 6. 10. 14. 18. 22. 26. 30. 34. (d) (b) (a) (d) (b) (a) (b) (c) (a) 3. 7. 11. 15. 19. 23. 27. 31. 35. (c) (b) (a) (a) (c) (b) (c) (c) (c) 4. 8. 12. 16. 20. 24. 28. 32. 36. (d) (c) (b) (c) (c) (b) (c) (c) (a,c,d) 37. 41. 45. 49. 53. 54. 55. 56. 58. (b,d) 38. (a,b,d) 39. (a,d) 40. (a,b,d) 42. (a,b) 43. (a,c) 44. (a) 46. (d) 47. (b) 48. (b) 50. (d) 51. (a) 52. A ® p, q, t B ® p, s, t C ® r, s D ® p. A ® p, q, s B ® q, r C ® q, r, s D ® q, r. Sodium potassium tartarate False 57. False (5) 59. (1) (a,b) (b) (a) (c) Hints & Solutions 3 1. Step I 2 KOtBu Br Step II O3/Me2S OHC Ozonolysis 1 Cyclohex-1, 3-diene CHO and OHC ¾CHO moves in the forward direction that results in the formation of acetal. Reaction involved is as follows : Et C In step-1 dehydrohalogenation reaction takes place. Here, hydrogen is eliminated from b-carbon and the halogen is lost from a-carbon atom. As a result diene is formed. H O H+/MeOH H Et OH C 3. It is an intramolecular aldol condensation reaction. This attack will result 7-membering. So, it will not proceed. Cyclohex -1, 3-diene on ozonolysis gives butane-1, 4- dial and ethane- 1, 2- dial. O OHC H OH –H2O HC 5 7 1 2 3 CH2 s Acidic H-atom OH– – H 2O O 5 O= =CH 2 4 1 3 s Less acidic H nucleophilic addition reactions. D For the reaction, HCl - 4 6 H 2. Key Idea Aldehydes are more reactive than ketones in Carbonyl compound + MeOH Acetal Rate of reaction is the highest for propanal as substrate and methanol in excess. Propanal is an aldehyde and more reactive than ketones. When MeOH is taken in excess then reaction O O s CH2 H CHO | CHO + (CH2)2 | | CHO CHO Me2S OMe H Acetal Cyclohex-1,3-diene O3 OMe C Excess OMe H – OtBu Br Et MeOH O (option-c) Major –H2O HO s O H 2O – OH O H More acidic 5 1 s b-H for b-elimination O 4 3 2 Aldehydes and Ketones –I O H 4. O b-elimination Cl s r /D tBuOK –HCl Acidic hydrogen H This benzylic –H activates p-position by +ve for ArSO2 H O H 3C O r r 1° 1°, but more stable carbocation aldol condensation reaction. As a result of this, b-hydroxyketone (A) is obtained which on hydrolysis followed by heating produces a , b-unsaturated ketone (B) r H (Conc. H2SO4) Electrophilic addition H O 6. The reactant in presence of dil × NaOH undergoes intramolecular H O r CH3 O H H 2O D –H2O HO H (B) O r O O O O ArSE 2 H H3C O dil. H NaOH s 2° 2°, but less stable carbocation because of –I effect of ArSE 2 O s O H ( A) 7. LiAlH4 acts as a nucleophilic reducing agent that can reduce H D –Hr ¾ COOH to ¾ CH2OH, ¾ C == O into ¾ CH ¾ OH and ½ ½ ¾ NO2 into ¾ NH2, but it cannot reduce isolated C==C H 6- membered ring (more stable) H D –Hr O OH O O C==C isolated OH LiAlH4 (excess) Major product 5- membered ring (option-d) (less stable) CH3 Minor product (option-c) NO2 NH2 OH 5. According to the given conditions, compound (d) neither reacts with neutral ferric chloride solution nor with Fehling solution. It however reacts with Grignard reagent and gives positive iodoform test. ● As the compound does not contain any phenolic ¾ OH group. Hence, it gives negative neutral FeCl 3 test. OH | CH3 Compound (b) 8. (A) ® Q; B ® (P) ; C ® (S), D ® (R) (A) Lysine (R=–(CH2)4–NH2) NH2 CH R Ninhydrin test (B) COOH OH Neutral C—Et O OH O Compound gives reaction with RMgX as it contains ¾ C ¾ Et. ½½ O OH | CH3 C—Et RMgX Grignard reagent O (C) C—Et O O NaOH+I2 Iodoform test CH2SO4 Violet colouration D Molisch Test (P) (Molisch reagent) CH2OH CH3I ¯ + Et Yellow ppts. O CHO (Benzaldehydel) (Benzyl alcohol) OH OH – dil. KMnO4/OH /Cold (R) (D) O–Na+ Ceric ammonium nitrate (s) [CAN] 50% CH3COOH Compound with CH3CH ¾ group undergoes iodoform test in ½ OH presence of NaOH and I2. OH | CH3 CHO + (Furfural) C2H5 OH O (B) OH | CH3 R Violet colouration (Ninhydrin), OH boil no reaction (FeCl3 test for phenolic group) FeCl3 ● O (Pink) (Styrene) H H H +Mn2+(Colourless) Aldehydes and Ketones CH3 9. NaBH4 is a selective reducing agent. It reduces carbonyl O ½ ½ C 12. (a) O Carbonyl group O CH2 C O OH OCH3 Ester group CH2 NaBH4 MeOH ½ group into an alcohol but cannot reduce an isolated C == C and an ester group too. O3 D Zn.H2O2 O O ½½ ½½ CH3 ¾ C ¾ CH2 ¾ CH2 ¾ CH2 ¾ C ¾ CH3 OCH3 C CH3 Heptan -2, 6 dione CH3 ½ ½ Isolated C C O3 D Zn.H2O2 (b) ½ 10. O O O [Ag (NH3)2]OH Esterification CH3OH, H+ Tollen’s reagent CHO COOH HO CH3 CH3 O ½½5 4 3 2 1 CH3 ¾ C ¾ CH2 ¾ CH2 ¾ CH ¾ CHO ½ CH3 CH3MgBr (Excess) 6 OH COOCH3 5-keto-2-methyl hexanal Before final product is formed, intermediate is HO CH3 CH3 ½ ½ O3 D Zn.H2O2 (c) ½ O CH3 CH3 O CH3 ½½ ½ CH3 ¾ C ¾ CH2 ¾ CH ¾ CH2 ¾ CHO 11. O O 5-keto- 3-methyl hexanal OH H – CH3 Carbanion ½ ½ ½ (d) CH3 H-atom at a carbon O3 D Zn.H2O2 O O O H—C—H H—C—H O CH3 ½½ ½ CH3 ¾ C ¾ CH ¾ CH2 ¾ CH2 ¾ CHO 5-keto-4- methyl hexanal 13. O O Crossed aldol CH2O Complete reaction sequence is as shown below: – H2 O CH2OH a-carbon has no H-atom hence, next reaction with HCHO is crossed Cannizzaro reaction HCHO, OH O PLAN This problem includes concept of nucleophilic addition reaction to carbonyl compound (ketone here) and intramolecular nucleophilic substitution reaction. O Cl Cl HCHO Acetal formation CH2OH + HCOO– (By oxidation) (By reduction) Nucleophilic addition reaction OMgX OH O CH3MgBr, dry ether, 0°C CH3 aq. acid intramolecular CH3 nucleophilic CH3 substitution reaction CH3 O CH3 + Mg X Cl Aldehydes and Ketones PCC 14. R — CH2OH ¾ ¾¾® R — CH == O CH2OH O ½ ½½ H ¾ C ¾ H + – CH ¾ CHO r ½ CH2OH Pyridinium chlorochromate is the mild oxidising agent which causes conversion of alcohol to aldehyde stage. While others causes conversion of alcohol to acid. CH2OH O- CH2OH ½ ½ ½ H 2O H ¾ C ¾ C ¾ CHO ¾® HOCH2 ¾ C ¾ CHO ½ ½ ½ H CH2OH CH2OH 15. The first step is cyanohydrin reaction O ½½ CH3 ¾ CH2 ¾ C ¾ CH3 + CN ¾® OH O½ ½ H2O CH3 ¾ CH2 ¾ C ¾ CN ¾¾® CH3 ¾ CH2 ¾ C ¾ CN ½ ½ CH3 CH3 (I) In the second step the ¾ CN of intermediate (I) is first hydrolysed and then dehydrated on heating in the presence of conc. H2SO4. OH OH ½ ½ H2SO4 CH3 ¾ CH2 ¾ C ¾ CN ¾¾¾® CH3CH2 ¾ C ¾ COOH ½ ½ CH3 CH3 (I) CH2OH O O½ ½ ½½ Step IV HOCH2 ¾ C ¾¾¾ C ¾ H + H ¾ C ¾ OH ½ ½ CH2OH H OH Cannizzaro reaction OH + HCOO– HO OH In the last step, formaldehyde is oxidised and the other aldehyde is reduced giving the desired products. O 17. D ¾® CH3 ¾ CH == C ¾ COOH ½ CH3 H O3 H Zn-H2O O E 16. The given reaction is an example of repeated aldol condensation followed by Cannizzaro reaction. H2O O ½½ H ¾ C ¾ H + CH2 ¾ CHO r OH O½ ½ H 2O H ¾ C ¾ CH2 ¾ CHO ¾® CH2 ¾ CH2 ¾ CHO ½ H CHO KOH - Step I CH3CHO + OH- ¾® CH2 ¾ CHO + H2O Intramolecular aldol condensation reaction 18. Three different oximes are formed out of which two are optically active i.e.exists as a pair of enantiomers while other is optically inactive. CH3 19. NaOH H 3C - I2/H2O O Step II HOCH2 ¾ CH2 ¾ CHO + HO r CHI3 + CH3CH2COONa H+ - HO ¾ CH2 ¾ CH ¾ CHO + H2O O ½½ H ¾ C ¾ H + CH ¾ CHO r ½ CH2OH CH3CH2COOH 20. X is (CH3CO)2 O and it is an example of Perkin’s reaction. 21. The reactivity of carbonyl compound towards nucleophilic addition of Grignard’s reagent depends on extent of steric hindrance at a-carbon. Greater the steric hindrance smaller the reactivity. Hence, reactivity order is OH O½ ½ H 2O CH2 ¾ CH ¾ CHO H ¾ C ¾ CH ¾ CHO ¾¾® ½ ½ ½ CH2OH H CH2OH Step III HOCH2 ¾ CH ¾ CHO + HO- r ½ CH2OH - HOCH2 ¾ C ¾ CHO + H2O ½ CH2OH CH3CHO > CH3 ¾ CO ¾ CH3 > Ph ¾ CO ¾ Ph II I III O 22. —CHO + H—C—H NaOH —CH2OH H2O + H— COONa This is an example of cross Cannizzaro reaction in which formaldehyde is always oxidised. Aldehydes and Ketones O 23. 32. Grignard’s reagent reacts with both aldehydes and ketones 33. Only ¾ CHO group is to be reduced to ¾ CH2OH. HO HO while other three reagents reacts only with aldehydes, not with ketones. CH2CH3 NH2–NH2 HO– CH3 It can be done using NaBH4 in C2H5OH. (Wolff-Kishner reduction) 24. O O O O C O Cl H I (a) LiAlH4 / (C2H5 )2 O reduces I, II and III into ¾ CH2OH, and IV into diol. (b) BH3 / THF show same properties as (a). (c) NaBH4 / C2H5OH reduces III into ¾ CH2OH. (d) Raney nickel, same as (a) and (b), thus (c) is correct reagent. O Cl—C—–CH Cl H 34. O HO– CH3 OH O ½ ½½ [O] 27. CH3 ¾ CH ¾ CH2 ¾ CH3 ¾¾® CH3 ¾ C ¾ CH2 ¾ CH3 catalyst CH3 O O O Highly stable hydrate 2- butanol Aldol type Condensation O CH3 Ethyl methyl ketone O O ½ OH group gives iodoform test : Acetone CH3 ¾ CH ¾ H ½ OH CH3 ¾ CH ¾ CH3 ½ OH Ethanol Isopropyl alcohol Above three compounds has the desired group for iodoform test. Diethyl ketone does not has the required group for iodoform test. 29. Aldehydes lacking presence of a-H undergo Cannizzaro reaction. When treated with aqueous base CH3CHO does not undergo Cannizzaro reaction because it has a-H and in the presence of aqueous base it undergoes aldol condensation. 30. CH3 ¾ CHO + Fehling’s solution ¾¾® Cu 2O ¯ Red O ½½ 31. For iodoform test, CH3 ¾ C ¾ group is required Plan When acetone reacts with Br2 in basic medium, bromoform is formed. Reaction I CH3COCH3 + 3Br2 + 4NaOH 1 mol 1 mol 3 3 mol 1 mol ¾® CH3COONa + CHBr3 + 3NaBr + 3H2O (T ) (U ) When CH3COCH3 and Br2 are in equimolar quantity, all the Br2 (limiting reactant) is converted into desired products and 2/3 mole of CH3COCH3 remains unreacted, being in excess. When acetone reacts with Br2 in acidic medium, there is monobromination of acetone. Reactions II 1 mol 1 mol (P ) CH3COCH3 and Br2 react in 1 : 1 mole ratio and (P) is formed. In reaction I, (U ) and (T ) are formed and acetone (reactant) remains unreacted. In reaction II, (P) is formed. + NaOH + I2 COONa 35. CH COOH CH3 2-pentanone O 3 CH3COCH3 + Br2 ¾¾¾¾® CH3COCH2Br + HBr O H 3C CH3 H+ D –H2O a, b-unsaturated ketone 28. Compounds that contain either CH3 ¾ CO or CH3 ¾ CH ¾ O ½½ CH3 ¾ C ¾ CH3 H COOH OH O ½ ½½ 25. CH2 == C ¾ CH3 + D2O r D3C ¾ C ¾ CD3 All a-H will be replaced by deuterium. 26. III II Highly resonance stabilised Cl3C—CHO + H2O O + H+ – H O IV O + CHI3¯ Yellow 36. All those carbonyl compounds containing a-H to sp2 carbon show keto-enol tautomerism. Aldehydes and Ketones 37. In both Friedel-Craft’s reaction and Reimer-Tiemann reaction 43. For base catalysed aldol condensation, there must be at least one a-H to carbonyl group. CH3 ¾ CH2 ¾ CHO, new carbon-carbon bond is formed. CH3 + CH3Cl Propionaldehyde (2 a -H) AlCl3 OH OH 2-methyl propionaldehyde (1 a -H) CHO + CHCl3 + NaOH H CH3 ½ CH3 ¾ C — CHO ½ CH3 CH3 ½ CH3 ¾ CH ¾ CHO , Friedel-Crafts' alkylation + 2, 2-dimethyl propionaldehyde (no a -H) Reimer-Tiemann’ reaction Passage 1 (For Q. Nos. 44-46) The given product is an ester, obtained by condensation of a hydroxy acid obtained through hydrolysis of a cyanohydrin : 38. All carbonyl compounds containing a-H or a-D undergo aldol condensation. In given example, benzaldehyde does not contain a-H to —CHO, hence does not undergo aldol condensation. H3C H3C 39. C == O + C6H5NH2 ¾® C == N ¾ C6H5 H3C H3C H3C + C6H5NHNH2 ¾® C == N ¾ NHC6H5 H3C CH3 OH CH3—C——CH OH OH ½ Dil. NaOH 40. 2CH3CHO ¾¾¾® CH3 ¾ CH ¾ CH2CHO H 3C H+ D C==O CH2 OH H3C HO O O Acid above is obtained by acid hydrolysis of cyanohydrin S as CH3 OH Aldol CH3 OH H3C—C——CH—CN CH3 O O ½ ½½ ½½ Dil. NaOH 2CH3 ¾ C ¾ CH3 ¾¾¾® CH3 ¾ C — CH2 ¾ C ¾ CH3 ½ OH H+ H2O H3C—C——CH—COOH CH2OH S CH2OH S is obtained by nucleophile addition of HCN on R, hence R is Aldol CH3 O Dil. NaOH 2HCHO ¾¾¾® CH3OH + HCOONa Cannizzaro reaction CH3 OH H3C—C——C—H + HCN Dil. NaOH C6H5CHO + HCHO ¾¾¾® C6H5CH2OH + HCOONa H3C—C——CH—CN CH2OH S CH2OH R Cannizzaro reaction S 2 N 41. CH3CH2Cl + KCN ¾® CH3CH2CN + KCl O O ½½ ½½ CH3 ¾ C ¾ Cl + KCN ¾® CH3 ¾ C ¾ CN + KCl O O ½½ ½½ C6H5 ¾ C ¾ H + KCN ¾® C6H5 ¾ C ¾ CH ¾ C6H5 ½ OH R is obtained by treatment of P and Q with aqueous K2CO3 through aldol condensation reaction as CH3 O CH3 CH3—CH—CHO + H—C—H P+Q OH– OHC—C—CH2OH CH3 R Passage 2 (For Q. Nos. 47-49) Benzoin O ½½ 42. Aldehyde, ketones containing CH3 ¾ C ¾ group gives iodoform test. Also alcohol containing CH3 ¾ CH ¾ group gives ½ OH O OH ½½ ½ iodoform test. CH3 ¾ C — CH3 C6H5 ¾ C ¾ CH3 ½ H Both gives positive iodoform test Esters and amides do not give iodoform test. CH3 O CH3 CH3MgBr H+/H2O CH3 CH3 Q P CH3 (gives positive iodoform test) Q CHO CH3 O3 Zn-H2O O H 3C OH– aldol O CH3 R S Aldehydes and Ketones Passage 3 (For Q. Nos. 50-52) 54. Column I O A. C6H5CHO 1. NaBH4 2. PBr3 H Br I (ii) CO2 J J COOH K C—Cl H H2 with AgNO3 . H C. CN - Forms AgCN with AgNO3. A nucleophile is involved in cyanohydrin formation. D. I- Gives AgI precipitate with AgNO3 and it is a nucleophile. C==C Pd/BaSO4 H5C2 CH2CH2CHO L Column I Column II Br A. O OH B. B. CH3 ¾ C ºº CH Gives precipitate (CH3 ¾ C ºº CAg) A nucleophile, undergo electrophilic attack. SOCl2 O 53. Gives phenyl hydrazone with 2, 4-dinitrophenyl hydrazine. Gives precipitate with AgNO3, Tollen’s test forms cyanohydrin. (i) Mg/ether (iii) H3O+ Column II CHO C. OH Br D. NO2 Undergo nucleophilic substitution of Br - . Undergo elimination of HBr. Does not undergo nucleophilic addition. Does not esterify with acetic anhydride, can be dehydrogenated. Undergo nucleophilic substitution with SOCl 2 , PCl 5 etc. Does not undergo elimination. Does not undergo nucleophilic addition. Undergo esterification with acetic anhydride. Undergo dehydrogenation to give C6 H5 CHO. Does not undergo nucleophilic substitution, there is no leaving group. Does not undergo elimination. Undergo nucleophilic addition at carbonyl carbon of —CHO. Undergo esterification with acetic anhydride. Does not undergo dehydrogenation. Undergo aromatic nucleophilic substitution (SN Ar) Does not undergo elimination, nucleophilic addition, esterification or dehydrogenation. 55. Sodium potassium tartarate CH3 O ½ ½½ H2 O 56. CH3MgI + CH3 ¾ C ¾ CH3 ¾® CH3 ¾ C — OH ½ CH3 Tertiary alcohol 57. Benzaldehyde, lacking a-H does not undergo aldol condensation, rather it undergoes Cannizzaro reaction. 58. Molecular weight of the ketone is 100 So, molecular formula = C6 H12O 12 =1 2 According to question, compound contains ketone group. Since, the compound which contain chiral centre lead to the formation of diastereomer while other produces enantiomers. Various isomers and their possible reduced product are as shown below. Degree of unsaturation = (6 + 1) - n-butyl–CH—CH3 (1) n-butyl–C—CH3 OH O (2) Iso-butyl–C—CH3 O (3) 3° butyl–C—CH3 Iso-butyl–CH—CH3 All are + racemic mixture NaBH4 OH 3° butyl–CH—CH3 OH O CH3 (4) CH3CH2—CH—C—CH3 * O R (2°butyl) CH3 (5) CH3CH2—CH—C—CH3 * O CH3 * CH3—CH2—CH—CH—CH3 * OH 2-alcohols (R,R) and (R,S) diastereomeric pair NaBH4 CH3 * CH3—CH2—CH—CH—CH 3 * S (2°butyl) (Here, (*) represents chiral centre) OH 2-alcohols (S,S) and (S,R) diastereomeric pair Aldehydes and Ketones (6) n-propyl–C—Et O Iso-propyl–C—Et (7) A= n-butyl–CH—Et (±) racemic OH mixture NaBH4 While in case of (4) and (5) they do not produce enantiomer due to the presence of stereogenic centre on ketone. 59. H3C CH3 ½ 60. CH2 == CH ¾ C — CH3 ½ CH3 Ph O CH3 Br2 ‘B’ CH3 + CO2 C O 64. C C OH– CH3 Aldol D CHO O Þ CH3 CH3 C= B= HCHO G MeMgBr H+ A (C6H10O) ¾¾¾® ¾¾® B O 61. H Þ O O O H+ Heat F O COONa + CHBr3 NaOH ‘A’ D CH3 ½ CH3 ¾ C — CHO ½ CH3 + Enantiomer = Racemic mixture O O CH3 ½ CH3 ¾ C == C — CH3 ½ CH3 C CH3 63. CH3 CH3 B CH3 Cl ½ ½ CH3 ¾ CH — C — CH3 ½ CH3 E CH3 CH3 CH3 Cl ½ ½ CH3 ¾ CH — C — CH3 ½ CH3 A O ½½ CH3 ¾ C ¾ CH3 H3C Ph II + H2/Pd Only product Y Ph C == C II Zn-H2O O Ph H OH– O3 CH3 + H O O Ph C == C I Iso-propyl–CH—Et (±)racemic OH mixture O Ph + CH3MgBr CHO H A CH3 HO CH3 H+ Cl Cl A B B COOH KMnO4 Ph 65. Ph Base O Br Cl O Br _ Enolisation 62. B + NaOH / I2 ¾® C6H5COONa + CHI3 Ph O CH2CH3 C Þ B= CH3 N2H4/OH– O Aldehydes and Ketones 66. (a) G is benzoic acid C6H5 ¾ COOH , B and C are two O 68. O stereomeric oximes which undergo Beckmann’s rearrangement on treatment with acid to give amides D and E. O ½½ OHC6H5 ¾ C ¾ NHCH3 ¾¾® C6H5COOH + CH3NH2 H2 O E (ii) C6H5CHO CH—Ph G (i) LiAlH4 O OH H5C6 H C==N CH—Ph (i) NaOH (ii) H+, heat + C6H5—C—NHCH3 H 3C E 69. Aldehyde A does not has any a-H but undergo ozonolysis to give two moles of compound B and benzaldehyde. Compound B on oxidation gives oxalic acid, so A is C O H5C6 H+ C==N H 3C O3 C6H5 ¾ CH == CH¾ C ºº C ¾ CHO ¾® CH3—C—NHC6H5 A C6H5 ¾ CHO + 2COOH ½ CHO D OH B KOH H2O CH3COOH + C6H5NH2 B F F + CH3COCl ¾® C6H5NHCOCH3 D O ½½ Þ A = C6H5 ¾ C ¾ CH3 + Ag B ¾¾® COOH ½ COOH 70. (b) CH3 ¾ CH2 ¾ C ºº CH O O ½½ B2 H6 (i) CH3 MgBr ¾¾¾® CH3CH2CH2 ¾ C ¾ H ¾¾¾¾¾® H2 O2 /OH- OH H5C6 C==N C6H5—C—CH3 + H2NOH H3C + (ii) H3 O OH ½ CH3CH2CH2 ¾ CH ¾ CH3 O [O] ½½ ¾® CH3CH2CH2 ¾ C ¾ CH3 H5C6 C==N + H3 C O OH CH2 2-pentanone. 71. (i) 67. (i) C6H5 ¾ CH ¾ C ºº C ¾ H (Wittig reaction) + Ph3P==CH2 ½ OH A O ½½ CH3O(ii) Cl ¾ CH2 ¾ CH2 ¾ CH2 ¾ C ¾ Ph ¾¾® + + H ¾¾® C6H5 ¾ CH ¾ C ºº C ¾ H - H2 O ·· OH2 Cl O ½½ + -H ¾® C6H5 ¾ CH== CH ¾ C ¾ H Ph—C— CH2 COPh O O (iii) C6H5 H+ + O H B HC e OH O CH2 C6 H5 ¾ CH == C == C ¾ OH tautomerisation ½ H (ii) Ph H 3C OH– O – H2C C==O Ph O H+/D O OH – H+ O Ph —Ph C Aldehydes and Ketones 76. CH3 ¾ CO ¾ CH2 ¾ COCH3 has highest enol content due to 72. CH3 CH3 Dil. KMnO4 HIO4 OH OH CH3 O O CH3 A resonance and formation of six membered ring through intramolecular H-bonding CH3 B O H 3C CH3 O B HO– 2 77. Iodoform reaction is an oxidation reaction in which hypoiodite OI- acts as oxidising agent : O ½½ R ¾ C ¾ CH3 + OI- ¾® R ¾ COO- + CHI3 Iodide (I- ) is a reducing agent, does not give iodoform reaction. 2 O 3 Zn CH CHO (only) 78. D ¾® ¾® 3 H2 O I is formed as racemic mixture. C = CH3 ¾ CH == CH ¾ CH3, O ½½ C2 H5 ONa 74. C6H5CHO + CH3 ¾ C ¾ OC2H5 ¾¾¾® O D = CH3 ¾ CH CH ¾ CH3 ½ ½ O ¾¾ O C2 H5 OH O ½½ C6H5 ¾ CH == CH ¾ C ¾ OC2H5 OH O ½ ½½ B = CH3 ¾ CH ¾ CH2CH3 and A = CH3 ¾ C ¾ CH2CH3 75. —CHO + HCHO 79. (i) Tollen’s reagent gives grey precipitate of Ag, acetone KOH does not. CH3 (ii) —CH2OH + HCOOK MeO— 80. (Cannizzaro reaction) In cross-Cannizzaro reaction, methanal is always oxidised. O O O H + H II is less stable than I because II is less substituted enol. Acetone has greater enol content than ethanal O OH ½½ ½ CH3 ¾ C ¾ CH3 s CH2 == C ¾ CH3 (more substituted) C D OH ½ CH2 == C — H + HCl NaOH —CH==CH—CHO 81. COOH [O] K2Cr2O7/H+ II O ½½ CH3 ¾ C ¾ H s Cl2 (excess) C6H5CHO + CH3CHO CHO H3C CCl3 hn Cross-aldol condensation O H B CH3 Therefore, CH3COCH2CHO forms next most stable enol II MeO— O Also, enol content depends upon the number of substituents on double bond, greater the number of substituents, greater the stability, higher the enol content. aldol OH O ½ ½½ CH3CH2 ¾ C ¾ H + HCN ¾® CH3CH2 ¾ C — CN ½ OH H ½ LiAlH4 I ¾¾® CH CH ¾ C — CH NH ½ H H 3C (I) C 2 H CH 3 (A) 73. KCN + H2SO4 ¾® KHSO4 + HCN 3 O O NaOH CaO/D NO2 HNO3 H2SO4 NO2 NH2 Fe/HCl NaNO2 HCl/0°C (less substituted) Therefore, overall enol-content order is D < C < B < A + N2Cl– KCN CuCN CN Aldehydes and Ketones O ½½ R 82. R ¾ C ¾ R + PhNHNH2 ¾® C ==N ¾ NHPh R Hydrazone In acidic medium, hydrazine reacts to form salt and hydrazone is hydrolysed back to aldehyde/ketone. OHheat 83. CH3CH2CHO ¾¾® CH3 ¾ CH2 ¾ CH == C ¾ CHO ½ CH3 OH O ½ ½½ HCN H+ 87. CH3 ¾ C ¾ CH3 ¾¾® CH3 ¾ C — CN ¾¾® H2 O ½ CH3 OH ½ CH3 ¾ C — COOH ½ CH3 D Br CH3 ½ ½ Þ C = CH3 ¾ C — COOH , B = CH3 ¾ CH — COOH ½ CH3 Aldol followed by dehydration 84. Steric hindrance at carbonyl carbon determine the reactivity towards nucleophilic addition reaction. Greater the steric hindrance, smaller the reactivity. Þ C2H5COCH3 < CH3COCH3 < CH3CHO < HCHO 85. Ca(OH) 2 H+ CH3CHO ¾¾® CH3COOH ¾¾® (CH3COO)2 Ca K 2 Cr 2 O7 O ½ ½ H2 NOH heat (CH3COO)2 Ca ¾® CH3 ¾ C ¾ CH3 ¾¾® H3C H3C C == N ¾ OH CH3 CH3 ½ ½ A = CH3 ¾ C== CH ¾ CH ¾ CH3 O ½½ E ther 88. H ¾ C ¾ H + CH3MgBr ¾¾® CH3CH2OMgBr H+ H2 O ¾¾® CH3CH2OH N 89. acetoxime 86. Cl 3C ¾ CHO + NaOH(aq) ¾® Cl 3C ¾ CH2OH Cannizzaro reaction + Cl 3C ¾ COONa 6CH2O + 4NH3 N N N Hexamine ‘‘(CH2)6N4’’