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CAK – 1 ALDEHYDES AND KETONES C1A Physical Properties : It has high dipole moment Boiling point are lower than alcohols due to their inability to form intermolecular H-bonding. B.Pt. are higher than corresponding alkanes due to dipole-dipole interaction. Carbonyl group can form H-bond with H2O hence they are soluble in water to varying extent. C1B Method of Preparation : 1. Oxidation : (a) (b) (c) This oxidation is called as oppenauer oxidation. 2. Rossenmund Reduction : 3. [LiAlH (O-t-C4H9)3 or Lithium tri-t-butoxy Aluminium hydride] 4. 5. Hydrolysis of Gem-Dihalide (a) (b) Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 2 (c) 6. Cl , heat H O 2 ArCH 3 2 ArCHCl 2 ArCHO Hydration of Alkynes : Hydration of alkynes gives ketones (except CH CH which gives CH3CHO) (a) H O, H CH CH 2 CH 3 CHO (b) HgSO 4 7. H O, H CH 3 C CH 2 CH 3 COCH 3 HgSO 4 Hydroboration Oxidation : Hydroboration of a non-terminal alkyne followed by oxidation of the intermediate yields a ketone but terminal alkyne yield aldehyde. 8. 9. Use of Grignard Reagent : (a) With HCN aldehyde is formed. (b) With RCN a ketone is formed. With Esters (a) (b) 10. Decarboxylation of calcium salts of carboxylic acids : 11. Oxo Process : 2 CH3CH = CH2 + CO + H2 CH3CH2CH2CHO + [COH(CO)4] Cobal + Carbonyl hydride Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 3 12. Reduction of acid chloride with organocopper compounds : 13. Friedel-Crafts Acylation : (a) (b) 14. From Cadmium and Lithium Salts : 1. 2RMgX + CdCl2 R – Cd – R + MgCl2 + MgX2 RCOR’ + RCdCl R – Cd – R + R’COCl R should be 10 alkyl or aryl (–C6H5) 15. Aldehyde/Ketone Ozonylysis of Alkene (Zn/H+) C2 Chemical Properties : 1. Nucleophilic Addition to the Carbon-Oxygen Double bond : The most characteristic reaction of aldehyde and ketone is nucleophilic addition to the carbon-oxygen double bond. General Reaction : Relative reactivity of Aldehydes versus Ketones : Aldehydes are more reactive than Ketones. There are two reasons for this, they are as follows : 1. 1. Steric Factor 2. Electronic factor Steric Factor : With one group being the small hydrogen atom, the central carbon of the tetrahedral product formed from the aldehyde is less crowded and the product is more stable. With ketones two alkyl groups at the carbonyl carbon causes greater steric crowding in the tetrahedral product and make it less stable. Therefore small concentration is present at equilibrium. 2. Electronic Factor : Because alkyl group are electron releasing therefore aldehydes are more reactive on electronic grounds as well. Aldehyde have one electron releasing alkyl group to stablise the partial positive charge on the carbon atom of the carbonyl group. Whereas ketones have two alkyl groups. Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 4 (a) Addition of cyanide : (b) Addition derivatives of ammonia : (i) (ii) HCHO reacts with NH3 differently forming UROTROPINE [hexamethylene tetraamine]. 6HCHO + 4NH3 (CH2)6N4 + 4H2O H2N – G (c) Product H2N – OH Hydroxylamine = N – OH oxime H2N – NH2 Hydrazine = N – NH2 Hydrazone H2N – NHC6H5 Phenylhydrazine = N – NHC6H5 Phenylhydrazone H2N – NHCOCH2 Semicarbazine = NNHCOCH2 Semicarbazone Addition of Alcohols : Acetal Formation (i) Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 5 (ii) Important : (d) Addition of Grignard reagents : Formaldehyde with Grignard Reagent gives 10alcohol, all higher aldehydes with grignard reagent give 20 alcohol and ketones with grignard reagent gives the 30 alcohol. C3 Other Reactions of Aldehyde and Ketones : (a) Oxidation : (i) RCHO or ArCHO Aldehydes (except) benzaldehyde reduce “Fehling’s Solution” (Cu2+ reduced to Cu+) which is an alkaline solution of Cu2+ ion complexed with tetrarate ion. (ii) Example : Tollen’s Test : CH 3 CHO 2Ag( NH 3 ) 2 3OH – CH 3 COO 2Ag 4NH 3 2H 2 O Colourless Solution Silver mirror Fehling Solution CH 3 CHO 2Cu 2 3OH CH 3 COO 2Cu 2H 2O Re d PPt . Tollen’s test is cheifly given by aldehydes. Tollen’s reagent does not attack carbon-carbon double bond. Aldehyde also reduce benedict’s solution (Cu2+ complexed with citrate ion) to Cu+ (b) (i) Ketones with strong oxidants and at high temperature undergo cleavage of C-C bond on either side of carbonyl group. Carbonyl Group after bond cleavage goes with that alkyl group which is of smaller size. (ii) Ketones are also oxidised from cleavage of bond by caro’s acid (H2SO5) or SO 5 peroxybenzoic acid (C6H5CO3H) to esters. RCOR H 2 RCOOR [Bayer’ss [O ] Villiger Oxidation] Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 6 (c) Haloform Test : (i) Methylketones : (ii) Hypohalite NaOX (NaOH + X2) cannot only halogenate can also oxidise alcohols (d) Reduction : (a) Reduction to alcohol (e) Reduction to hydrocarbons : (f) Reductive Amination (Discussed under Amines) (g) Cannizzaro reaction : In the presence of an concentrated base i.e. alkali, aldehydes containing no -hydrogens undergo self oxidation and reduction to yield a mixture of an alcohol. (i) (ii) Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 7 (h) (i) Crossed Cannizzaro reaction : Tischenko Reaction : All aldehyde in presence of aluminium ethoxide, Al(OC2H5)3 can be simultaneously oxidised (to acid) and reduced (to alcohols) to form ester. This is called Tischenko reaction and is thus like cannizaro reaction. (j) Distinction Aldehyde and Ketones S.No. Test RCHO RCOR 1. Schiffs reagent magenta colour restored by RCHO no. reaction 2. Tollen’s reagent is reduced by RCHO is not reduced 3. Fehling’s solution is reduced by RCHO (except C6H5CHO) is not reduced -hydroxy, ketones reudce Tollen’s re agent and Fehling’s solution Practice Problems : NaOH 1. Identify (Z) in the reaction series, (a) 2. 3. C2H 5I (b) CH 2 CH 2 HBr (X) Hydrolysis (Y) I ( Z) 2 ( excess ) C2H5OH (c) CHI3 (d) CH3CHO A compound (X) of the formula C3H8O yields a compound C3H6O on oxidation. To which of the following class of compounds could (X) belong (a) aldehyde (b) secondard alcohol (c) alkene (d) tert. alcohol Which statement is incorrect in the case of acetaldehyde and acetone (a) both react with hydroxylamine (b) both react with NaHSO3 Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 8 4. (c) both react with hydrazine (d) both reduce ammonical silver nitrate Which of the following undergoes Cannizzaro’s reaction (a) 5. 7. 8. (c) (CH3)2CHCHO (d) HCHO acetaldehyde (b) formaldehyde (c) acetone phenol (d) (a) Cannizzaro’s reaction (b) Clemmensen’s reduction (c) Rosenmund’s reaction (d) Tischenko reaction When acetaldehyde is treated with aluminium ethoxide, it forms (a) ethyl acetate (b) ethyl alcohol (c) acetic acid (d) methyl propionate (c) acetone (d) Chloretone is formed when chloroform reacts with formaldehyde (b) acetaldehyde benzaldehyde Which of the following reagent reacts differently with HCHO, CH3CHO and CH3COCH3 (a) 10. CH3CH2CHO Hydrocarbons are formed when aldehydes and ketones are reacted with amalgamated zinc and conc. HCl. The reaction is called (a) 9. (b) Urotropine is formed by the action of ammonia on (a) 6. CH3CHO HCN (b) NH2OH (c) C6H5NHNH2 (d) NH3 In the following sequence of reactions, the end product is 2 / H 2SO 4 O] )2 2O CaC2 H (A) Hg (B) [ ((C) Ca ( OH (D) heat (E) (a) 11. acetaldehyde (b) formaldehyde (c) acetic acid (d) acetone In the following sequence of reactions, the end product is 2 / H 2SO 4 [O] HC CH Hg (A) CH 3MgX ((B) (C) [ H 2O ] 12. 13. (a) acetaldehyde (b) isopropyl alcohol (c) acetone (d) ethyl alcohol HCN (a) CH3COOH (b) CH3CHOHCOOH (c) CH3CH2NH2 (d) CH3CONH2 A compound, C5H10O, forms a phenyl hydrazone and gives negative Tollen’s and iodoform tests. The compound on reduction gives n-pentane. The compound A is (a) 14. H O 2 In the following sequence of reactions, the end product is CH 3CHO ( A ) ( B) pentanal (b) pentanone-2 (c) pentanone-3 (d) amyl alcohol CHI3 (d) CH3CHO ethyl acetate (d) acetylene The product Z in the series is Na 2 CO 3 CH 2 CH 2 HBr X Hydrolysis Y I Z 2 ( excess ) (a) 15. (b) C2H5OH (c) If formaldehyde and KOH are treated together, we get (a) 16. C2H 5I methane (b) methanol (c) The correct order of reactivity in nucleophilic addition reaction CH3CHO, CH3COC2H5 and CH3COCH3 is (a) CH3CHO > CH3COCH3 > CH3COC2H5 (b) C2H5COCH3 > CH3COCH3 > CH3CHO (c) CH3COCH3 > CH3CHO > C2H5COCH3 (d) CH3COCH3 > C2H5COCH3 > CH3CHO Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 9 17. 18. 19. 20. 21. To distinguish between 2-pentanone and 3-pentanone which reagent can be used (a) NaOH/I2 (c) K 2Cr2O7/H + Tollen’s reagent (d) Zn–Hg, HCl CH3CH = CHCHO is oxidised to CH3 – CH = CHCOOH, using oxidising agent as (a) alkaline KMnO4 (b) K2Cr2O7/conc. H2SO4 (c) ammonical AgNO3 (d) dilute HNO3 m-chloro benzaldehyde on reaction with conc. KOH at room temperature gives (a) potassium m-chloro benzoate and m-hydroxy benzaldehyde (b) m-chloro benzyl alcohol and m-hydroxy benzaldehyde (c) m-chloro benzyl alcohol and m-hydroxy benzyl alcohol (d) m-chloro benzyl alcohol and potassium m-chloro benzoate The reagent which can be used to distinguish acetophenone from benzophenone is : (a) 2, 4-dinitrophenyl hydrazine (b) benedict reagent (c) I2 and Na2CO3 (d) aqueous solution of NaHSO3 Best starting material to synthesize 2-methyl-2-butenoic acid is (a) (b) (c) (d) 22. CH3CH2CH2CHO X and Y can be distinguished by : (a) 23. (b) silver-mirror test(b) iodoform test (c) both (d) CH3 C CCl3 (b) CH 3 C CH 2 I CH 3 C CH 2Cl (d) none Which of the following will give haloform test O || (a) (c) | OH all || O [Answers : (1) c (2) b (3) d (4) d (5) b (6) b (7) a (8) c (9) d (10) d (11) c (12) b (13) c (14) c (15) b (16) a (17) a (18) a (19) d (20) c (21) c (22) c (23) d] Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 10 SINGLE CORRECT CHOICE TYPE 1. 2. Which of the following reagent reacts differently with HCHO, CH3CHO and CH3COCH3 HCN (b) NH2OH (a) Benedict’s solution (c) C6H5NHNH2 (d) NH3 (b) Tollen’s reagent (c) Fehling’s solution (d) NaHCO3 In the following sequence of reactions, the end product is Condensati on Dehydratin g agent mild alkali heat (a) aldol (b) crotonaldehyde (c) paraldehyde (d) 4. 5. 9. metaldehyde Which of the following reaction is a condensation reaction 10. (a) Cannizzaro’s reaction (b) Aldol condensation (c) Claisen’s reaction (d) Tischenko reaction The reaction involving condensation of acetic anhydride with an aromatic aldehyde by a carboxylate ion is an example of HCHO Paraformaldehyde (b) CH3CHO Paraldehyde (a) aldol condensation (c) CH3COCH3 Mesityl oxide (b) benzoin condensation (d) CH2 = CH2 Polyethylene (c) Perkin reaction (d) Wurtz reaction A compound A has molecular formula C2Cl3OH. It reduces Fehling’s solution and on oxidation gives a monocarboxylic acid. A is obtained by the action of Cl2 on ethyl alcohol. The compound A is 11. Benzaldehyde when refluxed with aqueous alcoholic KCN forms (a) chloroform (a) benzoin (b) chloral (b) benzene (c) trichloro ethanol (c) phenyl cyanide (d) trichloro acetic acid (d) phenyl isocyanide 12. The product Z in the series is Hydrolysis CH 2 CH 2 X Y Na 2 CO 3 I Z 2 ( excess ) 7. Al ( OC H ) 5 3 CH3COOCH2CH3. This 2CH3CHO 2 reaction is called (a) HBr 6. Formic acid and formaldehyde can be distinguished by treating with (a) CH 3CHO (A ) (B) 3. 8. (a) C2H 5I (b) C2H5OH (c) CHI3 (d) CH3CHO 13. acetone (b) acetaldehyde (c) propanal (d) HCHO (a) Wurtz reaction (b) Cannizzaro’s reaction (c) Perkin reaction (d) Claisen reaction Anhyd .AlCl 3 (X) + HCl C6H6 + CO + HCl The compound (X) is Cyanohydrin of which of the following gives lactic acid on hydrolysis (a) Benzyl alcohol is obtained from benzaldehyde by 14. (a) C6H5CHO (b) C6H5COOH (c) C6H5CH2Cl (d) C6H5CH3 Which of the following is the weakest acid (a) benzene sulphonic acid In the reaction CH3CHO + HCN CH3CHOHCN, a chiral centre is produced. The product is (b) benzoic acid (c) benzyl alcohol (a) meso compound (d) phenol (b) laevorotatory (c) dextrorotatory (d) racemic mixture Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 11 15. In the Cannizzaro’s reaction given below OH 2Ph CHO Ph CH 2 OH PhCO 20. O || 2 (a) the slowest step is 16. 17. 18. (a) attack of OH— at the carbonyl group (b) the transfer of hydride to the carbonyl group (c) the abstraction of proton from the carboxylic acid (d) the deprotonation of Ph–CH2OH Which of the following will give haloform test (b) CH3 C CCl3 CH 3 C CH 2 I | OH (c) CH 3 C CH 2Cl || O (d) all The reagent(s) which can be used to distinguish acetophenone from benzophenone is (are) : (a) 2, 4-dinitrophenyl hydrazine (b) benedict reagent (c) I2 and Na2CO3 (d) aqueous solution of NaHSO3 21. cannizzaro (Y is alcohol, D is deuterium) X and Y X and Y will have structure : m-chloro benzaldehyde on reaction with conc. KOH at room temperature gives (a) potassium m-chloro benzoate and m-hydroxy benzaldehyde (b) m-chloro benzyl alcohol and m-hydroxy benzaldehyde (c) m-chloro benzyl alcohol and m-hydroxy benzyl alcohol (d) m-chloro benzyl alcohol and potassium m-chloro benzoate (a) (b) A natural compound (X), C4H8O2, reduces Fehling’s solution, liberates hydrogen when treated with sodium metal and gives a positive iodoform test. The structure of (X) is (a) CH3CHOHCH2CHO (b) HOCH2CH2CHO (c) CH3COCH2CHO (d) CH3COCH2CH2OH (c) (d) none is correct NaOH A 22. A is : (a) 19. (b) X and Y can be distinguished by : (a) silver-mirror test(b) iodoform test (c) both none (d) (c) (d) Einstein Classes, none is correct Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 12 23. 24. A carbonyl compound with molecular weight 86, does not reduce Fehling’s solution but forms crystalline bisulphite derivatives and gives iodoform test. The possible compounds can be (a) 2-pentanone and 3-pentanone (b) 2-pentanone and 3-methyl-2-butanone (c) 2-pentanone and pentanal (d) 3-pentanone and 3-methyl-2-butanone In the Cannizzaro’s reaction given below, 2Ph CHO OH PhCH 2 OH PhCOO the slowest step is (a) the attack of OH— at the carbonyl group (b) the transfer of hydride to the carbonyl group (c) the abstraction of proton from the carboxylic acid (d) the deprotonation of Ph–CH2OH Einstein Classes, ANSWERS (SINGLE CORRECT CHOICE TYPE) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. d b c b c b d d d c 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. a b a c b c d a c d 21. 22. 23. 24. Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 a a b b CAK – 13 EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE 4. Structure of K formed is Comprehension-1 H CrO (a) Cyclohexanol 2 4 A(C 6 H10O) acetone ( 1) CH MgI HA 3 B(C 7 H 14 O ) C(C 7 H 12 ) heat ( 2 ) H 3O (b) (1 ) O 3 (1) Ag 2O , OH ( 2 ) Zn , HOAc ( 2 ) H 3O D(C 7 H 12O 2 ) E(C 7 H 12O 3 ) 1. (c) The structure of compound B formed is (a) (b) (d) 5. (c) Thus L is (d) (a) 2. The D formed in the above sequence is (b) (a) (c) (b) (d) (c) (d) 3. 6. HOOC(CH2)4COOH CH3CH = CHCO2Et Thus M is (a) CH3CH2CH2CO2Et – D on treatment with Ag 2 O, OH and upon acidification forms (b) (a) (c) (d) (b) CH3CH2COOH Comprehension-3 (c) (d) HOOC(CH2)4COOH A B Comprehension-2 The following reaction sequence shows how the carbon chain of an aldehyde may be lengthened by two carbon atoms. (1) BrCH CO Et , Zn ( i ) NaOH Product(s) ? HA , heat Ethanal 2 2 K (C 6 H 12 O 3 ) ( ii ) H ( 2 ) H 3O H , Pt ( 1) DIBAL H L(C 6 H 10O 2 ) 2 M(C 6 H 12O 2 ) 7. ( 2 ) H 2O butanal 8. Einstein Classes, The missing reagent A is (a) OsO4, NaHSO3 (b) dilute KMnO4 (c) any one of these (d) O3/H2O2 The missing reagent B is (a) HIO4 (b) (c) any one of these (d) NaIO4 None Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 14 9. The product(s) formed is/are 8. (a) (c) (b) both of these (d) none MULTIPLE CORRECT CHOICE TYPE 1. 2. 3. 4. 5. 6. 7. A new carbon-carbon bond formation is possible in (a) Cannizoro reaction (b) Friedal-craft reaction (c) Clemensen reduction (d) Reimen-Tiemann reaction Which of the following undergoes Aldol Condensation (a) Acetaldehyde (b) propionaldehyde (c) Benzaldehyde (d) Trideuteroacetaldehyde 9. 10. (c) 2-Methyl propionaldehyde (d) 2, 2-Dimethyl-propionaldehyde On heating which of the following get decarboxylated ? (a) Methylmalonic acid (b) succinic acid (c) 2, 2-Dimethyl acetoacetic acid (d) Maliec acid Cannizaro reaction will be given by (a) (b) (c) (d) Which of the following is/are correct about an acetal ? (a) it is a condensation product of two molecules of aldehyde (b) it is the condensation product of two molecules of ketones (c) it is the condensation product of one molecular of aldehyde two molecules of Alcohols it is a diether Which of the following on reduction with LiAlH4 with give ethyl alcohol ? (a) (CH3CO)2O (b) CH3COCl (d) (c) CH3CO NH2 (d) CH3COOC2H5 Assertion-Reason Type Acetophenone is obtained when (a) Benzoylchloride treated with dimethyl cadmium (b) Acetylchloride is treated with dimethyl cadmium (c) Acetyl chloride is treated with benzene in presence of anhydrous AlCl3 (d) Benzoyl chloride is reduced with H2 in presence of Lindar’s catalyst Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Which of the following on oxidation with alk. KMnO4 followed by acidification with HCl gives benzoic acid (a) Methyl benzene (b) Ethyl benzene (c) O-xylene p-xylene (d) Which of the following statements about Benzaldehyde is/are true ? CCl3 CHO 1. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True STATEMENT-1 : H2CO is a planar molecule. STATEMENT-2 : In H2CO, C is sp2 hybridised 2. STATEMENT-1 : H 2 + PhCH 2 CH 2 COCl Pd / C (a) Reduces Tollen’s reagent S Ph CH2CH2CHO or quinoline (b) Undergoes Aldol condensation above reaction is correct. (c) Undergoes cannizaro’s reaction (d) Does not form an addition compound with NaHSO3 STATEMENT-2 : The reduction stops at the aldehyde stage because the catalyst is poinsoned by adding quinoline. Base-catalysed Aldol condensation occurs with (a) propionaldehyde (b) Benzaldehyde Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 15 8. 3. STATEMENT-1 : H SO 2 4 STATEMENT-2 : Carbonyl compounds cannot form hydrogen bonding with H2O. 9. STATEMENT-2 : In Pinacol-pinacolone rearrangement formation of the cyclopbtyl C+ is unlikely because the ring strain decreases. STATEMENT-1 : Boiling points of carbonyl compounds are higher than those of alkanes. STATEMENT-2 : Boiling points of carbonyl compound are higher because of hydrogenbonding. The above reaction is pinacol-pinacolone rearrangement. 4. STATEMENT-1 : Aldehydes are more reactive than carboxyl group. 10. STATEMENT-1 : When C 6 H 6 react with CO + HCl in the presence of CuCl/AlCl3, it form C6H5CHO. STATEMENT-1 : The reaction of aldehydes and ketones with LiAlH4 and NaBH4 is uncleophilic addition reaction. STATEMENT-2 : Aldol condensation reaction is not given by C6H5CHO. STATEMENT-2 : The electrophilic aromatic substitution take place in above reaction. 5. STATEMENT-1 : Under wollff-kishner reduction RCHO reduce to CH4. STATEMENT-2 : Under Clemmenson reduction pHCH2CHO reduce to pHCH2CH3 6. STATEMENT-1 : H2CO is always oxidised in the crossed cannizzaro reaction. STATEMENT-2 : H 2CO is the most reactive aldehyde, it exists in aqueous OH– solution mainly as the conjugate base of its hydrate, H2C(OH)O–. STATEMENT-1 : PCC oxidises 1 0 alcohol to aldehyde and 20 alcohol to ketone. 7. STATEMENT-2 : Cu dehydrates 30 alcohol to alkene. (Answers) EXCERCISE BASED ON NEW PATTERN COMPREHENSION TYPE 1. a 2. b 3. a 7. c 8. c 9. c 4. c 5. c 6. a 5. a, b 6. a, c 5. D 6. A MULTIPLE CORRECT CHOICE TYPE 1. b, d 2. a, b, d 3. a, b, d 4. a, c 7. a, c 8. a, c 9. b, c, d 10. c, d ASSERTION-REASON TYPE 1. B 2. A 3. C 4. A 7. B 8. C 9. C 10. B Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 16 INITIAL STEP EXERCISE (SUBJECTIVE) 1. Complete the following equations and write down the names of the products : A ketone (A) which undergoes haloform reaction gives compound (B) on reduction. (B) on heating with sulphuric acid gives compound (C), which forms monozonide (D). (D) on hydrolysis in presence of zinc dust gives only acetaldehyde. Identify (A), (B) and (C). Write down the reactions involved. 8. Show, by equations, how acetaldehyde may be converted into : . KCN / H 2SO 4 1 ? (i) =O (ii) i ) BD 3 ,THF CH 3CH 2 C CH ( ? 2. LiAlH 4 ( ii ) H 2 O 2 , OH 2. 7. Identify the (A), (B), (C) and (D) in the following reactions sequence : (i) CH 3 CH CH CHO NaBH 4 ( HCl (i) EtOH; (ii) CH3COCl; (iii) CHI3; (iv) MeCOEt; (v) CH3CO2Et; (vi) CH3CHOHCO2H; (vii) EtNH2; (viii) CH3CHBrCHBrCO2H. 9. Convert ClCH2CH2CHO into HOCH2CHOHCHO. 10. Compound (A) and (B) on reaction in ether medium and subsequent acidification and oxidation give 2, 5-dimethyl 3-hexanone. What are (A) and(B) ? 11. Identify (A) to (E) as reactant, reagent, product or name of the reaction in following : KCN (A) ZnCl 2 (B) (C) H (ii) ( alc.) Br2 CH 3CH 2CH 2 Br KOH (A) ( Hg 2 KOH ( alc.) (i) (B) followed by NaNH 2 (C) H (D) 2SO 4 ( dil.) (iii) Aldol condensation (ii) PCl5 Ag 2 O CH 3CHO (A) (B) Rosenmund’s reaction (iii) CH 3 MgBr Pd / BaSO 4 (C) (D) H 2O / H 3. What happens when (Give equation only) Chloral is heated with aqueous sodium hydroxide. 4. How will you differentiate between 5. 6. (C) CH 3COCl H 2 (D) NH 2 NH 2 H2 (i) A) 3(CH 3 ) 2 C O ( ( B) Acetaldehyde and Acetone. ( E ) C CH 3CH 2 CH 3 (F) 2 H 5 ONa 12. Write the structural formula of the main organic prudent formed when : the compound obtained by the hydration of ethyne is treated with dilute alkali. 13. How can we convert PhCH = CHCOCH3 to How would you bring the following conversions : (a) PhCH = CHCOOH (a) Acetone from acetaldehyde. (b) PhCH = CHCHOHCH3 (b) Methanal to ethanal (not more than 3 steps) (c) PhCH2CH2COCH3 (d) PhCH = CHCH2CH3 (c) Acetone from methane. (e) Ph(CH2)3CH3 (d) 2-Butanone from ethyl alcohol. (e) 3-Hexanone from n-propyl alcohol. 14. An organic compound (A) contains 40% carbon and 6.7% hydrogen. Its V.D. is 15. On reaction with a conc. solution of KOH, it gives two compounds (B) and (C). When (B) is oxidised, the original compound (A) is obtained. (C) give effervecence with NaHCO3. Write the structures of A, B, C and D and explain the reactions. 15. An organic compound (A), C4H9Cl on reacting with aqueous KOH gives (B) and on reaction alc. KOH gives (C) which is also formed on passing the vapours of (B) over heated copper. The compound Arrange the following : (i) CH3CHO, CH3COCH3, HCHO, C2H5COCH3 in decreasing order of nucleophilic addition. (ii) CH3COCH2CHO,CH3COCH3, CH3CHO, CH3COCH2COCH3 in increasing order of expected enol content. Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 17 (C) readily decolourizes bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to give (F) and the compound (E) reacts with NaOH to give an alcohol (G) and sod. salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg++ and H2SO4. Identify (A) to (H) with proper reasoning. 16. How would you convert : 17. An organic compound A, C8H6, on treatment with dilute H2SO4 containing HgSO4 gives a compound B which can also be obtained from the reaction of benzene with an acid chloride in the presence of anhydrous AlCl3. The compound B when treated with iodine and aq. KOH, yields C and a yellow compound D. Identify A, B, C and D with justification. Show how C and D is formed. 18. A certain hydrocarbon A was found to contain 85.7% C and 14.3 % H. This compound consumes 1 molar equivalent of hydrogen to give a saturated hydrocarbon B. 1.0 g of hydrocarbon A just decolourised 38.05 of a 5 percent s o l u t i o n (by weight) of Br 2 in CCl 4 . Compound A on oxidation with concentrated KMnO 4 gave compound C (molecular formulae C4H 8O) and acetic acid. Compound C could easily be prepared by action of acidic aqueous HgSO4 on 2-Butyne. Determine the molecular formula of A and deduce the structure of A, B and C FINAL STEP EXERCISE (SUBJECTIVE) 1. How would you bring the following conversions : (a) (b) (ii) Ethanal to 2-hydroxy-3-butenoic acid. 5. Compound (A), C10H12O gives off hydrogen on treatment with sodium metal and also decolourises Br2 in CCl4 to give (B), C10H12OBr2. (A) on treatment with I2 in NaOH gives iodoform and an acid (C) after acidification. Give structures of (A) to (C) and also of all the geometrical and optical isomers of (A). 6. A organic compound (A), C9H12O was subjected in a series of test in the laboratory. It was found that this compound Propanal to 2. Convert : (i) n-butanol into 2-ethylhexan-1-ol; (ii) bromocyclohexane into cyclopentancarboxylic acid. 3. Suggest what compounds the letters stand for in the following equations : (i) i ) NaNH 2 MeCOMe CH CH ( B ( ii ) A (ii) (i) Rotates the plane of polarised light. (ii) Evolves hydrogen with sodium (iii) Reacts with NaOH and I2 to produce a pale yellow solid compound. (iv) Does not react with Br2/CCl4. (v) Reacts with hot KMnO4 to form compound (B) C7H6O2 which can also be synthesised by the reaction of benzene and carbonyl chloride followed by hydrolysis. (vi) looses optical activity as a result of formation of compound (C) on being heated with HI and P. (vii) Reacts with Lucas reagent in about 5 min. Give structures of (A) to (C) with proper reasoning and draw fischer projections for (A). Give reactions for the steps wherever possible. NaNH 2 PhCOMe ClCH 2 CO 2 Et C (iii) BuOK ClCH 2 CO 2 Et t D t BuOH 4. Complete the following equations and give reasons for your answer. (i) H Me 2 C(OH)CH 2OH ? Einstein Classes, H PhMeC(OH)C(OH)MePh ? Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111 CAK – 18 7. 8. A neutral organic compound (A) on analysis gave C = 52.17%; H = 13.04%. On mild oxidation (A) gave (B) which on treatment with dilute NaOH gave another neutral compound (C) with double the molecular mass. Compound (C) readily loses water on heating to form (D) which on oxidation gave CH3CH = CHCOOH (crotonic acid). Identify (A), (B), (C) and (D) and explain the reactions. (a) 11. Identify A, B, in the following set up O3 A B OH (b) 9. 10. An alkyne with 5 carbon atoms per molecule when passed through dilute sulphuric acid containing mercuric sulphate gives a compound which forms an oxime, but has no effect on Fehling’s solution. The compound on oxidation gives dimethyl acetic acid. It reacts with sodamide to forms a hydrocarbon. What is the structure of the alkyne ? When 0.0088 g of compound (A) was dissolved in 0.50 g. of camphor, the melting point of camphor was lowered by 80C. Analysis of (A) gave 68.18% C and 13.16% H. Compound (A) showed the following reactions : (i) It reacted with acetyl chloride and evolved hydrogen with sodium, (ii) When reacted with HCl + ZnCl2, a dense oily layer separated out immediately. Compound (A) was passed over Al2O3 at 3500C to give compound (B). (B) on ozonolysis followed by hydrolysis gave two neutral compounds (C) and (D) which gave positive tests with carboyl reagents but only (C) gave a positive test with Fehling solution. Identify (A), (B), (C) and (D) with proper reasoning. Kf for camphor = 40 mol–1 kg–1. Suggest the appropriate structures for the missing compound. (The number of carbon atoms remain the same throughout the reactions) Identify A to G. 12. Wolff-Kishner reduction of the compound shown gave compound A. Treatment of compound A with m-chloroperoxy benzoic acid gave B which on reduction with LiAlH 4 gave C. Oxidation of compound C with chromic acid gave compound D (C9H14O). Identify A, B, C and D. 13. An optically active alkyne A contains 89.52% C and 10.48 % H. After hydrogenation over a Pd catalyst it is converted to 1-Methyl-4-propyl cyclohexane. When compound A reacts with CH3MgBr no gas is liberated. Hydrogenation of A over Lindlar catalyst, followed by ozonylysis and reaction with KMnO4 gives product B. Product B reacts with I2/NaOH and gives a yellow precipitate, which is filtered off. Acidification of the filterate gives an optically active product C. Give the structures of A, B and C and account for all observations. 14. Give the structure for compounds A – F : . CH 3 MgI / 2 CrO 4 C 6 H12 O H B 1 C HA A acetone 2. H 3 O .O3 ,OH D 1 E Ag 2O F 2. Zn / CH 3COOH 15. 2. H 3 O C6H12O 6 can be glucose or fructose. To decide nature, one can carry out following steps : dil . KMnO 4 A HIO 4 B HO C / HI 3O C6 H12O 6 HCN H P ? What can be the end product starting with glucose and fructose ? Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi – 110 018, Ph. : 9312629035, 8527112111