31 Aromatic Aldehydes, Ketones and Acids

31 Aromatic Aldehydes, Ketones and Acids Objective Questions I (Only one correct option) 1. The major products of the following reaction are OH 3. Compound A(C9 H10 O) shows positive iodoform test. Oxidation of A with KMnO4 / KOH gives acid B (C8 H6 O4 ). Anhydride of B is used for the preparation of phenolphthalein. Compound A is O (1) CHCl3/aq. NaOH CH2 (2) HCHO, NaOH (conc.) (3) H3 OH Cl O COOH and methanol Cl OH OH OH CH3 (d) CH3 O O OH and formic acid OH (d) and formic acid 4. The major product Y in the following reaction is Ph CH3 NaOCl Cl OH CH3 CH3 (c) (i) SOCl2 (ii) Aniline Y O Ph N (i) CrO3 (ii) SOCl2/D X O 2. The major product of the following reaction is HO CH3 CH3 and methanol (b) (c) H (b) OH COOH (a) C (a) O+ HN O (a) Ph (b) (iii) D HO O O NH2 (c) (b) (a) NH2 O (d) Ph HO O HO O Ph 5. Major products of the following reaction are O CHO (d) (c) + HCHO Cl Cl (i) 50% NaOH (ii) H3O+ Aromatic Aldehydes, Ketones and Acids 9. The major product obtained in the following conversion is (a) CH3OH and HCO2H COOH CH3 (b) CH3OH and O O Br2 (1 eqv.) MeOH CH2OH (c) HCOOH and CH3 and (d) O O COOH CH2OH CH3 O Br O OMe (a) 6. The major product of the following reaction is O (b) Br OMe O O O CH3 Br NaBH4 EtOH (c) (a) (d) (b) Br O O 10. The major product of the following reaction is OH OH (c) COCH3 (d) (i) KMnO4/KOH,D OEt (ii) H2SO4 (dil.) CH3 7. In the following reactions, the product S is COOH (a) H3C (i) O3 R (ii) Zn,H2O (i) NH3 H3C N (b) HOOC S HOOC H3C (d) HOOC OHC N 11. The major product formed in the following reaction is CH3 O (d) H3C H3C H H3C Dil. NaOH + 8. The major product of the following reaction is O CN (i) DIBAL-H (ii) H3O+ O CHO CHO (b) H3C OH O OH O (b) H3C O O OH (a) O (a) O CH (c) CHO NH H (c) H3C O (d) OH OH COCOOH COOH N N (c) COCH3 (b) (c) (a) O O O O OH CH3 O OH CHO (d) H OH H3C Aromatic Aldehydes, Ketones and Acids 12. The major product obtained in the following reaction is 16. Which of the following reactants on reaction with O conc. NaOH followed by acidification gives following lactone as the main product ? O DIBAL-H O COOH OH O OH (a) (b) CHO CHO COOH (c) COOCH3 CHO (d) CHO (b) COOH CHO COOH (d) CHO H / Pd 4 2 2 A ¾¾® B ¾¾® C Toluene ¾¾® BaSO 4 The product C is (a) C6 H5 COOH (c) C6 H5 CH2 OH to give CH3 14. Sodium phenoxide when heated with CO2 under pressure at COONa ; CH3CO2H (a) 125°C yields a product which on acetylation produces C. 125 ° COOH 17. 4-methyl benzene sulphonic acid reacts with sodium acetate (b) C6 H5 CH3 (d) C6 H5 CHO ; SO3 (b) + ONa + CO2 ¾ ¾ ¾® B ¾H¾® C 5 atm COOH (c) 13. In the following sequence of reaction, SOCl CHO CHO CHO KMnO COOH (a) Ac2O SO3Na CH3 OCOCH3 SO3H The major product C would be OCOCH3 OH COOH (a) (b) ; SO3 (c) COCH3 (d) ; NaOH CH3 CH3 Hg 2+ / H+ 18. Ph ¾ C ºº C ¾ CH3 ¾¾¾® A , A is COCH3 OH O OCOCH3 COOCH3 (c) Ph (d) O– Ph Ph (c) CHO (b) H3C OHC (i) NaOH/100°C 19. (ii) H+/H2O CH3 CH2Br CHO CH2Br (c) (d) CH3 CH3 OH (d) H 3C C2H5 Br H3C H 3C warmed with alcoholic AgNO 3 . Oxidation of (A) gives an acid (B), C8 H6 O4 . (B) easily forms anhydride on heating. Identify the compound (A). CH2Br O (b) COOH 15. Compound ( A ) , C8 H9 Br gives a white precipitate when (a) Ph (a) Major product OHC Aromatic Aldehydes, Ketones and Acids HOOC 22. m-chlorobenzaldehyde on reaction with conc. KOH at room COOH HOOC temperature gives (a) potassium m-chlorobenzoate and m-chlorobenzyl alcohol (b) m-hydroxy benzaldehyde and m-chlorobenzyl alcohol (c) m-chlorobenzyl alcohol and m-hydroxy benzyl alcohol (d) potassium m-chlorobenzoate and m-hydroxy benzaldehyde CH2OH HOOC COOH (a) (b) Objective Questions II (One or more than one correct option) 23. Compound P and R upon ozonolysis produce Q and S , COOH HOH2C O O respectively. The molecular formula of Q and S is C8H 8O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction. (i) O3/CH2Cl 2 (i) P ¾¾¾¾® (ii) Zn/H2O (c) (i) O3/CH2Cl 2 (ii) R ¾¾¾¾® Q (ii) Zn/H2O (C 8H 8O) S (C 8H8O) The option(s) with suitable combination of P and R, respectively, is(are) O H3 C O CH2OH CH3 CH3 HOH2C (a) CH3 and CH3 CH3 (d) CH2OH HOH2C and H3C (b) H3C CH3 20. In Cannizzaro’s reaction, the intermediate which is the best hydride donor is H ½ (a) C6 H5 ¾ C ¾ O½ OH H ½ (b) C6 H5 ¾ C ¾ O½O H ½ C ¾ O– ½ O– (c) CH3 H3C CH3 H3C MeO CH3 24. Positive Tollen’s test is observed for OH O H H ½ C ¾ O– ½ O– CH3 and (d) O2N (d) and (c) H3C CHO Ph O Ph H H Br (a) O (b) (c) Ph Ph (d) 25. The aldehydes which will not form Grignard product with 21. In the Cannizzaro’s reaction given below : one equivalent Grignard reagents are KOH 2Ph ¾ CHO ¾® Ph ¾ CH2OH + PhCO–2 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 ¾ CH2 OH CHO CHO (b) (a) HO2C CHO CHO (d) (c) HO3CO HOH2C Aromatic Aldehydes, Ketones and Acids 29. The compound S is Passage Based Questions Paragraph X Treatment of benzene with CO / HCl in the presence of anhydrous AlCl 3 / CuCl followed by reaction with Ac2 O / NaOAc gives compound X as the major product. Compound X upon reaction with Br2 / Na 2 CO3 followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H2 / Pd - C, followed by H3 PO4 treatment gives Z as the major product. (a) (b) NH2 HN 26. The compound Y is OH NH2 (c) COBr H N (d) Br (a) (b) O HO Passage Br COBr In the following reactions sequence, the compound J is an intermediate. (d) (c) Br 27. The compound Z is (a) (CH3 CO) 2 O (i) H2 , Pd/C CH3 COONa (ii) SOCl 2 (iii) anhyd. AlCl 3 I ¾¾¾¾¾® J ¾¾¾¾¾¾® K J (C9 H8 O2 ) gives effervescence on treatment with NaHCO3 and positive Baeyer’s test. (b) 30. The compound K , is O O (a) OH (b) O (c) (d) (c) (d) O An organic acid P ( C11 H12 O2 ) can easily be oxidised to a dibasic acid which reacts with ethylene glycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S . (1) H2/Pd-C (2) NH3/D (3) Br2/NaOH (4) CHCl3/KOH, D (5) H2/Pd-C P (1) H2/Pd-C (2) SOCl2 (3) MeMgBr,CdCl2 (4) NaBH4 Q (1) HCl (2) Mg/Et2O 31. The compound I, is O H (a) (b) O CH3 (c) (3) CO2(dry ice) (4) H3O+ OH H H H (d) R Fill in the Blanks 28. The Compound R is 32. The structure of the intermediate product formed by the HO2C (a) O O Paragraph A S O oxidation of toluene with CrO3 and acetic anhydride, whose hydrolysis gives benzaldehyde is …………… CO2H (b) True/False 33. Benzaldehyde undergoes aldol condensation in an alkaline (c) CO2H (d) CO2H medium. Aromatic Aldehydes, Ketones and Acids Subjective Questions Column 1 Column 2 Column 3 (I) Toluene (i) NaOH/ Br2 (P) Condensation A to E with molecular formula C8 H8 O2 were given for identification. Based on the following observations, give structures of the compounds : (II) Acetophenone (ii) Br2/hn (Q) Carboxylation (III) Benzaldehyde (iii) (CH3CO)2 O/ CH3COOK (R) Substitution (i) Both A and B form a silver mirror with Tollen’s reagent; also, B gives a positive test with FeCl 3 solution. (ii) C gives positive iodoform test . (iii) D is readily extracted in aqueous NaHCO3 solution. (iv) E on acid hydrolysis gives 1, 4-dihydroxy benzene. (IV) Phenol (iv) NaOH/CO 2 (S) Haloform 34. Five isomeric para-disubstituted aromatic compounds 35. An organic compound A, C8H4O3, in dry benzene in the presence of anhydrous AlCl3 gives compound B. The compound B on treatment with PCl5, followed by reaction with H2/Pd/(BaSO4) gives compound C, which on reaction with hydrazine gives a cyclised compound D (C14H10N2). Identify A, B, C and D. Explain the formation of D from C. 36. Explain, why o-hydroxy benzaldehyde is a liquid at room temperature while p-hydroxy benzaldehyde is a high melting solid? 37. The only CORRECT combination in which the reaction proceeds through radical mechanism is (a) (IV) (i) (Q) (b) (III) (ii) (P) (c) (II) (iii) (R) (d) (I) (ii) (R) 38. For the synthesis of benzoic acid, the only CORRECT combination is (a) (II) (i) (S) (c) (IV) (ii) (P) (b) (I) (iv) (Q) (d) (III) (iv) (R) 30. The only CORRECT combination that gives two different carboxylic acids is (a) (IV) (iii) (Q) (c) (I) (i) (S) (b) (II) (iv) (R) (d) (III) (iii) (P) Integer Type Questions Matching Type Questions 40. Among the following the number of reaction(s) that Answer Q. 22, Q. 23 and Q. 24 by appropriately matching the information given in the three columns of the following table. Column 1, 2 and 3 contain starting materials, reaction conditions, and type of reactions, respectively. produce(s) benzaldehyde is CHCl2 I. H 2O 100°C CO, HCl II. Anhydrous AlCl3 / CuCl CO2Me COCl H2 Pd-BaSO4 III. IV. DIBAL-H Toluene, -78°C H2O Answers 1. (d) 5. (c) 9. (b) 2. (b) 6. (a) 10. (a) 3. (c) 7. (a) 11. (c) 4. (b) 8. (d) 12. (a) 21. (b) 22. (a) 23. (b,c) 24. (a,b,c) 25. (b,d) 26. (c) 27. (a) 28. (a) 29. (b) 30. (c) 31. (a) 13. (d) 14. (a) 15. (d) 16. (c) 18. (a) 19. (b) 20. (d) 32. C6H5 ¾ CH(OAc)2 33. False 17. (a) 38. (d) 40. (4) 39. (b) 37. (a) Hints & Solutions 1. The major products of the given reaction are as follows: OH Further, heating of product leads to intramolecular cyclisation. OH COCl OH (1) CHCl3/aq. NaOH HO Cl HO Cl In step-I, substituted phenol undergoes Reimer-Tiemann reaction in presence of CHCl3 / aq. NaOH OH C==O D and HCOOH (2) HCHO, conc. NaOH (3) H3O+ 3. (i) C9H10O shows positive iodoform test thus, ¾ C ¾ CH3 || O group is present. OH (ii) C9H10O on strong oxidation (KMnO4 / KOH), gives acid (C8H6O4 ), indicating it can be a dicarboxylic acid. So, ‘A’ contains ¾COCH3 and one ¾CH3 group which get oxidised into ¾COOH and ¾COOH respectively. CHO (1) CHCl3/aq. NaOH Cl Cl The aldehyde obtained in above equation does not possess a-hydrogen. In presence of formaldehyde and conc. NaOH it undergoes Cannizaro reaction. In this reaction, one molecule of aldehyde is reduced to alcohol while another molecule is oxidised to salt of carboxylic acid. O–Na+ OH Conc. OH + HCOO–Na+ NaOH Cl CH3 C O C (i) [O] KMnO4/KOH r (ii) H3O A(C9H10O) Upon hydrolysis, following reaction takes place O–Na+ + CHI3 – + COONa Yellow ppts. (Iodoform test) O Cl CH3 I2/NaOH CH3 CHO + HCHO (iii) In the preparation of phenolphthalein from phenol, phthalic anhydride is used. So, ‘B’ can be phthalic acid (benzene-1,2dicarboxylic acid) which readily forms anhydride. Thus, the reaction sequence is as follows : C OH OH + HCOO–Na+ O H3O+ Phthalic acid B(C8H6O4) OH + HCOOH H D –H2O (2 mol)/ Conc. H2SO4 Cl Cl 2. The major product formed in the reaction is as follows : HO (2) SOCl2/D (3)D HO O C OH O –H2O ==O (1) CrO3 OH + CO2 OH C O Phthalic anhydride O HO Primary alcohol readily oxidised to corresponding carboxylic acid with oxidising agent, chromium trioxide (CrO3) in acidic medium. O COOH HO CrO3 HO HO ¾ OH group of carboxylic acid get substituted by ¾Cl in presence of SOCl2 (Thionyl chloride). COOH COCl SOCl2/D + SO2­ + HCl­ HO HO OH OH (Phenolphthalein an indicator) 4. NaOCl (sodium hypochlorite) is the reagent of haloform (chloroform formation) reaction. 2NaOH + Cl2 ¾® NaOCl + NaCl + H 2O Aromatic Aldehydes, Ketones and Acids 7. Key Idea Grignard reagent usually attacks on > C == O group The given reaction takes place as follows : O as : ( C == O + RMgX ¾® ' O NaOCl Ph—C—CH3 s r Ph—C—ONa + CHCl3 Hydration ( C ' OMgX ¾¾¾® ( C ' OH ' (R - Mg (OH) X ' ( R r H3O (It is missing in the statement of the question) Keto-methyl group (Ionic salt) O The question is related to above reaction only with the condition that the consumption of RMgX will be more than 1 equivalent in some of the given cases. Among the given compounds B, i.e. Ph—C—OH Benzoic acid (X) O O CHO SOCl 2 Ph—C—Cl Ph—C—OH – SO2 Benzoyl (X) – HCl chloride [Acyl SN2 pathway] –HCl H2N— HO2C Aniline and D, i.e. HOH2C contain additional groups which can give active hydrogens. Grignard reagents produce alkanes whenever come in contact with any group or compound which can give active hydrogen as: O —NH—C—Ph Benzanilide Dry ether ROH + R ¢ MgX ¾ ¾ ¾® R¢ H + Mg ' ( (Y ) 5. The given reaction is a crossed Cannizzaro reaction which is a redox reaction too. Oxidation number of carbon atom of the ¾CHO groups of Ph¾CHO and H¾CHO are +1 and zero respectively. So, HCHO is the stronger reducing agent than PhCHO. As a result, HCHO is oxidised to HCOONa (by donation of hydride, H- ) and PhCHO (H- acceptor) is reduced to PhCH 2OH. +1 0 50% NaOH Ph ¾ C HO + HC HO ¾¾¾¾® -1 +2 CHO ¯ H3 O CHO (i) + CHO Dry ether + RMg X HO2C RH + C d+ d– X MgO PhCH2OH + HCOOH O (Major products) CHO The reaction proceed via following mechanism. H2C= =O d+ d– – OH Slower More polar and less crowding for nucleophilic addition OH | s H—C—O | H X These reactions are equivalent to acid-base reactions. So, in both of these compounds more than one equivalent will be required to form Grignard products. Remember these compounds will give 2 type of products as: (i) from the >C == O group (ii) from the group which release active hydrogen The additional reactions involved are: - + Ph ¾ C H2OH + HC OONa ½ OR + RMg X (ii) Slowest Ph—CH= =O – H shift OH | H—C= =O + Ph—CH2—Os Ö Fast Os | H—C= =O + PhCH2OH 6. Reducing agents like LiAlH4, NaBH4, i.e. complex hydrides usually does not affect olifenic or p-bonds. Thus, if NaBH4 is O HOH2C CHO Dry ether + RH XMgOH2C 8. DIBAL-H (Diisobutylaluminium hydride) is a reducing agent with formula [l-Bu 2AlH ] . At ordinary temperatures, nitriles give imines which are readily converted in aldehydes by hydrolysis whereas lactones are reduced directly to aldehydes. H Al Al (DIBAL–H) (DIBAL–H) + H N + Al N CN C C H O O O O O then its ( C == O bond will be ' applied to a compound like O H 2O OH reduced only and we get as the final product. Thus, option (a) is correct answer. CHO CHO +H+ O OH H O –O CHO DIBAL–H (Hs) H O O Aromatic Aldehydes, Ketones and Acids 9. In presence of Br2/EtOH, the reactant containing double bond undergoes electrophilic addition reaction via the formation of ·· (iii) CH3—C—C6H5 bromonium ion. On further attack of ¾ O Me on bromonium CH2 ·· H—C ion gives the addition product. H3 C O Br2(1 eqiv) in MeOH O O COMe O (Protic solvent) MeOH Br O O O OMe Br O (Major) 10. In presence of alkaline KMnO4, vigorous oxidation of alkyl or O OH (Self Aldol condensation (iv) C6H5—C—CH2—C—C6H5 product of C H COCH ) 6 5 3 CH3 – O O H of base will prefer to attack on ¾CH3 group of CH3CHO for the formation of carbanion and as among the > C == O groups available, the > C == O group of CH3CHO is the best carbanion acceptor. Hence, self condensation product of CH3CHO will be the major product. 12. DIBAL-H (Di-isobutyl aluminium hydride) is a reducing agent with formula. This is generally used for the preparation of aldehydes. Using DIBAL ¾ H, Lactones are reduced directly to aldehydes. O acyl benzene takes place. During oxidation, aromatic nucleus remains intact but the entire chain is oxidised to —COOH group irrespective of the length of carbon chain. O==C—CH3 OH O DIBAL–H sÅ Oxidative cleavage in strong (MnO4– /OH–/D–) (Cross Aldol condensation of CH3CHO and CH3COC6H5 in which >C=O group of C6H5COCH3 is carbanion acceptor). OH CHO COOH COOH O == C—O K 13. Toluene undergoes oxidation with KMnO4, forms benzoic acid. KMnO4/KOH/D Strong oxidation, [O] sÅ CH3 COOK (also oxidisable) + H3O (dil. H2SO4) COOH In this conversion, alkyl part of toluene converts into carboxylic group. Further, benzoic acid reacts with thionyl chloride (SOCl 2) to give benzoyl chloride which upon reduction with H2/Pd or BaSO4 forms benzaldehyde (Rosenmund reduction) The conversion look like, O Cl CH3 COOH C KMnO4 SOCl2 + SO2 + HCl COOH 11. In aldol condensation, generally aldehydes react at a faster rate (Toluene) (Benzoic Acid) ‘A’ than ketones towards base. In the given case CH3CHO will lose O (Benzoyl chloride) ‘B’ H2/Pd BaSO4 O a-hydrogen faster than CH3 due to one more reason, i.e. conjugation between benzene ring and ‚ ƒC == O group. Along with sterically less hindered nucleophile of CH3CHO will also add to the major product formation. Following four products are possible in the reaction: OH H (ii) CH3—C—CH2—C—C6H5 OH + HCl (Benzaldehyde) ‘C’ 14. It is a Kolbe Schmidt reaction. + H (i) CH3—C—CH2—CHO H C C (Self Aldol condensation of CH3CHO) + ONa ½ COONa ½ OH + CO2 125°C H+/Ac2O COOH O ½ OC ¾ CH3 5 atm (Cross Aldol condensation of CH3CHO and CH3COC6H5 in which >C=O group of CH3CHO is carbanion acceptor). ‘B’ Aspirin (pain killer) ‘C’ The second step of the reaction is an example of acetylation reaction. Aromatic Aldehydes, Ketones and Acids 15. Compound A gives a precipitate with alcoholic AgNO3, so it 20. Dioxoanion is better hydride donor. Electron donating group at ortho/para position further promote H– transfer. must contains Br in side chain. On oxidation, it gives C8 H6O4, which shows the presence of two alkyl chains attached directly with the benzene nucleus. Since, compound B gives anhydride on heating, the two alkyl substituent must occupy adjacent (1, 2) position. Thus, A must be O – O C—H CH2 Br O– H3CO + C—R H O CH3 C and the reactions are as follows : CH2Br ‘A’ CH3 H3CO + AgBr CH3 + H— C —R H D O - COOH O Phthalic anhydride CHO COOH OH– Intramolecular Cannizzaro reaction CHO O O ½ ½½ 21. Ph ¾ C ¾ H + HO- ¾® Ph ¾ C ¾ H ½ OH I O½ Ph ¾ C ¾ H ½ OH CH2OH O O ½½ Slow C ¾ Ph ¾¾¾¾¾® hydride transfer ½ H + I H+ O O½ ½½ Ph ¾ C ¾ OH + Ph ¾ C ¾ H ½ H O Ester formation 17. O – O COOH 16. O– + CH2 OR Alcoholic AgNO3 Oxidation ‘B’ – H3C— —SO3H + CH3COONa Base 22. Cl Cl Cl 4-methyl benzene sulphonic acid H3C— + KOH —SO3Na + CH3COOH + CHO H+ 18. Ph — C ºº C— CH3 ¾® Ph — C==CH— CH3 Stable carbocation O ½½ H2O ¾¾® Ph — C — CH2CH3 23. (b) CH3 CH CH2 O3 Zn–H2O CH3 (P) OHC CHO CH2OH COOK Cannizzaro’s reaction + CHO+H2CO (R) Gives Cannizzaro but not haloform reaction O NaOH 19. O3 Zn–H2O C (Q ) CHO OHC H COOH HOH2C (c) + CH CH CH3+H2CO (S) Gives haloform reaction but not Cannizzaro reaction CH3 O3 Zn–H2O CHO+CH3CHO (R) P H2O CH3 CH2OH HOOC The above reaction is an example of intramolecular Cannizzaro reaction. CH3 Gives Cannizzaro but not haloform reaction Aromatic Aldehydes, Ketones and Acids 24. RCHO + OH - ¾ ¾¾® RCOOH + Ag2O 2Ag (Tollen' s reagent) Silver mirror Tollen’s test is given by all aldehydes and all reducing sugars as glucose, fructose and a-hydroxy ketones. OH For this question we require only reaction 1 to 4 written above. Let us explore them one by one. Reaction 1 It is called formylation or Gatterman Koch reaction. A ¾ CHO group is introduced to benzene ring through this reaction as CHO O CO+HCl In anhyd. AlCl3/CuCl — CH— C — a H O Benzene having an aldehyde group gives Tollen's test. H H The attacking electrophile isH ¾ C == O which is generated as (i) CO + HCl q Br H — C — Cl ½½ O O (c) CHO Benzaldehyde + + CH C OH a-hydroxy ketone give positive Tollen’s test. Aldehyde 25. (ii) H — C — Cl + AlCl 3 q H — C + AlCl -4 ½½ ½½ O O Reaction 2 It is Perkin condensation which results in a, b unsaturated acid as CH CHO CHCOOH O H3C H3C (i) O3 C—H CHO (ii) Zn– H2O H3 C CHO CH NH3 Note Besides CH3COO- Na+, quinoline, pyridine, triethylamine can also be used as bases in this reaction. NH (X) CH C CHCOOH H 3C N—H H NH2 CHBr CHBr COO–Na+ Br2 Na2CO3 OH O Na 2CO3, Reaction 3 It is simple addition of bromine to unsaturated acid formed through reaction 2. In the above reaction, NH3 prefer to attack at aliphatic aldehyde group than an less reactive aromatic aldehyde group. H3C + CH3COOH 180°C Cinnamic acid R Tautomerism Xs CH3COO–Na+ + (CH3CO)2O Cinnamic acid Na2CO 3 works as a base in the reaction to trap H+ to be released OH – H2O O H 3C CH—CH—Br N COO–Na+ by — COOH of the reactant. is also formed ‘S’ 26. Given, 1 CO+HCl Benzene anhy. AlCl3/CuCl X (Major) Br2 + Na2CO3 3 d – C NaOAc in the reaction as the minor product. Reaction 4 It is decarboxylation and dehydrohalogenation of product produced by reaction 3 as CHBr CHBr Moist KOH 473 K 4 H 3P O 4 6 Z Y (Major product) C COO–Na+ H 2 /P 5 2 Ac2O C Hence, Y is Moist KOH 437 K CH + Na2CO3 + KBr CH i.e., (c) is the correct answer. Aromatic Aldehydes, Ketones and Acids Thus P can be 27. Reaction 5 The Perkin condensation product X is CH COOH CHCOOH i.e., cinnamic acid (Ref. Q. 15) CH This compound on hydrogenation with H2 in the presence of Pd activated with charcoal (Pd - C ) gives CH CHCOOH + H2 C CH2CH2COOH CH3 CH3 IUPAC name : 4-(2-methyl) prop-l-enyl benzoic acid Now look for the reactions Pd-C COOH 3-Phenyl propanoic acid COOH Reaction 6 The product of reaction 5 on heating with H3PO4 dehydrates to give CH2CH2COOH CH2 Oxidation CH2 COOH C H3PO4 D (P) O COOH Z 28. (a) Given, (i) CH2 OH OH C O + O CHO CHO other oxidised products (P) 1. H 2 / Pd– C 2. SOCl 2 Given, P ¾¾¾¾¾¾¾® Q O (Ethylene glycol) O + C CHO Aliphatic Glyoxal ketone (Acetone) (An organic acid ‘P’) This indicates the presence of alkyl or alkenyl branch in P along with -COOH group. (ii) Dibassic acid produced by oxidation of P + CHO COOH CHO O Ozonolysis C11H12O2 ¾Oxidation ¾ ¾¾® Dibasic acid CH2 Terephthalic acid —C—O—(CH2)2—O— C 3. MeMgBr, CdCl 2 4. NaBH 4 So, Cl O Dacron COOH This indicates presence of benzene ring in P; as concluded from the structure of dacron given below. H2/Pd––C O O C C O CH2 CH2 COOH CO SOCl2 O MeMgBr, CdCl2 O Dicarboxylic acid component Ethylene glycol component (P) H Dacron Attachment of —COO group in dacron also confirm the para position of branch with respect to —COOH group in P. O Me C NaBH4 [Here, Me = CH3] Ozonolysis (iii) P ¾ ¾ ¾ ¾® Aliphatic ketone + other oxidised products. This reaction confirms the presence of multiple bonded branch i.e., alkenyl group in P. ( Q) Further, 1. HCl 2. Mg/ Et 2O Q ¾¾¾¾¾® R 3. CO2 (Dry ice) 4. H 3O+ C OH Me Aromatic Aldehydes, Ketones and Acids Me H Me C OH Me C H Cl – CMg H 2+ Cl In the second step, J on treatment with H2 / Pd / C undergo hydrogenation at olefinic bond only as : J + H2 / Pd ¾® C6 H5 ¾ CH2 ¾ CH2 ¾ COOH – The hydrogenated acid, on treatment with SOCl 2 gives acid chloride. Mg Et2O (Dry ether) HCl C6 H5 — CH2 — CH2 — COOH + SOCl 2 ¾® C6 H5 — CH2 — CH2 — COCl + HCl + SO2 (Q) Grignard's reagent CO2 (Carboxylation) C AlCl3 Me Me H In the final step, acid chloride formed above undergo intramolecular Friedel-Craft acylation as: C H COOH Cl COOMgCl O CH3 32. H3O+ Mg(OH)Cl + C ‘K’ CH(OAc)2 CrO3 (Hydrolysis) O (CH3CO)2O Intermediate CHO H2O (R ) H+ 29. Given (In connection with Q. 17) 1. H 2 / Pd– C 2. NH 3 / D 3. Br2 / NaOH 33. For aldol condensation, presence of at least one a-H is essential, 4. CHCl 3 / KOH, D 5. H 2 / Pd– C 34. which is not available to benzaldehyde. P ¾¾¾¾¾¾¾® S CHO So, COOH COOH Gives positive Tollen’s test but does not give FeCl3 test. CONH2 NH3/D H2/Pd–C CH2OH A OHC— —OH (P) Gives positive Tollen’s test and FeCl3 test. B CH3 N NH H2/Pd–C C NH2 CHCl3/KOH (Carbylamine) O Br2/NaOH (Hofmann bromamide) CH3—C— —OH C HOOC— —CH3 (S) HO— Passage Sol for (Q. Nos. 30 to 31) The first step of reaction is Perkin’s condensation. CHO + (CH3CO)2O I 35. —OCH==CH2 O + CH CH COOH J J being a carboxylic acid gives effervescence with NaHCO3 . Also, J has olefinic bond, it will decolourise Baeyer’s reagent. Can be extracted with NaHCO3. On hydrolysis give 1,4-dihydroxy benzene. O CH3COONa C6H5 Gives positive iodoform test. O A AlCl3 Aromatic Aldehydes, Ketones and Acids O O Br 2 37. C6H5 ¾ CH3 ¾® C6H5 ¾ CH2Br + HBr hv (Free radical bromination) PCl5 O || NaOH / Br2 38. C6H5 ¾ C ¾ CH3 ¾¾¾¾® C6H 5COONa + CHBr3 (P ) Cl—C HOOC N 2H 4 CH 3 COOK N Pd/BaSO4 Bromoform O O || || 39. C6H 5CHO+ CH3 ¾ C ¾ O ¾ C ¾ CH3 O H2 (i ) (II) O B ¾¾¾¾¾¾® C6H5 ¾ CH == CH ¾ COOH Perkin's condensation H—C (Cinnamic acid) N D O C Cinnamic acid shows cis-trans isomerism. 40. I. Gattermann-Koch reaction. 36. Intramolecular H-bonding in ortho hydroxy benzaldehyde is responsible for decrease in melting and boiling points. O H O II. CH(OH)2 CHCl2 H2O 100°C (SN1) CHO gem-diol –H2O H Intramolecular H-bonding p-hydroxy benzaldehyde molecules are associated by intermolecular H-bonding, has higher melting and boiling points. III. Rosenmund’s reduction. IV. Acid chloride, anhydride and ester undergo controlled reduction with di-iso-butylaluminium hydride (DIBAL-H) at - 78°C to give aldehydes.