29 Aromatic Compounds Containing Nitrogen

29 Aromatic Compounds Containing Nitrogen Objective Questions I (Only one correct option) 1. Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives —NH2 (a) 3. Which of the following is not a correct method of the preparation of benzylamine from cyanobenzene? (a) (b) (c) (d) H 2 / Ni (i) HCl/ H 2O (ii) NaBH 4 (i) LiAlH 4 (ii) H3O+ (i) SnCl2 + HCl(gas) (ii) NaBH 4 4. Aniline dissolved in dil. HCl is reacted with sodium nitrite —N==N — (b) at 0°C. This solution was added dropwise to a solution containing equimolar mixture of aniline and phenol in dil.HCl. The structure of the major product is H2N (c) —N==N— (d) —N==N—NH— —NH2 2. The increasing order of the pK b of the following compound is (a) N N (b) N N (c) N N (d) N N NH2 O OH F S (A) N CH3O N H H S (B) N NH 5. The major product obtained in the following reaction is NH2 N (i) CHCl3/KOH (ii) Pd/C/H2 O2N H (C) H 3C H S CN N N H H S O H NCH3 H NCH3 (b) (a) (D) (a) (b) (c) (d) N N H H (A) < (C) < (D) < (B) (C) < (A) < (D) < (B) (B) < (D) < (A) < (C) (B) < (D) < (C) < (A) CN CN O H NCHCl2 (c) OH H NCH3 (d) CN OH H2N OH Aromatic Compounds Containing Nitrogen 6. The major product in the following reaction is 9. The major product of the following reaction is O NH2 N N H N +CH3I N + NCH3 N N H N N CH3 N (c) N H OH (d) NH2 reaction ? N N (d) + N N N N 10. What will be the major product in the following mononitration NH2 N (ii) DIBAL- H (b) O CHO (c) N NHCH3 (i) Ni/H2 H N (b) N NH (a) NH2 NH2 (a) OEt CN Base H N H CH3 7. Coupling of benzene diazonium chloride with 1-naphthol in alkaline medium will give HNO3 Conc.H2SO4 NO2 (a) OH O2N (b) N H N H (a) (b) (c) N N OH N (d) O2N N N H N H NO2 11. An aromatic compound ‘A’ having molecular formula C7H6O2 on treating with aqueous ammonia and heating forms compound ‘B’. The compound ‘B’ on reaction with molecular bromine and potassium hydroxide provides compound ‘C’ having molecular formula C6H7N. The structure of ‘A’ is OH OHC N (c) CHO (a) (d) N (b) OH N N OH COOH (c) (d) CH==CHCHO OH 12. The increasing basicity order of the following compounds is 8. The increasing order of reactivity of the following compounds (A) CH3CH 2NH 2 CH 2CH3 ½ (B) CH3CH 2NH CH3 ½ (C) H3C ¾ N ¾ CH3 CH3 ½ (D) Ph ¾ N ¾ H towards reaction with alkyl halides directly is O NH2 NH2 (A ) CN O NH2 NH O (B) (a) (A) < (C) < (D) < (B) (b) (B) < (A) < (C) < (D) (c) (B) < (A) < (D) < (C) (d) (A) < (B) < (C) < (D) (C) (D) (a) (D) < (C) < (B) < (A) (b) (A) < (B) < (C) < (D) (c) (A) < (B) < (D) < (C) (d) (D) < (C) < (A) < (B) Aromatic Compounds Containing Nitrogen 13. The tests performed on compound X and their inferences are : 18. The major product of the following reaction is OH Test Inference (a) 2, 4- DNP test Coloured precipitate (b) Iodoform test Yellow precipitate (c) Azo-dye test No dye formation (i) NaNO2/HCl/0°C (ii) (aq. NaOH) NH2 Compound ‘X ’ is H 3C N CH3 H 3C N CH3 (a) (b) N N NH2 CH3 NH2 N N (b) NH2 OH (c) + COCH3 CHO (a) – ONa OH OH (c) CHO (d) OH (d) Cl 14. The major product of the following reaction is 19. In the reaction, NH2 C NH2 CH2CH3 (i) Br2/hn NaNO2/HCl 0.5°C (ii) KOH (dil.) E + N2 CH3 NH (a) CuCN/KCN D D (b) The product E is NH CH3 COOH CH3 (b) H3C (a) NH (c) CH3 NH (d) CH3 CN 15. Arrange the following amines in the decreasing order of CH3 basicity: (c) N (I) N N H (II) H (III) (a) I > II > III (b) III > II > I (c) I > III > II (d) CH3 (d) III > I > II 16. The correct decreasing order for acid strength is (a) FCH2COOH > NCCH2COOH > NO2CH2COOH > ClCH2COOH (b) CNCH2COOH > O2NCH2COOH > FCH2COOH > ClCH2COOH (c) NO2CH2COOH > NCCH2COOH > FCH2COOH > ClCH2COOH (d) NO2CH2COOH > FCH2COOH > CNCH2COOH > ClCH2COOH 20. Amongst the compounds given, the one that would form a brilliant coloured dye on treatment with NaNO 2 in dil. HCl followed by addition to an alkaline solution of b-naphthol is (a) (c) N(CH3)2 NH2 NHCH3 (b) CH2NH2 (d) H 3C 21. The species having pyramidal shape is (a) SO3 17. Which of the following compounds will give significant amount (b) BrF3 (c) SiO23 (d) OSF2 22. In the following reaction, of meta-product during mononitration reaction? OH (a) OCOCH3 (b) NH2 (c) O NHCOCH3 (d) N H Conc. HNO3 Conc. H2SO4 X Aromatic Compounds Containing Nitrogen The structure of the major product X is 25. Benzene diazonium chloride on reaction with phenol in weakly basic medium gives O (a) (a) diphenyl ether (c) chlorobenzene NO2 N H (b) p-hydroxy azobenzene (d) benzene 26. Examine the following two structures of anilinium ion and choose the correct statement from the ones given below : O2 N O (b) + NH3 NH3 N H + I O (c) N H NO2 O (d) N H O 2N 23. F NO2 27. Chlorobenzene can be prepared by reacting aniline with (CH3)2NH DMF,D (a) hydrochloric acid (b) cuprous chloride (c) chlorine in the presence of anhyd AlCl 3 (d) nitrous acid followed by heating with cuprous chloride (A) (i) Fe/HCl (ii) NaNO2/HCl/0°C (iii) H2/Ni (a) H2N N II (a) II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ions (b) II is not an acceptable canonical structure because it is non-aromatic (c) II is not an acceptable canonical structure because the nitrogen has 10 valence electrons (d) II is an acceptable canonical structure (B), B is Objective Questions II (One or more than one correct option) 28. Aniline reacts with mixed acid (conc. HNO 3 and conc. H 2SO 4 ) at 288 K to give P (51%),Q (47%) and R (2%). The major product(s) of the following sequence is (are) CH3 CH3 (1) Sn /HCl (2) Br2 /H 2 O (excess) (1) Ac2 O, pyridine (2) Br2 , CH 3 CO 2 H R ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾® S ¾¾¾¾¾¾¾¾¾® (b) H2N NH2 (c) O2N N (d) O2N NH2 (3) H 3 O+ (4) NaNO 2 , HCl /273-278 K (5) EtOH, D CH3 (3) NaNO 2 , HCl /273-278 K (4) H 3 PO 2 Major product(s) CH3 Br Br (a) 24. The most unlikely representation of resonance structures of _ + N O O _ + N (b) (a) O Br (b) Br Br p-nitrophenoxide ion is O Br Br _ Br (d) Br (c) Br Br Br Br Br Br O– O N (c) O– + N O (d) O– O O 29. The product(s) of the following reaction sequence is (are) NH2 (i) Acetic anhydride/Pyridine (ii) KBrO3/Br – O (iii) H3O+, Heat (iv) NaNO2/HCl, 273-278 K (v) Cu/HBr Aromatic Compounds Containing Nitrogen Br Br Br Br Br 32. p-chloroaniline Br Br and anilinium (a) Sandmeyer reaction (c) AgNO3 Br Br (b) (a) , NaOH W V 0°C (c) loss of aromaticity when Br + attacks at the ortho and para positions and not at meta position (d) easier loss of H+ to regain aromaticity from meta position OH (a) than from ortho and para positions (b) OH N= =N (b) NaHCO3 (d) Carbylamine test (a) the electron density on meta carbon is more than that on ortho and para positions (b) the intermediate carbonium ion formed after initial attack of Br + at the meta position is less destabilised OH NaNO2,HCl be the major product formed is m-bromonitrobenzene. Statements which are related to obtain the m-isomer, are 30. In the following reactions, the major product W is NH2 can 33. When nitrobenzene is treated with Br2 in the presence of FeBr3 , Br (d) (c) hydrochloride distinguished by Br N= =N Assertion and Reason HO OH (c) (d) N= =N N= =N 31. In the reaction shown below, the major product(s) formed is/are Read the following questions and answer as per the direction given below : (a) Statement I is correct; Statement II is correct; Statement II is the correct explanation of Statement I. (b) Statement I is correct; Statement II is correct; Statement II is not the correct explanation of Statement I. (c) Statement I is correct; Statement II is incorrect. (d) Statement I is incorrect; Statement II is correct. 34. Statement I Aniline on reaction with NaNO2/HCl at 0°C followed by coupling with b-naphthol gives a dark blue coloured precipitate. Statement II The colour of the compound formed in the reaction of aniline with NaNO2/HCl at 0°C followed by coupling with b-naphthol is due to the extended conjugation NH2 Acetic anhydride NH2 CH2Cl2 Product(s) O N H 35. Statement I In strongly acidic solution, aniline becomes more CH3 reactive towards electrophilic reagents. O (a) Statement II The amino group being completely protonated in strongly acidic solution, the lone pair of electrons on nitrogen is no longer available for resonance. NH2 + CH3COOH O NH2 Match the Column 36. Match the compounds in Column I with their characteristic H N (b) CH3 + CH3COOH O H N test(s)/reaction(s) given in Column II. Column I Column II (A) s Å H2N ¾ NH3 Cl (p) Sodium fusion extract of the compound gives Prussian blue colour with FeSO4 (B) O2N (q) Gives positive FeCl3 test O CH3 O (c) H N CH3 + H2O O O s r NH3CH2COO Å NHNH3Br NO2 H N (d) O CH3 O Aromatic Compounds Containing Nitrogen Column I NH3Cl HO (r) Reacts with aldehydes to form the corresponding hydrazone derivative Å (D) NHNH3Br O2N O Gives white precipitate with AgNO3 Å (C) 44. What would be the major product in the following reaction? Column II (s) NO2 Br2/Fe N— 45. How would you bring about the following conversion (in 3 steps)? Aniline ¾® Benzylamine 46. Complete the following reactions with appropriate reagents Fill in the Blanks O 37. The high melting point and insolubility in organic solvents of I —NH—C—CH3 sulphanilic acid are due to its…… structure. —NO2 38. In an acidic medium, ……… behaves as the strongest base. (nitrobenzene, aniline, phenol) I 47. Following reaction gives two products. Write the structures of the products. Subjective Questions Br 2 /Fe CH3 CONHC6 H5 ¾¾¾® 39. Convert NO2 NO2 48. Write the structure of foul smelling compound obtained when aniline is treated with chloroform in the presence of KOH. to 49. Complete the following with appropriate structure: OH in not more than four steps. Also mention the reaction conditions and temperature. NH2 NO2 40. Which of the following is more acidic and why? + + NH3 NH3 (i ) NaNO2 and HCl at 5°C ¾¾¾¾¾¾¾¾¾® A (ii ) Anisole NO2 50. How will you bring about the following conversions? “Benzamide from nitrobenzene” 51. How will you bring about the following conversions? I “4-nitrobenzaldehyde from benzene” F II 52. Outline a synthesis of p-bromonitrobenzene from benzene in two steps. 41. Convert (in not more than 3 steps) : 53. Write the structure of the major organic product expected from COOH COOH the following reaction. CH3 —N F 42. There is a solution of p-hydroxybenzoic acid and p-amino + HNO2 CH3 benzoic acid. Discuss one method by which we can separate them and also write down the confirmatory test of the functional group present. 54. How will you bring about the following conversion? 43. Write structures of the products A, B, C, D and E in the following 55. Complete the following with appropriate structures: “4-nitro aniline to 1, 2, 3-tribromobenzene.” scheme O —NH2 + CH2CH2CH3 Cl D CH2 H2 /Pd/C Cl2/FeCl3 – + CHCH2 ONa H2O E A C Na-Hg/HCl HNO3/H2SO4 B —COCl Base ? 56. How would you convert aniline into chlorobenzene? 57. State the conditions under which the following preparation is carried out. “Aniline from benzene” 58. State the equation for the preparation of following compound. Chlorobenzene from aniline (in 2 steps). Answers 1. 5. 9. 13. 17. 21. 2. 6. 10. 14. 18. 22. (c) (d) (b) (b) (c) (d) 3. 7. 11. 15. 19. 23. (c) (*) (b) (a) (a) (b) 4. 8. 12. 16. 20. 24. (b) (c) (c) (d) (c) (a) (a) (b) (d) (c) (c) (c) 25. 29. 33. 36. (b) 26. (c) 27. (b) 30. (a) 31. (a,b) 34. (d) 35. A ® r, s B ® p, q C ® p, q, r 28. (d) 32. (c,d) (d) (a) (d) D® p 37. zwitter ionic 38. Aniline Hints & Solutions 1. Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid undergoes coupling reaction and produces p-amino azobenzene. In this reaction, benzene diazonium chloride reacts with aniline in which aniline molecule at its para-position is coupled with diazonium salt to give p-amino azobenzene (yellow dye). + – N NCl + H 3. The preparation of benzylamine from cyanobenzene using given reagents are as follows : Ph ¾ C= N (Cyanobenzene) H2/Ni D Ph ¾ CH2 ¾ NH2 Benzylamine (i) LiAlH4 (ii) H Or NH2 (ii) NaBH4 3 Benzene diazonium salt Aniline H+ N N (i) SnCl2 NH2+Cl–+H2O p-amino azobenzene (yellow dye) 2. Key Idea pK b value is defined as the minus lagarithm of K b smallar the value of pK b stronger is the base and vice-versa. In the given options, +R effect is shown by —CH3 and —OCH3 group ( ¾ OCH3 > ¾ CH3 ). These group increases the electron density at o and p-positions. Groups such as ¾ F and—NO2 shows –R-effect ( ¾ NO2 > ¾ F). These group decreases the electron density at o and p- positions. Increase in electron density at p-position makes the unshared electron pair of ‘N ’ more available and decrease in electron density at p-position makes the unshared electron pair of ‘N ’ less available. Compound containing —OCH3 group act as strongest base and hence possess lowest value of pK b . So, the correct increasing order of pK b in the given compound is O2N S > N H N H (ii) HCl (g) r [Ph ¾ CH==NH2[2 SnCl 62bis (benzaldiminium ) hexachlorostannate (IV) HCl , H O NaBH 4 2 PhCN ¾¾¾® PhCOOH ¾¾® PhCH2OH –NH4Cl Benzoic acid Benzyl alcohol Thus, option (b) is incorrect. 4. Major product of the reaction is N N NH2 It is obtained by coupling reaction. Reaction of aniline with dil. HCl and sodium nitrite at 0°C is shown below : +– NººNCl NH2 + NaNO2 + Dil. HCl 0°C Diazonium salt Diazonium salt formed when added to equimolar mixture of aniline and phenol in dil. HCl then aniline couples with diazonium salt. Reaction is as follows : (C) + F S + NººNCl + H2N > N H N H In acidic H+ medium (A ) H3C N= =N S > N H N H (D) H3CO —NH2 + HCl S N H N H (B) For an effective coupling, the solution must be so alkaline that the concentration of diazonium ion is too low. It must not be so acidic that the concentration of free amine or phenoxide ion is too low. That is why amines couple fastest in mild acidic solution and phenols couple fastest in mild alkaline solution. Aromatic Compounds Containing Nitrogen s 5. [ Ar ¾ N ºº N ]+ OH 1° NH2 responds NH2 NººC (Carbylamine reaction) H+ NººC d- more acidic than that of C ¾ H bond in the ring. Again, sp2-N is more electronegative (s% = 33.3) than sp3-N (s% = 25) of the ¾ NH2 group. sp3 NH2 N N Non-aromatic H ring s NH2 sp2 N Base sp2 sp2 Aromatic ring sp3 (Base) N N s N Both aromatic rings Apparently B d+ d– CH3–I C – BH N In compound (D), lone pair of ‘N’ are available but in compound (C), the lone pair of ‘N’ are not readily available due to the presence of electron withdrawing group ( ¾CN) attach to benzene ring. In compound (B), i.e. pthalimide, ‘N’ is not nucleophilic due to the presence of two polar groups attached to it that pulls lone pair towards them. Hence, the correct order is: ( B ) < ( A ) < (C ) < (D ). 9. Ni/H 2 can reduce —C ºº N into ¾ CH2 ¾ NH2 (1°-amine) but cannot reduce an ester group ( ¾ CO2Et) whereas DIBAL-H, di-isobutylaluminium hydride, [(CH3 )2CH]2AlH reduces the ester group ( ¾ CO2Et) into —CHO (an aldehyde) and C2H5 OH. Ni/H2/D OEt CºN 1 N 7. Coupling of benzene diazonium chloride with 1-naphthol in alkaline medium will give the following coupling reaction. –H2O (Nucleophilic addition elimination reaction) 10. Here, in mononitration the electrophile produced from mixed + acid (HNO3 + conc.H2SO4 ) is N O2. OH A +R N PhN2+ OH – O –R B H N==N + In the presence of OH ion, Ar N 2 exists in equilibrium with an unionised compound (Ar ¾ N == N ¾ OH) and salts s DIBAL–H –EtOH CH NH2 2 3 4 OEt NH2 N CH3 + – NH2 s (Option-b) - + C NH2 –I N N s O (A) NH2 OH Does not couple O (SN2 attack) Note This C C¢ are missing in all the option given. + nucleophilic substitution reaction. More nucleophilic nitrogen is more reactive with alkyl halide. Compound (A) is benzamide and lone pair of ‘N’ of it, is not available in this compound. d+ 6. N is more electronegative than ‘C’. So, H of N ¾ H bond is N - Ar ¾ N == N ¾ O Na 8. The reaction of alkyl halide with amine or amides is a OH In step (i), 1° amine is converted into isocyanide, when reacts with CHCl3 / KOH. This reaction is known as carbylamine reaction. This reaction is used to detect primary amine in a compound. In step (ii), isocyanide is reduced to secondary amine and cyanide is reduced to primary amine, and carbonyl group to alcoholic group when treated with Pd/C/H2. N Does not couple (a) The solution must be so alkaline that the concentration of diazonium ion is too low. (b) It must not be so acidic that the concentration of free phenoxide ion is too low. NH—CH3 H 2N Ar ¾ N == N ¾ OH Following conditions are very useful for coupling reaction to take place. O Pd/C/H2 sp2 H+ NaOH CHCl3/KOH CN O NaOH - [ ArN == N ] ¾OH derived from it, which do not couple. Ring-(A) is activated, i.e. becomes more nucleophilic by the + R · · effect of the ¾ N H ¾ group and it becomes o / p-directing towards the electrophile, NO+2 in the ArS E 2 reaction. For Å mononitration, N O2 will preferably come at p-position, Aromatic Compounds Containing Nitrogen whereas the ring-(B) gets deactivated by -R- effect of the O ½½ ¾ C ¾ group, Aromatic amines (D) are always weaker bases than aliphatic amines, because of the conjugation of lp of electrons of N (+ R-effect) with the benzene ring. So, the correct order is (D) < (C) < (A) < (B). r O2N r NO2 C rds of ArSE2 NH H 13. O N It is a 3°-amino group (—NMe2) though aromatic which will not respond to diazotisation (NaNO2 + dil. HCl/5°C). So, formation of diazonium ion and its further reaction with alkaline phenol or b-napthol at 5-10°C to produce a precipitate of azo-dye will not take place. CH3 CH3 O O2N r C NH (b) Fast,–Hr It is a keto group C==O and so it responds to 2, 4-DNP test with Brady’s reagent (2, 4-dinitrophenyl hydrazine) to give a reddish orange precipitate. O CH3 —— C==O O O2N C It is a keto-methyl group —C—CH3 and hence, it responds to iodoform test to give a yellow precipitate of iodoform (CHI3). NH 11. Using retro-synthesis, to get the required compounds, A , B and C. Br ¬¾ ¾2 ¾ C6H7N Compound (c) is an alcohol and does not give positive 2, 4-DNP test. Hence, eliminated (a) and (d) does not have ¾ CH3CO group and does not give idoform test. Hence, eliminated. B KOH [An amide] (C ) Hoffmann (No.of C =7) [1°-amine] bromamide (No. of C = 6 1 carbon less) reaction (step down) NH3/D 14. In the given reaction, Responds to Hoffmann's bromamide degradation and gets converted into [¾NH2 +K2CO3] by Br2/KOH A [Carboxylic acid C 7 H 6 O2 ] So, the reactions can be shown as: COOH NH3/D –H2O (A ) [C7H6O2] Benzoic acid Br2/KOH –K2CO3 –KBr (B) Benzenamide C NH2 CONH2 NH2 (i) Br2/hn (C) [C6H7N] Aniline 1° 2° B C 2° 3° (B ) Å c H + H N H –Br Br –H+ Formation of 2° carbocation s by removing Br But, this order does not match with the options given. So, it has been asked on basicity of the amines in aqueous solution. When no phase is given, then basicity of amine is considered in aqueous solution as they are liquids. In aqueous solution, basicity of 2°-amines (aliphatic) is maximum because, of the thermodynamic stability of its conjugate acid. Et 2 NH + H 2O Br H N vapour phase), the order of basicity will be. A CH3 (ii) KOH (dil.) 12. If we consider Lewis basicity (basicity in aprotic solvents or in D Undergo benzylic free radical substitution with Br2/hn to give [¾CH¾CH3] CH2 Et 2 N H 2 È + OH Conjugate acid CH3 NH CH3 15. Key Idea Basicity µ Ease of donation of lone pair of electrons 1 1 µ µ % of s - character of N Electronegativity of N The % of s-character in the given amines are as follows: Å Et 2 N H2 is a sterically symmetric tetrahedral ion as it contains equal number (two) of bulkier Et-group and small size H-atoms. Here, two H-atoms give additional stability through hydrogen bonding with H2O (solvent) molecules. Et d+ – H d H O H N d+ d– N sp2-hybridised [s-character=33.3%] Pyridine (I) N H sp2-hybridised [s-character=33.3%] Pyrrole (II) N H sp3-hybridised [s-character=25%] Piperidine (III) Therefore, piperidine (III) having minimum Et H-bonding Aromatic Compounds Containing Nitrogen % s-character is most basic. Among the rest, pyridine (I) and pyrrole (II) the lone pair of electrons of N in pyrrole (II) is involved in delocalisation and follows (4 n + 2) p aromatic (n = 1) system. So, the N-atom of pyrrole (II) will show least basicity. Thus, the order of basicity is as follows: aromatic amine, on treatment with NaNO2 in dil. HCl forms the corresponding diazonium chloride salt. NaNO2 HCl H3C > > + – N2 Cl NH2 H 3C OH N N N H (III) (I) H (II) b-naphthol N CH3 N 16. All the given compounds are a-monosubstituted acetic acid OH 21. ¾ NO2 > ¾ CN > ¾ F > ¾ Cl. F F Thus, the correct decreasing order for acid strength is: NO2CH2COOH > NCCH2COOH > FCH2COOH S is sp3-hybridised O F Pyramidal SO3 is planar (S is sp2-hybridised), BrF3 is T-shaped and SiO23 is planar (Si is sp2-hybridised). > ClCH2COOH. 17. Aniline in presence of nitrating mixture (conc. HNO3 + conc. H2 SO4) gives significance amount (» 47%) of meta-product because in presence of H2SO4 its protonation takes place and anilinium ion is formed. NH2 S F—S==O — derivatives and the a-substitutions have been made by strong–I groups/atoms. More powerful the - I group, stronger will be the acid. -I power of different groups is as follows : O ½½ 22. N ½ H I NH3HSO4 Conc. HNO3 Conc. H2SO4 II O2N O ½½ H2SO4 Aniline N ½ H Anilinium ion Here, anilinium ion is strongly deactivating group and meta-directing in nature. So, it gives meta-nitration product. NH2 Ring-I is activated while ring-II is deactivated towards electrophilic aromatic substitution reaction. NH3HSO4 Conc.H2SO4 23. F +Conc.HNO3 NO2 (CH3)2NH DMF NO2 18. Diazo coupling occur at para-position of phenol. OH (i) Fe/HCl OH OH (ii) NaNO2 (iii) H2/Ni (i) NaNO2/HCl/0°C H3C N— NO2 N— NH2 H3C H3C H 3C (ii) NaOH + N2Cl– NH2 19. + - N2 Cl NH2 O N N CN NaNO2/HCl D 0-5°C diazotisation CuCN/KCN O N Nitrogen has five bonds and 10 valence electrons, not an acceptable resonance structure. 24. O – + CH3 4-methyl aniline CH3 D 4-methyl diazonium chloride CH3 E 25. C6H5N2Cl– + C6H5OH 4-methyl benzocyanide 20. As we know, benzenediazonium salt forms brilliant coloured dye with b-naphthol, the compound under consideration must be p-toluidine (c) as it is a primary aromatic amine. Primary —N==N— p-hydroxy azobenzene —OH Aromatic Compounds Containing Nitrogen Br So, 26. In structure II, nitrogen is associated with five bonds and 10 NO2 valence electrons, hence not acceptable. NH2 NH2 Br2 Sn/HCl H H2O (Excess) Br H—N—H Br (S) Br Br Br NaNO2+HCl 273-278K + Br Br N-has five bonds and 10 valence electrons Br CuCl 27. C6H5NH2 + HNO2 ¾® C6H5N2OH ¾® C6H5 ¾ Cl (51%) NH2 (47%) (2%) NH2 NH2 Ac2O/Pyridine + + KBrO3 + HBr ¾® Br2 NO2 Ac is CH3CO (acetyl), it protects ¾ NH2 group from being oxidised. NH2 NHAc NHAc NO2 P (51%) Its given R (2%) Q (47%) (i) Ac 2O, pyridine (ii) Br2 , CH 3COOH (iii) H 3O+ (iv) NaNO2 , HCl / 273-278 K (v) EtOH, D HN O C CH3 HN CH3 Br 30. O+ H3 (Hydrolysis) Br CuBr/HBr (Sandmeyer reaction) 273-278 K (Diazotisation) Br Br –CH3COOH N2Cl NaNO2/HCl NO2 Br2 CH3COOH (R ) NH2 C NO2 Ac2O Pyridine + CH3COOH Br O NO2 H3O+ Br2 water R ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾¾® S NH2 Br Br + NH2 – N2Cl b-naphthol NaOH NaNO2 –HCl 0°C V + – N2Cl EtOH D NO2 NaNO2/HCl 273-278K (Diazotisation) W 31. Br (S) N==N––Ph OH NH2 NO2 NO2 Br Br Now from S to major products its given. (i) Sn / HCl (ii) Br 2 / H2O (Excess) S ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾¾® Major product (iii) NaNO 2 , HCl / 273-278K (iv) H3PO 2 Br Br Hence, only (d) is the correct answer. NH 2 NHAc 29. NO2 Conc. HNO3+ Conc. H2SO4 – Major product 1, 3, 4, 5-tetrabromobenzene Then P, Q and R will be NH2 Br Br Br Conc. HNO3 + 28. Given, Aniline ¾ ¾ ¾ ¾ ¾¾ ® P + Q + R Conc. H 2SO4 N2+Cl H3PO2 PLAN This problem includes concept of acetylation reaction and regioselectivity of chemical reaction. Regioselectivity means which group will react selectivity in the presence of two or more than two functional groups. Here, among two functional group ¾ NH2 and ¾ CONH2, NH2 is more nucleophilic, hence NH 2 group will undergo reaction faster than CONH2. Aromatic Compounds Containing Nitrogen CH2—NH2 O Chloride salt gives white precipitate of AgCl on treatment with AgNO3. Hydrazone formation occur effectively at pH = 4.5. The reaction proceeds in that condition only when H+ concentration is just sufficient to activate the following enolisation. O + CH3—C—O—C—CH3 C—NH2 CH2Cl2 O O ½½ H+ ¾ C— l CH2—NH—C—CH3 O C—NH2 As H+ concentration rises sufficiently, a large number of molecules of hydrazine gets converted into hydrazinium ion which is not nucleophilic and reaction becomes impossible. Further low concentration of H+ (in the case of 2,4-dinitrophenyl hydrazinium bromide) is not effective to proceed elimination. O ¾ CH2 ¾ NH2 is more nucleophilic than ¾ C ¾ NH2. || O Hence, correct choice is (a). 32. + 37. Zwitter ionic + AgNO3 ¾® AgCl ¯ (C6H5 NH3Cl – ) anilinium hydrochloride + No such precipitate is formed with p-chloroaniline. Also, carbylamine test will not be given by anilinium hydrochloride but p-chloroaniline give this test. than from meta position. Also the s-complex formed from meta attack is less destabilised than from ortho/para positions. NO2 + + Br+ 38. Aniline It is a stronger base than either phenol or nitrobenzene. NO2 NO2 NO2 NH4HS HNO3 39. H2SO4 heat NO2 NH2 + H Br NO2 NO2 NO2 H Br NaOH NaNO2 Nitro group not destabilising the positive charge by resonance. NO2 —SO3 sulphanilic acid 33. Nitro group withdraws electrons more from ortho/para position NO2 – H3N— precipitate NO2 OH ½ ¾ C+ ½ HCl/0°C NO2 OH 40. II is more acidic due to - I effect of F. + + H 2O + N2Cl– COOH COOH + Br + Br H Br H HNO3 41. Sn-HCl H2SO4 Nitro group destabilises the positive charge directly by resonance, less stable s-complex. NO2 COOH COOH Similar phenomenon is observed with ortho attack. 0 °C 34. C6H5NH2 + NaNO2 / HCl ¾® C6H5N+2 Cl – NaNO2 OH b-naphthol F COOH N==N—C6H5 HBF4, heat NH2 COOH coloured dye 35. In strongly acidic medium, aniline is fully protonated, becomes 42. + deactivated for SE Ar reaction. + NH2 + H + NH3 OH NH2 dissolve in diethyl ether HOOC— Lone pair on nitrogen is not available for resonance. Positive charge makes the group strongly electron withdrawing. 36. Sodium fusion extract gives Prussian blue colouration, nitrogen and carbon both present in the compound. Phenolic group and salt of carboxylic acid gives FeCl 3 test. aq HCl HO— + – —NH3Cl(aq) —COOH (ether layer) Aromatic Compounds Containing Nitrogen COOH FeCl3(aq) violet colouration OH NH2 NC + CHCl3 + KOH COOH COOK foul smell O O C C Cl2 CH2CH2CH3 43. FeCl3 HCl Cl Cl Cl NO2 Cl Cl (A) (B) NO2 NO2 CH2CH2CH2CH3 HNO3 CH2CH2CH2CH3 Na-Hg CH2CH2CH3A H2C==CH—CH2ONa H2/Pd/C H2SO4 Cl Cl Cl Cl (C ) O—CH2CH2CH3 O—CH2—CH==CH2 (E) (D ) O 44. O N Br2 Fe __ NaNO N__ __Br LiAlH CuCN 4 2 45. C6H5NH2 ¾¾¾® C6H5N+2 Cl – ¾¾® C6H5 — CN ¾¾¾® C6H5 — CH2NH2 HCl/0 °C NO2 O 46. —NH—C—CH3 H+ H2O NaNO2 —NH2 HCl/0°C NO2 NaNO2(aq) + —N2Cl– Heat I2 Fe I O O NH—C—CH3 47. O NH—C—CH3 Br2 NH—C—CH3 + Fe Br Br 48. C6H5NH2 + CHCl 3 + KOH ¾® C6H5NC Isocyanide (foul smell) + N2Cl– NH2 NO2 49. NO2 (i) NaNO2/HCl 0°C NO2 NO2 Anisole N O2N NO2 N OCH3 I Aromatic Compounds Containing Nitrogen NH2 NaNO 2 NO2 Zn-HCl 50. COOH NH 3 CN H+/H O 2 CuCN CONH2 D HCl/0°C Benzamide 51. AlCl3 + CH3Cl Conc. HNO3 CrO3 Conc. H2SO4 (CH3CO)2O NO2 HNO3 Fe + Br2 H+ NO2 Br Br 52. CHO CH3 CH3 H2SO4 NO2 53. No reaction. Tertiary amine does not react with nitrous acid. NH2 54. N2Cl– NH2 + Br2 (excess) Fe Br Br NaNO 2 Br Br Br CuBr Br Br Zn—HCl HCl/0°C NO2 NO2 NO2 NO2 Br Br Br Br NaNO2 HCl/0°C NH2 O 55. NH2 + COCl Base NaNO2 /HCl CuCl 0 °C Heat NH—C— 56. C6H5 — NH2 ¾¾¾¾® C6H5N2+ Cl – ¾® C6H5 — Cl Zn – HCl 57. C6H6 + Conc. HNO3 / Conc.H2SO4 ¾® C6H5 — NO2 ¾¾¾® C6H5 — NH2 Aniline HCl CuCl 0 °C Heat 58. C6H5NH2 + NaNO2 ¾® C6H5N2+ Cl – ¾® C6H5 — Cl H3PO2 Br Br