CONTENTSS.NO. TOPIC TOPIC PAGE NO PAGE NO. 1. INTRODUCTION 2 2. PREPARATION OF AMINES 2 3. PHYSICAL PROPERTIES 6 4. CHEMICAL PROPERTIES 7 5. DIAZONIUM SALTS 10 1 Amine AMINE 1. INTRODUCTION (i) (ii) (iii) (iv) (v) (vi) Amines are called alkyl derivative of NH3. If a hydrogen atom of NH3 is replaced by an alkyl group then it is called primary amine and possess –NH2(amino) group. If two hydrogen atoms of NH3 are replaced then it is called secondary amine and it posses > NH (mino) group. If all hydrogen atoms of NH3 are replaced then it is called tert. amine and has a nitrilo N group. N is in sp3 hybridisation and tetrahedral geometry. Bond angle increases from ammonia to 3º amines. NH3 (107º) < RNH2 < R2NH < R3N 2. PREPARATION OF AMINES (i) Reduction of nitro compounds (ii)Ammonolysis of alkyl halides The free amine can be obtained from the ammonium salt by treatment with a strong base: – – R – NH3 X NaOH R – NH2 H2O Na X primary amine is obtained as a major product by taking large excess of ammonia. The order of reactivity of halides with amines is RI > RBr >RCl. (iii) Reduction of nitriles R –C N H 2 Ni Na (Hg )C3H 4OH R – CH2 – NH 2 (iv) Reduction of amides 2 Amine (v) Gabriel phthalimide synthesis (vi) Curtius Reaction : H2 O R N C O R NH 2 RCON N 3 Acyl azide CO 2 2 Alkyl isocyanate Alkyl amide Mechanism : RCOCl NaN3 RCON3 NaCl R C O O O N N C N R R N C N2 N N N R N N2 R—NCO alkyl isocyanate O R C H2 O N C OH R H2 O OH R C O OH O NH C N R — NH 2 CO2 OH – CO 2 3 O R OH NH R NH 2 CO 2 HO Mechanism : O O R R OH H2N + HCl Cl NH HO O OH R R NH HO Hydroxamic acid (keto form) N HO (enol form) 4 . The hydroxamic form. the acyl derivatives exist in two tautomeric form keto form called hydroxamic form and enol form called hydroximic acid. R – COOH + N 3H H2SO4 R – N2H + C2 O + H 2O Mechanism : R O OH C H R C OH OH OH H N3 R C OH H N N + N –H2O O C H2 R — N C O R N2 R isocyanate N –H 2O + N C O N N + H N N R — NH2 CO2 (viii) Lossen Reaction : Hydroxylamine on treatment with acid chloride gives acyl derivatives of hydroxyl amine.Amine (vii) Schmidt Reaction : Carboxylic acid reacts with hydrozoic acid in presence of concentrated H 2SO 4 to give isocyanates. Compound (A) has molecular formula C9H14NCl. nitration etc. it is resistant to oxidation. Further. heat. possible structure of (A) may be : + CH3 C6H5 – N – CH 3 – Cl. Suggest structure for (A). (A) gives an immediate precipitate with AgNO3. quaternary-salt CH3 (A) (C9H14NCl) N. nitration and oxidation by KMnO4. It is very resistance to bromination in either acid or alkaline solution.N. It is also resistance to heat. Therefore. it must be an ionic compound.N-Trimethylaniliniumchloride 5 .1 Sol. Mechanism: 2NaOH Br2 NaOBr NaBr H 2O O O R C – R NH2 OBr C N Br OH – H N-bromoacetone O R C O N R Br OH – –H 2 O H C N – Br H2O Rearrangement H O R — NH2 CO2 2 R — N C O (Isocyanate) Ex.Amine The hydroxamic form (keto form) forms o-acyl derivatives of hydroxamic form which on heating with bases forms isocyanates and finally amines upon hydrolysis H R C O N OH R C H R Cl O C N O O C R O O H2 O R NH2 R — N C O R CO2 C N O C R O (ix) Hoffmann bromamide degradation reaction : The reaction of amid with bromin in present of base to form primary amine. it must be a quanternary ammoinium salt. Since (A) (C9H14NCl) gives immediate precipitate with AgNO3. Like ammonia. Sn / HCl CH3CH2—NO2 + 6[H] CH3CH2NH2 + 2H2O Nitroethane (d) Ethanamine When acetaldehyde is treated with ammonia. 1º > 2º > 3º amine.Amine Ex. This is all due to H.bond with water. imine is formed whichonreduction gives ethanamine.2 Identify (A) through (E) in the following sequence H3 O – HCl NaOH EtBr PhSO2Cl EtNH2 (A) (B) (C) (D) (E) Sol.Bonding. CH3CH2CONH2 + Br2 + 4KOH CH3CH2NH2 + K2CO3 + 2KBr + 2H2O Ethanamine (b) When ethanamide is reduced with lithium aluminium hydride. Because H. weight. than amine is formed.bonding in amines is less pronounced in 1º and 2º than that in alcohols and carboxylic acids. Because nitrogen is less electronegative than oxygen. amines are much more soluble in water. Boiling points of 1º. 2º and 3º amines follow the order. 6 . ethanamine is LiAlH4 obtained. Because All amines form a stronger H.3 Sol. Boiling points of amines are lesser than alcohols and acids of comparable mol. 1º > 2º > 3º amine.bonding. CH3CONH2 + 4[H] CH3CH2—NH2+ H2O Ethanamide (c) Nitroethane can be converted into ethanamine by the reduction with Sn/HCl. Thus every question regarding boiling point can be answered on the basis of H . ethanamine is obtained. Solubility in water follow the order. amines are polar compounds and except 3º amines can form intermoleculer Hbonds that’s why they have higher boiling points. PHYSICAL PROPERTIES : (i) (ii) (iii) (iv) (v) Unlike other organic compounds. O3 l2 CH3 CH2 NH2 H2 O CH3CH2OH + NH3 A723K Ethanamine 3. PhSO2 N – Et PhSO 2 NHEt Ph SO2 NEt2 (B) (A ) (C) H 3O PhSO2OH Et2 NH2 (D) Ex. (E) How the ethylamine is prepared from : (a) Propanamide (b) Ethanamide (c) Nitroethane (d)Acetaldehyde Ethanol (on large scale) ? (e) (a) When propanamide is treated with Br2 and KOH. H H | | H 2 / Ni CH 3 — C O + NH3 CH — C NH CH 3 — CH 2NH 2 3 H2O 2[H] Ethanal (e) Acetaldimine Ethanamine When a mixture of ethanol and ammonia is passed over alumina at 723 K and high pressure. atom. R-NH2 + CH3COCl RNHCOCH3 + HCl R -NH2 + (CH3CO)2O RNHCOCH3 + CH3COOH (CH3)2N-COCH3 + HCl (CH3)2NH + CH3COCl Note : (a) Tertiary amines donot undergo this reaction because of absence of replacable H. (b) When Benzoyl chloride is used in place of acet yl chloride react ion is called ‘Schotten . CHEMICAL PROPERTIES : Alkylation : Amines undergo alkylation on reaction with alkyl halides RNH2 R1 X HX RNHR1 R2 X HX R3 X R -N -R 2 HX R2 R1 | N R2 | R3 X¯ (quartenary amm. Benzoylation : WhenAmines react with benzoyl chloride (C6H5COCl).NC 3KCl 3H 2 O Cl R R — NH2 CCl2 Mechanism : HN C Cl H -HCl R — N C R N + C Cl 7 .Baumann’ reaction.Amine 4.) Acylation : 1º and 2º amines react with acetyl chloride or acetic anhydride to form acetyl derivatives.NH 2 CHCl 2 3KOH R . salts. CH 3 NH 2 C6 H5COCl CH 3 NHCOC6 H5 HCl Methanamine NMethylbenzamide Benzoyl chloride Heat Carbylamine reaction : R . then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine. This can be done by protecting the -NH 2 group by acetylation with acetic anhydride. reacts with primary and secondary amines to form sulphonamides. it is soluble in alkali.N 2 Cl ] ROH N 2 HCl - NaNO2 2HCl C6 H5 . (iii) ter. N.6-tribromoaniline. (i) The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide. The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group.Amine does not react with Hinsberg’s reagent it is present above solution.N-diethylbenzenesulphonamide is formed. Hence.4.Amine Reaction with nitrous acid : - NaOH2 HCl R . 8 . If we have to prepare monosubstituted aniline derivative.NH 2 HNO 2 [R . (ii) In the reaction with secondary amine.NH 2 C6 H5 . which is also known as Hinsberg’s reagent. It is in soluble alkali.N 2 Cl NaCl 2H 2O Ani ln e Benzenedia monium Chloride Reaction with arylsulphonyl chloride : Benzenesulphonyl chloride (C6 H5 SO2 Cl). aniline active toward Electrophilic substitution (i) Bromination: Aniline reacts with bromine water at room temperature to give a white precipitate of 2. Electrophilic substitution : Due to +M effect of -NH 2 genrate electron dencity at ortho and para position hence. C6 Hl5NH 2 CHCl3 3KOH C6 H5 N C 3Kl 3H 2 O (A) C6 H5 N C 2H 2 O H C6 H5 NH2 HCOOH (A) (B) (C ) 9 . Sol.4 Consider the following reaction. However. Therefore.Amine The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below: Hence. by protecting the -NH 2 group by acetylation reaction with acetic anhydride. H / H2O C 6 H 5 NH 2 CHCl 3 KOH ( A) (B) (C ) Find the compounds (B) and (C). (ii) Nitration: Direct nitration of aniline is not possible because in the strongly acidic medium. the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. activating effect of -NHCOCH 3 group is less than that of amino group. the nitration reaction can be controlled and the p-nitro derivative can be obtained as the major product. (iii) Sulphonation: s Ex. aniline is protonated to form the anilinium ion which is meta directing. SO 3. Comment. HSO 4-. However. Sulphanilic acid exist as a Zwitter ion and exhibits strong dipole-dipole interactions. BF -3.Amine Ex. strongly basic hydroxyl ion can abstract a proton from —NH 3 to form soluble salt. it is insoluble in water. Br-. etc. thus sulphanilic acid is insoluble in acid. NH2 + NH3 + H2 O + NaOH SO3Na SO3- Sodium p-aminobenzene sulphonate Zwitter ion structure of sulphanilic acid Ex. Resonance of benzene diazonium ion is 10 . Sulphanilic acid is insoluble in water and acid but soluble in caustic alkali. on adding acid. DIAZONIUM SALTS : - - The diazonium salts have the general formula R N 2 X where R stands for an aryl group and X ion may be Cl -. + when alkali is added. Therefore.fails to accept H + ion.6 How will you carry out the following conversions? O (a) C NH2 NH NH2 NH2 (b) CH3 O C Sol. (a) NC NH2 CHCl3 KOH KOH NH2 Br2 / NH LIAlH4 CH3 - NH NaNO 2 (b) 2 H 2SO OH HCl BaO COOH COOH K Mn 4O / H 4 NH O t NH H2 / P 3 NH 2 5.5 Sol. N 2 BF4 Ar .Amine Method of Preparation of Diazonium Salts - 273-278K C6 H5 NH 2 NaNO2 2HCl C6 H5 N 2 Cl NaCl 2H 2O Chemical Reactions The reactions of diazonium salts can be broadly divided into two categories. chlorine or bromine can also be introduced byGatterman reaction. (i) Reactions involving displacement of nitrogen Diazonium group being a very good leaving group. namely (A) reactions involving displacement of nitrogen and (B) reactions involving retention of diazo group. (a) Replacement by halide or cyanide ion: The Cl-. CN. Alternatively. Br. This reaction is called Sandmeyer reaction. Br.which displace nitrogen from the aromatic ring. is substituted by other groups such as Cl-.. (b) (c) Replacement by iodide ion: Replacement byfluoride ion: - - Ar N 2 Cl KI ArI KCl N 2 - Ar N 2 Cl HBF4 Ar .nucleophiles can easily be introduced at the benzene ring in the presence of Cu(I) ion.and CN.and OH.F BF2 N 2 (d) - Replacement by H: Ar N 2 Cl H 3PO2 ArH N 2 H3PO3 HCl - Ar N 2 Cl CH 3CH 2OH ArH N 2 CH 3CHO HCl (e) Replacement by hydroxyl group: (f) Replacement by –NO 2 group: - Ar N 2 Cl H 2O ArOH N 2 HCl (ii) Reactions involving retention of diazo group coupling reactions 11 . I-. Ex.9 The end product (Z) of the following reaction is ? N2Cl Sol. we know high pH is not desirable for coupling reactions. NHR or NR 2 .7 Provide structures for the products of the reaction of PhN 2with (a) PhNMe2 (b) 2-naphthol (c) PhCH3. NH2 . H+ NH2 + H3 N Anilinium ion Aniline The positive charge on protonated amine exerts -I effect. Also.8 No reaction. Ph (a) N N NMe2 N. The substrate ring is insufficiently activated. Cu / KCN N2Cl Cu / KCN (X) H / H 2O (Y) NaOH (Z) CN 12 .Amine Ex. then protonation of aniline takes place. +H NH2 Benzene diazonium chloride pH 4-5 273-278 K N=N -HCl Aniline NH2 p-Aminoazobenzene If the high conc. Comment. thus coupling of amine with diazonium salt is not favoured. optimum pH for coupling reactions with amines is 4-5. A weakly acidic medium is provided for coupling of benzene diazonium chloride with aniline. of H+ ions are used during these reactions. at low pH (or high acid strength). Therefore.N-dimethylaminoazobenzene (butter yellow N N Ph OH (b) 1-Phenylazo-2-naphthol (c) Ex. + - N2Cl Sol. ArN 2 is a weak electrophile that undergoes diazo couplignonly with rings activated by OH. Sol. compound (A) must be aniline which is soluble in chloroform but insoluble in aq. an ionic compound (C). C8H7N.KOH and thus it must be acidic in nature whereas another compound is soluble in chloroform is either basic or neutral in nature. with chloroform followed by acidification. When treated with potassium cuprocyanide. 5 6 13 .KOH. (C). The organic layer containing compound (A). Identify the compounds (A). (E) liberates CO2 from aqueous sodium bircarbonate. ON hydrolysis (D) gives (E). (E) and write their structures. when heated with chloroform and then acidified give a mixture of two isomeric compounds (D) and (E) of molecular formula C7H6O2. (b) The compound (A) on heating with alcoholic KOH solution (chloroform already present) produces compound (C) (C7H5N) having unpleasant odour. C7H9N.3 Aneutral compound (A) C 8H 9ON on treatment with sodium hypobromite forms an acid soluble substance (B). (D). (F) on nitration yields two isomeric mononitro derivatives (G) and (H) having molecular formula. (a) One of the compound A and B is soluble in aq. (B). Mixture oftwo aromatic compounds (A) and (B) was separated by dissolving in chloroform followed by extraction with aqueous KOH solution. C7H7N2Cl is obtained.2 Sol. C 8H NO . The reactions are as follows : NH2 N C + CHCl3 + 3KOH Warm + 3KCl + 3H3O (carbylamine reaction) Aniline (A) (Soluble in chloroform) Phenyl isocyanide (C) OH OH OH CHO CHCl3/KOH + H+ Phenol (B) (Soluble in KOH) Salicylaldehyde (D) CHO p-Hydroxybenzaldehyde (E) Ex. On addition of aqueous sodium nitrite to a solution of (B) in dil. The compound (C) is Phenylisocyanide and. (E) on permanganate oxidation furnishes (F). the COOH is not strong enough to donate H to the weakly basic ArNH2 . (C) gives a red dye with alk -naphthol solution. when heated with alcoholic solution of KOH produced a compound (C) (C7H5N) associated with an unpleasant odour. as does paminobenzenesulfonic acid (sulfanilic acid) but p-aminobenzoic acid does not. HCl at 0 -5°C. Write the reaction involved in different steps. Sol. The alkaline aqueous on the other hand. (c) The compound (B) must be phenol as it is soluble in aqueous KOH and produces isomers ohydroxybenzaldehyde (D) and p-hydroxylbenzaldehyde (E) on heating.Amine SOL VED EXAMPLE Ex.1 Account for the fact that 2-aminoethanoic acid (glycine) exists as a dipolar ion. (C) yields a neutral substance (D). C8H8O2. but SO3H is a sufficiently strong acid to do so. H3 N — CH 2COO - p — H 2N — C 6H 4 — COOH p — H 3N — C6 H4 — SO-3 pA min obenzoic acid Sulfanilic acid Glycine Ex. therefore. The aliphatic NH 2 is sufficiently basic to accept an H from COOH. C8H6O4 . 6 Compound (A). Deduce the structures of (A) and (B). Is three any structural isomer of (A). Identify (A). COOH (D) : CH3 CH3 COOH COOH (F) : (E) : CH3 COOH (H) : (G) : COOH O2N COOH CH3 CN N2Cl NH2 COOH NO2 Ex. NH2 OH N=C OH OH CHO Sol. Theorganic layer containing compound (A). Compound (E) on heating with Zn dust again gives benzene. compound (B). C 6H 7N (as its sulphate). Identify the compounds (A). The alkaline aqueous layer on the other hand. Ex. (C) and (e) giving proper reactions? CN CO2H C6 H5 NH 2 dil.Amine CONH2 (A) : (C) : (B) : CH3 Sol. (A) : (E) : (B) : CHO Ex. (B). CHCH2CH2CH3 NH2 NH 2 CH 2 CH 2 CH 2 CH 2 CH 3 is positional isomer of (A).4 An optically active amine (A) is subjected to exhaustive methylation and Hoffman elimination to yield on alkene (B). Compound (C) with dilute H2SO 4/NaNO 2 in the cold followed by heating with water gives (E).5 A mixture of two aromatic compounds (A) and (B) was separated by dissolving in chloroform followed by extraction with aqueous KOH solution. (D) and (E). when heated with alcoholic solution of KOH produced a compound (C) (C 7H 5N) associated with an unpleasant odour. H2SO4 (NH 4 ) 2 SO 4 C Sol. C8H4O3 and on heating with sodalime gives benzene.H 2O CO2H C CO2H (C) H 2SO4 (C6 H5 NH3 ) 2 SO 4 O (D) O 14 . when heated with chloroform and then acidified give a mixture of two isomeric compounds (D) and (E) of molecular formula C7H 6O 2. C14H10N2O whenheated with dilute sulphuric acid gives ammonium sulphate. (C). C 6H6O. (B) NH (B) O (A) O CO2H C . (B) on ozonolysis gives an equimolar mixture of formaldehyde and butanol. C8H6O 4 and compound (C). (B). if yes draw its structure? (A) : H3C CH CH2CH2CH3 (B) : H2C Sol. Compound (B) on heating gives (D). Comment. 3-diol CH2OH CH2 PCl5 CH2OH (A) CH2 CH2Cl CH2 2KCN CH2Cl (B) CH2CH2NH2 CH2CH2NH2 -NH3 CH2CN CH2 Na + C2H5OH CH2CN (C) Nitrobenzene -6H N (D) H Piperidine (Hexahydropyridine) (E) Ex. Pyridine is obtained on treatment of (E) with nitrobenzene. therefore. -COOH group of glycine releases H+ ion which is accepted by -NH2 group. Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate Sol. Treatment of (A) with PCl5 gives (B) which reacts with KCN to form (C). H6H5NH2 + CH3I Aniline Methyliodide + [C6H5NHCH3]IN-Methylanilinium iodide 2[C6H5 NH CH3]I. PCl5 ROH (A) RCl KCN (B) Na+ C2H5OH R -C N (C) R CH2NH2 (D) -NH3 E Nitrobenzene Start with propane-1. -NH2 fails to accept a proton. H H + H2N .COOH H Glycine Zwitter ion of glycine NH2 COOH In anthranilic acid.Aminobenzoic acid) Ex. Give the structures of (A) to (E) with proper reasoning.9 Electron withdrawing -COOH and phenyl group reduces electron density of N of –NH 2 group.C .C . Anthranilic acid (2. (A) is an alcohol which gives characteristic red colour with ceric ammonium nitrate. The reduction of (C) with Na and C2H5OH produces (D) which on heating gives (E) with evolution of NH3. An organic compound (A) composed of C.8 Sol.Amine Ex. Thus glycine exist of a Zwitter ion. H and O gives a characteristic colour with ceric ammonium nitrate.7 Sol. Sol.+ Na2CO3 2C6 H 5 NHCH3 CO 2 2NaI NMethylaniline 15 .COOH H3N . Glycine exists as a Zwitter ion but anthranilic acid does not. Thus anthranilic acid can not form Zwitter ion. 10 What happens when (a) Aniline is treated with sodium nitrate and dil. Al O 2 3 R — NH2 + H2 O ROH + NH3 723K 1° alcohol (e) 1° alcohol Ethyl alcohol is formed NaNO2 + HCl NaCl + HNO2 Nitrous acid C2H5NH2 + HNO2 C2H5OH + N2 + H2O Ex. (a) Benzenediazonium chloride is formed. HCl at 273 K. H NH2 O H C H+ Benzaldehyde (c) N C Benzalaniline Schiff's base Soluble complex-Bis-(ethylamine) silver (I) chloride is formed AgCl(s) + 2C2H5NH2(aq) [Ag(C2H5NH2)2]Cl(aq) Bis-(ethylamine) sliver (I) chloride (soluble) (d) Primary amine is formed. HCl at 273 K ? Sol.Amine 2CH I3 2C6H5 NH CH3 2C6H5N(CH3)2 + CO2 + 2NaI Na 2CO 3 [C6H5NH(CH3)] +I - C6H5NHCH3 + CH3I N. (e) Sodium nitrite is added to a solution of ethylamine in dil. N NCl NH2 NaNO2 + HCl 273 K Aniline (b) Benzenediazonium chloride Benzal aniline is formed. a mixture of 1°.N-Trimethyl anilinium ioidide 2[2C6H5 N (CH3)3] I.+ Na2CO3 [C6 H5 N(CH3 )3 ]2 CO3 2NaI N. (d) Amixture of alcohol and ammonia is passed over heated aluminium oxide as catalyst. 3° amines and quaternary ammonium salt is formed. N. If ammonia is taken in excess.11 Give one chemical test to distinguish between the following pairs of compounds : (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines Ethylamine and aniline (iv)Aniline and benzylamine (iii) Aniline and N-methyl aniline (v) 16 . 2°.N. (b) Aniline is treated with Benzaldehyde. NTrimethylanilinium Carbonate Ex. (c) Ethylamine reacts withAgCl. gives carbylamine reaction while N-methyl aniline (sec. NH2 + CHCl3 + 3KOH(alc) NC + 3KCl + 3H2O Phenyl isocyanide (offensive smell) 17 . NH2 NaNO2/HCl 273 K N N Cl OH OH N N (iv) Benzyl amine on treatment with NaNO2 and HCl forms benzyl alcohol with the evolution of N2 gas while aniline froms benzene drazonium chloride which gives organe dye with alkaline b-naphthol. amine) will not give this rest. if forms yellow coloured oily compound N-nitrosoamine.Amine Sol. (i) Methyl amine. (v)Aniline.) CH3NH2 + CHCl3 + 3KOH (alc) CH 3 NC 3KCl 3H 2 O Methyl isocyanide (offensive smell) (ii) When secondary amine is treated with HNO2(NaNO2 + HCl). 1-phenylazo-2-naphthol. This test is not given by tertiaryanines. which on warming with a crystal of phenol and H2SO4 gives a green solution and on further addition of aq KOH red solution(Libermann nitroso test). (CH3)2 NH + HO—N (CH3)2 N—N O Secondary amine Nitric acid O + H2O N-Nitrosodimethylamine Phenol + H2SO4 KOH + Water Green colour Red colour (iii) When aniline is treated by NaNO2 + HCl at 273 K. benzenediazonium chloride is formed which on treatment with b-naphthol gives a bright orange dye. being a primary amine will give carbylamine reaction (offensive smell of isocyanide when treated with CHCl3 + KOH) while dimethyl amine will not show any reaction with CHCl3 + KOH (alc. being a primary amine. C 2H 5NH 2 (vi) In increasing order of solubility in water: C6H5NH2.6 Write one chemical reaction each to illustrate the following (i) Hofmann Bromanide reaction. (b) How would you convert methylamine into ethylamine ? Q. C 2H 5NH 2. (ii)Aryl cyanides cannot be formed by the reaction of aryl halides and sodium cyanide.11 Describe tests to distinguish between : Secondary amine and tertiary amine. 18 .1 Arrange the following : (i) In decreasing order of the pK values: b C2H5NH2. (iv) In decreasing order of basic strength in gas phase: C2H5NH2. (C2H5) 2NH. (ii) Garbriel Phthalimide reaction. Q. secondary and tertiary amines.Amine EXERCISE-I Q.10 Account for the following : (i)Aniline is weaker base than methylamine.3 Explain Hofmann Bromanide reaction with Mechanism ? Q.5 Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid. (C2 H5 )2 NH and C6 H5 NH2 (ii) In increasing order of basic strength: C6H5NH2.8 Illustrate the following with an example of reaction in each case : (i) Sandmeyer reaction (ii) Coupling reaction Q. (C 2H 5) 3N and NH 3 (v) In increasing order of boiling point: C2H5OH.2 Describe a method for the identification of primary. Q. (CH3) 2NH. Q.Also write chemical equations of the reactions involved.9 Write the chemical reaction equations for one example each of the following (a)Acoupling reaction (b) Hofmann's bromamide reaction (c)Acetylation Q. (C2H5)2NH and CH 3NH 2 (iii) In increasing order of basic strength: (a)Aniline. C6H5N(CH3)2. C6H5CH2NH2. p-nitroaniline and p-toluidine (b) C6H5NH2. C6H5NHCH3 . C 6H 5NHCH 3 .4 Why cannot aromatic primary amines be prepared by Gabriel phthalimidesynthesis? Q.7 Assign a reason for the following statements (a)Alkylamines are stronger bases than arylamines. (C2H5) 2NH. Q. Q. 15 Write the method of formation of benzene diazonium chloride ? Q.Amine Q. (iv)Although amino group is o. (ii) Methylamine solution in water reacts with ferric chloride solution to give a precipitate of ferric hydroxide. Q.6-tribromofluorobenzene (iii) Benzoic acid to aniline (vi) Chlorobenzene to p-chloroaniline (v) Benzyl chloride to 2-phenylethanamine (viii) Benzamide to toluene (vii)Aniline to p-bromoaniline (ix)Aniline to benzyl alcohol. Write the equation of curtius reaction with mechanism ? Complete the following reactions: (i) C6H5NH2 CHCl3 alc.14 Write the physical property of aniline ? Q. Q.18 Q. (iii) Ethyl amide to methylamine.13 State the reactions and reaction conditions for the following conversions (i) Benzene diazonium chloride to nitrobenzene. (v)Aniline does not undergo Friedel-Crafts reaction. (ii)Aniline to benzene diazonium chloride. (ii) Hexanenitrile into 1-aminopentane (iv) Ethanamine into methanamine (vi) Methanamine into ethanamine (viii) Propanoic acid into ethanoic acid ? (ii) Diazotisation (iv) Coupling reaction (vi)Acetylation Accomplish the following conversions (ii) Benzene to m-bromophenol (i) Nitrobenzene to benzoic acid (iv)Aniline to 2.20 Q.and p.17 How willyou convert: (i) Ethanoic acid into methanamine (iii) Methanol to ethanoic acid (v) Ethanoic acid into propanoic acid (vii) Nitromethane into dimethylamine Q.12 Account for the following observations : (i) pKb for aniline is more than that for methylamine.KOH (iii) C 6H5NH2 H2SO4 (conc.19 Q.4. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines. (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.16 Account for the following: (i) pKb of aniline is more than that of methylamine. 2 19 . (ii) Ethylamine is soluble in water whereas aniline is not.21 Write short notes on the following: (i) Carbylamine reaction (iii) Hofmann’s bromamide reaction (v)Ammonolysis (vii) Gabriel phthalimide synthesis. Q. aniline on nitration gives a substantial amount of m-nitroaniline.) (v) C6 H5NH2 + Br2(aq) (ii) C6H5N2Cl H3PO2 H 2O (iv) C6H5N2Cl C2H5OH (vi) C6H5NH2 + (CH3CO) 2O ( i)H B F4 (vii) C6H5N2 Cl (ii)NaNO / Cu.directing in aromatic electrophilic substitution reactions. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iii)Aniline does not undergo Friedel Crafts reaction. bromophenol (v) Nitrobenzene to 2. 20 .26 Why aniline do not give fridel craft reaction ? Q. Show the angle between two methyl groups. 6-tribromoaniline.27 What is Gabriel phthalimide synthesis ? For what purpose is it used ? Give equation only to explainyour answer. Q.29 How will you convert 4-nitrotoluene to 2-bromobenzoic acid ? Q.Amine Q.23 Write the reaction and conditions for the following conversions (i)Aniline to benzene (ii) Methylamine to methyl cyanide (iii) Propanenitrile to ethylamine (iv) m-Bromoaniline to m.24 Write the method of formation of zwitter ion ? Q.25 Explain nitration of aniline ? Q. Q.22 Give plausible explanation for each ofthe following: (i) Whyare amines less acidic than alcohols of comparable molecular masses? (ii) Whydo primaryamines have higher boiling point than tertiaryamines? (iii) Whyare aliphatic amines stronger bases than aromatic amine? Q. 4.28 Write the equation of carbyl amine reaction with mechanism ? Q.30 Draw the structure of trimethylamine and tell the shape of the molecule. (A). why ? Glycine exists as H3N+CH2COO.C6H4. is insoluble in both dilute acid and base and its dipole. (2) The amino group in ethylamine is basic whereas that in acetamide it is not basic.1 Aspartame. C6H4N2O4. the former boils at a lower temperature (3ºC) than the latter (49ºC).while anthranilic acid. when treated with nitrous acid yields an alcohol B. is a peptide and has the following structures : NH2 CH2C6H5 HOOC . p-NH2. compound (C) is formed which reacts with NaNO2 and HCl at 0 . When (B) is heated with soda-lime.COOCH3 (a) (b) (c) (d) Identify the four functional groups. B on careful oxidation yields a substance C of vapour density 36 which forms oxime. (3) Although trimethylamine and n-propylamine have same molkecular weight. It gives no reaction with benzene sulphonyl chloride or with NaNO2 and HCl at 0ºC.7 Explain with reason ? (1) Although trimethylamine and n-propylamine have same molecular weight. Silver chloride dissolves in aqueous solution of methylamine. it is soluble in alkali but insoluble in mineral acids. What is (A) ? Q.2 Compound ofA(molecular formula C9H11NO) gives a positive Tollen’s test and is soluble in dilute HCl. an artificial sweetener.6 Explain. C4H10O with the evolution of N2. (3) Dimethylamine is a stronger base than trimethylamine.CH . Q.CH2CH . (3) It is difficult to prepare pure amines by ammonolysis of alkyl halides. Identify compoundAand write the structural formulae of the isomeric compounds that behave with HNO2 in the same manner. (4) Sulphanilic acid although has acidic as well as basic group.4 An organic compound (A). Q. Q. B can react with NaHSO3 but does not reduce Fehling solution. (4) Q. Explain. Write the zwitterionic structure Write the structures of the amino acids obtained from the hydrolysis of aspartame. Which of the two amino acids is more hydrophobic ? Q. moment is zero. (2) Aweakly basic solution favours coupling with phenol.Amine EXERCISE-II Q. (4) 21 . upon oxidation with KMnO4 gives an acid (B).CONH .3 An organic compoundA.5ºC.5 Explain the following observations : (1) Aniline dissolves in aqueous HCl.COOH does not exist (1) as dipolar ion. Benzenesulphonic acid is a stronger acid than benzoic acid. (2) Dimethylamine is a stronger base than methylamine but trimethylamine is a weaker base than both dimethylamine and methylamine. Deduce the structure of (A). Explain. the former boils at a lower temperature (3ºC) than the latter (49ºC). 6-Dimethyl -N N-dimethylaniline. Q. Explain. (5) Comment. Complete the following reactions : (1) (2) (3) PCl5 C6H5COOH P2O5 CONH2 KOH 2. does not undergo coupling with benezenediazonium chloride.9 Q. Tertiary amines do not undergo acetylation.NH2 (S) T+U (V) Gattermann reaction 22 .10 Explain it with reason ? tert-Butylamine cannot be prepared by the action of NH3 on tert-butyl bromide. 4-Dinitroaniline (6) C6H6 Oleum H2/Hi [E] KBr + H I (ii) anisole SO3H C6H5CN G (i) NaNO2/HCl. (2) How will you explain the acidic nature of 1º and 2º nitroalkanes ? (3) Aniline does not undergo Friedal Craft’s reaction ? (4) Although borontrifluoride adds on trimethylamine.Amine Q. HCl AlCl3 Br2. (1) Isocyanides are hydrolysed by dilute acids but not by alkalies to form amine and formic acid.8 Explain it ? (1) An aqueous solution of ethylamine gives a red precipitate with ferric chloride. although has a free p-position. Comment. Fe N (R) P PhNH. Comment (2) (3) 2. it does not add on triphenylamine. In the following compounds : (4) O N N = (III) H (IV) N N (II) H (I) The order of basicity is I > III > II > IV. 5ºC NaOH (K) P2O5 [D] H+ — N(CH3)2 + HNO2 (5) (L) (J) NaOH heat (M) OH 1 (7) CHCl3/NaOH — SO3H fuming O H SO 2 Et2SO4 NaOH (9) Phenol (10) CH3CONHC6H5 (11) F EtNH2 + KCN + Br2 (4) (8) NH3 [C] C6H5N2Cl (A) HCl 4 (Q) (i) NaOH fuse (ii) H+ HCN. Explain. (i) On hydrolysis it gives an amine (B) and a carboxylic acid (C) (ii)Amine (B) reacts with benzene sulphonjyl chloride and gives a product which is insoluble in aqueous sodium hydroxide solution. Give structure of compounds (A) to (E) with proper reasoning. What is the structural formula of the unknown amine ? Q. which reacts with KCN to form (C).21 What happens with cyclopentanoen reacts with (a) CH3CH2 NH2 (1° amine) (b) (CH3CH 2 )2 NH (2° amine) Cyclohexylamineis a stronger base than aniline. the reduction of (C) with warm Na/ C 2 H5OH produces (D). give the structures of compounds (A) and (B). which on heating gives (E) with evolution of ammonia Pyridine is obtained on treatment of (E) with nitrobenzene.20 Q.v. H and O gives characteristic colour with ceric ammonium nitrate. the evolved nitrogen. Why? From analysis and molecular weight determination.15.25 g of an unknown amine was treated with nitrous acid. Why? How does the formation of 2° and 3° amines can be avoided during the preparation of 1° amines by alkylation? It is necessary to acetylate aniline first for preparing bromoaniline. Treatment of (A) with PCl5 gives (B).14 An organic compound (A) composed of C. I KCN CuCN B C Hot H2SO4 Hot aq.T. (D) and (E) respectively. E H F Q.16 Q. 2 moles u. corrected to S. Explain the reactions. B and C.13 The aqueous solution of a nitrogen and chlorine containing organic compound (A) is acidic towards litmus. KMnO4 G CH3OH at 30ºC D Cl2.p.Amine Q.11 Give structures for the compounds (A) to (I) : C8H11N (A) NaNO2/HCl 0-5ºC Heat to m. The alcohol isolated from the reaction mixture gave a positive iodoform reaction. Compound (B) on treatment with C6 H 5SO 2Cl in the presence of NaOH gives an insoluble product (C).P.12 When 2.19 Q. but not chlorine. (iii)Acid (C) on reaction with Tollen’s reagent gives a silver mirror when areA. the molecular formula of (A) is C3H 7 NO . containing nitrogen.17 Q. One mole of bromoderivative (A) and mole of NH 3 react to give one mole of an organic compound (B). C13H13 NO 2S. Both (B) and (C) react with HNO 2 to give compounds. (D) on oxidation and subsequent decarboxylation gives 2-methoxy-2-methyl propane. the compound gave following reactions. (B) reacts with CH 3I to give (C). Q. (A) on treatment with aqueous NaOH gives a compound (B). Q. Why? Dimethylamienis a stronger base than methylamine but trimethylamine is a weaker base than both dimethyl amine and methylamine. 23 . Q.18 Q. measured 560 ml. Give structures of compounds (A) to (E) with proper reasoning. it forms (C) having the molecular formula C8 H 7 N . (B) and (C) rearrange to give (D) and (E).22 An optically active amine (A) is subjected to exhaustive methylation and Hofmann elimination to yield an alkene (B). (B) dissolves in NaHCO3 and gives colour reaction with FeCl3 and can be prepared by the action of CCl 4 and NaOH on phenol. Compounds (B). C6 H 3 N 3O 7 . an oil (F) C6 H 7 N separated out.27 An organic compound (A) having M. When (B) is reacted with excess HNO3 .25 An aromatic compound (A). C7 H 6O 2 . Q.F C7 H 9 N on treatment with NaNO 2 and HClat room temperature forms another compound (B). Identify the compounds (A) to (G). (C). Deduce the structures of (A) and (B). which on halogenation forms only one monohalo derivative. which is isomeric to (C). Q. C7 H 8O . which condenses with benzene in the presence of anhydrous AlCl3 to give anthraquinone. (E) on boiling with alkali followed by acidification gives a white solid (G). on nitration.F. Hydrolysis of (C) followed by reaction with NaNO2 and HCl at low temperature and subsequent reaction with HCN in the presence of Cu(D). (D) on hydrolysis followed by oxidation gives a dibasic acid. having M.HCl gives (B) and (C). When (A) or (B) is treated with bromine water. (C) gives red colour with ceric ammonium nitrate and on oxidation gives an acid (D). (D) and (E)are all isomers of molecular formula C8 H 9 NO . (B) on ozonolysis gives an equimolar mixture of formaldehyde and butanal. (F) reacts rapidly with CH 3COCl togive back (D). Is there any structural isomer to (A).23 An aromatic compound (a) having molecular formula C7 H 7 NO 2 dissolves in NaHCO3 to evolve CO2 and when reacted with NaNO 2 / HCl forms (b). The alkaline solution thus obtained on acidification gives the precipitate of a compound (B) having molecular weight 136. Q. Compound (B) is unable to forma dye with -naphthol. C8 H8O on treatment with NH 2OH. which on reaction with NaNO 2 / HCl gives (C). C7 H 6O3 . On the other hand. (C) undergoes acetylation and decomposes NaHCO3 to evolve CO 2 . they form dibromo derivatives. it forms (C).26 An organic compound (A) of molecular weight 135. (E) an isomer of (A) on boiling with dilute HCl gives an acid (F). Identify compounds (A) to (D). When (A) is reacted with chloroform and alkali. Compound (C) reacts with cold HNO 2 to gives (D). if yes draw its structure. having molecular weight 136. On oxidation followed by heating. respectively on treatment with acid. Treatment of(A)with HNO 2 also yields (B). (F) gives an anhydride (G). which give red colour with ceric ammonium nitrate. Decarboxylation of (D) gives (e) which forms a single mononitro derivative (F).24 Compound (A) having M. Q. on boiling with NaOH evolves a gas which gives white denso fumes on bringing a rod dipped in Hcl near it. However.F C7 H5 NO 2Cl2 on reduction with Sn/HCl gives (B). On the other hand. Q. Identify the compounds (A) to (E). Give the structures of (A) to (F) with proper reasoning. C6 H 5 N 3O6 Cl which when reacted with water gives back (C).Amine Q. When (D) is boiled with alcoholic KOH. having equivalent weight 191. 24 . Give the structures of (A) to (G) with proper reasoning. whereas it treatment with Br2 / KOH gives (C). On reaction with PCl5 . (C) is converted to (D). (D) with conc. (C) gives a red dye with alkaline -napththol solution. C2 H 6O .Account for the above reactions and suggest how (A) may be synthesized. 25 . B gave an acid C whose ethyl ester gave (D) on the action of hydrogen and platinum. C 4 H 8O 3 capable of being resolved. Elucideate the reaction mechanism and suggest a synthesis of (C). C5H 8O5 . (F) onnitration yields two isomeric mononitro derivatives (G and H) having molecular formula C8 H5 NO 6 . F with HNO 2 gave G.29 An optically inactive acid (A). Both (A) react with NaNO2 and HCl. NH 3 gave E. C6 H 8O 4 while the latter yields (E). On heating with soda lime (A) yields (B) C 2 H 7 N . ON hydrolysis (D) gives E (C8 H8O 2 ) . E liberates CO 2 from aqueous NaHCO3 . On addition of aqueous NaNO2 to a solution of B in dilute HCl at 0-5°C an ionic compound (C) C7 H7 N 2Cl is obtained. the former yielding a compound (C) C3H 6O .28 An optically active compound (A). (E) on permanganate oxidation furnishes (E) C8 H 6O 4 . On action of sulphuric acid. Write the reactions involved in different steps. on being heated lost CO 2 to give an acid (B). When treated with potassium cuprocyanide (C) yields a neutral substance (D) C8 H7 N . C3H7O2 N forms a hydrochloride but dissolves in water to give a neutral solution. Q. C3H9 N . Q. C 4 H9OH which with Br2 and KOH solution gave (F). which on heating is converted to (D).Amine Q. (G) on mild oxidation gave H. BothAand H gave the iodoform reaction.30 Aneutral compound (A) C8 H 9OH on treatment with NaOBr forms an acid soluble substance C7 H 9 N. Dissolves in dilute HCl in the cold and is reprecipitated by the addition of alkali (a) A-3.4 The correct order of basic strength of in CCl4 (3) R2NH (1) NH3 (2) RNH2 Where R is CH3 group is (a) 3 > 2 > 1 > 4 (b) 2 > 3 > 4 > 1 (c) 3 > 2 > 4 > 1 (4) R3N (d) None of these Q. C-4. (a) II is not an acceptable canonical structure. because the nitrogen has 10 valence electrons (d) II is an acceptable canonical structure. Q. B-3.2 When nitrobenzene is treated with Br2 in presence of FeBr3. Forms a precipitate withAgNO3 in ethanol (B) Ethyl benzoate 3. when Br+ attacks at the ortho and para positions. H3CNH2 2. (CH3)3N 1. B-3. Insoluble in water. List II List I 1. (a) 1 and 3 (b) 2 and 4 (c) 3 and 4 Q. because it is non aromatic (c) II is not an acceptable canonical structure. B-3.3 Examine the following two structures for the anilinium ion and choose the correct statement from the ones given below. it gives (a) Sulphanilic acid (b)Aniline sulphate (c) o-aminobenzenesulphonic acid (d) m-aminobenzenesulphonic acid. Statements which are related to obtain m-isomer are: (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 least destablilzed (c) Loss of aromaticity. D-4 26 . B-2. C-1. Reduces Fehling’s solution (A) Propyne 2. because carbonium ions are less stable than ammonium ions (b) II is not an acceptable canonical structure. Selects the correct answer using the codes given below the list . H3C-NH-CH3 Select the correct answer using the codes given below. Match the compounds in list I with the appropriate test that will be answered by each one of them inlist II from the combinations shown. but dissolves in aqueous NaOH upon heating (C) Acetaldehyde (D) Aniline 4.7 4. D-4 (b) A-2. (1) Ethylamine (2) 2-aminoethanol (3) 3-aminopropan-1-ol (a) 1 > 3 > 2 (b) 1 > 2 > 3 (c) 2 > 1 > 3 (d) None of these Q.5 Place the following in the decreasing order of basicity. and not at meta position (d) Easier loss of H+ to regain aromaticity from the meta position than from the ortho and para positions. Q. the major product formed is mbromonitrobenzene. Q. C-1.Amine EXERCISE-II Q.1 When aniline is treated with fuming sulphuric acid at 475K.6 Which of the following will give a positive carbylamine test ? 3. D-4 (c) A-2. C6H5NH2 (d) 1 and 4. C-2. D-1 (d) A-1. 15 (c) Br2 NaOH (b) CS2 (a) Na Which of the following does nto give ethylamine on reduction (a) methyl cyanide (b) Ethyl nitrile (c) Nitro ethane (d) Water (d)Acetamide In the reaction. (a) (d) None of these NH2 NH2 (b) (c) (d) N N (d) None of these Cl Me H Q.13 .10 R-CH-CH3 | SO4– Na+ useful as (a)Afertilizer (b)An explosive (c)Adetergent The basic strength of amines (ethyl) and ammonia in H2O is (a) NH3 > p > s > t (b) p > s > t > NH3 (c) s > p > t > NH3 Which of the following will have highest Kb value. CH3 — NO2 Cl 2 NaOH is (a) ClCH 2 NO2 (b) Cl 2CHNO2 (c) Cl3CNO2 (d) CH3 NH 2 Q.11 Activation of benzene ring by — NH2 in aniline can be reduced by treating with (a) Dilute HCl (b) Ethyl alcohol (c)Acetic acid (d)Acetyl chloride Q.14 Q.17 Asequential reaction may be performed as represented below: SO 2 Cl 2 NH3 R — CH 2 CO 2 H R — CH 2 COCl R — CH 2 CONH 2 (1) ( 2) R — CH2 NH R — CO2 H R — CH2 OH (3) ( 4) (5) The appropriate reagent for step (3) is (a) NaBr (b) Bromine and alkali (c) HBr (d) P2 C5 27 . Q.Amine Q.9 Q.reaction sequence R-CH = CH2 + H2SO4 R-CH-CH3 | SO4H the end product would be NaOH (Where R = C14 H29) Q.12 Dipolar ion structure for amino acid is (a) H2N (b) H3N CH COOH R (c) H3N CH COO R CH COO (d) None of these. R Q. excess CH Cl 3 CH3 NH2 (X) (a) (CH3 )3 N (Y) (Z) the final product (Z) is (b) (CH3 )4 N Cl - (c) (CH3 )4 N OH - (d) (CH3 )2 NH Q.8 In the following 2.NH2 group shows acidic nature while reacts with reagent.16 The product not obtained in the following reaction. K Cr O .Amine Q.22 Diazonium coupling reaction with aniline should be carried out in (a) Weakly basic medium (b) Weakly acidic medium (c) Strongly basic medium (d) Strongly acidic medium Q.20 (b) NH2 H2N C (c) O (d) H2N H2N CH OH H2N Identify compound (A) in the following oxidation reaction. CH3COOH (a)All have same chemical property (c) All are basic Bromine in CS2 reacts with aniline to give Q.23 For CH3CHO.21 Aniline is a weaker base than ethyl amine because (a) Phenyl gp in aniline is a +R gp (b) Ethyl gp in ethyl amine decreases the electron density on nitrogen atom (c) The lone pair of electron on nitrogen atom in aniline is delocalized over aniline (d)Aniline is less soluble in water than ethylamine Q.19 NH2 (d) NH(CH3) The strongest base among the following is H2N H2N C (a) NH H2N C (b) H2N Q. nucleophillic. H S O 2 2 7 2 4 O (A) O OH NH2 (a) (b) OH (c) (d)All of these. CH3NO2.18 Which of the following amine form N-nitroso derivative when treated with NaNO2 and HCl? (a) H3C NH2 N(CH3)2 (c) Q. NH2 NH2 OH Q.25 (b)All have one common chemical behaviour (d) None of these (d) Both (a) and (b) Br (b) Electrophillic.24 NH2 (a) Br NH2 (b) RNC cannot undergo (a)Acidic hydrolysis (c) Base hydrolysis Br NH2 Br (c) Br Q. addition on carbon (d) Both (b) & (c) 28 . 4 only (d)All of these 29 .H (c) Cl (d) None of these Ethylamine undergoes oxidation in the presence of KMnO4 to give (a) CH3COOH (b) CH3CH2OH (c) CH3CHO (d) N-oxide Baker Mulliken’s test is used to detect the presence (a) -COOH gp (b) -NO2 (c) -OH (d) -NH2 t-amines with different alkyl gp has a chiral nitrogen atom still it is optically inactive because (a) Chiral N-atoms cannot rotates plane polarized light (b) The lone pair prevents the rotation of plane polarized light (c) Both of these (d) None of these In CH3NO2 we can observe (a) H-bonding (b) -halogenation reaction (c) Tautomerism (d)All of these The reaction: O || C NH O || C :N:K + n-BuBr C || O KOH C || O O || C N Bu-n 1.34 During the conversion of NH2 with HNO2 at high temperatures the following substances or entermediates are formed. OH 4.Amine NH2 + phosgene Q. N2O3 (a) 1 only 2. aq. HC CCH2NH2 1.NaOH 2. C6H5NH .30 Q. Here X is O NH . H2C = CHCH2NH2 The increasing order of basicity is (a) 3 < 1 < 2 (b) 3 < 2 < 1 (c) 2 < 1 < 3 (d) None of these Q.33 Consider the following compounds : 2.C .32 Cl N=C=O (a) Carbylamine reaction (c) Gabriel phthalimide synthesis The conjugate acid of HO(CH2)3 NH2 is + (a) H2O(CH3)3NH2 is called + + (b) HO(CH2)3NH3 (b) Hofmann reaction (d) Cope reaction .Cl (a) Q.N = O (c) 2. 1. CH3CH2CH2NH2 3.28 Q.26 X. N2 Cl - (b) 1.31 (b) CH . 4 only 3.29 Q.27 Q. H O+ n-BuNH2 3 C || O Q. - (c) O(CH2)3NH2 + (d) HO(CH2)3NH Q.C . 2. (a) A-2. Isopropylamine can be obtained by LiAlH4 (a) (CH3 )2 CHO NH 2OH ? H 2 / Ni (b) (CH3 )2 CHO NH3 ? H3C CHOH NH3 (c) (d)All of these.35 Match list I (condition of reaction of nitrobenzene) with list II (products formed ) and select thecorrect answer using the codes given below. N Br Q. The compound.39 (d) RH 2 N < RCH = NR < RCN. D-3 (d) A-1. C-3.44 (c) 2.6-tribromoaniline (d) 2.38 Q.41 Amines are highly soluble in: (a)Alcohol (b) Diethyl ether (c) benzene (d) Water. B-3. B-4. D-1 (c) A-1. B-1. B-3. C-2.37 (i ) N H 3 (A).Amine Q. (b) CH3 — CO — CONH2 (d) CH 3CH(NH 2 )COOH . 6-dibromoaniline. (B) Zn and NH4Cl Methanolic NaOMe Phenyl hydroxylamine (C) 3. RCN = NR and RHN2 is (a) RCN < RCH = NR < RH 2 N (b) RNH 2 < RCN < RCH = NR (c) RCH = NR < RNH 2 < RCN Q.43 The bromination of aniline produces (a) 2-bromoaniline (b) 4-bromoaniline Q.36 The increasing order of basicity of RCN. which on reaction with aqueous nitrous acid at low temperature produces an oily nitrosoamine is (a) Methyl amine (b) Ethylamine (c) Diethylamine (d) Triethylamine 30 . C-2.42 Which of the following reagents can convert benzene diazonium chloride into benzene? (a) Water (b)Acid (c) Hypophosphorous acid (d) HCl. List I List II Sn and HCl 1.4. Zn and KOH (D) Aniline 4.40 Reaction of RCONH 2 with a mixture of Br2 and KOH gives RNH2 as the main product. (A) Hydrazobezene Azoxybenzene 2. The intermediate involved in the reaction is O (a) R C O NHBr (b) R—NHBr (c) R C Br (d) R—C = N = O. Q. H3C Q. D-4 Q. What is (A) ? CH3 — CO — COOH ( ii ) H / Pd 2 (a) CH 3CONH 2 Q. Q. (c) CH 3CH 2CONH 2 How many isomeric amines with that formula C7 H9 N contain a benzene ring? (a) two (b) three (c) four (d) five. C-2. D-4 (b) A-4. 45 Carbylamine test is performed in alcoholic KOH by heating a mixture of (a) Chloroform and silver powder (b) Trihalogenated methane and a primary amine (c)An alkyl halie and a primary amine (d)An alkyl cyanide and a primary amine. Therefore. CH3 N+ N The reactivity of these ions towards azo coupling reactions under similar conditions is (a) I < IV < II < III (b) I < III < IV < II (c) III < I < II < IV (d) III < I < IV < II Q. The reactivity of diazonium salts towards coupling reactions is favoured by presence of electron withdrawing groups. the reactivity of 2. Diazonium salts also couple with phenols and aromatic amines to form coloured azo dyes.Amine Q. O2N N+ N III. OCH3 (by heating with CH3OH) Cl (by CuCl/HCl or Cu/HCl). Me2N N+ N II. diazonium salts undergo a number of substitution reactions in which the diazo group is replaced by a monovalent atom/group such as H(by H3 PO2 in presence of Cu+ ions.47 Which of the following diazonium salts when boiled with dil. CH3O N+ N IV. Br (by CuBr/HBr or Cu/HBr) I (by KI in presence of Cu+ ions).). 6-trinitrobenzenediazonium chloride is so high that it even couples with reactive hydrocarbons such as mesitylene. CH3 CH2OH. CN (by first neutralizing with Na2 CO3 and then reacting with KCN/CuCN). NaBH4 etc. 4. Obviously electron donating groups favour diazotisation by retarding the decomposition of diazonium salts to phenyl cation.46 benzene in presence of NaOH) etc. H2 SO4 gives the corresponding phenol most readily? OMe (a) (c) Me N + N (b) MeO N+ N (d) N+ N N+ N 31 . OH (by boiling in presence of mineral acids). The high reactively of arenediazonium salts isdue to the excellent leaving ability of the diazo group as N2 gas. Consider the following ions: I. F (by first converting into N 2F 4 followed by heating). Comprehension : Arenediazonium salts are more stable than alkanediazonium salts due to dispersal of the positive charge on the benzene ring. NO2 (by first neutralizing with Na2CO3 and then treating with NaNO2) phenyl or substituted phenyl (by treating with benzene or substituted Q. 2 With proper choice of alkyl halides. Gabriel-phthalimide synthesis can be used.54 A: R: 3° amine is proved to be less basic in aq. Q. Q. The reaction occurs through intermediate formation of acylnitrene.50 Q. Conjugate acid of 3° amine is poorly solvated in aq.58 A: R: Acetamide reacts with Br 2 in presence of methanoic CH 3ONa to form methyl N-methylcarbonate.57 A: 1° amides react with Br2 + NaOH to give 1° amines with one carbon atom less than the parent amide. Q. solution. 2° and 3° amines. Q. then mark (c) (d) If both Aand R are false. then mark (a) (b) If both Aand R are true but R is not the correct explanation of A. Chlorobenzene undergoes nucleophilic substitution by elimination-addition mechanism while 4-nitrochlorobenzene undergoes nucleophilic substitution byaddition-elimination mechanism. then mark (b) (c) If Ais true but R is false.49 NH2 NH2 (b) Cl NH2 (d) CH3 NH2 The product formed when bromobenzene reacts with benzenediazonium chloride in presence of NaOH is (a) Diphenyl (b) p-Bromodiphenyl (c) p. phthalmide synthesis can be used to prepare 1°. p´-Dibromodiphenyl (d) p-Bromoazobenzene Benzendiazonium chloride on reaction with phenol in weakly basic medium gives: (a) Diphenyl ether (b) p-Hydroxyazobenzene (c) Chlorobenzene (d) Benzene Assertion and Reason : Each of the questions given below consists of two statements. Select the number corresponding to the appropriate alternative as follows (a) If both A and R are true and R is the correct explanation of A. White precipitate of silver chloride get dissolved in NH4OH soln. solution. Nitro group has +M and -I effect. R.52 A. an assertion (A) and reason (R).48 Which of the following arylamines undergoes diazotisation most readily? (a) NO2 (c) CH3O Q. NH3 reacts withAgCl to form a solution complex with formula [Ag(NH3)2]Cl. Q.Amine Q. then mark (d) Q. Benzyl amine is more basic than aniline. R: Q. Positive inductive effect of phenyl group creates high electron density around N atom. R. Q. 32 .53 A: R: o-nitrophenol is more acidic than p-nitrophenol.51 A.56 A: R: 4-Nitrochlorobenzene undergoes nucleophilic substitution more readily than chlorobenzene. Methyl isocyanate is formed as an intermediate which reacts with methanol to form methyl Nmethylcarbamate.55 A: R: In order to convert R-Cl to pure R-NH . 4 Q.Ph CHO (10) T = CH3 .C .10 (1) (2) C = C6H5COCl D = C6H5CONH2 F = C6H5CN G = C6H5COOH E = C6H5CHNH2 N(CH3)2 (3) H = EtNHBr (4) I= NO NO2 -N = N - (5) J = O2 N (6) K = C6H5SO3H Me L = C6H5SO3Na M = C6H5OH OH (7) I = NaOH CHO N= SO3H (8) OH O= P= SO3 H OH OEt (9) Q = Ph .NH.3 N(CH3)2 CH3 C H C 2 H3 . CHO COOH N(CH3)2 Q.NH O + para isomer Br Cl (11) A = Cu Powder V= + N2 33 . The other isomers should be p-amines only | NH 2 NO2 Q.O .2 A= B= C= N(CH3)2 Q.Amine ANSWER KEY EXERCISE .II Q.Et S = EtO R= CH = N.1 The hydrolysed products are aspartic acid and phenylalanine. Amine NH2 Cl NH2 Q.14 (A) (B) CH2 OH CH2Cl H2 C H2 C H2 C CH2 OH CH2Cl (A) (B) Cl C CH2CH2Br H3 C C CH2 CH2NH2 C CH2CH2OH H3C OCH3 C CH2CH2NH-CH3 CH3 CH3 (B) (C) CH3 CH3 (D) (E) CH2 (E) CH2CH2NH2H3C C NH CH2 (D) OCH3 OCH3 H3C CH2 CN CH2 H2 C OCH3 CH3 (A) Q.16 a O CH3 CH2 NH 2 N Tauto mer ise N H b + O (CH3 CH 2 )2 NH N H H CH2CH3 CH2CH3 H - CH2CH3 N CH2CH3 24 .12 C2H5NH2 CH3-NH CH3 — N H 2 Cl - Q.15 H2 C (C) OCH3 CH2 CH2 CH2NH2 CH2 CN CH3 CH2 CH2 N CH3 CH2CH3 CH2CH3 Q.OMe O O C O (I) C O Q.13 Q.CH 3 O (H) COOH C .CH 3 Cl Cl (F) COOH D Et Et COOH (E) CN C E COOH COOH (G) C .11 A B Et COOH C . 24 NO2 (D) picric acid Picryl chloride OH C H3C (A) NO2 (C) N C C6H 5 (B) Acetophenone Acetanilide N C6H 5 OH (C) E-Acetophenone oxime Z-Acetophenone oxime O O C6H5 NH C (D) H3C CH3 C 6 H5 C NH CH3 (E) N-methyl benzamide C6H5 NH2 (F) Aniline C6H5 COOH (G) Benzoic acid 35 .23 (A) (B) Anthranilic acid salicylic acid C 6 H5 HC C O Q.18 Q.17 Q. Amino group. being activating group. Dimethyl amine is stronger base because of inductive effect.21 CH3 (B) HN CH3 C HCOOH NH2 (A) H3C CH3 Q. N dim ethyl formamide) (A) NOCH CH3 Q.22 (B) H2C CH3 Cl OH COOH COOH OH O2N NH2 NO2 O2 N NO2 Q. Use of excess ammonia reduces chances of reaction of 1° amine with alkyl halide to form 2° and 3° amines.19 Q.20 Aniline is a weak base than cyclohexylamine because of resonance.Amine Q. CH3 (N. Trimethyl amine is a weaker base because positive charge on nitrogen could be stabilized and due to crowding by alkyl groups around the nitrogen atom protonation cannot take place. activates bromination of aniline and forms tribromoaniline. 26 O (A) (C) (B) (D) CH3 (E) (F) CH3 (G) C O NH2 OH NC CN COOH CH3 CH3 CH3 CH3 COOH (A) (B) (C) (D) Q. Degree of unsaturation of B = 0.NH2 N CH3 CH2 -OH CaO (B) ( E) NH2 (A) NaNO2 HCl H3C CH COOH OH (C) H3 C2 O CO -2 H2 O CH COOH CH 3HC CHCH O OH P / Br 2 CH3CH2 — COOH H3C2 3 O NH3 H3C2 CH COOH CH COOH Br NH2 36 . Thus. which forms a cyclic diester on heating. a hydroxyl acid. It is likely to be amino acid as the molecular formula contains one N and 2 O. B is CH 3CH 2 — NH 2 . Since Aforms hydrochloride and dissolves in water to give a neutral solution.27 Q. B is a saturated amine. B reacts with NaNO2 and dilute HCl forming (E) C 2 H5OH .28 (E) Degree of unsaturation ofA= 2.atoms. it contains both a basic and an acidic functional group. Therefore. Aalso reacts with NaNO 2 and dilute HCl forming C. On decarboxylation it forms an amine B.All the reactions can be given as H3C aOH aNO 2 HC l CH COOHN CH3 CH2 .25 Cl (A) Cl (B) Cl (C) Cl (D) Cl (E) Cl (F) CH2CONH2 CH2COOH CH2NH2 CH2OH O CONH2 COOH C Q.Amine Cl Cl CH2NO2 Cl Cl Cl COOH CH2OH CH2NH2 Cl NO2 Q. 55 d c d d d d c c Q.54 b c a b a a a a Q. b) d d b d c c c a Q.16 Q.34 Q.4 Q.26 Q.44 Q.33 Q.18 Q.58 (a.42 Q.19 Q.11 Q.14 Q.37 Q.7 Q.6 Q.40 Q.52 c a b c c b b a Q.Amine Q.28 Q.10 Q.47 Q.20 Q.41 Q.13 Q.12 Q.2 Q.46 Q.29 A = H3C CO2H C CH2 CH2OH CH2OH B = H3C CO2H CO2H C = H3C C C CO2H H D (CH 3 )2 CHCO2 C2 H 5 E (CH3 )2 CHCONH 2 F (CH3 )2 CHNH 2 G (CH3 )2 CHOH H CH 3COCH 3 CONH2 A= Q.50 Q.45 Q.1 Q.9 Q.56 b b c b b c b b 37 .39 Q.29 Q.21 Q. H are CO2H O2 N CO2H NO2 EXERCISE .49 Q.38 Q.17 Q.53 d d d c b a b d Q.31 Q.15 Q.30 NH2 B= N2 Cl C= CH3 CH3 CH3 CN CO2H COOH E= D= F= CH3 CH3 CO2H COOH CO2H G.III Q.35 Q.27 Q.3 Q.43 Q.32 Q.36 Q.5 Q.25 Q.8 Q.51 Q.57 a c c b d a c b a Q.24 Q.23 Q.22 Q.30 Q.48 Q.