27 Aliphatic Compounds Containing Nitrogen

27 Aliphatic Compounds Containing Nitrogen Objective Questions I (Only one correct option) (a) (b) (c) (d) OH ½ Ethyl formate (1 equiv.) CH3 CH CH2CH2NH2 ¾¾¾¾¾¾¾¾® Triethylamine OH ½ (a) CH3 CHCH2CH2NHCHO 5. Which of the following amines can be prepared by Gabriel phthalimide reaction? 1. The major product of the following reaction is n - butylamine triethylamine t-butylamine neo -pentylamine 6. In the following compounds, the decreasing order of basic (b) CH3CH== CH ¾ CH2NH2 O OH ½ (c) CH3 ¾ CH ¾ CH == CH2 O (d) H CH3CHCH2CH2NH2 strength will be (a) (b) (c) (d) C2H5NH2 > NH3 > (C2H5 )2 NH (C2H5 )2 NH > NH3 > C2H5NH2 (C2H5 )2 NH > C2H5NH2 > NH3 NH3 > C2H5NH2 > (C2H5 )2 NH 7. The major product of the following reaction is 2. Ethylamine (C2H5 NH2) can be obtained from (a) NaBH4 (b) NH2NH2 O H3 C N-ethylphthalimide on treatment with (c) H2O (d) CaH2 NH2 O 3. Hinsberg’s reagent is (a) SOCl 2 (b) C6H5COCl (c) C6H5SO2Cl (d) (COCl)2 4. The major products A and B for the following reactions are, (i) NaNO2/H+ (ii) CrO3/H+ (iii) H2SO4 (conc.), D O (a) CH3 O (b) HO O respectively O O I KCN DMSO [A] H2/Pd [B] (c) HO (d) O O OH O CN CH2NH2 ; (a) HO CN HO CH2 I O O CN CH2NH2 ; (c) HO CN I (d) NH2 H ; (b) HO O CH3 ; CH2 NH2 I O 8. A compound ‘X’ on treatment with Br2 / NaOH, provided C3 H9N, which gives positive carbylamine test. Compound ‘X’ is (a) CH3COCH2NHCH3 (b) CH3CH2CH2CONH2 (c) CH3CON(CH3 )2 (d) CH3CH2COCH2NH2 9. The major product of the following reaction is O CH3N NaBH4 Aliphatic Compounds Containing Nitrogen OH COCH3 OH (a) CH3N (a) (b) CH3NH OH (b) NH2 NHCOCH3 OH OCOCH3 (c) CH3NH OH (c) NHCOCH3 (d) CH3N 10. The correct structure of product ‘P’ in the following reaction is Cl Asn -Ser + (CH3CO)2 O ¾® P N H O N H (b) H3C O Cl OH (a) OCOCH3 OCOCH3 O N H (c) H3C OCOCH3 O H N O Ol Cl N H n O NH2 OH NH2 NH2 N H O O 14. The increasing order of basicity of the following compounds is NH2 I. NH II. NH2 IV. III. NHCH3 NH OH OCOCH3 11. The major product formed in the reaction given below will be NH2 NH2 Cl HN NHCOCH3 O NH O O (d) n O (d) H3C n (c) O O (b) HN OH NHCOCH3 O O Cl NH2 N H O O N H O n O O (i) Et3N (ii) Free radical polymerisation O O O (a) H3C NH2 Cl + H 2N (Excess) O NH2 13. Major product of the following reaction is NEt 3 NH2 H N (d) NaNO2 (a) (I) < (II) < (III) < (IV) (b) (II) < (I) < (III) < (IV) (c) (II) < (I) < (IV) < (III) (d) (IV) < (II) < (I) < (III) 15. The order of basicity among the following compounds is Aq. HCI, 0-5°C NH2 O (a) NO2 (b) NO2 (c) OH OH (d) H3C NH2 N (I) (a) II > I > IV > III (c) IV > I > II > I NH (II) HN N H2N (III) NH (IV) (b) I > IV > III > I (d) IV > I > II > III 16. In the Hofmann-bromamide degradation reaction, the number of 12. The major product obtained in the following reaction is OH NH2 (CH3CO)2O/pyridine (1 eqv.) room temperature moles of NaOH and Br2 used per mole of amine produced are (a) four moles of NaOH and two moles of Br2 (b) two moles of NaOH and two moles of Br2 (c) four moles of NaOH and one mole of Br2 (d) one mole of NaOH and one mole of Br2 Aliphatic Compounds Containing Nitrogen 17. Considering the basic strength of amines in aqueous solution, 23. The correct order of basicities of the following compounds is which one has the smallest pK b value? (a) (CH3 )2 NH (c) (CH3 )3 N ethanolic potassium hydroxide, the organic compound formed is (b) an alkanediol (d) an alkyl isocyanide 19. The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was (a) (b) (c) (d) O ½½ (CH3 )2 NH, H3C ¾ C ¾ NH2 4 3 (a) 2 > 1 > 3 > 4 (b) 1 > 3 > 2 > 4 (c) 3 > 1 > 2 > 4 (d) 1 > 2 > 3 > 4 (a) N, N-dimethylaniline (c) N-methyl-o-methylaniline 25. p-chloroaniline anilinium (a) Sandmeyer reaction (c) AgNO3 O C and (b) 2, 4-dimethylaniline (d) p-methylbenzylamine hydrochloride (b) NaHCO3 (d) Carbylamine test mixture of —CH2Cl (ii) Br— O O C N—CH2— (a) chloroform and silver powder (b) trihalogenated methane and a primary amine (c) an alkyl halide and a primary amine (d) an alkyl cyanide and a primary amine 27. Acetamide is treated separately with the following reagents. —Br Which of these would give methylamine? (a) PCl 5 (c) NaOH + Br2 C O O (b) Sodalime (d) Hot, conc. H2SO4 28. The compound which on reaction with aqueous nitrous acid at low temperature produces an oily nitrosamine, is C (b) N— (a) methylamine (c) diethylamine —CH2Cl C (b) ethylamine (d) triethylamine Objective Question II (Only one more than one correct option) O O C (c) 29. The major product of the reaction is N H3C CO2H C O —CH2— (a) C NH2 H3 C 0°C (b) H 3C CH3 OH N C O— NaNO2, aq. HCl CH3 NH2 —Br O (d) (c) —CH2Cl H3 C 21. CH3NH2 + CHCl3 + KOH ® Nitrogen containing compound + KCl + H2O. Nitrogen containing compound is - + (b) CH3NHCH3 + - (d) CH3 ¾ N ºº C 22. Benzamide on treatment with POCl3 gives (a) aniline (c) chlorobenzene (b) benzonitrile (d) benzyl amine CO2H CH3 OH CO2H (d) NH2 H 3C CH3 OH CH3 OH (c) CH3 ¾ N ºº C be 26. Carbylamine test is performed in alc. KOH by heating a (i) KOH NH (a) CH3CN can distinguished by 20. The major product of the following reaction is (a) 2 1 24. A positive carbylamine test is given by methyl isocyanate methylamine ammonia phosgene C , CH3 ¾ CH2 ¾ NH2 , NH2 18. On heating an aliphatic primary amine with chloroform and (a) an alkanol (c) an alkyl cyanide NH H3C ¾ C (b) CH3NH2 (d) C6H5NH2 30. A positive carbylamine test is given by (a) N, N-dimethyl aniline (c) N-methyl-o-methyl aniline (b) 2, 4-dimethyl aniline (d) p-methyl benzyl amine Fill in the Blanks 31. (CH3OH+2 ) is …… acidic than (CH3NH+3 ). Aliphatic Compounds Containing Nitrogen 38. Arrange the following in increasing order of basic strength: Match the Columns methylamine, dimethylamine, aniline, N-methylaniline. 32. Match each of the compounds in Column I with its characteristic reaction(s) in Column II. 39. Give a chemical test and the reagent used to distinguish between Column I Column II the following : “Ethylamine and diethylamine”. (A) CH3 CH2 CH2 CN (p) Reduction with Pd - C/ H2 (B) CH3 CH2 OCOCH3 (q) Reduction with SnCl 2 / HCl 40. For nitromethane molecule, write structures (i) showing significant resonance stabilisation (ii) indicating tautomerism (C) CH3 CH == CHCH2 OH (r) Development of foul smell on treatment with chloroform and alcoholic KOH. (D) CH3 CH 2 CH2 CH2 NH2 (s) Reduction with diisobutylaluminium hydride (DIBAL-H) 41. State the equation for the preparation of following compounds : n-propyl amine from ethyl chloride. Passage Based Questions Treatment of compound O with KMnO 4 / H+ gave P, which on heating with ammonia gave Q. The compound Q on treatment with Br2 / NaOH produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound T.4 (2016 Adv.) (t) Alkaline hydrolysis Subjective Questions 33. NaNO2 , HCl – N2 ¾¾¾¾® Y(tertiary alcohol + other products) C5H13N optically active X Find X and Y. Is Y optically active ? Write the intermediate steps 34. Give reasons for the following in one or two sentences. Dimethylamine is a stronger base than trimethylamine. (O) 42. The compound R is 35. Following reaction gives two products. Write the structures of O the products. NH2 (CH CO) O, heat (a) 3 2 CH3CH2NH2 ¾¾¾¾¾¾® Br (b) Br NH2 36. Give the structure of A. ‘A (C3H9 N) reacts with benzenesulphonyl O chloride to give a solid, insoluble in alkali’. O O 37. A basic volatile nitrogen compound gave a foul smelling gas when treated with chloroform and alcoholic KOH. A 0.295 g sample of the substance dissolved in aq. HCl and treated with NaNO2 solution at 0°C, liberated a colourless, odourless gas whose volume corresponds to 112 mL at STP. NHBr (c) (d) NHBr O After evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave a yellow precipitate. Identify the original substance assuming that it contains one N-atom per molecule. NBr O 43. The compound T is (a) glycine (c) valine (b) alanine (d) serine Answers 1. 9. 17. 25. (a) (c) (a) (c) 2. 10. 18. 26. (b) (a) (d) (b) 3. 11. 19. 27. (c) (*) (a) (c) 32. A ® p, q, s, t; B ® p, s, t; C ® p; D ® r 4. 12. 20. 28. (c) (b) (a) (c) 42. (a) 5. 13. 21. 29. (a) (d) (d) (c) 43. (b) 6. 14. 22. 30. (c) (c) (b) (b,d) 7. 15. 23. 31. (c) (d) (b) (more) 8. (b) 16. (c) 24. (d) Hints & Solutions 1. The mechanism of the given reaction is as follows: Gabriel phthalimide reaction. This method produces only primary amines without the traces of secondary or tertiary amines. In this method, phthalimide is treated with ethanolic KOH, it forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis forms corresponding primary amines. O OH H—C—OEt CH3—CH—CH2CH2—NH2 [–NH2 is stronger nucleophile than —OH] Et3N —EtOs OH r CH3—CH—CH2CH2—NH—CH= =O H The basic mechanism of the reaction is acyl S N 2 because the nucleophile, CH3 (OH)CH2CH2 NH2 attacks the sp2 carbon of the ester (H ¾ CO2Et) and gets substituted. 2. The reaction for the production of ethylamine from N -ethylphthalimide can be takes place as follows: H HN N—C2H5 O O (An aliphatic 1°-amine) O This reaction is the second step of Gabriel phthalimide synthesis for the preparation of aliphatic 1°-amines and amino acid. In this step concentrated alkali can also be used in place of hydrazine. 3. Hinsberg’s reagent is C6H5SO2Cl (benzene sulphonyl chloride). This reagent is used to distinguish between primary, secondary and tertiary amines. O O I COO–Na+ NaOH(aq.) + n Bu—NH2 – + COO Na CO N—R CO 1° amine N-alkyl phthalimide Triethylamine, t-butylamine and neo-pentylamine cannot be prepared by Gabriel phthalimide reaction. alkanamine and ammonia with proton. NH | + C2H5—NH2 NH Ethylamine 4. N –K + CO 6. Basic strength can be compared by the reaction of an Hydrazinolysis NH2—NH2 (Hydrazine) H HN (N-ethylphthalimide) CO NH+ Alc. KOH –H O 2 nBu—X CH3—CH—CH2CH2—NH—CHO O CO CO EtOs OH 5. n-butylamine (CH3CH2CH2CH2NH2 ) can be prepared by CN Step I KCN DMSO [A] H | R—N + H+ | H H | R—N+—H | H H | H—N + H+ | H H | H—N+—H | H Basicity of an amine in aqueous solution depends upon the solubility of ammonium cation formed by accepting proton from water. The stability of ammonium cation depends upon the following factors. (a) + I effect (b) Steric effect (c) Solvation effect In first case, on increasing the size of alkyl group + I effect increases and the positive charge of ammonium cation gets dispersed more easily, Therefore, order of basicity is 2º amine > 1º amine > ammonia. In second case, substituted ammonium cation is also stabilised by solvation with water molecules. Greater the size of ion, lesser will be the solvation and less stabilised is the ion. OH2 Step II H2/Pd O CH2NH2 [B] Step I involves the nucleophilic substitution reaction in which I (Iodine) is substituted by ¾ CN group. In step II, H2/Pd reagent is used for reduction process. Here, ¾ CN group reduces itself to ¾CH2NH2. H C2H5 | C2H2—N+—H----OH2 > | C2H5 H + H----OH2 N H----OH2 2° amine OH2 1° amine Considering both steric and solvation effect, it can be concluded that greater the number of H-atoms on the N-atom, greater will be the H-bonding and hence more stable is the ammonium Aliphatic Compounds Containing Nitrogen cation. But in case of ¾C2H5 group, + I effect predominates over H-bonding. Therefore, order is 2º amine > 1° amine > ammonia i.e. (C2H5 )2 NH > C2H5NH2 > NH3. 7. or CH3 CH(CONH2 )CH3 (b). Carbylamine test given by (a) and (b) D CH3 CH2 CH2 NH2 + CHCl 3 + 3KOH ¾¾ ® CH3 (CH2 )2 NC + 3KCl + 3H2 NH2 ½ D ® CH3 ¾ CH ¾ CH3 + CHCl 3 + 3KOH ¾¾ Key Idea The reaction involves: A. Deamination in step (i) B. Oxidation in step (ii) C. Hydroysis in step (iii) NC ½ CH3 ¾ CH ¾ CH3 + 3KCl + 3H2 The complete pathway of reactions is as follows: H 3C O NH2 (i) O H 3C NaNO2/H+ OH Deamination H 3C As (b) is not among the given options So, it is ruled out and compound (X) is CH3CH2CH2CONH2. O 9. NaBH4 is a selective reducing agent. It can reduce O O OH (ii) O H 3C O and ¾ OH (alcohol) groups only, but not with ‘¾ NH2’ part of O ½½ ¾ C ¾ NH2 (amide) groups, because the lp of nitrogen in Conc. H2SO4/D Cyclisation (iii) O H isolated Keto C C group 10 Acetylation by Ac2O/Et3 N is possible with ¾ NH2 (1°-amine) O Hydrolysis HO OH amide group is involved in resonance with Intramolecular dehydration O NH2 NH2 C O 8. The molecular formula, C3H9 N refers to four structural isomers of amines H2 N (b) CH3–CH–CH3 CH C NH CH2 C NH2 CH2 Primary amine Asn (Asparagine) CH3 COOH OH Ser (Serine) Ac2O or (CH3CO)2O (2 moles) Et3N –2 AcOH (d) CH3–N–CH3 Tertiary amine Here, only a and b (primary amines) can be prepared from their respective amides by Hoffmann bromamide (Br2/NaOH) method and can give positive carbylamine test. CH O NH2 Secondary amine O part. r C Thus, option (c) is the correct answer. (c) CH3–NH–CH2CH3 C Os O Final product (a) CH3CH2CH2NH2 NH H3 C N-methyl imine OH H O OH NaBH4 H 3C O Oxidation HO H3C O N O CrO3/H+ C==O group into alcohol, N-methyl imine group, Me ¾ N == CH ¾ into 2°-amine but cannot reduce an isolated C == C. O CH3 C O NH CH CH2 C NH CONH2 CH CH2O COOH C CH3 O Þ CH3CH2CH2—C—NH2 Br2 CH3CH2CH2—NH2 NaOH –Na2CO3 H2N—C Þ CH3—CH—CH3 NH2 Br2/NaOH –Na2CO3 CH3—CH—CH3 So, ‘X ’ can be CH3 CH2 CH2 ¾ CONH2 (a) 11. No option is the correct answer. Amines in presence of NaNO2 + dil. HCl undergoes diazotisation. The diazotised product readily loses nitrogen gas with the formation of carbocation. The resulting carbocation rearranges itself to give the expanded ring. Aliphatic Compounds Containing Nitrogen Decrease in the delocalisation of p-bonds because of which it undergoes free radical addition (chain growth) polymerisation NH2 NaNO2+dil. HCl 0-5°C (Diazotisation) (1°-amine) r s N2Cl Cl Cl Aliphatic diazonium salt (unstable) 5 6 1 2 –Hr H2O –H absorbed by Et3N 4 4 3 –N2 Cl Free radical polymerisation –Cl – O n O C NH r Ring expansion leads to give thermodynamically more stable carbocation 5 3 6 Cl O NH2 r 2 O NH2 1 NH2 O O OH NH2 NH2 14. Key Idea Among the given compounds the basic nature depends upon their tendency to donate electron pair. (Major) 12. Rate of acetylation : ¾ NH2 > ¾ OH because N-bases are stronger than O-bases. Size of N-atom is larger than O-atom and at the same time, N atom is less electronegative than O-atom. ·· So, N-atom of the ¾ N H2 group can donate its lone pair of electrons (Lewis basicity) more easily than that of O atom of the ·· ¾ O H group. OH OH NH Among the given compounds in , Nitrogen is sp2-hybridised. This marginally increases the electronegativity of nitrogen which in turn decreases the electron donation tendency of nitrogen. Thus making compound least basic. NH2 Among the rest NH is totally different from others as in this compound lone pair of one nitrogen are in conjugation with p bond i.e. As a result of this conjugation the cation formed after protonation becomes resonance stabilised s CH3 CH3COO H 2N Protonation HN—COCH3 C + HN O—COCH3 H NH2 H2N NH2 + NH2 H2N Equivalent resonance s O—C—CH3 OH CH3COOH + NH2 NHCOCH3 Absorbed by pyridine N N NH2 most This equivalent resonance in cation makes HN basic among all. Categorisation is very simple between rest two as CH3 (primary amine) is less basic than (secondary amine) Hence, the correct order is (II) < (I) < (IV) < (III) i.e. option (c) is correct. H 13. The analysis of both the substrates : Vinylic centre Cl (+R) 15. IV is most basic as conjugate acid is stabilised by resonance of two ¾ NH2. r NH2 NH O Cl NH2 H+ Acyl SN2 centre NH2 H2N NH2 H2 N Acts as H2N nucleophile NH2 (1°-amine) O NH2 + R This N-atom can not act as nucleophile Amide group r NH2 H2N r H2N NH2 III is least basic as H N N H+ H N So, the reaction can take place as follows: Destablised by -I-effect of sp2-carbons. r N H Aliphatic Compounds Containing Nitrogen 16. Hofmann-bromamide degradation reaction is given as: RCONH2 + 4NaOH + Br2 ¾® RNH2 (1° amine) : Bromine is not substituted in the above reaction as it is in resonance with benzene ring giving partial double bond character to C—Br bond, hence difficult to break. – + ClCH2 Br: Br ClCH2 : : + Na2 CO3 + 2NaBr + 2H2O Hence, four moles of NaOH and one mole of Br2 are used. 17. This problem can be solved by using the concept of effect of steric hindrance, hydration and H-bonding in basic strength of amines. Order of basic strength of aliphatic amine in aqueous solution is as follows (order of K b ) ·· ·· ·· ·· (CH3 )2 NH > CH3 NH2 > (CH3 )3 N > C6H5 NH2 ·· So, (CH3 )2 NH will have smallest pK b value. In case of phenyl amine, N is attached to sp2-hybridised carbon, hence it has highest pK b and least basic strength. 18. This reaction is an example of carbylamine test which includes conversion of amine to isocyanide. C2 H 5OH + KOH isocyanide O ½½ POCl 3 22. C6H5 ¾ C ¾ NH2 ¾¾® C6H5 ¾ C ºº N + H2O 23. 4, (acetamide) is least basic because lone pair of nitrogen is involved in delocalisation. - O O ½½ · · ½ + CH3 ¾ C ¾ NH2 ¬® CH3 ¾ C == NH2 ‘1’ is most basic due to formation of resonance stabilised conjugate acid. – R ¾ NH2 + CHCl 3 ¾¾¾® R— N ºº C: 1° amine + Alkyl isocyanide NH NOTE The mechanism of this reaction included rearrangement of nitrene in which migration of alkyl group from carbon to nitrogen takes place. NH2 + H+ H3C—C H3C—C NH2 NH2 (1) 19. Methyl isocyanate CH3 ¾ N == C == O (MIC) gas was leaked NH2 H3C—C from the storage tank of the union carbide plant in Bhopal gas tragedy. 20. It is the first step of Gabriel's phthalimide synthesis. The hydrogen bonded to nitrogen is sufficiently acidic due to two a-carbonyls. – 3 (secondary amine) is stronger base than 2 (primary amine). Hence, overall order of basic strength is : 1>3>2>4 N– H2O + O 24. H3C— O O – O O O – The conjugate base formed above act as nucleophile in the subsequent step of reaction. As shown above, the nucleophile exist in three resonating form, one may think of oxygen being the donor atom in the nucleophilic attack. However, nitrogen act as donor as it is better donor than oxygen. O N–+ O CH2 Br SN 2 Cl O N CH2 O Br KOH —CH2NH2 + CHCl3 p-methyl benzylamine (a primary amine) H3C— —CH2—NC (isocyanide) N N + NH2 O O N¾H + OH - POCl 3 brings about dehydration of primary amide. As we know, pK b = - log K b ·· + 21. CH3NH2 + CHCl 3 + KOH ¾® CH3 ¾ N ºº C + KCl + H2O Carbylamine test is not given by secondary or tertiary amine. + + 25. C6H5 NH3Cl - + AgNO3 ¾® C6H5 N H3NO-3 + AgCl (s) ¯ anilinium hydrochloride No such precipitate of AgCl(s) would be formed with p-chloroaniline. KOH 26. R ¾ NH2 + CHCl 3 ¾¾® R ¾ NC ‘Carbylamine test’ isocyanide (foul smell) The above test is characteristic of primary amine and used to distinguish primary amine from other amines. O ½½ NaOH 27. CH3 ¾ C ¾ NH2 + Br2 ¾¾® CH3NH2 acetamide methylamine (Hofmann’s bromamide reaction) Aliphatic Compounds Containing Nitrogen 28. A secondary amine, on treatment with aqueous nitrous acid at low temperature produces oily nitrosamine. (C2H5 )2 NH + HNO2 ¾® (C2H5 )2 N ¾ NO + H2O nitrosamine 29. Reaction proceeds via diazonium salt with neighbouring group participation. R HNO2 COOH H R H OH O N¾ ¾N r H Or COOH r –H H2O OH (I) Conformer of (I) C O ¾ N¾ N r H CH3 H NH2 HO H OH COOH R or C2H5 ½ CH3 ¾ CH ¾ CH2NH2 * (X ) ¾R CH3 H ½ 33. CH3 ¾*C ¾ NH2 ½ CH ¾ CH3 ½ CH3 ¾O H R 30. Carbylamine test is given by primary amines only. NH2 CH3 CH3 CH3 N H CH3 ½ ½ NaNO2 * CH3 ¾ C ¾ NH2 ¾¾¾® CH3 ¾ CH2 ¾ C ¾ CH3 HCl ½ ½ OH CH ¾ CH3 Y -optically inactive ½ CH3 Above reaction proceeds via carbocation intermediate, hence rearrangement takes place. 34. Conjugate acid of dimethylamine is more stable than conjugate base of triethyl amine due to exhaustive H-bonding with water. O ½½ (CH3 CO) 2 O 35. CH3CH2NH2 ¾¾¾¾® CH3CH2 ¾ NH ¾ C ¾ CH3 Heat + CH3COOH 36. A must be a secondary amine : CH3 N,N-dimethylaniline (3°-amine) 2,4-dimethylaniline (1°-amine) NHCH3 CH3 H 3C N-methyl-o-methylaniline (2°-amine) CH2NH2 p-methyl benzylamine (1°-amine) 31. More : CH3NH2 is stronger base than CH3OH. Column I 32. (A) (B) CH3 CH2 CH2 CN : Column II Gives amine with Pd-C/ H2 CH3CH2NHCH3 + benzene sulphonylchloride ¾® CH3 ½ NaOH CH3CH2 ¾ N ¾ SO2 ¾ C6H5 ¾¾® Insoluble 37. Starting compound is a primary amine. R ¾ NH2 + CHCl 3 + KCl ¾® R ¾ NC foul smell R ¾ NH2 + HNO2 ¾® R ¾ OH + N2 112 Moles of N2 = = 5 ´ 10- 3 22500 Also, Q One mole of N2 is obtained from 1.0 mole of R ¾ NH2, mole of R ¾ NH2 = 5 ´ 10- 3 Gives aldehyde with SnCl 2 / HCl Gives amide with diisobutyl- aluminium hydride. Gives carboxylic acid on alkaline hydrolysis. Þ CH3 CH2 OCOCH3 : Reduced to alcohol with ester Pd - C/ H2 Reduced with diisobutylaluminium hydride into aldehyde. Undergo alkaline hydrolysis. Þ (C) CH3 CH == CHCH2 OH Reduced to butanol when : treated with Pd-C/ H2 . (D) CH3 CH 2 CH2 CH2 NH2 A primary amine, gives carbylamine test. 5 ´ 10- 3 mol R ¾ NH2 weigh = 0.295 g 1 mole of R ¾ NH2 will weigh = 0.295 = 59 g 5 ´ 10- 3 In R ¾ NH2, ¾ NH2 has molar mass = 16 R— has molar mass = 43 Also, the alcohol R—OH gives iodoform test; it must has the following skeleton : Þ R ¢¾ CH ¾ CH3 ½ OH R ¢— has molar mass = 15 (CH3 ) Þ CH3 ½ Original compound is CH3 ¾ CH ¾ NH2 Aliphatic Compounds Containing Nitrogen 38. Aniline > N-methylaniline < methylamine < dimethylamine. O 39. Carbylamine test given only by primary amine. NH2 NH2 KOH CH3CH2 ¾ NH2 + CHCl 3 ¾¾® CH3 ¾ CH2 ¾ NC primary amine O oily nitrosamine + (ii) CH3—N O O + NH + CH3—N O O O– O O O – S N—K : Tautomerism Br O COOC2H5 –KBr O LiAlH NaCN SN 2 KOH O – OH + CH2==N O D – CH3—N O– R Q (CH3CH2 )2 NH + HNO2 ¾® (CH3CH2 )2 N ¾ NO + NH2 foul smell Nitrosamine test is given only by secondary amine. 40. (i) CH3—N NH2 Br2/NaOH D 4 41. CH3CH2Cl ¾¾® CH3CH2CN ¾¾¾® CH3CH2CH2NH2 O S1 N 42. (a) COOC2H5 O S2 43. Explanation O O OH – MnO4 /H+ D OH OP NH3 D OH H 3O + OH O (P) NH2 + COOH (T )