EG_Chemistry-12_(22-07-2022) a

revIsIon chart Physical Properties Nomenclature Preparation • These are highly polar molecules with boiling • In the IUPAC system, aldehydes are • Aldehydes were prepared by mild points higher than the comparable molecular named by replacing ‘e’ of the parent oxidation of primary alcohols, using mass hydrocarbons or weakly polar molecules hydrocarbon chain with ‘al ’. reagents like Cu/573 K, PCC, etc. like ethers. • Lower members are highly soluble in water Aldehydes • Dehydrohalogenation of alcohols due to their ability to form H-bond however These are (by Cu at 573 K), will give aldehydes. with increase in hydrocarbon chain length, carbonyl Mild selective reduction of acyl there is less extent of H-bonding and compounds halides, selective reduction of therefore, higher members are insoluble in in which the nitriles, esters by DIBAL-H and water, but soluble in organic solvents. carbonyl group Stephen’s reduction also produce ( C = O) is aldehydes. Chemical Properties directly bonded • Nucleophilic substitution reactions occur to an alkyl/aryl • Aromatic aldehydes are prepared by chain and a oxidation of Methyl benzene using in aldehydes at carbonly carbon atom on H-atom. (CErtOa3r;d C rreOa2cCtli2o fno)ll, oowr bedy barye hnyedsr uoslyinsigs (rteoa cfotiromn walictohh HoClsN),, aNlacHohSoOl3s, G(troig fonramrd a receatgael)n.t CO gas and HCl in the presence • Ahaldloefhoyrdme rse wacitthio n(C. H3—CO) group undergo o( fG aatntehrymdarno-uKso cAhl Crel3a catsio cna).talyst. • Aldehydes with a-H-atom undergo aldol condensation in the presence of dilute alkali. • Aldehydes which do not have a-H atom on treatment with conc. alkali form alcohol and carboxylic acid. Nomenclature Carboxylic Acids: Preparation • According to IUPAC system, these • Carboxylic acids were prepared by These are the organic were named by replacing ‘e’ of the compounds having carboxyl oxidation of primary alcohols, aldehydes hydrocarbon chain with -oic acid,. and alkenes, by hydrolysis of nitriles O by reaction of CO2 gas with Grignard ( C OH) functional group. reagent and by hydrolysis of acid chlorides and anhydrides. • Aromatic carboxylic acids can be prepared by side chain oxidation of Chemical Properties alkyl benzenes. • Presence of electron donating groups decreases the acidic strength while electron withdrawing group enhances their acidic strength. • Carboxylic acids with mineral acids form anhydrides, on Physical Properties reaction with PCl5, PCl3, SOCl2, etc. they give acyl chlorides • Lower members of carboxylic acids are soluble in water, and ammonium salt formation takes place with NH3. but as the size of hydrocarbon chain increases, the • CB2aHrb6,o oxry liitc u ancdidesr gfoorems daleccoahrbolosx iynl atthioe np rteos egnivcee (oofn Lei ACl-Hle4 sosr) solubility of carboxylic acid decreases. hydrocarbon. • Carboxylic acids are higher boiling liquids than • They also undergo Hell-Volhard-Zelinsky reaction and aldehydes, ketones, alcohols of comparable molecular aromatic acids can undergo electrophilic substitution masses. This is due to extensive intermolecular reaction at o/p-position. H-bonding in carboxylic acids. 300 Chemistry-12

revIsIon chart Nomenclature Preparation • In the IUPAC system, ketones are named • Ketones are prepared by mild oxidation of secondary alcohols, by replacing ‘e’ of the parent hydrocarbon hydration of alkynes (using HgSO4/H2SO4), by reaction of acyl chain with ‘one’. chloride with dialkyl cadmium, by dehydrohalogenation of sec. alcohols, by ozonolysis of alkenes, or by treating nitriles with Grignard reagent followed by hydrolysis. Ketones •A romatic ketones can be conveniently prepared by Friedel-Crafts acylation Chemical Properties Analogous to reaction. • Ketones too like aldehydes can undergo aldehydes they also Physical Properties nucleophilic substitution reactions, but contain carbonyl • Ketones too like aldehydes are polar its tendency is low due to the presence of group ( C = O), two electron-releasing alkyl/aryl groups. but the carbonyl molecules with boiling points higher than • Ketones do not react with alcohols like carbon is directly hydrocarbons and ethers of comparable aldehydes, but with powerful oxidising bonded to two alkyl molecular mass. agents it form carboxylic acids. In case or aryl chains. • Like aldehydes lower members of ketones of unsymmetrical ketones, cleavage are soluble in water but higher members occurs such that keto group remains are insoluble. with smaller alkyl group. This is known as Popoff’s rule. • KNei tcoantaelsy sutn tdoe grigvoe r(2e°d)u aclctioohno wl. Aitlhs oL iwAiltHh4 /ZNna-BHHg4/ HoCr lH o2r wHiIt/hR ePdt /PP do/r NH2-NH2/KOH they reduce to hydrocarbons. • Ketones like aldehydes can undergo haloform reactions and the one with a-H atom can undergo aldol condensation (self or cross) reactions. • To distinguish between aldehydes and ketones Tollens’ test and Fehling’s test can be used. Aldehydes, Ketones And CArboxyliC ACids 301

chapter trend—Based on past Years’ cBse exams ➣ It has been observed from this chapter that the weightage of topics ‘Aldehydes and Ketones’ (especially, their chemical properties) is maximum. Hence, these are most important topics. ➣ From this chapter , generally 1 mark questions were asked from the topic ‘Aldehydes and Ketones’. ➣ Most of the 2 and 3 marks questions asked from this chapter belong to the topic ‘Aldehydes and Ketones’. Questions For practice Very Short Answer Type Question (1 Mark) 1. What type of aldehydes undergo Cannizzaro’s reaction? [CBSE 2017 (C)] Short Answer Type Questions-I (2 Marks) 2. Identify A to D in the following reactions. [Delhi 2017 (C)] 3. Give reasons: (i) Benzaldehyde does not undergo aldol condensation (ii) Sodium bisulphite is used to separate carbonyl group compound from non-carbonyl compounds. [Delhi 2017 (C)] 4. Write the reactions involved in the following: (i) Etard reaction (ii) Stephen’s reaction Short Answer Type Questions-II (3 Marks) 5. mAheocaolt–m1i.npDgoeugdanuvdcee‘A‘Cs’t(’r.Cu‘2BcHt’ u6oOrne)sooxnoifodAxai,tdiBoan,tiCownaitnwhditKwhMrPinCteOCt4hggeaivvreeeas‘Bcat’ iwomnhosenrioenovbnoalstvirecedac.tamrbeonxtywlicithacaiqd. alkali and subsequent with molar mass 60 g 6. (i) Distinguish between the following: [AI 2015] (a) CH3COCH2CH3 and CH3CH2CH2CHO, (b) Ethanal and Ethanoic acid (ii) Write the structure of 4-chloropentan-2-one. 7. (i) Write the chemical reaction involved in Cannizzaro reaction. (ii) Draw the structure of the semicarbazide of ethanal. (iii) How can you distinguish between propanal and propanone? 8. (i) Write the product(s) in the following reactions: [Delhi 2017] O (a) + HCN (b) (ii) Give simple tests to distinguish between the following pair of organic compounds: Butanal and Butan- 2-one Long Answer Type Question (5 Marks) 9. Write the structures of A, B, C, D and E in the following reactions. C6H6 CH3COCl Zn-Mg/conc. HCl B (i) KMnO4/KOH, ∆ C Anhyd. AlCl3 (ii) H3O+ A NaOI D+E 302 Chemistry-12

assIgnment Time: 45 Minutes M.M.: 25 Multiple Choice Questions (MCQs) (1 Mark) 1. Name the alkene which on reductive ozonolysis gives acetone as the only product. (i) 2,3-Dimethylbut-2-ene (ii) Hex-3-ene (iii) Propene (iv) 2-Methylbut-2-ene 2. Which of the following substituent increases the acidic strength of carboxylic acid to the maximum extent? (i) Cl (ii) F (iii) CN (iv) NO2 Assertion-Reason Type Questions (1 Mark) Note: In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (i) Assertion and reason both are correct statements and the reason is correct explanation of assertion. (ii) Assertion and reason both are correct statements but reason is not correct explanation of assertion. (iii) Assertion is correct statement but reason is wrong statement. (iv) Assertion is wrong statement but reason is correct statement. 3. Assertion: Cross canizzaro reaction between formaldehyde and benzaldehyde gives benzyl alcohol and formate ion. Reason: Formaldehyde is a better hydride donor than benzaldehyde. 4. Assertion: The a-hydrogen atom in carbonyl compounds is less acidic. Reason: The anion formed after the loss of a-hydrogen atom is resonance stabilised. Very Short Answer Type Questions (1 Mark) 5. Why benzoic acid is more acidic than acetic acid? 6. Distinguish between benzoic acid and methylbenzoate Short Answer Type Questions-I (2 Marks) 7. How is Benzaldehyde prepared from: (i) Toluene, (ii) Benzoic acid? 8. Discuss some important use and test for carboxylic acids? 9. Write short note giving example of each of following reactions: (i) Wolff Kishner reduction (ii) HVZ reaction 10. Identify A, B, C and D Short Answer Type Questions-II (3 Marks) 11. What happens when: (i) 2-butanone is treated with Zn/Hg and conc HCl (ii) 2 molecules of benzaldehyde are treated with conc. NaOH. (iii) acetone is treated with semicarbazide? 12. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compound A, B and C. Long Answer Type Question (5 Marks) 13. Carry out the following conversions: (ii) Methanal to Ethanal (i) Formaldehyde to Ethanol (iv) Methyl cyanide to Ethanal (iii) Acetylene to Acetaldehyde (v) Acetyl chloride to Acetone Answers 1. (i) 2. (iv) 3. (i) 4. (iv) Aldehydes, Ketones And CArboxyliC ACids 303

9 Amines Topics covered 9.1 Amines: Classification, Nomenclature and Preparation 9.2 Physical and Chemical Properties of Amines 9.3 Diazonium Salts C hapter map AMINES Classification Preparation Physical properties Chemical properties Structure • By reduction of nitro • Boiling point Basicity of amines Nomenclature compounds/nitriles/amides • Solubility • Alkylation/Acylation reaction • Ammonolysis of alkyl halides • Carbylamine reaction • Gabriel phthalimide synthesis • Reaction with nitrous acid • Hoffmann's bromamide • Reaction with Hinsberg reagent • Ring substitution reactions in degradation aromatic amines • Diazotisation reaction Benzene diazonium salts • Displacement of N2 – by Cl– or Br– ion (Sandmeyer reaction) – by cyanide ion (Gattermann reaction) – by I– ion/F– ion/H/–NO2 group/–OH group • Coupling reactions Topic 1. Amines: Classification, Nomenclature and Preparation Classification of Amines Amines are regarded as the derivatives of ammonia in which one, two or all three hydrogen atoms are replaced by alkyl or aryl groups. 304

Structure of Amines sp3-hybridised orbital, thus reducing the C–N–R bond angle (where R is C or H) less than 109.5°. • Nitrogen atom is trivalent in amines and carries lone pair of electrons. Nomenclature • Commonly amines were named as alkylamine. • The nitrogen atom is sp3-hybridised and three of • But in IUPAC system, they are named as sp3-hybridised orbitals overlaps with the orbitals alkanamines. This rule is based on the replacement of hydrogen and carbon atoms, depending upon of ‘e’ of the corresponding hydrocarbon with the composition of amines. ‘amine’. • The lone pair of electrons is present in the fourth Amines Common Name IUPAC Name Primary Amines (1° Amines) CH3NH2 Methylamine Methanamine CH3–CH2– NH2 n–Propylamine Propan-1-amine CH3 — CH — CH2 Isopropylamine Propan-2-amine | NH2 1 23 Allylamine Prop-2-en-1-amine H2N — CH2 — CH = CH2 Secondary Amines (2° Amines) CH3NHCH3 Dimethylamine N-methylmethanamine CH3NHC2H5 Ethylmethylamine N-methylethanamine CH3 — CH — CH3 N-Methylisopropylamine N-Methylpropan-2-amine | NHCH3 Tertiary Amines (3° Amines) (CH3)3N Trimethylamine N, N-Dimethylmethanamine 1234 C2H5 — N — C H2 — C H2 — C H2 — C H3 | N, N-Diethylbutylamine N, N-Diethylbutan-1-amine C2H5 Diamines Hexamethylenediamine Hexane-1, 6-diamine H2N(CH2)6NH2 Aromatic Amines NH 2 Aniline Benzenamine CH3 o-Toluidine 2-Methylaniline NH2 NHCH3 N-methylaniline N-Methyl benzenamine N(CH3)2 N, N-Dimethyl benzenamine N, N-Dimethylaniline Amines 305

Preparation of Amines ã It should be noted that this method is not suitable for the preparation of aryl • By reduction of nitro compounds: Nitro amines because aryl halides are relatively compounds can be catalytically reduced by less reactive than alkyl halides towards passing hydrogen gas in the presence of Raney- nucleophilic substitution reactions. Ni, finely divided Pt or Pd metals as catalyst at room temperature. • By reduction of nitriles: Nitriles can be reduced to amines using H2/Ni, LiAlH4 or Na(Hg)/ R – NO2 + 3H2 Ni, Pt or Pd R – NH2 + 2H2O; C2H5OH as reagents. Ar – NO2 + 3H2 Ni, Pt or Pd Ar – NH2 + 2H2O • By reduction of amides: Amides are reduced ã Nitro compounds can also be reduced with active metals such as Fe, Sn, Zn, etc. with conc. HCl. R – NO2 + 3H2 Sn/HCl R – NH2 + 2H2O; to corresponding amines by LiAlH4 as follows: or Fe/HCl O Sn/HCl Ar – NO2 + 3H2 or Fe/HCl Ar – NH2 + 2H2O R–C–NH2 (i) LiAlH4 R–CH2 – NH2 (ii) H2O • Ammonolysis of alkyl halides: Reaction of alkyl halides with an ethanolic solution of • By Gabriel phthalimide synthesis: This ammonia in a sealed tube at 373 K forms a method is used for the preparation of primary mixture of primary, secondary and tertiary amines and finally quaternary ammonium salt. amines. Phthalimide on treatment with ethanolic Process of cleavage of C-X bond by ammonia is called ammonolysis. potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the RNH2 RX R2NH RX R3N RX +– corresponding primary amine according to the R4NX reaction as follows: (1°) (2°) (3°) Quaternary OO ammonium salt ã The free amine can be obtained from the C N—H alc. KOH C N– K+ R–X C C∆ ammonium salt by treatment with a strong base as follows: NH3 + RX RNH+3 X – NaOH O O RNH2 + H2O + Na+ X– Phathalimide Potassium phthalimide (1° amine) O O C N—R NaOH (aq) R2NH+2 X – NaOH C (hydrolysis) C O– Na+ + R–NH2 R2NH + H2O + Na+ X– RNH2 + RX C O– Na+ (1° amine) (2° amine) O O +– N-Alkylphthalimide R2NH + RX R3NH X NaOH It should be noted that aromatic primary amines cannot be prepared by this method because aryl R3N + H2O + Na+ X– halides do not undergo nucleophilic substitution with potassium phthalimide due to double bond (3° amine) character between C—X bond, arising as a result of resonance effect. ã Order of reactivity of halides is: RI > RBr • By Hoffmann bromamide degradation > RCl reaction: Primary amines can be prepared from amides by treatment with Br2 and KOH. Amines ã Larger the size of halogen atom easier is the which are prepared using this method, contains one carbon atom less than the parent amide. cleavage of R–X bond O ã Limitations of this method is that it gives a mixture of amines which are difficult to separate in a laboratory. ã When ammonia is taken in excess, primary amine is formed as the main product. ã When alkyl halide is used in excess, R C NH2 + Br2 + 4NaOH R – NH2 + Na2CO3 + 2NaBr + 2H2O quarternary ammonium salt is formed as the main product. ExErCiSE 9.1 Multiple Choice Questions (MCQs) (1 Mark) (ii) Triethylamine 1. Which of the following is a 3° amine? (iii) tert-butylamine (i) 1-methylcyclohexylamine (iv) N-methylaniline 306 Chemistry-12

2. The correct IUPAC name for CH2 CHCH2NHCH3 Very Short Answer Type Questions (1 Mark) is 10. Write the structure of zwitter ion of sulphanilic (i) Allylmethylamine acid. (ii) 2-amino-4-pentene 11. Mention the chief use of quaternary ammonium salts derived from long chain amines. (iii) 4-aminopent-1-ene (iv) N-methylprop-2-en-1-amine 12. How will you convert an amide into an amine having the same number of carbon atoms? 3. Which of the following reagents would not be a good choice for reducing an aryl nitro compound 13. Give reason for the following: to an amine? Gabriel phthalimide synthesis is preferred for (i) H2 (excess)/Pt (ii) LiAlH4 in ether synthesising primary amines. [NCERT] (iii) Fe and HCl (iv) Sn and HCl 14. Write short note on Gabriel phthalimide synthesis. 4. Which of the following reagents would not be a [Delhi, AI 2012][NCERT] good choice for reducing an aryl nitro compound to an amine? Short Answer Type Questions-I (2 Marks) (i) H2 (excess)/Pt (ii) LiAlH4 in ether 15. Write chemical equations for the following (iii) Fe and HCl (iv) Sn and HCl reactions: 5. The source of nitrogen in Gabriel synthesis of (i) Reaction of ethanolic NH3 with C2H5Cl. amines is .................... . (ii) Ammonolysis of benzyl chloride and reaction of amine so formed with two molecules of (i) Sodium azide, NaN3 (ii) Sodium nitrite, NaNO2 CH3Cl. [NCERT] (iii) Potassium cyanide, KCN 16. What is Hoffmann bromamide reaction? (iv) Potassium phthalimide, C6H4(CO)2 N–K+ [Delhi 2012] 6. The best reagent for converting 2–phenyl- 17. Write chemical equations for the following propanamide into 2-phenylpropanamine is conversions: .......................... . (i) CH3CH2Cl into CH3CH2CH2NH2 (ii) C6H5CH2Cl into C6H5CH2CH2NH2 [NCERT] (i) excess H2 (ii) Br2 in aqueous NaOH 18. Write structures and IUPAC names of (iii) iodine in the presence of red phosphorus (i) The amide which gives propanamine by (iv) LiAlH4 in ether Hoffmann bromamide reaction. 7. Hoffmann Bromamide Degradation reaction is (ii) The amine produced by the Hoffmann shown by .................... . degradation of benzamide. [NCERT] (i) ArNH2 (ii) ArCONH2 19. Classify the following amines as primary, (iii) ArNO2 (iv) ArCH2NH2 secondary or tertiary: Assertion-Reason Type Questions (1 Mark) (i) NH2 (ii) Note: In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (iii) (C2H5)2CHNH2 (iv) (C2H5)2NH [NCERT] (i) Assertion and reason both are correct statements 20. Write the IUPAC names of this following and the reason is correct explanation of assertion. compounds: (ii) Assertion and reason both are correct statements but reason is not correct explanation of assertion. (i) (CH3)2CHNH2 (ii) H2C = CHCH2CH2NH2 [NCERT] (iii) Assertion is correct statement but reason is wrong statement. 21. Give the IUPAC names of the following compounds: (iv) Assertion is wrong statement but reason is correct (i) p-Anisidine (ii) Acetanilide [NCERT] statement. 22. Why cannot aromatic primary amines be prepared 8. Assertion: Only a small amount of HCl is by Gabriel phthalimide synthesis? [NCERT] required in the reduction of nitro compounds with 23. What are arylamines? Give some examples. iron scrap and HCl in the presence of steam. Short Answer Type Questions-II (3 Marks) RHeCal sdounri:nFgeCthl2efroeramcetidonge. ts hydrolysed to release 24. Convert the following: 9. Assertion: C6H5CH2NH2 is an arylamine. (i) Benzyl chloride to 2-phenylethanamine Reason: In arylamines, aromatic ring is directly [Delhi 2012] attached to nitrogen atom. (ii) Benzoic acid to aniline. (iii) Chloroethane to N,N-dimethylethanamine Amines 307

25. (i) Write structures of different isomeric amines 32. Write the IUPAC names of the following corresponding to the molecular formula, compounds: C4H11N. (i) (ii) Write IUPAC names of all the isomers. (ii) (iii) (iii) What type of isomerism is exhibited by (iv) different pairs of amines? [NCERT] (v) 26. How will you convert: 33. Write the IUPAC names of the following compounds: (i) Benzene into aniline (i) (ii) Benzene into N, N-dimethylaniline (iii) Cl—(CH2)4—Cl into hexane-1,-6-diamine? [NCERT] 27. How will you convert: (i) Ethanoic acid into propanoic acid (ii) Methanamine into ethanamine (iii) Nitromethane into dimethylamine [NCERT] 28. How will you convert: (i) Acetic acid to ethylamine (ii) Ethylamine from methanol (in 3 steps) (iii) n-Butylamine from propene (in 3 steps) 29. Complete the following equations: (i) LiAlH4 /ether (i) CH3CH2NO2 (ii) H2 O (ii) H3C N C CH2CH3 (ii) X Thermal decomposition (CH3)3 N + CH3 OH CH3 C2H5 (iii) C6 H5 CONH2 + Br2 + KOH (iii) [Delhi 2009, 11] (iv) (CH3)3C—NH2 (v) CH3CH2CH2N(CH3)2 Long Answer Type Questions (5 Marks) 34. Give the IUPAC names of the following compounds. 30. Write the IUPAC names of the following (i) Ethylenediamine (ii) p-Toluidine compounds: (iii) Acetanilide (iv) tert-Butylamine (i) (CH3)2CHNH2 (ii) CH3—NH—C2H5 (v) Benzylamine (iii) C6H5NH2 (iv) C6H5N CH3 35. Write structures of different isomers corresponding C2H5 to the mofotlheecuislaormfeorrsmwuhliac,hCw3Hill9Nlib. eWrartietenIitUroPgAeCn + names (v) C6 H5 (CH3) 3 I– gas on treatment with nitrous acid. [NCERT] N 31. Write the IUPAC names of the following 36. Write IUPAC names of the following compounds compounds: and classify them into primary, secondary and (i) CH3CH2N CH3 tertiary amines. CH2CH3 (i) CH3(CH2)2NH2 (ii) CH3NHCH(CH3)2 (iii) (CH3)3CNH2 (iv) C6H5NHCH3 (ii) H2C = CHCH2CH2NH2 [Delhi 2010] (v) m-BrC6H4NH2 [NCERT] NH2 37. How will you convert: (iii) C2H5 — N CH3 (iv) (i) Ethanoic acid into methanamine CH3 Br (ii) Hexanenitrile into 1-aminopentane (iii) Methanol into ethanoic acid (v) (iv) Ethanamine into methanamine (v) Methanamine to dimethylamine 1. (ii) 2. (iv) 3. (ii) Answers 8. (ii) 9. (iv) 4. (iii) 5. (iv) 6. (iv) 7. (ii) 10. 308 Chemistry-12

11. As detergents (cleansing agent), RCONH2 + Br2 + 4KOH Heat CH3(CH2)15N+ (CH3)3Cl–, Acid amide ∆ i.e. cetyl trimethyl ammonium chloride. R—NH2 + 2KBr + K2 CO3 + 2H2 O 12. Reduction of 1°, 2° and 3° amides into the 1° Amine corresponding amines having the same number 17. (i) CH3 CH2 Cl NaCN (alc), CH3 CH2 CN of carbon atoms takes place by reduction using T,(–NaCl) Propanenitrile LiAlH4 /ether followed by hydrolysis as follows: Na/alcohol or LiAlH4 /ether CH3 CH2 CH2 NH2 (Reduction) Propan - 1 - amine C6 H5 CONH2 i, LiAlH4 /Ether C6 H5 CH2 NH2 ii. H2 O (ii) C6 H5 CH2 Cl KCN(alc) C6 H5 CH2 CN Benzamide Benzyla min e Benzyl chloride T,–KCl Benzyl cyanide 13. Gabriel phthalimide reaction gives pure 1° amine without any impurity of 2° or 3° amines. Na/Alcohol C6 H5 CH2 CH2 NH2 Therefore, it is preferred for synthesising 1° or LiAlH4 /ether amines. 18. (i) Propanamine contains three carbon atoms, 14. Gabriel Phthalimide Synthesis: It is a therefore the amide molecule undergoing synthetic pathway to obtain 1°-amines specifically Hoffmann bromamide reaction must contain without any impurities of 2°, 3°-amines. It four carbon atoms. Thus, the structure and involves treatment of phthalimide with KOH IUPAC name of the starting amide with four solution followed by reaction with alkyl halides carbon atoms is: and finally alkaline hydrolysis. OO O C C CH3CH2CH2 — C — NH2 (Butanamide) NH KOH NK RCl The complete reaction is: C C CH3 CH2 CH2 CONH2 + Br2 + 4KOH T OO Bu tan amide OO CH3 CH2 CH2 NH2 + 2KBr + K2 CO3 + 2H2 O C H2O C—ONa Propana min e OH– + RNH2 N—R (ii) Benzamide is an aromatic amide containing C—ONa seven carbon atoms. Hence, the amine C formed from benzamide is an aromatic primary amine containing six carbon atoms, O O i.e. 15. (i) C2 H5 Cl NH3 C2 H5 NH2 C2 H5 Cl (alc.) Chloroethane Ethana min e (1° amine) C2 H5 NHC2 H5 C2 H5 Cl The complete reaction is: N - Ethylethanamine C6 H5 CONH2 + Br2 + 4KOH T (2° amine) Benzamide C2H5 C2 H5 Cl C6 H5 NH2 + 2KBr + K2 CO3 + 2H2 O Aniline C2H5 — N — C2H5 19. (i) Primary (1°), (ii) Tertiary (3°), (iii) Primary N,N-Diethyl ethanamine (1°), (iv) Secondary (2°) (3° amine) 20. (i) Propan-2-amine (ii) But-3-en-1-amine (C2 H5)4 N+ Cl– 21. (i) 4-methoxybenzenamine Tetraethyl ammonium chloride (ii) N-Phenylethanamide (Quaternary ammonium salt) 22. The preparation of aniline by Gabriel phthalimide (ii) C6 H5 CH2 Cl NH3 C6 H5 CH2 NH2 reaction requires the treatment of potassium Benzyl chloride Benzyla min e phthalimide with haloarene (C6H5Cl or C6H5Br). Since, the cleavage of C—X bond in haloarenes is 2CH3 Cl C6 H5 CH2 N (CH3)2 quite difficult, due to double bond character and –2HCl hence the reaction is not possible. N,N - Dimethylphenylmethanamine 23. Arylamines are amines in which aromatic ring like benzene is directly attached to nitrogen atom. 16. The conversion of aonneamcairdbeon(—atCoOmNleHs2s) group For example: into an amine with by the Amines 309 treatment of amide with bromine and aqueous KOH solution is called Hoffmann bromamide reaction or Hoffmann degradation reaction. The reaction takes place as follows:

NH2 CH3 , H (c) CH3 — C — NH2 N Aniline CH3 Diphenylamine 2-Methylpropan-2-amine 24. (i) (d) CH3CH2 C|HCH3 NH2 (Butan-2-amine) (e) Secondary amines: CH3CH2CH2NHCH3 (N-Methylpropanamine) (f) (CH3 )2 CHNHCH3 (N-Methylpropan-2-amine) COOH CONH2 (g) CH3CH2NHCH2CH3 (N-Ethylethanamine) (ii) NH3 Benzamide NH2 (h) Tertiary amines: ∆ Benzoic acid (CH3 )2 NCH2CH3 (N,N-Dimethylethanamine) Br2/NaOH (iii) Chain isomerism: (a) and (b), (a) and (c), (b) ∆ and (d), (c) and (d). Position isomerism: (a) and (d), (e) and (f) Aniline Metamers: (e) and (g), (f) and (g) Functional isomers: All 8 isomers are (ii) C2H5Cl NH3 → C2H5NH2 C2H5Cl → functional isomers. C|2H5 NC2H5 NO2 NH2 C2H5NHC2H5 C2H5Cl → C2H5 26. (i) conc. HNO3 6[H] Sn + HCl 25. (i), (ii) (a) Primary amines: Benzene conc. H2SO4 Aniline CH3 CH2 CH2 CH2 NH2 (Nitration) Nitrobenzene Butan - 1 - amine NH2 N(CH3)2 CH3 (ii) (i) Nitration 2CH2I (ii) 6[H] Heat,–2HI (b) CH3 — CH — CH2NH2 (Sn+HCl) 2-Methylpropan-1-amine Benzene Aniline N, N-Dimethyl- aniline (iii) CH2 — CH2 — CH2 — CH2 2KCN (alc) CH2 — CH2 — CH2 — CH2 LiAlH4 H2 N— (CH2)6 —NH2 T CN CN (Reduction) Cl Cl Hexane - 1, 6 - diamine 1, 4-Dichlorobutane 27. (i) CH3 COOH LiAlH4 CH3 CH2 OH P, I2 CH3 CH2 I KCN (alc.) CH3 CH2 CN (Reduction) or PI3 Ethanoic acid Ethanol Iodoethane Propanenitrile H+,H2 O CH3 CH2 COOH Hydrolysis Propanoic acid (ii) CH3 NH2 HNO2 CH3 OH P + I2(PI3) CH3 I KCN (alc.) CH3 CN Na/alcohol CH3 CH2 NH2 [H] Methana min e Methanol Iodomethane Ethanenitrile Ethana min e (iii) CH3 NO2 Sn/HCl CH3 NH2 CHCl3, KOH, T CH3 NC Na/C2 H5 OH CH3 NHCH3 Nitromethane (Reduction) (Carbylamin e (Reduction) Methana min e Methyliso- Dimethyla min e reaction) cyanide 28. (i) CH3 COOH LiAlH4 CH3 CH2 OH SOCl2 CH3 CH2 Cl NH3 CH3 CH2 NH2 Reduction T T or CH3 COOH (i) NH3 CH3 CONH2 P2 O5 CH3 CN Na/C2 H5 OH CH3 CH2 NH2 (ii) Heat Reduction 310 Chemistry-12

(ii) CH3 OH SOCl2 CH3 Cl KCN CH3 CN Na/C2 H5 OH CH3 CH2 NH2 T T Reduction (iii) CH3 CH == CH2 HBr/H2 O2 CH3 CH2 CH2 Br KCN CH3 CH2 CH2 CN T T Na/C2 H5 OH Reduction CH3 CH2 CH2 CH2 NH2 29. (i) CH3 CH2 NO2 (i)LiAlH4 /ether (iii) N-Phenylethanamide (ii) H3 O+ (iv) 2-Methylpropan-2-amine (v) Phenylmethanamine CH3 CH2 NH2 + 2H2 O 35. C3H9N has four structural isomers. These are: (ii) [(CH3)4 N]+ OH– Thermal (∆) (i) CH3 CH2 CH2 NH2 CONH2 decomposition Propan - 1 - amine (1° amine) (CH3)3 N + CH3 OH (ii) CH3 — CH — CH3 (iii) + Br2 + 4KOH ∆ NH2 Benzamide NH2 Propan-2-amine + 2KBr + K2CO3 + 2H2O (1° amine) Aniline (iii) C2 H5 — NH — CH3 30. (i) Propan-2-amine N - Methylethanamine (ii) N-Methylethanamine (2° Amine) (iii) Benzenamine (Aniline) (iv) N-Ethyl-N-methylbenzenamine CH3 (v) Trimethyl phenyl ammonium iodide (iv) CH3 — N — CH3 31. (i) N-Ethyl-N-methylethan-1-amine (ii) But-3-en-1-amine N, N-Dimethylmethanamine (iii) N, N-Dimethylethanamine (3° Amine) (iv) 3-Bromobenzenamine (v) N, N-Dimethylbenzenamine Only 1° amine will react with HNO2 to liberate N2 gas. 32. (i) 4-Nitro-N, N-dimethylbenzenamine (ii) N-Phenylbenzenamine CH3 CH2 CH2 NH2 + HNO2 (iii) Phenylmethanamine CH3 CH2 CH2 OH + N2 + H2 O (iv) 2-Phenylethanamine (v) 3-Amino-2-chlorobutanamide Propan – 1 – ol 33. (i) 4-Ethyl-2-methylbenzenamine CH3 — CH — CH3 + HNO2 (ii) N, N-Dimethylpentan-3-amine (iii) 4-Methoxybenzenamine NH2 (iv) 2-Methylpropan-2-amine (v) N, N-Dimethylpropan-1-amine Propan-2-amine 34. (i) Ethane-1, 2-diamine CH3 — CH — CH3+ N2 + H2O (ii) 4-Methylbenzenamine OH Propan-2-ol 36. (i) Propan-1-amine (1°) (ii) N-Methylpropan-2-amine (2°) (iii) 2-Methylpropan-2-amine (1°) (iv) N-methylbenzenamine (2°) (v) 3-Bromobenzenamine or 3-Bromo-aniline (1°) 37. (i) CH3 COOH NH3 CH3 COONH4 Heat CH3 CONH2 Br2 /KOH CH3 NH2 or (Excess) (–H2 O) Ethanoic acid Ammonium acetate Acetamide Methanamine CH3 COOH SOCl2 CH3 COCl NH3 CH3 CONH2 Br2 /KOH CH3 NH2 (Excess) Ethanoic acid Ethanoyl chloride Ethanamide Methanamine (ii) CH3 (CH2)4 CN H3 O+ CH3 (CH2)4 COOH SOCl2 CH3 (CH2)4 COCl (Hydrolysis) –SO2, –HCl Hexanenitrile Hexanoic acid NH3 (excess) CH3 ^CH2h4 CONH2 Br2 /KOH CH3 (CH2)4 NH2 –NH4 Cl (Hoffmann bromamide 1 - Aminopentane reaction) (iii) CH3 OH PCl5 CH3 Cl KCN (alc.) CH3 CN H3 O+ CH3 COOH (Hydrolysis) Methanol Ethanenitrile Ethanoic acid Chloromethane Amines 311

(iv) CH3 CH2 NH2 HNO2 CH3 CH2 OH K2 Cr2 O7 /H2 SO4 CH3 COOH NH3(excess) Ethana min e Ethanol (Oxidation) Ethanoic acid CH3 COONH4 Heat CH3 CONH2 Br2 /KOH CH3 NH2 (–H2 O) Ethanamide (Hoffmann bromamide reaction) Methanamine (v) CH3 NH2 + CH3 Cl \" CH3 NHCH3 + HCl Methana min e N,N - Dimethylamine Topic 2. Physical and Chemical Properties of Amines Physical Properties of Amines or secondary amines. This can be explained on the basis of following factors: • Solubility: Lower aliphatic amines are soluble in water because they are able to form hydrogen ã Solvation effect: Greater is the stability of bond with water. Solubility decreases with the substituted ammonium cation formed, increase in molar mass of amines due to increase stronger is the corresponding amine as in size of the hydrophobic group. a base. Tertiary ammonium ion is less hydrated than the secondary ammonium • Boiling point: Among the isomeric amines, ion, which is further less hydrated than the primary and secondary amines have higher primary amines. Thus, tertiary amines has boiling point because they can form hydrogen least tendency to form ammonium ion and bonds. consequently they are least basic. Tertiary amines cannot form hydrogen bond O| H2 due to the absence of hydrogen atom attached H to electronegative nitrogen atom. Hence order of | R N+ H–OH2 boiling point of isomeric amines follows the order: R— N+ —H—OH2 R H–OH2 Primary > Secondary > Tertiary | (2°) Chemical Properties of Amines H Less hydration • Basic character of amines: Amines have an | unshared pair of electrons on nitrogen atom, OH2 (1°) R Maximum hydration R N+–H–OH2 due to which they behave as a Lewis base. Basic R character of amines can be better understood in (3°) terms of their Kb and pKb values as follows: R – NH2 + H2 O 5 + OH– RNH3 + K= + [OH –] or K[H2O] = [R–NH3] [OH –] Minimum hydration [R–NH3] [R–NH2] On the basis of solvation effect, order of [R–NH2] [H2 O] basicity of aliphatic amines should be: or Kb = + [OH –] Primary amines > Secondary amines > Tertiary amines [R–NH3] [R–NH2] pKb = –log Kb the pKb ã Steric factor: As the crowding of alkyl group GinrdeiacatetresththeaKt bb avsaeluise or smaller value increases from primary to tertiary amines stronger. hindrance to hydrogen bonding increases which eventually decreases the basic • Comparison of basic strength of aliphatic strength. Thus, there is a subtle interplay amines and ammonia: Aliphatic amines are of the inductive effect, solvation effect and stronger bases than ammonia due to + I effect of steric hinderance of the alkyl group which alkyl groups leading to high electron density on decides the basic strength of alkylamines in the aqueous state. the nitrogen atom. H NH R NH When the alkyl group is small like CH3, there is no steric hindrance to hydrogen bonding. HH In this case, order of basicity in aqueous medium is: • Comparison of basic strength of primary, secondary and tertiary amines: The order of (CH3)2NH > CH3NH2 > (CH3)3N > NH3 basicity of amines in the gaseous phase follows the When alkyl group is ethyl, the order of basicity in aqueous medium is: expected order on the basis of + I effect: tertiary amines > secondary amines > primary amines >terNtiHar3y. Ianmainqueseoaures solution, it is observed that (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 less basic than either primary 312 Chemistry-12

ã Comparison of basic strength of arylamines e.g. pyridine which removes HCl formed in the and alkanamines: Generally, arylamines are reaction and shifts the equilibrium towards the considerably less basic than alkylamines. product side. For example, ethylamine is more basic than aniline. In aniline lone pair of electrons on R – NH2 + RCOCl Base RNHCOR + HCl nitrogen is less available for protonation Acid chloride O Substituted amide because of resonance effect. Resonating O structures of aniline are as follows: R′–NH2 + R–C–O–C–R Base N+ H2 N+ H2 Acid anhydride R′NHCOR + RCOOH NH2 Substituted amide R2 NH + RCOCl Base R2 NCOR + HCl It should be noted that since tertiary amines do not contain any replaceable hydrogen atom. Therefore, they do not undergo acylation reaction. I II III N+ H2 • Carbylamine Reaction: Only aliphatic and NH2 aromatic primary amines on heating with choloroform and ethanolic potassium hydroxide form isocyanides or carbylamines. IV V R – NH2 + CHCl3 + 3KOH Heat R – NC + 3KCl + 3H2 O Less basicity of aniline can also be explained by comparing the relative stability of aniline It should be noted that secondary and tertiary amines do not give this test. and anilinium ion obtained by accepting a • Reaction of primary amines with nitrous acid: Primary aliphatic amines on reaction with proton. Greater the number of resonating snaitltr,ouwshiacchidbe(HinNg Ou2n) sftoarbmle aliphatic diazonium structures, greater is the stability of that decomposes to form species. alcohol and evolves nitrogen gas. Aniline is a resonance hybrid of five resonating structures whereas anilinium ion R –NH2 NaNO2 + HCl [R – N+2 Cl–] (diazotisation) has only two resonating structures, as shown 273 - 278 K below: N+ H3 N+ H3 H2 O ROH + N2 (-) + HCl (Hydrolysis) While primary aromatic amines on reaction with dniiatrzoounsiuamcidsa(lHt.NO2) in cold (273–278 K) forms NaNO2 + 2HCl I II C6 H5 –NH2 273 – 278 K Thus aniline has less tendency to accept a proton to form anilinium ion. Aniline (diazotisation) • Effect of substituent on basic character of amines: Electron donating or electron releasing C6 H5 –N+2 Cl– + NaCl + 2H2 O group/groups (EDG) strengthen the electron density on nitrogen atom thus, increases the basic Benzenediazonium strength of amines. chloride Electron withdrawing group (EWG) decreases the electron density on nitrogen thus, decreases the • Reaction with benzenesulphonyl chloride: basic strength of amines. Hinsberg’s reagent, i.e. benzenesulphonyl cshecloorniddaery(Ca6Hm5iSnOes2Ctol) reacts with primary and Chemical reactions of Amines form sulphonamides. The hydrogen atom attached to nitrogen atom in • Acylation Reaction: Aliphatic and aromatic sulphonamides formed by primary amines is primary and secondary amines (which contains strongly acidic in nature due to the presence of replaceable hydrogen atom) react with acid a strong electron withdrawing sulphonyl group. chlorides, anhydrides and esters to form Hence, it is soluble in alkalies. substituted amides. The process of introducing O an acyl group (R—CO—) into the molecule is called acylation. The reaction is carried out in S Cl + H N R the presence of a stronger base than the amines, O H 1° amine O (Hinsberg reagent) S N R + HCl OH N-Alkylbenzenesulphonamide (soluble in alkali) Amines 313

Since sulphonamides formed by secondary amines NH do not contain any hydrogen atom attached to 2 nitrogen atom, so it is not acidic. Hence, it is insoluble in alkalies. HNO3, H2SO4, 288 K O NH NH NH 2 2 2 + NO S Cl + H N R 2 OR + 2° amine NO 2 O NO (47%) (2%) 2 S N R + HCl (51%) OR It should be noted that aromatic amines cannot N, N-Dialkylbenzene be nitrated directly because tHhNe On3itbreoinagmainsetrsonsgo sulphonamide oxidising agent oxidises (insoluble in alkali) formed, to form a black mass of various oxidation • Bromination: Aniline reacts with bromine water products. However, nitration by protecting the at room temperature to give a white precipitate –anNhHy2drgirdoeucpanbybeaccoentytrlaoltlieodntoreoabcttaioinnpwaritah-isaocmeteirc as the major product. of 2, 4, 6-tribromoaniline: NH2 NH2 Br Br + 3HBr NHCOCH3 + 3Br2 H2O NH2 Aniline Br (CH3CO)2O HNO3,H2SO4,288 K Pyridine (Nitration) 2, 4, 6–Tribromoaniline Acetanilide To stop the reaction at monosubstitution product, NHCOCH3 NH2 activating effect of A–cNetHy2l group is reduced by OH– or H+ acetylation reaction. group is removed later (Hydrolysis) by hydrolysis. O NH2 H N C CH3 NO2 NO2 p-Nitroacetanilide p-Nitroaniline • Sulphonation: Aniline reacts with conc. H2SO4 (CH3CO)2O Br2 to form anilinium hydrogen sulphate which on Pyridine CH3 COOH heating at 453-473 K produces p-aminobenzene sulphonic acid, commonly known as sulphanilic Aniline N-Phenylethanamide (Acetanilide) acid, as the major product. O +– H N C CH3 NH2 NH2 NH3SO4 OH– or H+ H2SO4 453-473 K (Hydrolysis) ∆ Br Br Anilinium + (Major) 4-Bromoaniline hydrogensulphate NH3 NH2 • Nitration: In strongly acidic medium, aniline gets protonated first to form the anilinium ion, which is a deactivating group and meta directing in nature. Hence, meta-nitroaniline is also formed SO3H – in 47% yeild along with ortho and para substitutes SO3 products. Sulphanilic acid Zwitter ion of sulphanilic acid ExErCiSE 9.2 Multiple Choice Questions (MCQs) (1 Mark) (i) CH3NH2 (ii) NCCH2NH2 1. Amongst the following, the strongest base in (iii) (CH3)2NH (iv) C6H5NHCH3 aqueous medium is .................... . 314 Chemistry-12

2. Which of the following is the weakest Brönsted C6H5NH2, C2H5NH2, (C2H5)2NH, NH3 base? [AI 2014, 2008, Delhi 2011][NCERT] NH 12. Write reactions of the final alkylation product 2 of aniline with excess of methyl iodide in the presence of sodium carbonate solution. (i) (ii) N H NH [NCERT] 2 (iv) CH3NH2 13. Write chemical reaction of aniline with benzoyl (iii) chloride and write the name of the product 3. Amongst the following, the strongest base in obtained. [NCERT] aqueous medium is .................... . 14. Mention two important uses of sulphanilic acid. (i) CH3NH2 (ii) (CH3)3N 15. Why amines have lower boiling point than those (iii) (CH3)2NH (iv) C6H5NHCH3 of corresponding alcohols or carboxylic acids? 4. Name the best reagent used for the separation of 16. Identify (A) and (B) in the following chemical 1°, 2° and 3° amines. equation: (i) C6H5SO2Cl (ii) (CH3CO)2O C6 H5 CH2 Cl NH3 (A) 2CH3Cl (B) (iii) C6H5NH2 (iv) C6H5CH2Cl Short Answer Type Questions-I (2 Marks) Assertion-Reason Type Questions (1 Mark) 17. What is carbylamine reaction? [Delhi 2012] Note: In the following questions, a statement of 18. Explain mwhetyhpyKlabmvainlue?e for aniline is greater than that for [Delhi 2011] assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. OR (i) Assertion and reason both are correct statements Why aniline has weaker alkaline nature than and the reason is correct explanation of assertion. aliphatic amines? (ii) Assertion and reason both are correct statements 19. State reasons for the following: but reason is not correct explanation of assertion. (i) Primary amines have higher boiling point than tertiary amines. [Delhi 2011] (iii) Assertion is correct statement but reason is wrong statement. (ii) Ethylamine is soluble in water whereas aniline is insoluble in water. [Delhi 2011] (iv) Assertion is wrong statement but reason is correct statement. 20. Carry out the following conversions: (i) Methanamine to N-Methylbenzamide 5. Assertion: Acylation of amines gives a monosubstituted product whereas alkylation of (ii) Aniline to sulphanilic acid amines gives polysubstituted product. 21. Complete the following given chemical equations: Reason: Acyl group sterically hinders the (i) C6 H5 NH2 + CHCl3 + alc. KOH approach of further acyl groups. (ii) C6 H5 NH2 + Br2 (aq) [Delhi, AI 2012] 6. Assertion: Hoffmann’s bromamide reaction is given by primary amines. 22. Complete the following acid-base reactions and name the products: Reason: Primary amines are more basic than secondary amines. (i) CH3CH2CH2NH2 + HCl 7. Assertion: N-Ethylbenzene sulphonamide is (ii) (C2H5)3N + HCl [NCERT] soluble in alkali. 23. Write the products of the following reactions: Reason: Hydrogen attached to nitrogen in sulphonamide is strongly acidic. (i) C6 H5 NH2 + (CH3 CO)2 O 8. Assertion: N, N-Diethylbenzenesulphonamide is (ii) CH3 NH2 NaNO2 + HCl [NCERT] insoluble in alkali. 24. Write the reactions of: Reason: Sulphonyl group attached to nitrogen atom is strong electron withdrawing group. (i) aromatic and Very Short Answer Type Questions (1 Mark) (ii) aliphatic primary amines with nitrous acid. 9. Name the best reagent used for the separation of 1°, 2° and 3° amines. [NCERT] 10. Give one reaction that can be used as a test for 25. Give the structure of A (C3H9N) if it reacts with benzene sulphonyl chloride to form a compound primary amines. [Foreign 2011] which is insoluble in alkali. 11. Arrange the following in decreasing order of their 26. Give the structure of the compound formed when basic strength: 3-chlorobutanamide is treated with Br2 and KOH. Amines 315

27. Why aniline does not give Friedel-Crafts reaction? insoluble in aqueous sodium hydroxide [CBSE 2022] solution. 28. Aniline is a weaker base than methylamine. (iii) Acid (C) on reaction with Tollens’ reagent Explain. gives silver mirror test. 29. Why aniline is not as basic as ammonia? Explain. What are the name and molecular structures of Short Answer Type Questions-II (3 Marks) (A), (B) and (C)? Explain the reactions with the help of equations. 30. Complete the following given chemical equations: 35. Arrange the following in increasing order of their (i) C6 H5 NH2 Na2 Cr2 O7 /H2 SO4 basic strength: (ii) C6 H5 N (CH3)2 + HNO2 (i) (CC2H2H5N5)2HN2H,C6H5NH2, NH3, C6H5CH2NH2 and (ii) C2H5NH2,(C2H5)2 NH, (C2H5)3N,C6H5NH2 (iii) C2 H5 NH2 + C6 H5 SO2 Cl (iii) CC6HH35NCHH22,N(HC2H3)2 NH, (CH3)3N, C[N6HC5ENRHT2], [Delhi 2009, 2011] 36. Give plausible explanation for each of the following: 31. Identify (A), (B) and (C) in the following chemical reactions: (i) (A) Br2 /KOH (B) CHCl3 /KOH(alc) Heat (C) H2/Pt CH3 NHCH (CH3)2 (i) Why are amines less acidic than alcohols of comparable molecular masses? (ii) (A) SOCl2 (B) NH3 (ii) Why do primary amines have higher boiling (C) LiAlH4 H2 NCH2 CH2 CH2 NH2 point than tertiary amines? (iii) C6 H5 COOH NH3 (A) Br2 /KOH (iii) Why are aliphatic amines stronger bases than Heat aromatic amines? [NCERT] (B) conc. H2 SO4 (C) 37. Complete the following reactions: 453 - 473 K (i) C6 H5 NH2 + CHCl3 + (alc) KOH 32. Identify (A), (B) and (C) in the following reactions: (ii) C6 H5 NH2 + H2 SO4 (conc.) (i) CH3I (A) CH3I (B) HNO2 (C) (iii) C6 H5 NH2 + Br2 (aq) Long Answer Type Questions (5 Marks) (ii) Br2/KOH (A) Br2/(excess) 38. What happens when: (B) NaNO2 + HCl (C) (i) Aniline solution is added to an aqueous 273K solution of bromine or aniline is treated with bromine water. O NO2 (ii) Aniline is heated with conc. H2SO4 at 475 K. (i) A (iii) Aniline is treated with benzoyl chloride in (iii) NH — C — CH3 (ii) B the presence of aqueous NaOH. (iii) C (iv) Aniline is treated with acetyl chloride. 33. Identify (A), (B) and (C) in the following chemical equations: (v) Aniline is treated with sodium nitrite and NH2 dilute HCl at 273 K. 39. Write the chemical reaction when: (i) CH3COCl (A) HNO3 (B) H2O/H+ (C) (i) Benzamide reacts with bromine and aqueous KOH. H2SO4, 288 K (ii) Aniline is treated with benzaldehyde. NH2 (iii) Aniline reacts with CHCl3 in the presence of alcoholic KOH. (ii) conc. H2SO4 (A) Heat (B) 453 K (C) Heat (iv) Aniline reacts with 2 moles of CH3I. 34. From analysis and molecular weight determination, (v) Aniline is treated with acetic anhydride in the molecular formula of a compound (A) is found tfoorbtehCis3Hco7mNOpo.uTnhde:following information is given the presence of conc. H2SO4. (i) On hydrolysis it gives an amine (B) and a 40. A mixture of two aromatic compounds (A) and carboxylic acid (C). (B) was separated by dissolving in chloroform (ii) Amine (B) reacts with benzene sulphonyl followed by extraction with aqueous KOH chloride and gives a product which is solution. The organic layer containing compound (A) when heated with alcoholic solution of KOH, produced a compound (C) (C7H5N), which is an 316 Chemistry-12

associated molecule and has an unpleasant odour. (iv) In decreasing order of basic strength in gas phase: The alkaline aqueous layer on the other hand when heated with chloroform and then acidified C2H5NH, (C2H5)2 NH, (C2H5)3N and NH3 gives a mixture of two isomeric compounds (D) (v) In increasing order of solubility in water: and (E) of molecular formula, C7H6O2. Identify the compounds (A), (B), (C), (D) and (E) and write the C2H5NH2, (C2H5)2NH, C6H5NH2 [NCERT] structures of (A), (B), (C), (D) and (E). 44. (i) Accomplish the following conversions: [NCERT Exemplar] (a) Nitrobenzene to Benzoic acid 41. Give one chemical test to distinguish between the (b) Benzene to m-bromophenol following pairs of compounds: (ii) An aromatic compound ‘A’ on treatment with aqueous ammonia and on heating forms KcoOmHpofuonrmd s‘Ba’ wcohmichpoounnhde‘aCt’inwgitwhitmhoBlerc2ualnadr (i) Ethylamine and Dimethylamine [Delhi 2013(C)] [Delhi 2010] faonrdmIuUlaPCA6CHn7Na.mWesriotfe down the structures the compounds A, B (ii) Secondary and Tertiary amines (iii) Ethylamine and Aniline and C. [NCERT] [Delhi AI 2011, 10, 09, 14; Foreign 2009, 08] 45. How will you convert: (iv) Aniline and Benzylamine (i) Chlorobenzene to p-Chloroaniline [Delhi AI 2011, 10; Foreign 2009, 08] (ii) Aniline to p-Bromoaniline [Delhi 2010] (iii) Benzamide to Toluene (v) Aniline and N-methylaniline? (iv) Aniline to Benzyl alcohol [NCERT] [Delhi AI 2010] [Delhi 2010] [NCERT] (v) Benzyl chloride to 2-Phenylethanamine 42. Account for the following: 46. Give the structures of A, B and C in the following reactions: (i) pKb of aniline is more than that of methylamine. (i) CH3 CH2 I NaCN A OH– (Partial hydrolysis) (ii) Ethylamine is soluble in water whereas aniline is not. B NaOH + Br2 C (iii) Methylamine in water reacts with ferric (ii) C6 H5 N2 Cl CuCN A H2 O/H+ B NH3 C chloride to precipitate hydrated ferric oxide. T (iii) CH3 CH2 Br KCN A LiAlH4 B HNO2 C (iv) C6 H5 NO2 273 K C [Delhi AI 2008] NaNO2 + HCl Fe/HCl 273 K (iv) Although amino group is o-and p-directing in A nature in electrophilic substitution reactions, B H2 O/H+ T yet aniline on nitration gives substantial amout of m-nitroaniline. [CBSE 2022] (v) CH3 COOH NH3 A NaOBr T (v) Aniline does not undergo Friedel-Crafts B NaNO2/HCl C reaction. [NCERT] [AI 2014] 43. Arrange the following: HNO2 (vi) C6 H5 NO2 Fe/HCl A 273 K (i) In decreasing order of the pKb values: C2H5NH2,C6H5NHCH3, (C2H5)2NH and B C6H5OH C C6H5NH2 [NCERT] (ii) In increasing order of basic strength: 47. Compound a(Aci)d, iCc 5sHol1u1NtiOon.isWnhoetnso(Alu)bilse in cold, C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and alkaline or refluxed CH3NH2 in NaOH solution, a gas (B) is evolved and a salt (iii) In increasing order of basic strength: (C) is formed. Acetyl chloride reacts with (B) to give (D), yCe4lHlo9wNOoi.l (B) reacts swtrituhctHurNeOs 2ofin(Ac)oltdo (a) Aniline, p-nitroaniline and p-toluidine to give a (E). Give (b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2 (E) with suitable reactions. 1. (iii) 2. (i) Answers 3. (iii) 4. (i) 5. (iii) 6. (iii) 9. Hinsberg’s reagent (Benzene sulphonyl chloride, 7. (i) 8. (ii) C6H5SO2Cl). 10. Primary amine reacts with aryl sulphonyl chloride gives N-alkylbenzene sulphonyl chloride which is soluble in alkali, Amines 317

O O methylamine due to electron donating –CH3 || || group, the electron density increases on N-atom S—Cl + H—N—R→ S—N—R + HCl || | || | which increases its basicity and decreases its pKb OH OH value. (Soluble in alkali) 19. (i) The high boiling point is due to intermolecular H-bonding in primary amines while there is 11. aDipnhusetternuotyronnlg+gaerIrroeeubfpasfetsorceofttnaohgnfaeitnlrhineCbeaa2Hslekex5syNelrtHtghsr2ao–anunIpNde, Hfb(fCoe3t.2cHhtH5aoo)fn2wNtdheHtvehemiirss,, results in less availability of electrons on the no H atom for H-bonding in tertiary amines. (ii) Ethylamine is soluble in water due to the formation of H-bond with water molecules nitrogen atom for protonation. Thus, aniline is whereas due to bulkier phenyl group (larger a weaker base than aliphatic amines aCn2dH5NNHH32. hydrocarbon part) aniline is not being able to H>CeN6nHHc5e3N,>tHhC2e6Hor5CNdHeH3rI2fol[lCow6 Hs:5(NC+ H2H25C)2HN3H] I>– form H-bond and thus it is insoluble in water. 12. Aniline n - Methylanilinium iodide 20. (i) CH3 NH2 + C6 H5 COCl Methanamine Benzoylchloride + CH3 NHCOC6 H5 + HCl N - Methylbenzamide NH2 2:C6 H5 CH3DI– + Na2 CO3 \" NH2 NH+3 [HSO4]– 2C6 H5 NHCH3 + CO2 + 2NaI (ii) (conc.) H2SO4 Heat (D) 455 K (–H2O) C6 H5 NHCH3 CH3 I C6 H5 N^CH3h2 CH3 I Aniline Na2 CO3 Na2 CO3 :C6 H5 N+ ^CH3h3D2 CO32– + 2NaI NHSO3H NH2 O Rearrangement 455 K NH2 O NH — C — C6H5 SO3H 13. + C6H5 — C — Cl NaOH(aq) Sulphanilic acid Benzoyl chloride 21. (i) C6 H5 NH2 + CHCl3 + 3KOH (alc.) T C6 H5 N == C + 3KCl + 3H2 O Aniline N-Phenylbenzamide Phenyl isocyanide (Benzanilide) 14. Sulphanilic acid is mainly used for manufacturing (ii) of dyes and sulpha drugs. 15. Electronegativity of nitrogen is less than that of oxygen (i.e., N-H bond is less polar than O-H bond) thus extent of hydrogen bonding and boiling point is lower in amines. 16. (A) C6 H5 CH2 NH2; (B) C6 H5 CH2 —N (CH3)2 22. (i) CH3 CH2 CH2 NH2 + HCl Benzyla min e N, N - Dimethylbenzylamine n - Propylamine 17. Aliphatic and aromatic primary amines on heating with choloroform in the presence of + CH3 CH2 CH2 Cl – alcoholic solution of KOH form isocyanides or NH3 n - Propyl ammonium chloride carbylamines, which has offensive smell. This (ii) (C2 H5)3 N + HCl (C2 H5)3 N+ HCl– reaction is known as carbylamine reaction. Triethyl amine Triethyl ammonium R—NH2 + CHCl3 + 3KOH (alc.) Heat chloride ∆ Alkyl amine(1°) 23. (i) C6 H5 NH2 + (CH3 CO)2 O conc. H2 SO4 R—NC + 3KCl + 3H2 O Aniline Acetic or Pyridine (Acetylation) Alkyl isocyanide Anhydride C6 H5 NH2 + CHCl3 + 3KOH (alc.) Heat C6 H5 NHCOCH3 + CH3 COOH Aniline ∆ Ace tan ilide C6 H5 N =\" C + 3KCl + 3H2 O (ii) CH3 NH2 HNO2 Phenyl isocyanide _NaNO2 + HCli 18. Due to resonance in aniline, lone pair of electrons CH3 + Cl – H2 O CH3 OH + N2 + HCl on nitrogen gets delocalised over the benzene ring and N-atom acquires some +ve charge, which N2 decreases the basicity of aniline and increases its pKb value, whereas in aliphatic amines like 24. (i) Aromatic primary amines react with nitrous acid at 273-278 K to form aromatic diazonium salts. 318 Chemistry-12

NH2 29. Aniline, C6H5NH2 has five resonance structures in which the lone pair of electrons on the nitrogen + NaNO2 + 2HCl 273-278 K atom is delocalised on the ring, thus making the N+ NCl– nitrogen atom less available to accept a proton. Aniline In an ammonia molecule the lone pair of electrons on the nitrogen atom is localised and + 2H2O + NaCl is fully available for coordination with a proton. Benzenediazonium Therefore, aniline is not as basic as ammonia. chloride NH2 O (ii) Aliphatic primary amines also react with 30. (i) Na2Cr2O7 / H2SO4 HdiNazOo2naiut m273s-a2l7ts8 K to form unstable aliphatic which decomposes even at Aniline ∆ low temperatures readily to form alcohols O (which generally predominates) and N2 gas p-Benzoquinone is evolved. N(CH3)2 N(CH3)2 CH3 CH2 NH2 + NaNO2 + 2HCl 273–278 K (ii) + HONO (–2H2 O, –NaCl) Ethyla min e + N,N-Dimethyl- /NCl– H2 O aniline NO CH3 CH2 —N Ethanediazonium chloride p-nitroso-N,N-dimethylaniline (Unstable) (iii) C6 H5 SO2 Cl + H HNC2 H5 ∆ C6 H5 SO2 NHC2 H5 + HCl CH3 CH2 OH + N2 + HCl Ethyl alcohol 25. Secondary (2°) oraNm-mi neteh,yleCth2aHn5aNmHineC H 3 31. (i) (A) (CH3)2CHCONH2; (B) (CH3)2CHNH2; (ethylmethylamine) (C) (CH3)2CHNC CH3 NHC2 H5 + C6 H5 SO2 Cl OO OO CH3 (ii) (A) HOCCH2COH; (B) ClCCH2CCl; C6 H5 SO2 N–C2 H5 KOH Insoluble OO 26. (C) H2NCCH2CNH2 (iii) (A) C6H5CONH2; (B) C6H5NH2; (C) p-H2N—C6H4—SO3H + 2KBr + KCl NHCH3 N(CH3)2 N(CH3)2 32. (i) (A) (B) (C) 27. Electron pair owfhNicHh2isgraoLuepwiins aniline attacks on NO AlCl3 catalyst acid to form a salt. N2+Cl– As a result, it reduces the electron density on the NH2 Br NH2 Br Br Br benzene ring and hence aniline does not undergo (ii) (A) Friedel-Crafts reaction. (B) (C) H Cl N Al Cl Br Br (iii) (A) NaNO2 + HCl; (B) HBF4; (C) NaNO2, Cu; H Cl NHCOCH3 NHCOCH3 NH2 28. In aniline, the lone pair of electrons on the 33. (i) (A) (B) (C) nitrogen atom is involved in conjugation with the p-electrons of the ring. Therefore, nitrogen atom NH3+HSO–4 NO2 NO2 becomes positively charged or electron deficient NHSO3H NH2 and is not in a position to donate the electron pair (ii) (A) to the acid so easily as compared to methylamine (B) (C) in which there is no such conjugation infact, methyl group due to its +I effect increases the SO3H electron density on nitrogen atom. Thus, aniline is a weaker base than methylamine. Amines 319

O pair of electrons gets delocalise over the aromatic ring and thus, not readily available 34. (A) is H — C — N(CH3)2 ; (B) is (CH3)2 NH ; for donation while in aliphatic amines it N, N-Dimethylformamide Dimethylaniline is localised on nitrogen atom and easily available for donation. (C) is HCOOH 37. (i) C6 H5 NH2 + CHCl3 + 3KOH (alc.) Heat Formic acid Aniline Reactions: C6 H5 N == C + 3KCl + 3H2 O (i) HCON (CH3)2 + H2 O (A) Phenyl isocyanide HCOOH + (CH3) 2 NH (ii) C6 H5 NH2 + H2 SO4 (conc.) (C) (B) 2° amine + HSO4– (ii) (CH3)2 NH + C6 H5 SO2 Cl C6 H5 NH3 (B) Aniline hydrogen sulphate (CH3)2 NSO2 C6 H5 + HCl (Insoluble in NaOH) NH2 NH2 Br Br (iii) HCOOH + 2 7Ag^NH3h2A+ + 2OH – (iii) + Br2 (aq.) + 3HBr (C) 2Ag + CO2 + 2H2 O + 4NH3 Aniline Silver mirror Br 35. (i) C< 6(HC52NHH5)22N<HNH3 < C6H5CH2NH2 < C2H5NH2 2, 4, 6-Tribromoaniline (ii) C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH (iii) <C6(HC5HN3H)2N2 <HC6H5CH2NH2 < (CH3)3N < CH3NH2 38. (i) 36. (i) Loss of proton from amines can form an amide NH2 NH2 ion, whereas loss of proton from alcohols forms an alkoxide ion. R—NH2 R—NH– + H+ ; (ii) + conc. H2SO4 475 K A min e Amide ion Heat, ∆ Aniline R—OH RO– + H+ SO3H Alcohol Alkoxide ion Sulphanilic acid Since O is more electronegative than N, O therefore, RO– can accommodate the negative charge more easily than RNH–. Thus, alcohols NH2 NH C C6H5 are more acidic than amines. O (ii) aPtroimmasrbyoanmdeindews (iRth—tNheHN2) have two hydrogen (iii) + C6H5 C Cl aq. NaOH atom, so they can Heat, D Benzoyl chloride undergo extensive intermolecular H-bonding. Aniline Benzanilide (N-Phenylbenzamide) R RR O N—H N— H N— H NH2 NH C CH3 H HH O aTteormtisarbyonadmeidnewsit(hR3tNhe) do not have hydrogen (iv) + CH3 C Cl Pyridine N-atom and therefore ∆ they do not undergo H-bonding. As a Aniline Acetanilide (N-Phenylacetamide) result, primary amines have higher boiling point than tertiary amines of comparable molecular masses. For example, boiling (v) point of n-butylamine (351 K) is much higher than that of tertiary butylamine (b.p. 319 K) although the molecular masses of both of them are quite the same. (iii) Aniline and other aromatic amines are far 39. (i) less basic than NH3 and aliphatic amines. It can be explained on the basis of resonance + 2KBr + K2CO3 + 2H2O stability in aromatic amines. The lone 320 Chemistry-12

H (ii) Secondary amine reacts with benzene (ii) NH2 + O C ZnCl2 sulphonyl chloride and forms a solid ppt. ∆ which is insoluble in alkali. Tertiary amines do not react with benzene sulphonyl chloride. Aniline Benzaldehyde H (iii) Aniline reacts with nitrous aaciddia(fzroonmiuNmaNsaOlt2/, HCl) at 273-278 K to form NC which gives a scarlet azo dye with b-naphthol in NaOH. Benzylidene aniline (Schiff's base) Ethylamine reacts with nitrous acid to produce ethyl alcohol and nitrogen gas (as (iii) seen through bubbles): C2 H5 NH2 + ONOH C2 H5 OH + N2 + H2 O (iv) C6 H5 NH2 + CH3 I CH3 I C6 H5 N (CH3)2 (iv) Aniline gives dye test and forms a scarlet C6 H5 NHCH3 azo-dye. NHCOCH3 NH2 Benzylamine reacts with nitrous acid to form benzyl alcohol and nitrogen gas (as seen through bubbles). (v) conc. H2SO4 + (CH2CO)2O ∆ + CH3COOH C6H5CH2NH2 + ONOH → C6H5CH2OH + N2 + H2O Aniline Acetanilide (v) Aniline reacts with nitrous acid to form 40. The compound (C) having molecular formula, diazonium salt which gives azo dye test. iCs7Hob5Ntaipnoesdsebsyseths eanactuinonpleoaf saalncothoodliocuKr OanHd it on N-Methylaniline forms nitrosoamine (yellow chloroform layer of compound (A), so it should be phenyl isocyanide. Thus, the compound (A) is oily liquid) on treatment with nitrous acid which is stable at room temperature. aniline. Carbylamine reaction is exhibited by (A) However, on reaction with HCl in ether and as follows to form compound (C): alcohol, the nitroso (—NO) group migrate NH2 towards the para-position. + CHCl3 + 3KOH(alc.) NC (A) Aniline + 3KCl + 3H2O (C) Phenyl isocyanide 42. (i) In methylamine due to +I effect the electron density on the N-atom increases. In contrast, Alkaline aqueous layer of (B) when heated with in aniline, the lone pair of electrons present chloroform and then acidified give two compounds on the N-atom is delocalised over the benzene (D) iasnpdh(eEn)owl,hwichhicahreunthdeerisgoomeseRrse(imC7eHr6-OT2i)e. mThaunsn, ring and hence the electron density on the (B) nitrogen atom decreases. Therefore, aniline is reaction as follows: a weaker base than methylamine and so, its pKb value is higher than that of methylamine. OH (ii) Ethylamine is soluble in water due to + CHCl3 (i) Heat OH intermolecular H-bonding as shown below: (ii) Acidification (B) Phenol OH CHO + HH (D) CHO H— N H O H — N H—O o-Hydroxy- benzaldehyde (E) p-Hydroxy- C2H5 H C2H5 H benzaldehyde 41. (i) E t h y l a m i n e ( P r i m a r y a m i n e ) g i v e s However, in aniline, due to larger hydrocarbon carbylamine test. It gives a foul odour when part (hydrophobic part) the extent of heated with choloroform and alcoholic KOH. H-bonding decreases considerably and Dimethylamine (secondary amine) does not hence aniline is almost practically insoluble give this test. in water. Amines 321

(iii) Methylamine being more basic than water, (v) Aniline being a Lewis base reacts with Lewis accepts a proton from water, liberating OH– acid AlCl3 to form a salt as follows: ions. C6 H5 NH2 + AlCl3 C6 H5 + AlCl3– or CH3 NH2 + H—OH CH3 + OH– Lewis base Lewis acid NH2 NH3 CH3 + + OH – NH3 These —OH– ions react with Fe3+ ions of FfeerCrilc3 forming a brown precipitate of hydrated hydroxide. As a result, the nitrogen atom of aniline becomes positively charged and is not in a 2Fe3+ + 6OH- 2Fe (OH)3 or Fe2 O3 .3H2 O position to activate the ring for electrophilic reddish (Brown ppt.) substitution reactions. Consequently aniline (FeCl3) does not undergo Friedel-Crafts (alkylation and acylation) reaction. (iv) HItmno2-SdsftiOorrreo4)mcn,tgimnlaygonasaitcnliioddnfidictuehmamecetadiinivouianlmtiinn((eNcgogiHnnect3n.s+)aHptWNruorOtheo3,in/cwcahohtnieilcdes. —NH2 group in aniline is o, p-directing and 43. (i) (C2H5)2NH > C2H5NH2 > C6H5NHCH3 > an activating group. That is why besides the C6H5NH2, is the decreasing order of basicity but a stronger base has lower pKb value, o, p-derivatives, significant amount of meta therefore pKb values decreases in the reverse derivative is also formed. order as follows:C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH NH2 (ii) C6H5NH2 < C6H5N(CH3)2 < CH3NH2 < Aniline conc. HNO3 NH2 NH2 (C2H5)2NH conc. H2SO4 NO2 (iii) (a) p-nitroaniline < aniline < p-toluidine NH2 (b) C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2 ++ (iv) (C2H5)3 N > (C2H5)2NH > C2H5NH2 > NH3 (v) C6H5NH2 < (C2H5)2NH < C2H5NH2 NO2 (47%) m- NO2 (2%) o- (51%) p- NO2 NH2 N2+Cl– CN COOH 44. (i) (a) (i) Sn/HCl NaNO2 + HCl CuCN/KCN H2O, H+ (ii) NaOH 273-278 K (Sandmeyer's (Hydrolysis) (Diazotisation) reaction) Nitrobenzene Aniline Benzenediazonium Phenyl cyanide Benzoic acid chloride NO2 NO2 NH2 (b) conc.HNO3 Br2/Fe Sn/HCl conc.H2SO4, 333 K (Bromination) (Reduction) Benzene Br Br Nitrobenzene m-Bromonitro– m-Bromoaniline benzene N2+Cl– OH NaNO2 + HCl (273-278 K) Boiling (Diazotisation) H2SO4(dil) Br Br (ii) 'A' NH3 'B' Br2 /KOH 'C' m-Bromophenol T T (C6 H7 N) CSrei6anHcc5teNi,oHtnh2tea(Ackonemsilpipnolaeu)cneadna‘dsCf’(oBilsl)oaiwsnsC:a6mHi5nCeO(N—HN2H(B2)e.nSzoam‘Bi’dme)u. sHt ebnecean‘Aa’ misidCe6H(—5CCOOONHH(2B).eTnhzoeircefaocried)(.CT)hies NH3 C6 H5 COOH (T) C6 H5 CONH2 Br2 /KOH C6 H5 NH2 (A) (B) (C) Benzene (Benzamide) Benzena min e Carboxylic acid (Aniline) 322 Chemistry-12

Cl Cl Cl 45. (i) conc. HNO3/conc. H2SO4 Sn/HCl ∆, (Nitration) (Reduction) Chlorobenzene NO2 NH2 NH2 NH2 p-Chloronitro p-Chloroaniline benzene NHCOCH3 NHCOCH3 (ii) (CH3CO)2O/Pyridine Br2/CH3COOH H2O/OH– (Acetylation) (Bromination) (Hydrolysis) Aniline Acetanilide Br Br CONH2 COOH p-Bromoacetanilide p-Bromoaniline CH3 (iii) H2O/H+ NaOH/CaO CH3Cl –NH3 ∆ Anhyd. AlCl3 Benzamide Benzoic acid Benzene Toluene NH2 N2+Cl– CH3 (iv) NaNO2 + HCl H3PO2 – H2 O/Cu CH3Cl 273-278 K ∆ , (Reduction) AlCl3 (anhyd.) Aniline (F.C. alkylation) Benzene Toluene CH2Cl CH2OH Cl2(1 mole) KOH (aq.) hν, 383 K (Hydrolysis) Benzylchloride Benzyl alcohol NH2 NH2 N+ 2Cl– N+ 2BF4– F Br Br Br Br Br Br Br Br (v) Br2 water NaNO2/HCl HBF4 ∆ 273-278 K –N2,–BF3 Aniline Br Br Br Br 46. (i) CH3 CH2 I NaCN CH3 CH2 CN OH– CH3 CH2 CONH2 2, 4, 6 - Tribromo- (–NaI) (Partial fluorobenzene Iodoethane Propanenitrile (A) Hydrolysis) Propanamide (B) NaOH + Br2 CH3 CH2 NH2 Ethanamine (C) (ii) C6 H5 N+2 Cl– CuCN C6 H5 CN H2 O/H+ C6 H5 COOH NH3 C6 H5 CONH2 (Sandmeyer T Benzenediazonium reaction) Cyanobenzene Benzoic acid Benzamide chloride (A) (B) (C) (iii) CH3 CH2 Br KCN CH3 CH2 CN LiAlH4 CH3 CH2 CH2 NH2 HNO2 CH3 CH2 CH2 OH (Reduction) 273 K Bromoethane Propanenitrile (A) Propan - 1 - amine (B) Propan - 1 - ol (C) (iv) C6 H5 NO2 Fe/HCl C6 H5 NH2 NaNO2 + HCl C6 H5 + / NCl– H2 O/H+ C6 H5 OH (Reduction) 273 K (Hydrolysis),T Nitrobenzene Aniline (A) N Phenol (C) Benzenediazonium chloride (B) (v) CH3 COOH NH3 CH3 CONH2 NaOBr CH3 NH2 NaNO2 /HCl CH3 OH T (Hoffmann bromamide Ethanoic acid Ethanamide Methanamine Methanol (A) reaction) (B) (C) NO2 NH2 + NCl– N (vi) Fe/HCl HNO2 C6H5OH N==N OH 273 K (Coupling reaction) Aniline Benzene p-Hydroxyazobenzene (orange dye) (A) diazonium chloride (C) (B) Amines 323

47. The given information indicates that the Also it reacts with acetyl chloride to form a compound (A) is a substituted amide. ecoxmplpaoinuenddthDrowugithhchmeomleiccaullaerqufaortimonuslaasCfo4Hllo9NwOs: O CH3CH2 — C — N(CH3)2 NaOH O O Reflux (A) CH3 NH + Cl—C—CH3 –HCl CH3 N—C—CH3; N, N-Dimethyl propanamide CH3 CH3 Acetyl chloride (D) (CH3)2 NH + CH3CH2COONa (B) (B) (C) CH3 NH + HONO CH3 CH3 CH3 Dimethylamine (gas) (Salt) N—N == O (B) Compound (B) gives yellow oily liquid on reaction with HNO2; thus it should be a secondary amine. (E) (Yellow oily liquid) N, N-Dimethly-N-nitrosoamine Topic 3. Diazonium Salts are more stable than the alkyldiazonium salts. Stability of Diazonium Salts This can be accounted on the basis of +R effect in aryldiazonium salts, due to which the positive • They have a general formula: R2 NX+ 2– X– , in which charge over N-atom in N+ 2 form is delocalised over R is an alkyl or aryl group and may be Cl–, the benzene ring as shown below: Br–, HBF4–, HSO4– etc. Arenediazonium salts + +– +– +– + NN NN NN NN NN ++ + Preparation of Diazonium Salts R–NH2 + NaNO2 + 2HX 273 – 278 K R–N+ 2 X– + NaX + 2H2O (Here R = alkyl/aryl group) • They are prepared by the treatment of alkyl or This process of converting an amine into its corresponding diazonium salt is called diazotisation. aryl aatm2i7n3e–s2w7i8thK,HaNccOo2rd(ainmg tixottuhree roefaNctaioNnO: 2 + HX) reactions of Benzene Diazonium Chloride • Reactions involving displacement of nitrogen: CuCl/HCl ArCl + N2 CuBr/HBr ArBr + N2 CuCN/KCN Cu/HCl ArCN + N2 Cu/HBr ArCl + N2 + CuCl KI +– HBF4 ArBr + N2 + CuCl ArN2Cl H3PO2 + H2O H2O ArI + KCl + N2 CH3CH2OH +– ∆ Ar – F + BF3 + N2 Ar – N2BF4 ArH + N2 + H3PO3 + HCl ArOH + N2 + HCl ArH + N2 + CH3CHO + HCl +– N2BF4 NO2 HBF4 NaNO2 + N2 + NaBF4 • Reactions involving retention of diazo Cu, ∆ group, coupling reactions: Diazonium ion acts as an electrophile because there is a positive uasndNe2r+g.oTeshecroeufpolriengbernezaecntieondiwaiztohnieulemctrcohnlorriidche charge on the terminal nitrogen atom and it exists compounds like phenol and aniline to form azo- compounds. 324 Chemistry-12

Azo compounds contains —N = N— bond and the • Uses of Diazonium Salts: reaction is known as coupling reaction. ã Diazonium salts, (eg. Benzene diazonium chloride) can be used for the production of +– – dyes and are thus used in paints and pigment N NCl + H industry. OH OH ã They are also very useful for synthesising N=N – many reagents (eg. phenolphthalein) and OH + Cl + H2O organic compounds like aryl halides and aryl p-Hydroxyazobenzene (orange dye) cyanides, which otherwise require very harsh + conditions for their preparation. +– N NCl + H NH2 H N=N NH2 + C–l + H2O p-Aminoazobenzene (yellow dye) ExErCiSE 9.3 Multiple Choice Questions (MCQs) Reason: Diazotisation of aniline belongs to +– electrophilic aromatic substitution. 1. The reaction ArN2Cl Cu/HCl ArCl + N2 + CuCl is named as .................... . 6. Assertion: Chlorobenzene can be formed from diazonium salt using Gatterman reaction. (i) Sandmeyer reaction Reason: Yield in Sandmeyer reaction is better (ii) Gatterman reaction than that of Gatterman reaction. (iii) Claisen reaction Very Short Answer Type Questions (1 Mark) (iv) Carbylamine reaction 7. What is diazotisation? 2. Which of the following compound will not undergo 8. Give an example of Sandmeyer reaction. azo coupling reaction with benzene diazonium chloride. 9. Write short note on diazotisation. (i) Aniline (ii) Phenol 10. Write short note on coupling reaction. (iii) Anisole (iv) Nitrobenzene 11. What is Gattermann’s reaction? [Delhi 2009] 3. Which of the following can be prepared by Sandmeyer’s reaction? 12. Give an example of Balz-Schiemann reaction. (i) Bromobenzene (ii) Iodobenzene 13. H o w i s i o d o b e n z e n e o b t a i n e d f r o m benzenediazonium chloride? [Delhi 2008] (iii) Fluorobenzene (iv) All of these 14. What type of compounds are formed in coupling 4. The reagent that can be used to convert reaction between an aryl diazonium salt and benzenediazonium chloride to benzene is phenol (or its derivatives)? (i) SnCl2/HCl (ii) CH3CH2OH 15. N a m e t h e r e a g e n t u s e d t o c o n v e r t (iii) HBF4 (iv) LiAlH4 benzenediazonium chloride to benzene. Assertion-Reason Type Questions 16. Give reason for the following: Note: In the following questions, a statement of Diazonium salts of aromatic amines are more assertion followed by a statement of reason is given. stable than those of aliphatic amines. [NCERT] Choose the correct answer out of the following choices Short Answer Type Questions-I (2 Marks) on the basis of the above passage. 17. How will you convert: [Delhi 2012] (i) Assertion and reason both are correct statements (i) Aniline into Benzyl alcohol and the reason is correct explanation of assertion. (ii) Nitrobenzene to Benzoic acid (ii) Assertion and reason both are correct statements but reason is not correct explanation of assertion. [Delhi 2012] [NCERT] (iii) Assertion is correct statement but reason is wrong 18. Convert: statement. (i) Aniline to Benzonitrile (iv) Assertion is wrong statement but reason is correct (ii) Aniline to Bromobenzene statement. 19. Complete the following reactions: 5. Assertion: Coupling reaction of aryldiazonium (i) C6 H5 N2+Cl–+ H3 PO2+ H2 O salts belongs to electrophilic aromatic substitution. [Delhi 2011, 2012, 2015 (C)] Amines 325

(ii) C6 H5 N+2 Cl– + C2 H5 OH Long Answer Type Questions (5 Marks) [Delhi 2010] [NCERT] 23. Convert: 20. Write the products of the following reactions: (i) Aniline into benzonitrile (i) C6 H5 N2+ Cl– (i) HBF4 (ii) Benzene into m-bromoiodobenzene (ii) NaNO2 /Cu,T (iii) Aniline into p-dinitrobenzene NaNO2 /HCl C6 H5 NH2 (ii) C6 H5 NH2 (A) OH– (B) (iv) Aniline into phenol [Delhi 2010] (v) Benzene into m-bromophenol [NCERT] 21. Convert: 24. What happens when benzenediazonium chloride reacts with: (i) 3-Methylaniline into 3-nitrotoluene (ii) Aniline into 1,3,5-tribromobenzene (i) Hypophosphorous acid [NCERT] (ii) Cuprous chloride/HCl 22. Convert: (i) 3-Methylaniline into 3-nitrotoluene (iii) KCN/Cu (ii) Aniline into 1, 3, 5-tribromobenzene [NCERT] (iv) Boiling water /H+ (v) Fluoroboric acid, heat Answers 1. (i) 2. (iv) 3. (i) C6H5N2+Cl– HCl/Cu C6H5Cl + N2 + CuCl 4. (ii) 5. (iii) 6. (ii) 7. The process of conversion of aromatic primary 12. ArN+2 X– + HBF4 –HCl ArN+2 BF4— Heat ∆ amines into arenediazonium salts with NaNO2 / Aryldiazoniumfluoroborate HCl at 273-278 K is called diazotisation. ArF + N2 + BF3 Arylfluoride 8. C6 H5 N+2 Cl– CuCl/HCl C6 H5 Cl + N2 (-) or 1.13. C6 H5 N+2 Cl– + KI Warm C6 H5 —I + N2 + KCl C6 H5 N+2 Cl– CuBr/HBr C6 H5 Br + N2 (-) 9. Diazotisation: The chemical reaction in which 14. Coloured azo dyes are formed in coupling reaction Nipnr+i≡mcNoalXdry– (b2ao7rno3dm–2ias7t8iccaKallle)kdytoldiafaomzroimntiesdsaitraieozaonc.tsawlitt,hhHavNinOg2 as follows: + NCl– + OH OH– (273–278 K) N (Coupling reaction) NH2 Phenol + NaNO2 + 2HCl 273-278 K N==N OH + NCl– p-Hydroxyazobenzene (orange dye) N + NaCl + 2H2O 15. H3PO2/H2O or alkaline SnCl2/C2H5OH followed by boiling. 10. Coupling Reaction: It is the chemical reaction 16. The diazonium salts of aromatic amines are involving formation of –N=N– bond in azo dyes by more stable than those of aliphatic amines due treating diazonium salt with phenols/amines. The to dispersal of the positive charge (at o-and –N=N–bonds are joined together at para position p-positions) of the benzene ring as shown: wrt. the phenolic (–OH)/amine (–NH2) group. + +– +– NN NN NN + NCl– + OH ++ N N=N OH + +– + NN NN 11. The process of introducing a halogen atom This type of resonance stability is not possible in (chlorine or bromine) in the benzene ring by case of aliphatic amines. reacting the arenediazonium salt with the corresponding halogen acid, i.e. HCl and HBr in the presence of powdered copper metal as catalyst is called Gattermann reaction. 326 Chemistry-12

NH2 N2+Cl– CN COOH NH2 CH2OH NH2 Br Br 17. (i) NaNO2+ HCl CuCN H3 O+ (ii) Reduction (LiAlH4) Br2/H2O NaNO2/HCl CN ∆ 273–278 K (Diazotisation) N2+Cl– COOH CH2OH Br CuCN H3 O+ N2+Cl– N+ 2Cl– Reduction NO2 (LiAlH4) Br Br Br Br NH2 CN COOHH3PO2/H2O ∆ (ii) Sn/HCl NaNO2 + HCl CuCN/ H3 O+ N2+Cl– Reduction KCN NH2 Br Br N2+Cl– H2 Aniline COOH HBF4 NaNO2 + HCl CN e CH3 H3 O+ 22. (i) NaNO2 + HCl NO2 273-278 K CuCN/ N2+ Cl– KCN CH3 (Diazotisation) 3-Methylaniline NH2 CN N2+BF4– 18. (i) NaNO2 + HCl CuCN/KCN NaNO2 / Cu 273–278 K D or K3[Cu(CN)4] Aniline Benzonitrile CH3 CH3 NH2 N2+ Cl– Br 3-Nitrotoluene (ii) NaNO2 + HCl NH2 NH2 273–278 K CuBr/HBr Br Br Aniline Bromobenzene (ii) Br2 NaNO2 + HCl 19. (i) C6 H5 N+ 2 Cl– + H3 PO2 + H2 O Aniline 273-278 K (Diazotisation) Benzene diazonium Br chloride C6 H6 + N2 + HCl + H3 PO3 2, 4, 6-Tribromoaniline Benzene N+ Cl – 2 (ii) C6 H5 N+2 Cl– + C2 H5 OH Br Br Br Br Benzenediazonium H3PO2 chloride C6 H6 + CH3 CHO + N2 + HCl H2O Benzene Ethanal Br Br + + 20. (i) C6 H5 Cl – HBF4 C6 H5 BF4– 1, 3, 5- Tribromobenzene N2 (–HCl) N2 Benzene diazonium fluoroborate NaNO2 /Cu C6 H5 NO2 + N2 + BF3 + NaF 23. (i) T Nitrobenzene (ii) (A) C6 H5 N+2 Cl– NO2 NO2 NH2 (B) NH2 N+=NBF4– (ii) HNO3/H2SO4 Br2/FeBr3 Sn/HCl Di (Nitration) (Reduction) N NO2 N2+Cl– NH2 Br Br I 21. (i) NaNO2/HBF4 NaNNOO2/C2u 273–278 K ∆ 3-Methyl Aniline CH3 H3N-MOe3/tHhy2Sl O4 CH3 Br2/FeBr3 Sn/HCl Diazotisation KI Benze(Nneitdriaatzioonni)um (Reduction) NaNO2/HCl ∆ fluoroborate Br Br (0-5° C) Br Br NO2 NH2 NHCOCH3 NHCOCH3 NH2 (iii) (CH3CO)2O HNO3 H2O/H+ (Diazotis (Acetylation) H2SO4 NaNO2 + 3-Nitro- CH3 Acetanilide (0–5°C toluene NO2 NO2 Amines 327

22 COCH3 NHCOCH3 NH2 N2+Cl– NO2 HNO3/H2SO4 Br2/Fe Sn/HCl NO2 373 K HNO3 (i) HBF4 N+2 Cl– Br H2SO4 (ii) NaNO2/Cu (Nitration) OH NO2 NH2 H2O/H+ (Diazotisation) (Diazotisation) Boiling NaNO2 + HCl NO2 NaNO2/HCl dil. H2SO4 Br (0–5°C) Br Br NO2 + 24. (i) C6 H5 / NCl– + H3 PO2 + H2 O N C6 H6 + N2 (-) + H3 PO3 + HCl Benzene NH2 N2+Cl– OH + (ii) C6 H5 / NCl– CuCl/HCl Diazotisation H2O/H+ N NaNO2/HCl ∆ (iv) (iii) C6 H5 + / NCl– + KCN C6 H5 Cl + N2 (-) (0–5°C) N Cchulorobenzene + C6 H5 CN + N2 (-) + KCl NO2 NO2 (iv) C6 H5 N / NCl– + H2 O Heat C6 (v) HNO3/H2SO4 Br2/Fe Sn/HCl H5HO+H + N2 (-) + HCl 373 K N+2 Cl– Br (v) C6 H5 + / NCl– + Phenol (Nitration) OH N HBF4 NH2 C6 H5 N2+BF4– Heat C6 H5 F + N2 (-) + BF3 ∆ Fluorobenzene Case based questions(Diazotisation) Boiling Br NaNO2/HCl dil. H2SO4 1. Case StudyBr Br (iii) T h e b e s t r e a g e n t f o r c o n v e r t i n g , 2-phenylpropanamide into Reliable N-alkylation of secondary amines have been developed. By using DIAD and TPP (or 1-phenylethanamine is PS-TPP) a variety of secondary amines can be (a) excess H2/Pt (b) NaOH/Br converted to the corresponding tertiary amines in good to excellent yields with diverse alkyl (c) NaBH4/methanol (d) LiAlH4/ether OR halides, no formation of quaternary amine salts are observed. The nature of amine is (a) basic (b) acidic (c) neutral (d) None of these (Reference: Michio Kurosu, Sevendy Sekhar (iv) A compound ‘X’ on reaction with Ni/H2 gives Day, Dean C. Crick, Efficient Synthesis of compound C2H7N, which on reaction with Tertiary Amines from Secondary Amines, 2006, CH3Cl gives compound ‘Y’. Compound ‘Y’ on Tetrahedron Letter, 47 (28), 4871–4875) reaction with CH3Br in presence of DIAD and TPP gives compound ‘Z’. What is compound Z? The following questions are multiple choice questions. Choose the most appropriate (a) CH3CH2N(CH3)2 (b) (CH3CH2)2NCH3 answer: (c) (CH3CH2)3N (d) (CH3)3N 2. Case Study (i) Which of the following is a secondary amine? Aromatic amines 1-amino-4-phenoxybenzene (a) CH3—NH—CH3 (b) CH3—NH2 H (A-1A), 2-(4-aminophenoxy) naphthalene (A-2A), and 1-(4-aminophenoxy) naphthalene (A-3A) were synthesized by the reduction of corresponding nitroaromatics with hydrazine monohydrate and (c) CH3—N—CH3 (d) CH —C—NH Pd/C 5% (w/w). The compounds were highly active 32 in protecting DNA against hydroxyl free radicals. CH CH DNA interaction and antioxidant results back up 3 3 each other indicating that these compounds have (ii) Give product of the reaction: potential to be used as cancer chemopreventive DIAD, TPP agents. Additionally, one compound (A-1A) (CH3)2NH + CH3CH2Cl THF, r.t. ? showed significant antibacterial and antifungal (a) (CH3)2N+CH2CH3 (b) (CH3)3N activity as well. (c) (CH3)4N (d) (CH3CH2)2NCH3 Aromatic amines were synthesized by the following procedure. Mixture of corresponding 328 Chemistry-12

nitroaromatic compound, hydrazine monohydrate, synthetic chemistry. Recently, electrochemical and Pd/C 5% (w/w) was refluxed in ethanol for 18 C–N bonding methods have attracted attention hours in 250 mL flask equipped with a magnetic due to several advantages such as their simplicity stirrer under an inert atmosphere created by and possibility of low-cost scale up. Meanwhile, purging dry nitrogen. due to severe environment problems such as global warming and air pollution associated with N2H4 + R – O NO2 Ethanol, Pd/C (5%) non-sustainable synthesis of chemicals and fuels is indispensable. Electrochemical C–N bonding Reflux, 18h reactions (e.g., amino acid synthesis from biomass, athmeidenesvifrroonmmeCnOta2)llayrbeenmiugnchadnveaednetadgbesecoafutsheesoef R–O NH2 methods. However, there are still some future aspects on which study should continue so as to (Source: Hammad Ismail, Bushra Mirza, improve the efficiency and selectivity of these Ihsan-ul Haq, Muhammad Shabbir, Zareen reactions. Akhter, and Amina Basharat, 2015, Synthesis, (Reference: Jeong Eun Kim, Seungwoo Choi, Characterization, and Pharmacological Mani Balamurugan, Jun Ho Jang, Ki Tae Nam, 2020, Electrochemical C–N bond formation for Evaluation of Selected Aromatic Amines, Journal sustainable amine synthesis.) of Chemistry, Volume 2015, Article ID 465286, 9 In these questions, a statement of assertion Pages) followed by a statement of reason is given. Choose the correct answer out of the The following questions are multiple choice following choices. questions. Choose the most appropriate (a) Assertion and reason both are correct answer: statements and reason is correct explanation (i) The product of reaction, when nitrobenzene for assertion. is reduced in the presence of Pd/C is (b) Assertion and reason both are correct (a) Benzene (b) Aniline statements but reason is not correct explanation for assertion. (c) Cyclohexane (d) None of these (c) Assertion is correct statement but reason is (ii) The product of reaction: wrong statement. N2H4 + Ethanol, Pd/C (5%) (d) Assertion is wrong statement but reason is correct statement. O NO Reflux, 18h 2 (i) Assertion: Amine derivatives have various applications in biology and synthetic (a) O NH chemistry. 2 Reason: Proteins, vitamins, hormones (b) O NO in biology and drugs, dyes, polymers in 2 synthetic chemistry are some important examples of amine derivatives. (c) O NO 2 (ii) Assertion: Simplicity and possibility of low-cost scale up are advantages of (d) O CH electrochemical C–N bond formation. 3 Reason: Conventional methods to prepare (iii) The substance which p1roves to be active in C–N bond are more environment friendly. prevention of DNA against free radicals is (iii) Assertion: The electrochemical C–N bonding (a) hydrazine monohydrate methods have various advantages over (b) Pd/C methods used conventionally in industries. (c) 1-nitro-4-phenoxybenzene (d) 2-(4-aminophenoxy)naphthalene Reason: This new electrochemical C–N OR bonding methods are perfect and there is no chance for further improvement in this The compound which shows significant method. antibacterial and antifungal properties is (iv) Assertion: Gabriel phthalimide synthesis is an electrochemical C–N bonding reaction. (a) 1-amino-4-phenoxybenzene (b) 2-(4-aminophenoxy)naphthalene Amines 329 (c) 1-(4-aminophenoxy)naphthalene (d) 1-nitro-4-phenoxybenzene (iv) N2H4 is acting as a (a) Reducing agent (b) Oxidising agent (c) Catalyst (d) Solvent 3. Case Study C–Nbond-formingreactionsrepresentanimportant synthetic class because amine derivatives play crucial roles in various applications in biology and

Reason: Formation of amides from carbon In these questions, a statement of assertion dioxide is an electrochemical C–N bonding followed by a statement of reason is given. reaction. Choose the correct answer out of the following choices. OR (a) Assertion and reason both are correct Assertion: Amino acids prepared from biomass can be used to prepare proteins. statements and reason is correct explanation for assertion. Reason: Proteins are the polymers of amino acids. (b) Assertion and reason both are correct statements but reason is not correct 4. Case Study explanation for assertion. Transition metal-catalyzed substitution of (c) Assertion is correct statement but reason is alcohols by N-nucleophiles (or N-alkylation of wrong statement. amines and related compounds with alcohols) avoids the use of alkylating agents by means (d) Assertion is wrong statement but reason is of borrowing hydrogen (BH) activation of the correct statement. alcohol substrates. Water is produced as the only by-product, which makes the ‘BH’ processes (i) Assertion: Alkaline earth metal (M) is used atom-economic and environmentally benign. to catalyse the substitution of alcohols by Diverse types of homogeneous organometallic N-nucleophiles. and heterogeneous transition metal catalysts, and substrates such as N-nucleophiles including Reason: Metal, M first reduces to MH2 amines, amides, sulfonamides and ammonia, and then oxidise back to metal M in the and various alcohols, can be used for this substitution reaction. purpose, demonstrating the promising potential of ‘BH’ processes to replace the procedures using (ii) Assertion: The new method is named as ‘BH’ traditional alkylating agents in pharmaceutical process i.e., borrowing hydrogen activation of and chemical industries. ‘BH’ strategy provides alcohol substrates. new principles for establishing green processes to replace the relevant traditional synthetic methods Reason: In the reaction, metal, M first takes for C–N bond formation. hydrogen from alcohol and then return the hydrogen to form the final N-substituted R OH [M ] R NHR' product. [M] (iii) Assertion: The ‘BH’ process is considered atom economical and environmentally [MH2] benign. R O H2NR' R NR' Reason: The by-product of the reaction is H2O only water molecule. (Source: Qin Yang, Qingfu Wang and Zhengkun Yu, (iv) Assertion: Amines can be converted into 2015, Substitution of alcohols by N-nucleophiles alcohols using HNO2. via transition metal-catalyzed dehydrogenation, Reason: The by-products in the reaction are Chemical Society Reviews, 44(8), 2305-2329) nitrogen gas and water molecule. OR Assertion: Different types of homogeneous and heterogeneous catalysts are used in ‘BH’ process. Reason: Different types of substrates like N-nucleophiles and some alcohols are used in the ‘BH’ process. 1. (i) (a) (ii) (a) (iii) (b) or (a) (iv) (a) Answers (ii) (c) (iii) (c) (iv) (d) or (b) 2. (i) (b) (ii) (a) (iii) (d) or (a) (iv) (a) (ii) (a) (iii) (a) (iv) (a) or (b) 3. (i) (a) 4. (i) (d) Analogy based questions (ii) Aliphatic Amines : Weaker base :: Ammonia 1. Which of the following analogies is correct? : Stronger base (i) Aromatic Amines : Weaker base :: Ammonia : Stronger base 330 Chemistry-12

(iii) Aryl Fluoride : Prepared by direct halogenation (i) A : Reacts with Hinsberg’s reagent :: Aryl Iodide : Cannot be prepared by direct B : Do not react with Hinsberg’s reagent halogenation (ii) A : Gives Carbylamine reaction (iv) Aniline with bromine water : Trisubstituted B : Do not give Carbylamine product :: Acetanilide on bromination : Meta- substituted product (iii) A : Shows H-bonding among their molecules B : Do not show H-bonding among their 2. Complete the following analogy. molecules Primary Amines : A :: Secondary Amines : B (iv) A : Colourless B : Coloured Answers 1. (i) 2. (ii) Matching type questions 1. Match the reactions given in Column I with the statements given in Column II. Column I Column II A. Ammonolysis (1) Amine with lesser number of carbon atoms B. Gabriel phthalimide synthesis (2) Detection test for primary amines C. Hoffmann Bromamide reaction (3) Reaction of phthalimide with KOH and R X D. Carbylamine reaction (4) Reaction of alkylhalides with NH3 (ii) A (4) B (3) C (1) D (2) Code: (i) A (4) B (2) C (3) D (1) (iii) A (4) B (3) C (2) D (1) (iv) A (1) B (3) C (4) D (2) 2. Match the compounds given in Column I with the items given in Column II. Column I Column II A. Benzenesulphonyl chloride (1) Zwitter ion (2) Hinsberg reagent B. Sulphanilic acid (3) Dyes (4) Conversion to alcohols C. Alkyl diazonium salts (ii) A (2) B (1) C (4) D (3) D. Aryl diazonium salts (iv) A (2) B (4) C (1) D (3) Code: (i) A (1) B (2) C (4) D (3) (iii) A (3) B (1) C (4) D (2) Answers 1. (ii) 2. (ii) Quick revision notes • Like ammonia, N-atom carries an unshared • Amines are considered as derivatives of ammonia, electron pair (lone pair) in amines also, by the e.g. CH3NH2, CH3CH2NH2. virtue of which they behave as Lewis bases. • They are derived by replacing one, two or all the three H-atoms of ammonia by alkyl/aryl group • Amines are named as per IUPAC rules by forming primary (1°), secondary (2°) and tertiary replacing ‘e’ of the corresponding hydrocarbon (3°) amines, respectively. with ‘amine’, e.g. CH3NH2 is methanamine, • Among secondary and tertiary amines, if the CH3CH2NH2 is ethanamine. alkyl groups attached to the N-atom is the same then they are considered as simple amines and • Amines have sp3-hybridised N-atom and their if the alkyl groups are different, then they are geometry is pyramidal as only three constituent considered as mixed amines. atoms are bonded with the N-atom and the fourth Amines 331

sp3-hybridised orbital is occupied by lone pair of electron withdrawing group (eg. –NO2) decreases electrons. the basicity. • Amines are generally prepared by reduction of • With chloroform and alcoholic KOH solution, nitro compounds, Hoffmann bromamide method, amines react to form compounds having ‘R– by reduction of amides, nitriles and by treating NC’ moeity, where R is an alkyl or aryl group ammonium salts with a strong base, like NaOH, associated with amines. Such a reaction is and by ammonolysis of alkyl halides. known as ‘carbylamine reaction’. Isocyanides are offensive smelling compounds. This reaction • The lower aliphatic amines are gases with fish is only given by primary amines. Therefore used like odour. to distinguish 1° amines with 2° and 3° amines. • Primary amines with more than two carbon atoms • Reaction with nitrous acid (NaNO2 + HCl) in are liquids. cold helps in distinguishing between 1°, 2° and 3° amines. • Primary amines with more than 12 carbon atoms are solids. • With nitrous acid, primary amines form alcohols with liberation of N2 gas while secondary • Amines, particularly aromatic amines are usually amines form N-nitrosoamines (Liebermann’s colourless, however they develop colour, on nitroso reaction) and tertiary amines from alkyl keeping in air for a long time due to oxidation. ammonium nitrite salts. • The electronegativity of N-atom and H-atom is • Hinsberg test involves treatment of 1°, 2° and quite different and this is responsible for the 3° amines with benzenesulphonyl chloride, (i.e. extensive intermolecular H-bonding in amines Hinsberg Reagent) followed by aqueous KOH (primary and secondary). solution and is used to distinguish between the three types of amines (1°, 2° and 3°). • However the extent of H-bonding in amines is less than that of the corresponding alcohols • Amines can undergo electrophilic substitution and carboxylic acids, since O-atom is more reactions with Br2, HNO3 (nitration), H2SO4 electronegative than N-atom. (sulphonation) at ortho and para positions, in case of aromatic amines, since NH2 is an o, p-directing • The boiling point will increase with increase in group. However, direct nitration of aniline gives the molecular weight of amines, however primary 47% m-nitroaniline. and secondary amines have higher boiling point than tertiary amines of corresponding molecular • Amines can also undergo alkylation and acylation masses, due to more effective H-bonding. Boiling with alkyl halides; acyl chlorides, or anhydrides point decreases with increase in branching. in the presence of a base, like pyridine but cannot undergo Friedel-Crafts reaction. • In a similar manner, the solubility of amines in water and stability of ammonium cations follows • Diazonium salts can be obtained by treatment the same order: 1° > 2° > 3°. of alkyl/aryl amines with HNO2 in cold, i.e. By treating aniline with a mixture of NaNO2 + HCl • Aromatic amines are practically insoluble in (i.e. nitrous acid) in cold at 273–278 K. water, due to larger hydrophobic aromatic ring present in it, e.g. C6H5NH2 is insoluble in water. • Due to their instability, diazo salts needs to be immediately used. • All amines are however soluble in organic solvents like ethers, benzene, alcohol, etc. • They are very important for synthesising a large number of organic compounds. • Among alcohols, ethers and amines, amines are the most basic, due to less electronegativity of N • Coupling reaction of aryl diazonium salts with than O-atom. phenol or arylamines form azo dyes. • Alkyl amines are more basic than ammonia • The lower mass aliphatic amines and aniline are followed by aromatic amines because alkyl group used for the synthesis of many organic compounds is electron releasing whereas phenyl group is and in the manufacture of sulpha drugs. electron withdrawing. • In aromatic amines, electron donating groups (eg. –CH3 group) increases the basicity while ImportAnt reActIons 1. C2 H5 Cl + NH3 –HCl C2 H5 NH2 C2 H5 Cl (C2 H5)2 NH C2 H5 Cl (C2 H5)3 N –HCl –HCl Chloroethane Ethana min e N - Ethylethanamine N,N - Diethyl ethana min e C2 H5 Cl (C2 H5)4 N+ Cl– Tetra ethyl ammonium chloride 332 Chemistry-12

OH NH2 O 2. + NH3 anhyd. ZnCl2 + H2O NH2 O NH—C—C6H5 Phenol Aniline 13. + C6 H5 —C—Cl Pyridine + HCl (Benzenamine) NO2 NH2 NHCOCH3 NH2 3. + 6[H] Sn/HCl + 2H2O 14. + CH3CO O Pyridine + CH3COOH or Fe/HCl CH3CO or H2 /Pd 15. C2 H5 NH2 + HNO2 C2 H5 OH + N2 + H2 O 16. Diazotisation: 4. CH3 C / N + 4 [H] H2 /Ni CH3 CH2 NH2 NH2 Na(Hg)/C2 H5 OH or LiAlH4 5. Hoffmann’s Bromamide Reaction: + NaNO2 + 2HCl 0–5°C O N+2 Cl– –NaCl, –2H2 O ã CH3 —C—NH2 + Br2 + 4KOH OH CH3NH2 + K2CO3 + 2KBr + 2H2O H2O + N2 + HCl CONH2 ã + Br2 + 4KOH ã (C2 H5)2 NH + HNO2 NH2 (C2 H5)2 N–N = O + H2 O + 2KBr + K2CO3 + 2H2O (Yellow oily compound) N - Ethyl - N - nitroso ethanamine O ã (C2 H5)3 N + HNO2 (C2 H5)3 N+ HNO2– 17. Carbylamine reaction: ]Colourlessg 6. C6 H5 —C—NH2 ]i g LiAlH4 C6H5CH2NH2 + H2O ã CH3 NH2 + CHCl3 + 3KOH (alc.) ]iig H2 O CH3 N =\" C + 3KCl + 3H2 O 7. CH3 N \"= C + 4 [H] Na/C2H5OH CH3 NHCH3 8. Gabriel Phthalimide synthesis: ã C6 H5 NH2 + CHCl3 + 3KOH (alc.) O C6 H5 N =\" C + 3KCl + 3H2 O C NK + C2H5Cl 18. Hinsberg reagent test: C O O ã C2 H5 NH2 + C6 H5 SO2 Cl Potassium phthaliamide ]1° amineg Benzene sulphonyl chloride C NaOH^aqh C N—C2H5 C6 H5 SO2 NHC2 H5 KOH O N - ethyl benzene sulphonamide COONa C6 H5 SO2 NK+ – C2 H5 + C2H5NH2 COONa Soluble 9. CH3 NH2 + HCl ? CH3 + Cl– ã (C2 H5)2 NH + C6 H5 SO2 Cl NH3 ]2° amineg 10. C6 H5 NH2 + HCl ? C6 H5 + Cl– C6 H5 SO2 N (C2 H5)2 KOH Insoluble O NH3 11. C2 H5 NH2 + CH3 —C—Cl Pyridine ã (C2 H5)3 N + C6 H5 SO2 Cl No reaction O ]3° amineg NH2 CH3 —C—NH—C2 H5 + HCl 19. + CH3I O NHCH3 N(CH3)2 NH2 NH—C—CH3 Aniline O + CH3I 12. + CH3 —C—Cl Pyridine + HCl N-Methyl aniline N, N-Dimethylaniline Amines 333

NH2 ã N+ ∫NCl– + OH OH– 20. + H2 SO4 N=N OH + Cl– + H2O conc. Cl p-Hydroxyazobenzene Aniline +NH3HSO4– NH2 + 26. Sandmeyer’s Reaction: NH3 heat 453 K SO3H SO3– Cu2Cl2/HCl + N2 (Sulphanilic acid) (Zwitter ion) NH2 N2+Cl– Br 21. + HNO3 conc. H2 SO4 Cu2Br2/HBr + N2 conc. C∫N 288 K NH2 NH2 NH2 + NO2 + NO2 CuCN/KCN + N2 NO2 47% 2% 27. Gattermann’s Synthesis: 51% NH2 Cl 22. + (CH3CO)2O Pyridine N2+Cl– Cu/HCl Aniline NHCOCH3 NHCOCH3 Br + HNO3 conc. H+ or OH– Cu/HBr conc. H2 SO4 NNHO22 28. Balz-Schiemann Reaction: N2+Cl– N2+BF4– + CH3COOH + HBF4 Heat F NO2 + BF3 + N2 NH2 pa-nNiliitnreo N2+BF4– NO2 29. Cu/NaNO2 23. + CH3COCl Pyridine NHCOCH3 + N2 + NaBF4 NHCOCH3 ∆ Br2 OH– or H+ CH3 COOH N2+Cl– I NH2 30. KI + N2 + KCl + CH3COOH Br Br N2+Cl– H3PO2 + H2O + N2 + HCl + H3PO3 NH2 p-Bromoaniline 31. O N2+Cl– 24. + 3Br2(aq) NH2 CH3CH2OH + CH3—CH—CHl + Br Br OH + + 3HBr Br N2 (2,4,6-Tribromoaniline) 25. Coupling Reaction: 0–5°C 32. + H2O + N2 + HCl ã N+ ∫NCl– + H+ NH2 N=N NH2 + HCl p-Aminoazobenzene 334 Chemistry-12

common errors errors correctIons (i) Students usually done direct bromination of In aniline, +–RNHef2fegcrtouthpuiss, a strong activating group aniline to prepare p-bromoaniline as product. due to its results in 2,4,6-tribromo aniline on direct bromination. So acetylation of (ii) Students commit mistakes in writing about the aniline decreases the +R effect due to electron basic nature of amines in gaseous phase and in withdrawing carbonyl group attached to nitrogen aqueous phase. atom, which gives only para-bromo isomer as major product. In gaseous phase, only inductive and resonance effect of groups attached to nitrogen will play their role while in aqueous phase, solvation effect and steric hindrance of alkyl groups also decides the basicity of amines. Amines 335

Amines revIsIon chArt Physical Properties: • Lower aliphatic amines are water soluble, whereas higher amines are not. • Boiling point of isomeric amines follows the order: Primary amines > Secondary amines > Tertiary amines Due to decrease in hydrogen bonding • Considered as derivatives of NH3. Basic Character: • Classification: On the basis of no. of H-atom in NH3- • Like ammonia, all the three types of amines too being replaced: behave like Lewis bases, as they also contain a lone (i) Primary (1°) amines-RNH2 pair of electrons. (ii) Secondary (2°) amines-R2NH • Alkyl amines are stronger bases than ammonia, (iii) Tertiary (3°) amines-R3N electron releasing groups increases the basic • On the basis of type of hydrocarbon moeity attached character, while electron withdrawing group decreases the basic character. to N-atom: • In alkanamines, a combination of electron releasing,  Aliphatic (R–N) amines steric and solvation effect influences the stability of  Aromatic (Ar–N) amines the substituted ammonium cations in polar protic solvents and thus it affect the basic nature of amines. • structure: N-atom in amines is sp3-hybridised like CN-HN3- Ran-bdo cnodn asnisgtsle o ifs ale lsosn teh apna 1ir0 o9f. 5e°l einc tarmonins,e sso. the Chemical Properties: • Acylation reactions: Aliphatic and aromatic • They are tetrahedral in shape. primary and secondary amines react with acid • nomenclature: Commonly known as alkyl/aryl chlorides, esters and anhydrides to form amides. amines. • Carbylamine Reaction: Only 1°-aliphatic and aromatic amines can react with CHCl3 and ethanolic • iUPAC system: Alkanamine, where alkane represents KOH to form isocyanides or carbylamines. hydrocarbon moeity. • Reaction with Benzene sulphonyl chloride (Hinsberg reagent): Only 1° and 2° amines react to • Simplest aryl amine is aniline having –NH2 group form sulphonamides while 3° amines do not. directly bonded to benzene ring. • Reaction with nitrous acid: 1°-aliphatic amines form unstable diazonium salt which decomposes • Aniline is also the accepted IUPAC name of to give alcohol and N2 gas. 1°-aromatic amines form benzenamine and derivatives of aniline were named benzene diazonium chloride, only at 0-5°C. This as per IUPAC norms. reaction is used to distinguish between 1°, 2° and 3°-amines. • Aromatic amines can undergo ring substitution reactions like Bromination, Nitration and Sulphonation under suitable conditions. Preparation: Amines are usually prepared by reduction Arenediazonium salts are usually obtained by of nitro compounds, nitriles and amides, ammonolysis diazotisation reaction and they react with variety of of alkyl halides, By Gabriel Phthalimide synthesis and compounds to form aryl halides, cyanides, phenols and from amides (By Hoffmann’s degradation method). arenes, which are otherwise difficult to obtain. 336 Chemistry-12

Chapter trend—Based on past Years’ CBSe exams ➣ It has been observed from this chapter that the weightage of topics ‘Physical and Chemical Properties of Amines’ and ‘Diazonium Salts’ are maximum. Hence, these are most important topics. ➣ From this chapter , generally 1, 2 and 3 marks questions were asked from the topic ‘Physical and Chemical Properties of Amines’. Questions For practice Very Short Answer Type Questions (1 Mark) 1. Give reason for the following: Although –NH2 is o/p directing group yet aniline on nitration gives sufficient amount of m-nitroaniline. [Delhi 2017] 2. Arrange the following in the increasing order of their basic strength: CH3NH2, (CH3)2NH, C6H5NH2, C6H5CH2NH2 [AI 2017] 3. Write one chemical test to distinguish between aniline and ethylamines. [AI 2017] 4. How will you convert the following: Aniline into N-phenyl ethanamide (Write the chemical equations involved.) [Delhi 2014] Short Answer Type Questions-I (2 Marks) 5. Give reasons for the following: (i) Acetylation of aniline reduces its activation effect. (ii) CH3NH2 is more basic than C6H5NH2. [Delhi 2017] 6. Write the structure of the main products when benzene diazonium chloride reacts with the following reagents: (i) CuCN (ii) CH3CH2OH [AI 2017] 7. Write the chemical equations involved when aniline is treated with the following reagents: (i) Br2 water (ii) HCl [Delhi 2015] Short Answer Type Questions-II (3 Marks) 8. Write the structures of compounds A, B and C in the following reactions: (i) CH3 COOH NH3/∆ A Br2/KOH(aq) B CHCl3 + alc. KOH C [AI 2017] (5 Marks) (ii) CH3 N+2 BF4– NaNO2/Cu A Fe/HCl B CH3COCl/Pyridine C Long Answer Type Questions 9. Write the structure of A, B, C, D and E in the following reactions: [AI 2017] C6H5NO2 Sn/HCl A (CH3 CO)2 O B HNO3 + H2 SO4 C OH– or H+ D Pyridine 288 K Ω Ø H2SO4 E 10. An aromatic compound ‘A’ of molecular formula C7H7ON undergoes a series of reactions given below. Write the structures of A, B, C, D and E in the following reactions: (C7H7ON) Br2 + KOH C6H5NH2 NaNO2 + HCl B CH3 CH2 OH C CHCl3 + NaOH D KI E [Delhi 2015] ‘A’ 273 K 11. Write the structures of A, B, C, D and E in the following reaction: CH3 COCl Br2 /CH3 COOH B H+ C [Delhi 2015] Pyridine . .C6H5NH2 A CH3 Cl + KOH HNO3 + H2 SO4 'D' 'E' Amines 337

AssIgnment Time: 45 Minutes M.M.: 25 (1 Mark) Multiple Choice Questions (MCQs) 1. The correct IUPAC name for CH2 CHCH2NHCH3 is (i) Allylmethylamine (ii) 2-amino-4-pentene (iii) 4-aminopent-1-ene (iv) N-methylprop-2-en-1-amine 2. Which of the following is the weakest Brönsted base? NH 2 NH 2 (i) (ii) NH (iii) (iv) CH3NH2 3. Which of the following compound will not undergo azo coupling reaction with benzene diazonium chloride. (i) Aniline (ii) Phenol (iii) Anisole (iv) Nitrobenzene Assertion-Reason Type Questions (1 Mark) Note: In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (i) Assertion and reason both are correct statements and the reason is correct explanation of assertion. (ii) Assertion and reason both are correct statements but reason is not correct explanation of assertion. (iii) Assertion is correct statement but reason is wrong statement. (iv) Assertion is wrong statement but reason is correct statement. 4. Assertion: Only a small amount of HCl is required in the reduction of nitro compounds with iron scrap and HCl in the presence of steam. Reason: FeCl2 formed gets hydrolysed to release HCl during the reaction. 5. Assertion: Acylation of amines gives a monosubstituted product whereas alkylation of amines gives polysubstituted product. Reason: Acyl group sterically hinders the approach of further acyl groups. (1 Mark) Very Short Answer Type Questions 6. Why aliphatic amines are soluble in water while aromatic amines are insoluble? 7. What happens when benzenediazonium chloride reacts with aniline? Short Answer Type Questions-I (2 Marks) 8. Write giving chemical equations about the following: (i) Hoffmann’s bromamide reaction (ii) Carbylamine reaction (iii) Gabriel Phthalimide reaction (iv) Sandmeyer reaction 9. Carry out the following conversions: (i) Benzene into m-dichlorobenzene (ii) Benzene into p-nitrobenzaldehyde Short Answer Type Questions-II (3 Marks) 10. Account for the following: (i) Boiling point of ethyl alcohol is more than that of ethylamine. (ii) Tertiary amines does not undergo acylation reaction. (iii) Before nitration, aniline is converted into acetanilide. 11. How would you distinguish between the following: (i) Ethylamine and Dimethylamine, (ii) Ethylamine and Acetamide, (iii) Aniline and Benzylamine. 12. How would you distinguish between primary, secondary and tertiary amines by Hinsberg’s test? Long Answer Type Question (5 Marks) 13. atArlnecoaohtrmogleaicnntKicwOaiHrtohtmodaifltoirHcmCcoalmnitpogobiunvnoedxsia‘oAuw’swastimtehrelsmlionlouglebcclouemlcapormofuopnromdun‘uCdl’a.‘‘BCA’6’.Hr‘Ae7aN’ catilsssowspriteaharcibntesgnlwzyeistnhoeluCsbuHllepChiln3oniwnylaptcreherlsoe‘rAnid’c.eeOtonof form alkali soluble compound ‘D’. ‘A’ reacts with NaNO2 and conc HCl at 0–5°C to form ‘E’ which on reaction with phenol forms an orange red dye ‘F’. Elucidate the structure of organic compounds ‘A’ to ‘F’. Answers 1. (iv) 2. (i) 3. (iv) 4. (ii) 5. (iii) 338 Chemistry-12

10 Biomolecules Topics covered 10.2 Proteins 10.4 Nucleic Acids 10.1 Carbohydrates 10.3 Vitamins C hapter map BIOMOLECULES Carbohydrates Proteins Vitamins Nucleic Acids Classification: Amino Acids Classification Nucleoside Aldoses, Ketoses of Vitamins and Nucleotide Peptide bond Monosaccharides Sources and DNA and RNA (Glucose, Fructose) Polypeptides Deficiency Diseases Structure elucidation Structure of Proteins of glucose Primary, Secondary, Open Chain & Ring Tertiary and Structure of Glucose Quaternery Structure and Fructose Denaturation Preparation and of Proteins Reactions of Enzymes Glucose, Fructose Hormones D,L-Configurations: Mutarotation Haworth projections of Glucose, Fructose Oligosaccharides (Sucrose, Maltose, Lactose) Haworth Projections Polysaccharides (Starch, Importance of Carbohydrates Cellulose, Glycogen) 339

Topic 1. Carbohydrates • Oligosaccharides: These are the carbohydrates which gives 2 to 10 monosaccharide units on Definition hydrolysis. Example: Sucrose, Lactose, Maltose, Optically active polyhydroxy aldehydes or polyhydroxy Raffinose. ketones or the compounds which produce such units on hydrolysis, e.g. glucose, fructose, sucrose, maltose, etc. • Polysaccharides: These are the carbohydrates which on hydrolysis gives large number of Classification monosaccharide units. Example: Cellulose, • Monosaccharides: They are the simplest Starch, Glycogen. carbohydrates and cannot be hydrolysed into simpler compounds. Example: Glucose, Mannose, Galactose, Ribose, Deoxyribose. Monosaccharides • Preparation of Glucose: ã From sucrose: C12 H22 O11 + H2 O H+ C6 H12 O6 + C6 H12 O6 (Hydrolysis) Glu cos e Fructose Sucrose ã From starch: (C6 H10 O5) n + nH2 O H+ nC6 H12 O6 Starch or Cellulose 393 K; 2–3 atm Glu cos e (Hydrolysis) • Structure Elucidation of Glucose: CHO (CHOH)4 HI H3C—CH2—CH2—CH2—CH2—CH3 shows the presence of 6 carbons atoms linked in a CH2OH n-Hexane linear chain. CHO OH (CHOH)4 HCN H—C—CN shows the presence of carbonyl ( C = O) CH2OH (CHOH)4 group (open chain structure). CH2OH Glucose cyanohydrin CHO (CHOH)4 NH2OH CH = N – OH + H2O shows the presence of carbonyl ( C = O) group (open chain structure). CH2OH (CHOH)4 shows that the carbonyl CH2OH group is an aldehydic (–CHO) group. Glucose monoxime CHO CHO (CHOH)4 Br2 water (CHOH)4 CH2OH CH2OH CHO Gluconic acid (CHOH)4 + 2Cu(OH)2 + NaOH COONa + Cu2O + 3H2O shows that Glucose is CH2OH a reducing sugar Red ppt. (open chain structure). (CHOH)4 CH2OH Sodium salt of Gluconic acid 340 Chemistry-12

CHO (CHOH)4 + 2 [Ag(NH3)2]+OH– COOH + 2Ag + 4NH3 + H2O shows that Glucose is CH2OH a reducing sugar Silver (open chain structure). (CHOH)4 mirror CHO CH2OH Gluconic acid COOH (CHOH)4 HNO3 (CHOH)4 This indicates the presence of one primary CH2OH alcoholic group (R–OH) and one aldehyde CHO COOH group (–CHO) in Glucose. Saccharic acid CHO (CHOH)4 5(CH3CO)2O (CHOCOCH3)4 + 5CH3COOH Shows the presence of 5 ZnCl2 –OH group. CH2OH CH2OCOCH3 Glucose pentaacetate • Unusual reactions of glucose: (Presence of ring structure) CHO CH3OH H C OCH3 H3CO C H shows the dry HCl H C OH H C OH H OH HO C H ⇒ presence HO H H C OH O + HO C H O of ring HC H C OH H OH structure H OH HC and 2 anomers. CH2OH CH2OH CH2OH D-Glucose Methyl α-D Glucoside Methyl β-D Glucoside ã Glucose does not give 2,4-DNP test Schiff’s base test and also does not give additon product with sodium bisulphite. ã Glucose pentaacetate does not react with hydroxyl amine. This shows that is absence of free —CHO group and hence there is presence of ring structure. • Cyclic Structure of Glucose: O H C OH H 1C H OH C H H 2C OH H C OH HO 3C H H C OH H 4C OH HO C H O H 5C O H HO C H O H C OH H C OH 6CH2OH HC HC D–(+)–Glucose CH2OH CH2OH α–D–(+)–Glucopyranose β–D–(+)–Glucopyranose Anomers: The optical isomers which differ in configuration only around yCr1a-antoosme are called anomers, e.g . a-D-glucop and b-D- glucopyranose (a-glucose and b-glucose) Epimers: The pair of optical isomers which differ in the configuration around C-atom other than C1 atom are called epimers, e.g. D-glucose and D-mannose are C2 epimers. biomoleCules 341

• Cyclic Structure of Fructose: HO 2C C1H2OH 1CH2OH 1 2C O HO 3C H O HO 3C H HOH2C 2C OH O H 4C OH HO 3C H H 5C O H H 4C OH H 4C OH 6CH2OH H 5C H 5C D–(–)–Fructose 6CH2OH 6CH2OH β–D–(–)–Fructofuranose α–D–(–)–Fructofuranose • Mutarotation: The spontaneous change in Œ Haworth projection of Fructose: specific rotation of an optically active compound is called mutarotation, e.g. a-D-glucose (m.p. = 1 2C OH 6 O 1CH2OH +1gl41u69c°o.C2s°)e.h(Wmash.pes.np=eec1iit5fhi0ce°rrCoo)tfhattahiseosnsep, efaocDrifm=ics+ri1os1tad1ti°siosanonl,dv(eabd)-DDin=- water and is allowed to stand, the specific rotation HOH2 C HOH2 C 2 of the solution changes slowly and reaches to a HO 3C H O5 OH constant value of +52.5°. H OH H 4C OH H 3 H 5C 4 6CH2OH OH H a–D–(–)–Fructofuranose a–D–(–)–Fructofuranose D–D–glucose Open chain form E–D–glucose HO 2C 1CH2OH 6 O DD = 111° DD = + 52.5° DD = + 19° OH HOH2C • Haworth Projection: Representation for ring HO 3C H O 2 structure: 5 H 4C OH C1H2OH ã 6 membered ring is called pyranose ring. H OH H 5C H ã 5 membered ring is called furanose ring. 3 4 Œ How to write the projection: Groups 6CH2OH OH H projected to the right in the Fischer b–D–(–)–Fructofuranose b–D–(–)–Fructofuranose projection are written below the plane Œ Glycosidic Linkage: The oxide linkage formed by the loss of a water molecule and those to the left are written above the when two monosaccharide units are joined plane. together through oxygen atom is called Œ Haworth projection of Glucose: glycosidic linkage. H 1C OH 6CH2OH H 2C OH O Disaccharides HO 3C H 5 O H • Sucrose (Invert Sugar): Sucrose is dextrorotatory HH but on hydrolysis it gives dextrorotatory and H 4C OH 41 laevorotatory forms and the mixture on the whole H 5C HO OH H OH is laevorotatory which is named as invert sugar. 3 2 6CH2OH H OH C12 H22 O11 + H2 O H+ C6 H12 O6 + C6 H12 O6 D–glu cos e a–D–(+)–Glucopyranose a–D–(+)–Glucopyranose Sucrose [a]D = + 52.5° D–fructose [a]D = + 66.5° [a]D = –92.4° HO 1C H 6 Sucrose is a non-reducing sugar because the H 2C OH CH2OH O two monosaccharide units are held together by HO 3C H O H 5 OH a glycosidic linkage between C1 of a-glucose and C2 of b-fructose. Since the reducing groups of 4 H 1 both glucose and fructose units are involved in H C OH HO OH H H glycosidic bond formation, therefore sucrose is a 4 32 H 5C 6CH2OH H OH non-reducing sugar. This is due to the absence of b–D–(+)–Glucopyranose b–D–(+)–Glucopyranose free aldehyde group. 342 Chemistry-12

ã Haworth Projection of Maltose: H1 OH O 2 1CH2OH 6CH2OH H 6CH2OH H H2 O HO 3H 5O 5O HO 3 H 4 OH O H4 H H HH HH H5 5 OH 4 1 O 4 1 6CH2OH HO OH H Glycosidic OH H OH β–D–Fructose linkage 3 2 3 2 6CH2OH H OH H OH α–D–Glucose α–D–Glucose α–D–Glucose (C-1) (C-2) • Lactose (Milk Sugar): It is composed of b-D- • Haworth Projection of Sucrose: galactose and b-D-glucose units. The linkage is 6CH2OH there between C–1 of galactose and C–4 of glucose. 5O O Hence it is also a reducing sugar. HH H HOH2C1 H 4 1 O 2 5 HO 1 H 1H H 2 OH HO OH H Glycosidic H OH CH2OH O H 2 OH linkage HO 3 H O HO 3 H O 3 2 3 4 6 H4 HO 4 H H OH OH H H5 H5 6CH2OH α–D–Glucose β–D–Fructose 6CH2OH β–D–Galactose • M a l t o s e i s a n o t h e r s u g a r h a v i n g t w o β–D–Glucose monosaccharide units held together by glycosidic (C-1) (C-4) loifnaknaogteh,eirngwluhciochseCu1noift.oTneheglfureceosaeldisehliyndkiecdgtroouCp4 scaolnutbieonpraondducitedshaotwCs1 of second glucose unit in ã Haworth Projection of Lactose: reducing properties, so it is a reducing sugar. 6CH2OH 6CH2OH 5O H1 O HO 5 O OH H 2 OH HH HO 3 H O H 1 OH 4 H 1O 1 H 2 OH HH 4 H 4 OH HO 3 H O H OH 2 Glycosidic H H5 H4 OH H H5 3 linkage 6CH2OH OH 32 6CH2OH H α–D–Glucose H OH α–D–Glucose (C-1) β–D–Galactose β–D–Glucose (C-4) Polysaccharides • Starch: It is a polymer of a-glucose units and it consists of two components—Amylose and Amylopectin. Amylose Amylopectin It is the water soluble component. It is the water insoluble component. It is a long unbranched chain linear polymer. It is a branched chain polymer. It contains about 200 – 1000 a-D-(+)-glucose units It is a branched chain polymer of a-D-glucose units in held by a-glycosidic linkage involving C1–C4 glycosidic which the chain is formed by C1–C4 glycosidic linkage, linkage. whereas branching occurs by C1–C6 glycosidic linkage. It constitutes about 15–20% of starch. It constitutes about 80–85% of starch. CH2OH 6 CH2OH O O CH2OH H H H HH 5 O HH H OH H H OH H 4 14 H 14 1 O O OH O O H OH 32 H OH H OH α-Link α-Link Amylose biomoleCules 343

CH2OH CH2OH HH O H H OH H OH H1 4 14 α-Link O OH HO H OH H OH O Branch at C6 CH2OH 6CH2 CH2OH O H HH 5 OH H OH H H 4 14 H 1 OH H 14 O O OH HO HO OH H OH H OH H OH α-Link α-Link Amylopectin ã Cellulose: It is a straight chain polysaccharide stored in animal body and its resemblance of b-D-glucose units in which C1 of one with the structure of amylopectin. It is highly glucose unit and C4 of the next glucose unit branched and is present in liver, muscles is linked together by glycosidic linkage. and brain. It is broken down to glucose by the enzymes to provide energy to the body. HOH2C Importance of Carbohydrates OO • They acts as a fuel to provide energy in the living HOH2C OH systems: OO OH C6H12O6 + 6O2 3 6CO2 + 6H2O + Energy • They are also used as storage molecules in plants HOH2C OH OO and animals OH • Cellulose can act as a structural material in the OH β-links cell wall of plant cells. O • They can also provide raw materials for many OH industries. Cellulose • D-Ribose and Deoxyribose sugar are present in RNA and DNA molecules, respectively. ã Glycogen: It is called animal starch as ExErCIsE 10.1 Multiple Choice Questions (MCQs) (1 Mark) 4. Which of the following pairs represents anomers? 1. Which of the following statements is not true (NCERT Exemplar) about glucose? (NCERT Exemplar) CHO CHO (i) It is an aldohexose H OH HO H (i) HO (ii) On heating with HI it forms n-hexane H HO H H OH H OH (iii) It is present in furanose form OH H OH H (iv) It does not give 2,4-DNP test CH OH CH2OH 2. a–D–(+)-glucose and b–D–(+)–glucose are 2 CHO (i) Anomers (ii) Epimers CHO (iii) Enantiomers (iv) Geometrical isomers H OH HO H (ii) HO 3. Glycolysis is HH OH H OH HO H (i) oxidation of glucose to pyruvate H OH HO H (ii) conversion of glucose to haem CH OH CH OH 2 2 (iii) oxidation of glucose to glutamate (iv) conversion of pyruvate to citrate 344 Chemistry-12

H OH HO H (i) Fructose (ii) Saccharic acid H OH HO HO (iii) Galactose (iv) Ribose (iii) HO H H OH O H 9. Which of the following is the C-2 epimer of H HO glucose? H OH H (i) Fructose (ii) Mannose CH OH CH OH (iii) Galactose (iv) Ribose 2 2 H OH HO H 10. Which of the following is the product formed when glucose reacts with H2/Ni? H OH H OH O (i) Hexane (ii) Hexanol (iv) HO O HO H OH (iii) Sorbitol (iv) Hexan-2-ol H H H H OH H 11. Gluconic acid and glucose when oxidized with nitric acid yields a dicarboxylic acid, saccharic CH OH CH OH acid. This indicates the presence of which 2 2 substance in glucose? 5. Which of the following reactions of glucose can be explained only by its cyclic structure? (i) Secondary – OH group (NCERT Exemplar) (ii) Primary – CHO group (i) Glucose forms pentaacetate. (iii) Primary – OH group (ii) Glucose reacts with hydroxylamine to form an oxime. (iv) Primary – COOH group (iii) Pentaacetate of glucose does not react with 12. How many enantiomeric forms does glyceraldehyde hydroxylamine. have? (iv) Glucose is oxidised by nitric acid to gluconic (i) 1 (ii) 2 acid. (iii) 3 (iv) 4 6. Three cyclic structures of monosaccharides are 13. The pentaacetate of glucose does not react with given below which of these are anomers? hydroxylamine. What does that indicate? (NCERT Exemplar) (i) Presence of free-CHO group (ii) Absence of free-COOH group HO H (iii) Absence of free-CHO group H OH HO H HO H (iv) Absence of free-OH group H OH H OHO HO HO Assertion-Reason Type Questions (1 Mark) O HO H HO H H OH Note: In the following questions, a statement of assertion followed by a statement of reason is given. H OH H OH HO H Choose the correct answer out of the following choices on the basis of the above passage. HH H CH2OH CH2OH CH2OH (i) Assertion and reason both are correct statements (I) (II) (III) and the reason is correct explanation of assertion. (i) I and II (ii) II and III (iii) I and III (ii) Assertion and reason both are correct statements but reason is not correct explanation of assertion. (iv) III is anomer of I and II (iii) Assertion is correct statement but reason is wrong 7. Optical rotations of some compounds along with statement. their structures are given below. Which of them (iv) Assertion is wrong statement but reason is correct have D configuration? (NCERT Exemplar) statement. CHO CHO CH2OH 14. Assertion: The process of change in the optical H OH H OH rotation of either form of sugar, glucose (or HO H CO fructose) in the aqueous solution due to the CH2OH H OH HO H equilibrium between two anomers is called H OH mutarotation. H OH H OH (+) rotation CH2OH CH2OH Reason : Either a-glucose or b-glucose when dissolved in water and allowed to stand then the (I) (+) rotation (–) rotation specific rotation of the solution reaches a constant value of +52.5°. (II) (III) (i) I, II, III (ii) II, III (iii) I, II (iv) III 15. Assertion: D(+)-Glucose is dextrorotatory in nature. 8. Which of the following monosaccharide gives the same osazone as given by glucose? Reason: ‘‘D” represents its dextrorotatory nature. biomoleCules 345

16. Assertion: Glucose has an open structure. 35. Write any two reactions of glucose which cannot be Reason: Glucose on reacting with HI gives explained by the open chain structure of glucose n-hexane. molecule. [Delhi 2012] 17. Assertion: Carbon atoms in glucose are linked together in a straight chain. 36. Glucose and sucrose are soluble in water but cyclohexane and benzene (simple six membered Reason: Glucose on complete reduction with HI ring compounds) are insoluble in water. Explain. gives n-hexane. [NCERT] 18. Assertion: Glucose does not give 2, 4-DNP test. Reason: Glucose reacts with hydroxylamine to 37. Classify the following as monosaccharides and form an oxime. disaccharides: Ribose, 2-deoxyribose, Maltose, Galactose, Very Short Answer Type Questions (1 Mark) Fructose and Lactose. [NCERT] 19. Name the polysaccharide which represents the 38. What are the products obtained on hydrolysis of polymeric structure of b-D-glucose units. (i) sucrose, (ii) lactose? [A.I. 2013] [NCERT] 20. What are monosaccharides? Short Answer Type Questions-II (3 Marks) [A.I. 2010, Delhi 2010] 39. How do you explain the absence of aldehydic group 21. What is invert sugar? in the pentaacetate of D-glucose? [NCERT] [Foreign 2012, 2010, Delhi 2010] 40. What is glycogen? How it is different from starch? 22. Write a chemical reaction which shows that all [Foreign 2012] [NCERT] the carbon atoms in glucose are linked together 41. What is the basic structural difference between in a straight chain. [A.I. 2012; Delhi 2012] starch and cellulose? [Foreign 2012][NCERT] OR 42. What happens when D-glucose is treated with the following reagents? [NCERT] How can you explain the presence of all the six carbon atoms in glucose in a straight chain? (i) HI, (ii) Bromine water, [NCERT Exemplar] (iii) HNO3 [A.I. 2008; Delhi 2008] [NCERT] 23. Give one example each of a reducing and a non- Long Answer Type Questions (5 Marks) reducing disaccharide. 43. What happens when? 24. How is oxygen replenished in our atmosphere? 25. What is meant by mutarotation? (i) Glucose is treated with HCN. 26. What is meant by pyranose structure of glucose? (ii) Glucose is treated with hydroxylamine. [Delhi 2012, 2011 A.I. (C)] [Delhi 2011, 12] (iii) Glucose is allowed to react with ammoniacal 27. What are reducing sugars? [A.I. 2010] [NCERT] silver nitrate solution. 28. What is meant by the term glycosidic linkage? (iv) Glucose is treated with acetic anhydride. Short Answer Type Questions-I (2 Marks) (v) Glucose is oxidised with nitric acid. 29. In the human system, starch can act as a nutrient [Delhi 2012] but cellulose does not. Why? 44. (i) Classify the following as monosaccharides, disaccharides and polysaccharides: 30. What are the products obtained on hydrolysis of sucrose? Why sucrose is not a reducing sugar? Ribose, Glycogen, Maltose, Deoxyribose, Lactose, Fructose, Glucose, Cane sugar, [A.I. 2013, 10, Delhi 2014] Starch, Cellulose. 31. What are polysaccharides? Give one example. (ii) G i v e o n e e x a m p l e f o r e a c h o f t h e [Foreign 2014] [Delhi 2010] following: (i) Reducing sugars and (ii) Non- reducing sugars. 32. How are carbohydrates stored in the liver of animals? (iii) Name the monomer unit of starch. 33. Mention two main functions of carbohydrates in (iv) Name the monomer unit of cellulose. plants. [A.I. 2008] [Delhi 2010(C)] 34. Name two components of starch. [Delhi 2013(C)] 1. (iii) 2. (ii) 3. (ii) Answers 4. (ii) 5. (iii) 6. (ii) 7. (ii) 8. (ii) 9. (ii) 19. Cellulose 10. (iii) 11. (iii) 12. (ii) 20. They are the monomer unit of carbohydrates, i.e. 13. (iii) 14. (ii) 15. (iii) 16. (ii) 17. (i) 18. (ii) polyhydroxy aldehydes (aldoses) or polyhydroxy ketones (ketoses), which cannot be broken further into lower sugars upon hydrolysis. They have the 346 Chemistry-12

general ftohramtuvlaa,riCensHf2rnoOmn, when n is a positive 29. Human body does not contain enzymes which can integer 3 to 9 representing reduce or hydrolyse cellulose into glucose. Thus, cellulose cannot act as a nutrient. the number of carbon atoms. The common monosaccharides are glucose, fructose ribose, etc. 30. C12 H22 O11 + H2 O H+ C6 H12 O6 + C6 H12 O6 21. Equimolar mixture of a-D-glucose and b-D- Sucrose D(+) glucose D(–) fructose fructose obtained by hydrolysis of sucrose. Sucrose is not a reducing sugar, since both glucose 22. Glucose on complete reduction with HI and red phosphorus at 383 K gives n-hexane (a straight (oCth1 e–rat)harnodugfrhutchtoesier (rCed2 –ucbi)nagrececnotnrnee, cit.ee.dstuoceraocshe chain hydrocarbon). This indicates that all the carbon atoms in glucose are linked in a straight does not contain any aldehydic or ketonic group. chain. 31. Polysaccharides are carbohydrates which contains CH2 OH (CHOH)4 CHO HI/Red P large number of monosaccharide units joined Glucose ' (Reduction) together. On hydrolysis they give a large number of monosaccharide units. For example, starch or CH3 (CH2) 4 CH3 cellulose. n - hexane 32. It is stored as glycogen. 23. Reducing sugar—Maltose or Lactose; 33. (i) It is used as structural material for cell walls Non-reducing sugar—Sucrose. of plant cells. 24. Oxygen is replenished in our atmosphere (ii) It is used as a reserve food material. through photosynthesis process in plants. During 34. Amylose and Amylopectin photosynthesis, CO2 gas from air and produced as a result of combustion of fuels, breathing of 35. (i) Glucose on heating with methyl alcohol in the animals and human beings is consumed a(wndhiOch2 presence of dry HCl gas forms two isomeric gas is generated along with carbohydrates methyl glucosides (a and b-forms). is being used by plants as a food material). (ii) Glucose raenacotxwimithe,hbyudtrogxlyulcaomseinpee(n—taNaHce2OtaHte) Sunlight to form xCO2 + yH2 O Chlorophyll Cx (H2 O) y + xO2 does not react with hydroxylamine. This Carbohydrate 25. The process which involves the change in the optical rotation of either form of sugar, glucose shows that free —CHO group is not present (or fructose) in the aqueous solution to that of the in glucose. equilibrium mixture is called mutarotation. (iii) Glucose does not give, 2, 4-DNP test and Schiff’s base test, despite having a —CHO 26. The six membered cyclic structure of glucose, group. (i.e. pryinragniossefosrtmruecdtubreetwofegelnucCo1seanindaCn5a)loisgycawlliethd the 36. Glucose and sucrose are soluble in water due to the presence of large number of –OH groups (five pyran, a heterocyclic compound. in case of glucose and eight in case of sucrose), as they can form H-bond with water molecules (being 27. Carbohydrates which have free aldehyde group polar nature). Benzene and cyclohexane (being and reduces Tollens’ reagent to metallic silver non-polar) does not contain –OH groups and hence do not form H-bond with water molecules. That or Fehling’s solution to red ppt. of aCrue2Oreodrucgiinvge is why they are insoluble in water. red ppt with Benedict’s solution sugars. Examples are D–(+)–glucose, D–(+)– mannose, D(+)–galactose, D–(+)–fructose, etc (all monosaccharides) and disaccharides such as 37. Monosaccharides: Ribose, 2-Deoxyribose, Galactose and Fructose. D–(+)–maltose, D–(+)–lactose, etc. Disaccharides: Maltose and Lactose. 28. The oxygen atom (ethereal) linkage through which two monosaccharide units are joined together 38. Sucrose on hydrolysis gives equimolar mixture resulted by the loss of a water molecule to form a of D-(+)-glucose and D-(–)-fructose but lactose molecule of disaccharide is called the glycosidic on hydrolysis gives one molecule each of b-D- linkage. ‘For example, sucrose (a disaccharide) is galactose and b-D-glucose as follows: bjo-ifnruecdtotosegeththroeur gthhrtohuegghlyCco1soifdiac-lgilnukcaogseeaasnsdhCow2 nof: (i) C12 H22 O11 + H2O H+ C6 H12 Sucrose or Invertase O6 6 O6 + C6 H12 CH2OH D–(+)–Glucose D–(–)–Fructose 5O 1O C12 H22 O11 + H2 O H+ H HOH2C C6 H12 HH H (ii) Lactose or Lactase 4 OH 1 O 2 5 O6 + C6 H12 O6 HO 3 H H HO C6 H2OH 4 2 Glycosidic 3 b–D–Glucose b–D–Galactose H OH linkage OH H 39. When glucose (a or b form) is treated with acetic α–D–Glucose β–D–Fructose anhydride it forms a penta–acetyl derivative Sucrose biomoleCules 347

which does not contain a free –OH group at C-1, unit is attached to C4 of the other glucose unit through b-glycosidic linkage. so it cannot get hydrolysed in aqueous solution to give the open-chain aldehydic form and hence 42. (i) CH2 OH (CHOH)4 CHO HI/Red P, ' D - Glucose (Reduction) glucose pentaacetate does not react with NH2OH (ii) CH2 OH (CHOH)4 CHO to form glucose oxime. CH3 (CH2) 4 CH3 n–Hexane Br2 –H2 O (Oxidation) CH2 OH (CHOH)4 COOH D–Gluconic acid (iii) CH2 OH (CHOH)4 CHO HNO3 (Oxidation) COOH— (CHOH)4 COOH D–Saccharic acid OH CHO H—C—CN 43. (i) (CHOH)4 + HCN (CHOH) 4 CHOCOCH3 CH2 OH CH2 OH H OCOCH3 CH2 OH (CHOH)4 CHO + H2 NOH CH3OCO H O NH2OH (ii) Glucose Hydroxyl No oxime amine formation H OCOCH3 CH2 OH (CHOH)4 CH == NOH Glucose oxime H CH2OCOCH3 (iii) C6 H12 O6 + 2 [Ag (NH3)2]+ OH– Glucose Penta-acetyl derivative COOH (having no free (CHOH)4 + 2Ag + 4NH3 + H2 O aldehyde group) Silver mirror CH2 OH 40. Glycogen is more highly branched chain polymer (than amylopectin) of a–D–glucose units and CHO its molecular weight is very high (as it contains ~105 glucose units). It acts as the reserve food (iv) (CHOH)4 + 5 (CH3 CO)2 O for animals and is also known as animal starch. It is present in liver, muscles and brain. When CH2 OH CHO the body needs glucose, enzymes break down the (CHOCOCH3)4 + 5CH3 COOH glycogen to form glucose. Glycogen is also found in yeast and fungi. CH2 OCOCH3 While starch is the main storage polysaccharide Glucose pentaacetate found in plants and it is a mixture of two (A mixture of a- and b-) units, namely (i) amylose (15–20%) and (ii) amylopectin (80–85%). It is water insoluble on (v) CH2 OH (CHOH)4 CHO + 3 [O] HNO3 the whole. Amylose is the water soluble fraction Glucose oxidation and it is a linear polymer of a–D–glucose units linked by C1 – C4 glycosidic linkage. HOOC (CHOH)4 COOH + H2 O 41. Starch consist of two components amylose and Saccharic acid amylopectin. Amylose is a linear polymer of a–D– wglhuicloesceeullnuiltossejoiisnaedlinbyeaCr1p–olCy4mgelyrcoofsbid–iDc–lignlukcaogsee, 44. (i) Monosaccharides: Ribose, Deoxyribose, auoCnmn4eioytgfsloltujshoceeio,nsoCeetd1huoebnfryiot-gngalelunycdgcoloCussec4idoousifncetihlutienntnhkiteraxoigsuteaggbtlhuteacatcow-hsgeeeleyduncntooCisti1t.dhIoinecf Glucose, Fructose. linkage. However, in cellulose, C1 of one glucose Disaccharides: Maltose, Lactose, Cane sugar Polysaccharides: Glycogen, Starch, Cellulose (ii) (a) Reducing sugar: Glucose, (b) Non-reducing sugar: Cane sugar (sucrose) (iii) a-D-Glucose (iv) b-D-Glucose 348 Chemistry-12

Topic 2. Proteins R—CH—COO– OH– R—CH—COOH – Amino Acids R—CH—COO These are the compounds which contains both carboxyl group, –COOH and amino group, –NH2 in NH2 NH2 + NH3 its molecules. + • Depending on the relative positions of –COOH and –NH2 groups they are classified as a, b, g, d H and so on. R—CH—COOH R CH COOH + NH3 NH2 where R – Any alkyl/aryl side chain. • Isoelectric Point: The pH at which the dipolar ion can exists as neutral due to equal +ve and – ve charges and does not migrate to either of the electrode, cathode or anode is called isoelectric point. Except glycine, all other a-amino acids are Proteins optically active. Proteins are the polymers of a-amino acids. a-Amino • Most naturally occurring amino acids have L– acids are connected to each other by peptide bond or Configuration, i.e. –NH2 group is on left hand side. peptide linkage in proteins. A polypeptide with more COOH COOH than hundred amino acid residues, having molecular H NH2 H2N H mass greater than 10,000 u is called a protein. RR • Peptide Linkage: Peptide linkage is an amide linkage formed by condensation reaction between (D) (L) • Classification of amino acids: –COOH group of one amino acid and –NH2 group of another amino acid. ã Essential Amino Acids: Those amino acids which cannot be synthesised in the H2N—CH—COOH + H2N—CH—COOH –H2O body and are necessarily required for proper functioning of the body and hence must be R1 R2 O obtained through regular diet, are known as essential amino acids. Example: Valine, H2N—CH— C—NH —CH—COOH Leucine. R1 R2 ã Non-essential Amino Acids: The amino acids, which can be synthesised in the body Peptide linkage and need not necessarily to be taken through regular diet, are known as non-essential • Primary Structure of Proteins: The structure amino acids. Example: Glycine, Alanine. representing the sequence of amino acids in a protein is said to be the primary structure of ã Zwitter ion form of Amino Acids: Amino that protein. acids behave like salts rather than simple amines or carboxylic acids. This behaviour is • Secondary Structure of Proteins: It refers to due to the presence of both acidic (carboxyl the shape in which the long polypeptide chain of a group) and basic (amino group) groups within protein can exist. Two different types of secondary the same molecule. structures are: In aqueous solution, the carboxyl group can ã a-Helix: lose a proton and amino group can accept Œ It exists when R-group (or hydrocarbon a proton, thus giving rise to a dipolar ion part) is large. known as zwitter ion. Œ Polypeptide chain twisted into right It is neutral and contains both positive and handed screw with the NH2 group of each negative charges in equivalent amount. amino acid residue is H–bonded to the C=O group of adjacent turn of the helix. In zwitter ionic form, amino acids show amphoteric behaviour as they can react both ã b-pleated sheet: with acids as well as bases: Œ It exists when R group (or hydrocarbon part) is small. Œ In this conformation all the peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by hydrogen bonds. biomoleCules 349