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Class 12 Notes 2016 : Chemistry-AMINES (R. N. Podar School CBSE, Santacruz West, Mumbai)

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ANSWER KEY Class: 12 Month: JULY Worksheet Number: 12/chem/13/AK Subject: CHEMISTRY Chapter 13: AMINES Name: ____________________________________________ 1. Section A Each question carries 1 mark : What happens when Chloroform is treated with aniline in presence of alc.KOH. Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. 2. Write a chemical reaction in which iodide ion displaces diazonium group from a diazonium salt. Iodine is not easily introduced into the benzene ring directly, but, when the diazonium salt solution is treated with potassium iodide, iodobenzene is formed. 3 Classify the following amines as primary, secondary or tertiary: (i) (ii) (iii) (C2H5)2CHNH2 (iv) (C2H5)2NH Primary (I )&(iii) secondary (iv) tertiary (ii) 4. Each question carries 2 marks : What is Hinsberg s reagent? Give the reactions of Hinsberg s reagent with primary and secondary amines. Benzenesulphonyl chloride (C6H5SO2Cl) is Hinsberg s reagent. It reacts with primary and secondary amines to form sulphonamides. The reaction of benzenesulphonyl chloride with primary amine yields Nethylbenzenesulphonyl amide. In the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed. 5. 1 Starting from benzene diazonium chloride, how will you prepare the following compounds: Iodo benzene, Benzene, p-Amino azobenzene, Phenol 1 ANSWER KEY Class: 12 Month: JULY Worksheet Number: 12/chem/13/AK Subject: CHEMISTRY Chapter 13: AMINES 6. i) Write short note on : Gabriel phthalimide synthesis 7 Each question carries 3 marks : Answer the following giving reaction conditions and a complete chemical equation in each case: a) How would you prepare ethyl amine from acetaldehyde? NaBH4 8 SOCl2 NH3 Acetaldehyde Ethanol Ethyl chloride C2H5NH2 b) How can you get benzonitrile from aniline? Convert Aniline to benzene diazonium chloride followed by CuCN/KCN. c) How is ethyl isonitrile obtained from ethyl amine? Carbylamine reaction. Describe the following giving a chemical equation: a) Hoffmann bromamide reaction Hoffmann bromamide is a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. The amine so formed contains one carbon less than that present in the amide. b) Coupling reaction Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form p-hydroxyazobenzene. This type of reaction is known as coupling reaction. This is an example of electrophilic 2 2 ANSWER KEY Class: 12 Month: JULY Worksheet Number: 12/chem/13/AK Subject: CHEMISTRY Chapter 13: AMINES substitution reaction. c) Sandmeyer reaction 9 10 11 The Cl , Br and CN nucleophiles can easily be introduced in the benzene ring in the presence of Cu(I) ion. This reaction is called Sandmeyer reaction. Section B Each question carries 1 mark : Give the IUPAC name of H2N-CH2-CH2-CH=CH2 . But-3-en-1-amine Rearrange the following in an increasing order of their basic strengths: C6H5NH2,C6H5N(CH3)2, (C6H5)2NH and CH3NH2. C6H5N(CH3)2 <C6H5NH2 <(C6H5)2NH < CH3NH2 Each question carries 2 marks : Explain the following : (a)Like ammonia, amines are good nucleophiles. All amines contain an active lone pair of electrons on the very electronegative nitrogen atom. It is these electrons which are attracted to positive parts of other molecules or ions.Hence good nucleophiles. (b)Aromatic amines are less basic than aliphatic amines. It is because in aniline or other arylamines, the -NH2 group is attached directly to the benzene ring. The unshared electron pair on nitrogen atom is in conjugation with the benzene ring and thus making it less available for protonation. 3 3 ANSWER KEY Class: 12 Month: JULY Worksheet Number: 12/chem/13/AK Subject: CHEMISTRY Chapter 13: AMINES Less stability of anilinium ion. Aniline has five resonating structure while anilinium ion obtained by accepting a proton has only two resonating structures. greater the number of resonating structures, greater is the stability. 12 Explain the following : (a)Boiling point of methylamine is lower than that of methanol. O is more electro negative than H. Stronger H bonds in methanol. (b) Primary amines have higher boiling points than tertiary amines. Due to hydrogen bonding in primary amines. No H bonding in tertiary amines. 13 How will you convert the following: (a)Aniline to phenol (b) p- Toluidine to 2-bromo-4-methylaniline (a) (b) 14 (a) NaNO2 + HCl Aniline H2O benzene diazonium chloride Phenol How will you convert the following: (a)An alkyl halide to a quaternary ammonium salt (b) Aniline to benzonitrile Ammonolysis RNH2 (b) 15 4 NaNO2 + HCl Aniline CuCN/KCN benzene diazonium chloride benzonitrile Each question carries 3 marks : Complete the following equations: (i) C2H5NH2 + C6H5SO2Cl C2H5NH SO2C6H5 + HCl (ii) C6H5 NH2 + CH3COCl C6H5 NH COCH3 4 ANSWER KEY Class: 12 Worksheet Number: 12/chem/13/AK Subject: 16 Month: JULY CHEMISTRY Chapter 13: AMINES (iii) C6H5NH2 + (CH3CO)2O Complete the following equations: 4 , H 2O RCONH 2 LiAlH (i) RCH2NH2 + CH3COOH (ii) C6H5NH2 + Br2(aq) 17 (iii) C6H5N2Cl + H3PO2 + H2O Benzene Complete the following chemical equations: (i) C6H5N2Cl + C6H5NH2 (ii) C6H5N2Cl + CH3CH2OH Benzene (iii) RNH2 + CHCl3 + KOH 18 Explain how ethylamine be prepared from each of the following: a)Ethyl bromide b) Nitroethane c) Acetonitrile NH3 a) CH3CH2Br C2H5NH2 Sn/HCl b)CH3CH2 NO2 C2H5NH2 H2/Ni c)CH3 CN 19 a b c 20 5 C2H5NH2 Explain how ethylamine be prepared from each of the following: a) Acetaldehyde b) Acetamide c) Propionamide Refer Q11(a) LiAlH4/H2O CH3CO NH2 C2H5NH2 CH3CH2 CO NH2 + Br2 + NaOH C2H5NH2 + NaBr + Na2CO3 + H2O Account for the following: 5 ANSWER KEY Class: 12 Worksheet Number: 12/chem/13/AK Subject: 21 Month: JULY CHEMISTRY Chapter 13: AMINES a) CH3NH2 is a stronger base than NH3. b) Silver chloride dissolves in aqueous methylamine. a) Due to presence of electron donating alkyl group. b) Due to formation of a complex. [Ag(CH3NH)2]Cl In the following cases rearrange the compounds as directed : (i) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2,(C2H5)2NH ,and CH3NH2 (ii) In decreasing order of basic strength : Aniline ,p-Nitro aniline and p-Toluidine (iii) In increasing order of pKb values: C2H5NH2 ,C6H5NHCH3, (C2H5)2NH and C6H5NH2 (i)C6H5N(CH3)2 is more basic than C6H5NH2 due to the presence of the +I effect of two CH3groups in C6H5N(CH3)2. Further, CH3NH2 contains one CH3 group while (C2H5)2NH contains two C2H5 groups. Thus, (C2H5)2 NH is more basic than C2H5NH2. Now, C6H5N(CH3)2 is less basic than CH3NH2 because of the R effect of C6H5 group. Hence, the increasing order of the basic strengths of the given compounds is as follows: C6H5NH2 < C6H5N(CH3)2 < CH3NH2 < (C2H5)2NH ii) In p-toluidine, the presence of electron-donating CH3 group increases the electron density on the N-atom. Thus, p-toluidine is more basic than aniline. On the other hand, the presence of electron-withdrawing NO2 group decreases the electron density over the N atom in p-nitroaniline. Thus, pnitroaniline is less basic than aniline. Hence, the increasing order of the basic strengths of the given compounds is as follows: p-Nitroaniline < Aniline < p-Toluidine 22 6 iii)In C2H5NH2, only one C2H5 group is present while in (C2H5)2NH, two C2H5 groups are present. Thus, the +I effect is more in (C2H5)2NH than in C2H5NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2. Also, both C6H5NHCH3 and C6H5NH2 are less basic than (C2H5)2NH and C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3 and C6H5NH2, the former will be more basic due to the +T effect of CH3 group. Hence, the order of increasing basicity of the given compounds is as follows: C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH We know that the higher the basic strength, the lower is the pKb values. C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH i. Give plausible reasons for each of the following (i)Why are amines less acidic than alcohols of comparable molecular masses 6 ANSWER KEY Class: 12 Month: JULY Worksheet Number: 12/chem/13/AK Subject: CHEMISTRY Chapter 13: AMINES (ii) Why do primary amines have higher boiling point than tertiary amines (iii) Why are aliphatic amines stronger bases than aromatic amines Due to R- effect electrons on the N- atoms are less available in case of aromatic amines. Therefore the electrons on the N-atom cannot be donated easily. Hence aliphatic amine are stronger bases than aromatic amines 23 Each question carries 5 marks : Give reasons for the following: (a) Aniline does not undergo Friedel-Crafts reaction. Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst (b)Methylamine in water reacts with ferric chloride to give a precipitate of ferric hydroxide. Due to the +I effect of CH3 group, methylamine is more basic than water. Therefore, in water, methylamine produces OH ions by accepting H+ ions from water. Ferric chloride (FeCl3) dissociates in water to form Fe3+ and Cl ions. Then, OH ion reacts with Fe3+ ion to form a precipitate of hydrated ferric oxide. (c) pKb value for aniline is more than that for methylamine. because aniline is a weaker base and due to electron withdrawing effect of benzene ring (d)The observed Kb order is Et2NH Et3N EtNH2 in aqueous solution. Due to the inductive effect, solvation effect and steric hinderance of the alkyl group 7 7 ANSWER KEY Class: 12 Worksheet Number: 12/chem/13/AK Subject: 24 Month: JULY CHEMISTRY Chapter 13: AMINES which decides the basic strength of alkyl amines in the aqueous state the above order is observed. (e)Aniline is a weaker base than phenol. Arrange the following compounds in an increasing order of basic strengths in their aqueous solutions : a) NH3,CH3NH2,(CH3)2NH,(CH3)3N (CH3)2NH > CH3NH2 > (CH3)3N > NH3 The inductive effect, solvation effect and steric hinderance of the alkyl group decides the basic strength of alkyl amines in the aqueous state. b) C2H5NH2, (C2H5)2NH, (C2H5)3N, CH3NH2 (gas phase) CH3NH2 < C2H5NH2 <(C2H5)2NH< (C2H5)3N In gas phase only inductive effect determines the basic character. More R groups in tertiary amines means more electron donating groups . So more electron donating power. c) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 (C2H5)2NH > C2H5NH2 > NH3 > C6H5NH2 Reasoning as above+ resonance for aniline. d)Aniline, p-Nitroaniline, and p-Toluidine. p-Nitroaniline < Aniline < p-Toluidine In case of substituted aniline, it is observed that electron releasing groups like OCH3, CH3 increase basic strength whereas electron withdrawing groups like NO2, SO3, COOH, X decrease it. 25 e)Aniline, phenol Phenol < Aniline Explain the following : a) Sulphanilic acid is soluble in dil.NaOH and in dil.HCl but insoluble in water. Sulphanilic acid is a zwitter ion so it is soluble in acids and bases. b) Aniline gets coloured on standing in air for a long time. Due to aerial oxidation. c) Carbon-nitrogen bond length in aromatic amines shorter than in aliphatic amines. Due to partial double bond character. This is due to resonance between benzene ring and nitrogen of amine group. d) Electrophilic substitution in case of aromatic amines takes place more readily than in benzene. This is because of the electron donating nature of the amine group which increases the electron density on the benzene ring and hence increasing the activity of the benzene ring towards electrophilic substitution. 8 8 ANSWER KEY Class: 12 Month: JULY Worksheet Number: 12/chem/13/AK Subject: CHEMISTRY Chapter 13: AMINES e) Tertiary amines do not undergo acylation reaction. Due to absence of acidic H. 26 State reasons for the following : (a) Ethyl amine is soluble in water whereas aniline is not soluble in water. (b) Primary amines have higher boiling points than tertiary amines. (c)Before nitration, aniline is converted into acetanilide (a) Like ethanol, ethylamine (CH3CH2NH2) is small, polar and easily forms hydrogen bonds with water, and is therefore miscible with water. Aniline, does exhibit hydrogen bonding with water, although that is offset by the larger ring structure that does not lend itself to water solubility H bonding Aniline is highly activated towards electrophilic aromatic substitution. Aniline is also a base .the anilium ion formed is a meta directing group. The nitrogen basicity makes direct nitration difficult. How will you convert the following: a) Benzene into aniline b)Benzene into N, N-dimethyl aniline c)Cl-(CH2)4-Cl into hexane-1, 6-diamine d)Aniline into 1, 3, 5-tribromobenzene e) 3-Methylaniline into 3-nitrotoluene f)Benzyl chloride into 2-phenylethanamine a) Convert benzene to nitrobenzene and reduce nitro benzene with Sn/HCl to get aniline. b) Convert Benzene into aniline and then react with excess methyl bromide. c) Substitute both Cl groups by CN using KCN. Reduce CN groups by H2/Ni d) Refer29(iii) e) Refer29(ii) f) Substitute Cl group by CN using KCN. Reduce CN group by H2/Ni (b) (c) 27 28 9 How will you convert the following : (i) 4- Nitrotoluene to 2- bromobenzoic acid (ii) 3- Methylaniline to 3- nitrotoluene (iii) Aniline to 1,3,5-tribromobenzene (iv)Benzene diazonium chloride to nitrobenzene (v)Aniline to benzene diazonium chloride 9 ANSWER KEY Class: 12 Month: JULY Worksheet Number: 12/chem/13/AK Subject: CHEMISTRY Chapter 13: AMINES i. ii. Convert 3- Methylaniline to its diazonium form using NaNO2+HCl at 273-275K , then convert diazonium group to nitro group.(ref iv) iii. Treat 2,4,6-tribromoaniline with NaNO2+HCl at 273-275K followed by H3PO2 iv. v. 29 a 10 Convert aniline to benzene diazonium chloride using NaNO2+HCl at 273-275K Give a chemical test to distinguish between (i)Ethyl amine and aniline. (ii)Aniline and N-methylaniline (iii)Aniline and Benzylamine (iv)Methylamine and Dimethylamine (v)CH3CH2NH2 and CH3NHCH3. Azo dye test Reagents used: Nitrous acid (prepared in situ by reacting NaNO2+HCl) followed by phenol at 273-275K Observations: Aniline form benzene diazonium chloride which gives a yellow orange dye with phenol Ethylamine no orange dye. 10 ANSWER KEY Class: 12 Month: JULY Worksheet Number: 12/chem/13/AK Subject: CHEMISTRY Chapter 13: AMINES b Carbylamine Test Reagents: Chloroform and ethanolic potassium hydroxide Observations: N-methylaniline gives no offensive gas. Aniline forms benzonitrile which is a foul smelling substance. c Azo dye test or Carbylamine Test Refer a, b part for details. Hinsberg reagent d Benzenesulphonyl chloride (C6H5SO2Cl) is Hinsberg s reagent. It reacts with primary and secondary amines to form sulphonamides. The reaction of benzenesulphonyl chloride with primary amine yields Nethylbenzenesulphonyl amide. In the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed. e Hinsberg reagent 30 How will you convert the following: (a)Methyl bromide to ethyl amine (b)Aniline to Chlorobenzene (c) Aniline to nitrobenzene (d) Ethanamine to N- ethylethanamide (e) Chloro ethane to propan-1- amine (a) (i)CH3Br with KCN gives CH3CN (ii)Reduce CH3CN with H2/Ni to get ethyl amine (b) (i)Convert aniline to benzene diazonium chloride using NaNO2+HCl (ii) benzene diazonium chloride is converted to chloro benzene using CuCl/HCl (c) (i)Convert aniline to benzene diazonium chloride using NaNO2+HCl (ii) (d) 11 11 ANSWER KEY Class: 12 31 Worksheet Number: 12/chem/13/AK Subject: (e) Month: JULY CHEMISTRY Chapter 13: AMINES Same steps as (a). Section C Each question carries 1 mark : Before reacting aniline with HNO3 for nitration, it is converted into acetanilide. Why is this done and how is nitroaniline obtained subsequently? Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. To prevent this amine group is protected by forming an acetanilide. 32 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution. Aniline reacts with methyl iodide to produce N, N-dimethylaniline. With excess methyl iodide, in the presence of Na2CO3 solution, N, N-dimethylaniline produces N, N, N trimethylanilinium carbonate. 33 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained. Amines also react with benzoyl chloride (C6H5COCl). This reaction is known as benzoylation. 12 12 ANSWER KEY Class: 12 Month: JULY Worksheet Number: 12/chem/13/AK Subject: CHEMISTRY Chapter 13: AMINES Reaction is similar to above. Replace methyl of methanamine with Ph group. 34 35 Each question carries 2 marks : Write chemical equations for the following reactions: (i) Reaction of ethanolic NH3 with C2H5Cl (ii) Ammonolysis of benzyl chloride and reaction of amine so formed with 2 moles of CH3Cl. Write structures and IUPAC names of (i) The amide which gives propanamine by Hoffman bromamide reaction (ii) The amine produced by the Hoffman degradation of benzamide (i) (ii) 36 37 38 39 13 Propanamine contains three carbons. Hence, the amide molecule must contain four carbon atoms. Structure and IUPAC name of the starting amide with four carbon atoms are given below: Butanamide Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is aromatic primary amine containing six carbon atoms. Aniline. Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate N2 gas on treatment with HNO2. Do it on your own. Each question carries 3 marks : An organic compound, A has M.F C2NH3. A on reduction with Na and alcohol gives B, C2H7N and on acid hydrolysis gives C, C2H4O2, B on treatment with HNO2 gives D, C2H6O, C can also be obtained from D by oxidation with K2Cr2O7/ H+. Write structures for A, B, C, D & write all the reactions. A= CH3CN B= CH3CH2NH2 C= CH3COOH D= CH3CH2OH Equations on your own. Refer text book. An organic compound A has M.F. C2NH5O2. A on reduction with tin and conc.HCl gives a primary amine B which on treatment with HNO2 gives ethyl alcohol. Further B when warmed with chloroform and KOH forms a compound C with offensive odour. C on reduction gives ethylmethylamine. What structure will you assign to A, B and C? A=CH3CONH2 B=CH3CH2NH2 C= CH3CH2NC An organic compound A, C2H7NO2 on heating looses water and gives compound B, C2H5NO. B when distilled with P2O5 gives C, C2H3N. C on acid hydrolysis gives acetic acid. Write structures for compounds A,B and C. A= CH3COONH4 B= CH3CONH2 13 ANSWER KEY Class: 12 41 42 43 14 Worksheet Number: 12/chem/13/AK Subject: 40 Month: JULY CHEMISTRY Chapter 13: AMINES C= CH3CN An aromatic compound A on treatment with aqueous ammonia and heating forms compound B which on heating with Br2 and KOH forms compound C of molecular formula C6H7N.Write the structure and IUPAC names of compounds A,B,C A= C6H5NH2 B= C6H5CONH2 C= C6H5 COOH Section D Ram & Shyam are good friends, from 2 different localities. Ram used to go to cake point which used a colour light as insect attractants, but Shyam used to go to a way side XYZ bakery . Shyam was affected with poor health and ulcer in the stomach and admitted in the hospital.While Ram stayed healthy.The Doctor diagnosed and found Shyam s poor health due to poor quality of the food . a) How do you distinguish primary and secondary amine? b) Which organic compound is used as insect attractants? c) What is the lesson learnt by Shyam from Ram? Creatinine is a break-down product of creatine phosphate in muscle, and is usually produced at a fairly constant rate by the body. It belong to amines. Its presence can be tested in our blood and urine. Does this test help us to be aware of our health? Explain. Write the value associated with it. The amount of creatinine tells us about proper functioning of kidneys. Also tells about ailments like hypertension. Values: Should take proper care of our health, should have knowledge about indicators which can tell us about diseases. A mother brought her 2 year old child to a clinic with a complaint that the child would not stop crying and was profusely vomiting. The doctor noticed the child shirt collar colour was faded. a) Why dyes are coloured? Give 2 reasons? b) What is coupling reaction? Give equation c) What are the values involved in this? 14

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