{"id":37431,"date":"2022-02-23T11:00:34","date_gmt":"2022-02-23T05:30:34","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=37431"},"modified":"2022-02-23T11:25:02","modified_gmt":"2022-02-23T05:55:02","slug":"ncert-exemplar-problems-class-12-chemistry-amines","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/ncert-exemplar-problems-class-12-chemistry-amines\/","title":{"rendered":"NCERT Exemplar Class 12 Chemistry Chapter 13 Amines"},"content":{"rendered":"

NCERT Exemplar Class 12 Chemistry Chapter 13 Aminese are part of NCERT Exemplar Class 12 Chemistry<\/a>. Here we have given NCERT Exemplar Class 12 Chemistry Chapter 13 Amines. https:\/\/www.cbselabs.com\/ncert-exemplar-problems-class-12-chemistry-amines\/<\/p>\n

NCERT Exemplar Class 12 Chemistry Chapter 13 Amines<\/h2>\n

Multiple Choice Questions<\/strong><\/p>\n

Single Correct Answer Type<\/strong><\/p>\n

Question 1. Which of the following is a 3\u00b0 amine?<\/strong>
\n (a) 1 -Methylcyclohexylamine (c) tert-Butylamine<\/strong>
\n Solution:<\/strong> (b)
\n\"NCERT<\/p>\n

Question 2. The correct IUPAC name for CH2<\/sub> = CHCH2<\/sub>NHCH3<\/sub> is<\/strong>
\n (a) allylmethylamine (b) 2-amino-4-pentene<\/strong>
\n (c) 4-aminopent-l-ene . (d) N-methylprop-2-en-l-amine.<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 3. Amongst the following, the strongest base in aqueous medium is .<\/strong>
\n (a) CH3<\/sub>NH2<\/sub> (b) NCCH2<\/sub>NH2<\/sub> (C) (CH3<\/sub>)2<\/sub>NH (d) C6<\/sub>H5<\/sub>NHCH3<\/sub><\/strong>
\n Solution:<\/strong> (c) 2\u00b0 amine is more basic than 1\u00b0 amine, i.e., (CH3<\/sub>)2<\/sub>NH is more basic than CH3<\/sub>NH2<\/sub>. Due to -I effect of CN group, NC-CH2<\/sub>NH2<\/sub> is less basic than CH3<\/sub>NH2<\/sub>. Further C6<\/sub>H5<\/sub>NHCH3<\/sub> is less basic than both CH3<\/sub>NH2<\/sub> and (CH3<\/sub>)2<\/sub>NH due to delocalization of lone pair of electrons present on the nitrogen atom into benzene ring. Hence, the decreasing order of amines is:
\n(CH3<\/sub>)2<\/sub>NH > CH3<\/sub>NH2<\/sub> > C6<\/sub>H5<\/sub>NHCH3<\/sub> > NC – CH2<\/sub>NH2<\/sub><\/p>\n

Question 4. Which of the following is the weakest Bronsted base?<\/strong>
\n\"NCERT
\n Solution:<\/strong> (a) Due to delocalization of lone pair of electrons on the N-atom into the benzene ring, C6H5NH2 is the weakest base.
\nResonating Structure of Aniline
\n\"NCERT<\/p>\n

Question 5. Benzylamine may be alkylated as shown in the following equation: C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub> + R – X > C6<\/sub>H5<\/sub>CH2<\/sub>NHR<\/strong>
\n Which of the following alkyl halides is best suited for this reaction through SN1 mechanism? .<\/strong>
\n (a) CH3<\/sub>Br (b) C6<\/sub>H5<\/sub>Br (c) C6<\/sub>H5<\/sub>CH2<\/sub>Br (d) C2<\/sub>H5<\/sub>Br<\/strong>
\n Solution:<\/strong> (c) SN1 reaction occurs in two steps. In first step R – X bond is broken to produce a carbocation which is attacked by nucleophile. The greater the stability of carbocation, the greater will be the rate of reaction. Benzylic halides show high reactivity towards SN1 reaction.
\n\"NCERT<\/p>\n

Question 6. Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?<\/strong>
\n (a) H2<\/sub> (excess)\/Pt \u00a0 \u00a0(b) LiAlH4<\/sub> in ether<\/strong>
\n (c) Fe and HCl \u00a0 \u00a0 (d) Sn and HCl<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 7. In order to prepare a 10 amine\u2018from an alkyl halide with simultaneous addition of one CH2<\/sub> group in the carbon chain, the reagent used as source of nitrogen is<\/strong>
\n (a) sodium amide, NaNH2<\/sub><\/strong>
\n (b) sodium azide, NaN2<\/sub><\/strong>
\n (c) potassium cyanide, KCN<\/strong>
\n (d) potassium phthalimide, C6<\/sub>H4<\/sub>(CO2<\/sub>)N K<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 8. The source of nitrogen in Gabriel synthesis of amines is .<\/strong>
\n (a) sodium azide, NaN3<\/sub><\/strong>
\n (b) sodium nitrite, NaNO2<\/sub><\/strong>
\n (c) potassium cyanide, KCN<\/strong>
\n (d) potassium phthalimide, C6<\/sub>H4<\/sub>(CO2<\/sub>)N’KT<\/strong>
\n Solution:<\/strong> (d) Potassium phthalimide is the source of nitrogen in Gabriel\u2019s synthesis.
\n\"NCERT<\/p>\n

Question 9. Amongst the given set of reactants, the most appropriate for preparing 2\u00b0 amine is .<\/strong>
\n (a) 2\u00b0 R – Br + NH3<\/sub><\/strong>
\n (b) 2\u00b0 R – Br + NaCN followed by H2<\/sub>\/Pt<\/strong>
\n (c) 10 R – NH2<\/sub> + RCHO followed by H2<\/sub>\/Pt<\/strong>
\n (d) 1\u00b0 R – Br + (2 mol) + potassium phthalimide followed by H3<\/sub>O+\/heat<\/strong>
\n Solution:<\/strong> (c)
\n\"NCERT<\/p>\n

Question 10. The best reagent for converting 2-phenylpro-panamide into-2-phenyl- propanamine is .<\/strong>
\n (a) excess H2<\/sub><\/strong>
\n (b) Br2<\/sub> in aqueous NaOH<\/strong>
\n (c) iodine in the presence of red phosphorus<\/strong>
\n (d) LiAlH4<\/sub> in ether<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 11. The best reagent for converting-2-phenylpro-panamide into 1-phenylethana- mine is .<\/strong>
\n (a) excess H2<\/sub>\/Pt (b) NaOH \/Br2<\/sub><\/strong>
\n (c) NaBH4<\/sub>\/methanol (d) LiAlH4<\/sub>\/ether<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 12. Hoffmann bromamide degradation reaction is shown by .<\/strong>
\n (a) ArNH2<\/sub> (b) ArCONH2<\/sub> (c) ArNO2<\/sub> (d) ArCH2<\/sub>NH2<\/sub><\/strong>
\n Solution:<\/strong>
\n\"\"<\/p>\n

Question 13. The correct increasing order of basic strength for the following compounds is<\/strong>
\n\"NCERT
\n Solution:<\/strong>(d)
\n\"NCERT
\nElectron withdrawing group decreases the basic strength while electron releasing groups increases the basic strength of aniline.<\/p>\n

Question 14. Methylamine reacts with HN03 to form .<\/strong>
\n (a) CH3-0-N = 0 (b) CH3-0-CH3<\/strong>
\n (c) CH3OH (d) CH3CHO<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 15. The gas evolved when methylamine reacts with nitrous acid is .<\/strong>
\n (a) NH3<\/sub> (b) N2<\/sub> (C) H2<\/sub> (d) C2<\/sub>H6<\/sub><\/strong>
\n Solution:<\/strong> (b) Chemical reaction takes place during reaction of methylamine with nitrous acid is as follows
\n\"NCERT<\/p>\n

Question 16. In the nitration of benzene using a mixture of cone. H2<\/sub>SO4<\/sub> and cone. HNO3<\/sub>,<\/strong>
\n the species which initiates the reaction is .<\/strong>
\n (a) NO2<\/sub> (b) NCf (c) NO2<\/sub> (d) NH2<\/sub><\/strong>
\n Solution:<\/strong> (c)
\nNO2<\/sub> (Nitronium ion) electrophile initiates the process of nitration. It is obtained as:
\n\"NCERT<\/p>\n

Question 17. Reduction of aromatic nitro compounds using Fe and HCl gives .<\/strong>
\n (a) aromatic oxime (b) aromatic hydrocarbon<\/strong>
\n (c) aromatic primary amine (d) aromatic amide<\/strong>
\n Solution:<\/strong> (c)
\n\"NCERT<\/p>\n

Question 18. The most reactive amine towards dilute hydrochloric acid is<\/strong>
\n\"NCERT
\n Solution:<\/strong> (b) The greater will be the strength of base, the greater will be its reactivity towards dilute HCl. Hence, (CH3<\/sub>)2<\/sub>NH has the highest basic strength as it has the highest reactivity.<\/p>\n

Question 19. Acid anhydrides on reaction with primary amines give .<\/strong>
\n (a) amide (b) imide<\/strong>
\n (c) secondary amine (d) imine<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 20.<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 21. Best method for preparing primary amines from alkyl halides without<\/strong>
\n changing the number of carbon atoms in the chain is<\/strong>
\n (a) Hoffmann bromamide reaction (b) Gabriel phthalimide reaction (c) Sandmeyer reaction (d) reaction with NH3<\/sub><\/strong>
\n Solution:<\/strong> (b) Gabriel phthalimide synthesis is used to get primary amine from alkyl halide without any change in number of carbon atoms.
\n\"NCERT<\/p>\n

Question 22. Which of the following compound will not undergo azo coupling reaction with benzene diazonium chloride?<\/strong>
\n (a) Aniline (b) Phenol (c) Anisole (d) Nitrobenzene<\/strong>
\n Solution:<\/strong> (d) Diazonium cation is a weak electrophile and hence reacts with electron rich compounds containing electron donating groups such as -OH, -NH2<\/sub> and -OCH3<\/sub> groups and not with compounds containing electron withdrawing groups such as -NO2<\/sub>, etc.<\/p>\n

Question 23. Which of the following compounds is the weakest Bronsted base?<\/strong>
\n\"NCERT
\n Solution:<\/strong> (c) Amines (a, b) have stronger tendency to accept a proton and hence are stronger Bronsted bases than phenol (c) and alcohol (d). Since phenol is more acidic than alcohol, therefore, phenol (c) has the least tendency to accept a proton and hence it is the weakest Bronsted base.<\/p>\n

Question 24. Among the following amines’ the strongest Bronsted base is<\/strong>
\n\"NCERT
\n Solution:<\/strong> (d) Aniline is weaker base than NH3<\/sub> due to delocalization of lone pair of electrons on the N-atom into the benzene ring. Pyrrole (c) is not at all basic because the lone pair of electrons on N-atom is donated towards aromatic sextet formation. Therefore, pyrrolidine (d) has a strong tendency to accept a proton and is hence, the strongest base.<\/p>\n

Question 25. The correct decreasing order of basic strength of the following species is<\/strong>
\n H2<\/sub>O, NH3<\/sub>, OH , NH;<\/strong>
\n (a) NH3<\/sub> > OH > NH3<\/sub> > H2<\/sub>O (b) OH > NH2<\/sub> > H2<\/sub>O > NH3<\/sub><\/strong>
\n (c) NH3<\/sub> > H2<\/sub>O > NH2<\/sub> > OH (d) H2<\/sub>O > NH3<\/sub> > OH > NH2<\/sub><\/strong>
\n Solution:<\/strong> (a) NH2<\/sub> > OH > NH3<\/sub>\u00a0> H2<\/sub>O. Due to higher electronegativity of O than N atom, O – H bond is more polar than N – H bond. Hence, O – H is more acidic in nature than N – H bond. Now, NH2<\/sub> and OH have negative charge due to which they are more basic than NH3 and H20.<\/p>\n

Question 26. Which of the following should be most volatile?<\/strong>
\n\"NCERT
\n Solution:<\/strong> (b) 1\u00b0 and 2\u00b0 amines have higher boiling points due to intermolecular H-bonding (and hence less volatile than 3\u00b0 amines and hydrocarbons of comparable molecular mass).<\/p>\n

Question 27. Which of the following methods of preparation of amines will not give same number of carbon atoms in the chain of amines as in the reactant?<\/strong>
\n (a) Reaction of nitrile with LiAlH4<\/strong>
\n (b) Reaction of amide with LiAlH4 followed by treatment with water.<\/strong>
\n (c) Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis<\/strong>
\n (d) Treatment of amide with bromine in aqueous solution of sodium hydroxide.<\/strong>
\n Solution:<\/strong> (d) Aliphatic and arylalkyl primary amines can be prepared by the reduction of the corresponding nitriles with LiAlH4.
\n\"NCERT
\nHeating alkyl halide with primary, secondary and tertiary amine can be prepared by reduction of LiAlH4 followed by treatment with water.
\n\"NCERT
\nHeating alkyl halide with potassium salt of phthalimide followed by hydrolysis produces primary amine. This process is known as Gabriel phthalimide reaction. The number of carbon atoms in the chain of amines of product is same as reactant.
\n\"NCERT<\/p>\n

More than One Correct Answer Type<\/strong><\/p>\n

Question 28. Which of the following cannot be prepared by Sandmeyer\u2019s reaction?<\/strong>
\n (a) Chlorobenzene (b) Bromobenzene<\/strong>
\n (c) Iodobenzene (d) Fluorobenzene<\/strong>
\n Solution:<\/strong> (c, d) Chloro and bromo arenes are easily prepared by Sandmeyer\u2019s reaction. Iodoarenes are prepared by simply warming the diazonium salt solution with aqueous KI solution.
\nFluoroarenes are prepared by Balz-Schiemann reaction.
\nAll other reagents give aniline.<\/p>\n

Question 29. Reduction of nitrobenzene by which of the following reagents give aniline?<\/strong>
\n (a) Sn\/HCl (b) Fe\/HCl (c) H2<\/sub>-Pd (d) Sn\/NH4<\/sub>OH<\/strong>
\n Solution:<\/strong> (a, b, c)
\n\"NCERT<\/p>\n

Question 30. Which of the following species are involved in the carbylamine test?<\/strong>
\n (a) R-NC (b) CHCl3<\/sub> (c) COCl2<\/sub> (d) NaN02<\/sub> + HCl<\/strong>
\n Solution:<\/strong> (a, b) Carbylamine reaction: Amine on reaction with a mixture of CHCl3<\/sub> and KOH produces alkyl isocyanate.
\nR-NH2<\/sub> + CHCl, + 3KOH -> RNC + 3KCl + 3H2<\/sub>O Only RNC and CHCl3<\/sub> are involved in carbylamine reaction.<\/p>\n

Question 31. The reagents that can be used to convert benzene diazonium chloride to<\/strong>
\n benzene are<\/strong>
\n (a) SnCl2<\/sub>\/HCl (b) CH3<\/sub>CH2<\/sub>OH<\/strong>
\n (c) H3<\/sub>PO2<\/sub> (d) LiAlH4<\/sub><\/strong>
\n Solution:<\/strong> (a, b)
\n\"NCERT<\/p>\n

Question 32. The Product of the following reaction is .<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 33. Arenium ion involved in the bromination of aniline is.<\/strong>
\n\"NCERT
\n Solution:<\/strong> (a,b,c)
\nArenium ion involved in the bromination of aniline are as follows
\n\"NCERT
\n\"NCERT<\/p>\n

Question 34. Which of the following amines can be prepared by Gabriel synthesis?<\/strong>
\n (a) Isobutyl amine (b) 2-Phenylethylamine<\/strong>
\n (c) N-Methylbenzylamine (d) Aniline<\/strong>
\n Solution:<\/strong> (a, b)
\nOnly primary aliphatic amines such as (a) (CH3<\/sub>)2<\/sub>CH-CH2<\/sub>NH2<\/sub> and C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub> (b) can be prepared by Gabriel synthesis. 2\u00b0 amines, i.e., C6<\/sub>H5<\/sub>CH2<\/sub>NHCH3<\/sub>\u00a0(C) and 1\u00b0 amine, C6<\/sub>H5<\/sub>NH2<\/sub> (d), however, cannot be prepared.<\/p>\n

Question 35. Which of the following reactions are correct?<\/strong>
\n\"NCERT
\n Solution:<\/strong> CH3CH2NH2 + NH4C1
\n\"NCERT<\/p>\n

Question 36. Under which of the following reaction conditions, aniline gives p-nitro derivative as the major product?<\/strong>
\n (a) Acetyl chloride\/pyridine followed by reaction with cone. H2<\/sub>SO4<\/sub> + cone.<\/strong>
\n (b) Acetic anhydride\/pyridine followed by cone. H2<\/sub>SO4<\/sub> + cone. HNO3<\/sub><\/strong>
\n (c) Dil. HCl followed by reaction with cone. H2<\/sub>SO4<\/sub> + cone. HNO3<\/sub><\/strong>
\n (d) Reaction with cone. HNO3<\/sub> + cone. H2<\/sub>S04<\/sub><\/strong>
\n Solution:<\/strong> (a, b)
\nAniline or reaction with acetyl chloride or acetic anhydride in the presence of pyridine produces N-acetyl aniline which is a ortho, para directing group which on further reaction with nitrating mixture (cone. HN03<\/sub> + cone. H2SO4<\/sub>) produces p-nitroaniline preferentially as shown below:
\n\"NCERT<\/p>\n

Question 37. Which of the following reactions belong to electrophilic aromatic substitution?<\/strong>
\n (a) Bromination of acetanilide<\/strong>
\n (b) Coupling reaction of aryldiazonium salts<\/strong>
\n (c) Diazotisation of aniline<\/strong>
\n (d) Acylation of aniline<\/strong>
\n Solution:<\/strong> (a, b) Acylation is a nucleophilic substitution reaction in which H atom of -NH2<\/sub> is replaced by acyl group. Diazotisation is also a nucleophilic substitution reaction.<\/p>\n

Short Answer Type Questions<\/strong><\/p>\n

Question 38. Which is the role of HNO3<\/sub> in the nitrating mixture used for nitration of<\/strong>
\n benzene?<\/strong>
\n Solution:<\/strong> HNO3<\/sub> acts as a base in the nitrating mixture (HNO3<\/sub> + H2<\/sub>SO4<\/sub>) and provides the electrophile.
\n\"NCERT<\/p>\n

Question 39. Why is NH2<\/sub> group of aniline acetylated before carrying out nitration?<\/strong>
\n Solution:<\/strong> Direct nitration of aniline is not possible on account of oxidation of -NH2<\/sub> group. However, nitration can be carried after protecting the -NH2<\/sub> group by acetylation to give acetanilide which is then nitrated and finally hydrolysed to give o- and p-nitroanilines.
\nThe acetyl group being electron withdrawing attracts the lone pair of electrons of the N-atom towards carbonyl group.
\n\"NCERT
\nAs a result, the activating effect -NH2<\/sub> group is reduced i.e., the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of \u2014NHCOCH3<\/sub> group is less than that of \u2014NH2<\/sub> group.<\/p>\n

Question 40. What is the product when C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub> reacts with HN02?<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 41. What is the best reagent to convert nitrile to primary amine?<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 42. Give the structure of \u2018A\u2019 in the following reaction.<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 43. What is Hinsberg reagent?<\/strong>
\n Solution:<\/strong> Hinsberg reagent is benzene sulphonyl chloride (C6<\/sub>H5<\/sub>SO2<\/sub>Cl).<\/p>\n

Question 44. Why is benzene diazonium chloride not stored and is used immediately after its preparation?<\/strong>
\n Solution:<\/strong> Benzene diazonium chloride is very unstable. Therefore, it cannot be stored and is used immediately after its preparation.<\/p>\n

Question 45. Why does acetylation of -NH2<\/sub> group of aniline reduce its activating effect?<\/strong>
\n Solution:<\/strong> Direct nitration of aniline is not possible on account of oxidation of -NH2<\/sub> group. However, nitration can be carried after protecting the -NH2<\/sub> group by acetylation to give acetanilide which is then nitrated and finally hydrolysed to give o- and p-nitroanilines.
\nThe acetyl group being electron withdrawing attracts the lone pair of electrons of the N-atom towards carbonyl group.
\n\"NCERT
\nAs a result, the activating effect -NH2<\/sub> group is reduced i.e., the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of \u2014NHCOCH3<\/sub> group is less than that of \u2014NH2<\/sub> group.<\/p>\n

Question 46. Explain why is MeNH2<\/sub> stronger base than MeOH?<\/strong>
\n Solution:<\/strong> Nitrogen is less electronegative than oxygen, therefore, lone pair of electrons on nitrogen is readily available for donation. Hence, MeNH2<\/sub> is more basic than MeOH.<\/p>\n

Question 47. What is the role of pyridine in the acylation reaction of amines?<\/strong>
\n Solution:<\/strong> Pyridine and other bases are used to remove the side product i.e., HCl formed during reaction mixture.<\/p>\n

Question 48. Under what reaction conditions (acidic\/basic), the coupling reaction of aryl diazonium chloride with aniline is carried out?<\/strong>
\n Solution:<\/strong> Coupling reaction of aryl diazonium chloride with aniline is carried out in mild acidic condition (pH = 4-5).<\/p>\n

Question 49. Predict the product of reaction of aniline with bromine in non-polar solvent such as CS2<\/sub>.<\/strong>
\n Solution:<\/strong>
\n\"NCERT
\nIn non-polar solvent (such as CS2<\/sub>) the activating effect of -NH2<\/sub> group is reduced (due to resonance) and hence, mono substitution occurs only at o- and p-positions.<\/p>\n

Question 50. Arrange the following compounds in increasing order of dipole moment: CH3<\/sub>CH2<\/sub>CH3<\/sub>, CH3<\/sub>CH2<\/sub>NH2<\/sub>, CH3<\/sub>CH2<\/sub>OH.<\/strong>
\n Solution:<\/strong> CH3<\/sub>CH2<\/sub>CH3<\/sub> < CH3<\/sub>CH2<\/sub>NH2<\/sub> < CH3<\/sub>CH2<\/sub>OH
\nSince O is more electronegative than N, therefore, dipole moment of ethyl alcohol is higher than that of ethyl amine. Propane however, has the least dipole moment since it is almost a non-polar molecule.<\/p>\n

Question 51. What is the structure and IUPAC name of the compound, allyl amine?<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 52. Write down the IUPAC name of<\/strong>
\n\"NCERT
\n Solution:<\/strong> N, N-dimethylbenzenamine.<\/p>\n

Question 53. A compound Z with molecular formula C3<\/sub>H9<\/sub>N reacts with C6<\/sub>H5<\/sub>SO2<\/sub>Cl to give a solid, insoluble in alkali, identify Z.<\/strong>
\n Solution:<\/strong> Compound \u2018Z\u2019 with molecular formula C3<\/sub>H9<\/sub>N is an aliphatic amine which on treatment with C6<\/sub>H5<\/sub>SO2<\/sub>Cl gives a solid, insoluble in alkali. Therefore, the product does not have any replaceable hydrogen on the nitrogen atom. In other words, the amines (Z) must be secondary amine, i.e., Z is ethyl methylamine (C2<\/sub>H5<\/sub>NHCH3<\/sub>).
\n\"NCERT<\/p>\n

Question 54. A primary amine, RNH2<\/sub> can be reacted with CH3<\/sub>-X to get secondary amine, R NHCH3<\/sub> but the only disadvantage is that 3\u00b0 amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where RNH2<\/sub> forms only 2\u00b0 amine?<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 55. Complete the following reaction.<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 56. Why is aniline soluble in aqueous HCl?<\/strong>
\n Solution:<\/strong> Aniline forms anilinium chloride salt with aqueous HCl which is soluble in water.
\n\"NCERT<\/p>\n

Question 57. Suggest a route by which the following conversion can be accomplished.<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 58. Identify A and B in the following reaction.<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 59. How will you carry out the following conversions?<\/strong>
\n (i) toluene –> p-toIuidine<\/strong>
\n (ii) p-toluidine diazonium chloride –> p-toluic acid<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 60. Write following conversions:<\/strong>
\n (i) nitrobenzene –> acetanilide<\/strong>
\n(ii) acetanilide –> p-nitroaniline<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 61. A solution contains 1 g mol each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. To this 1 g mol of alkaline solution of phenol is added. Predict the major product. Explain your answer.<\/strong>
\n Solution:<\/strong> This reaction is an example of electrophilic aromatic substitution. In alkaline medium, phenol forms phenoxide ion which is more electron rich than phenol and hence more reactive for electrophilic attack. The electrophile in this reaction is arydiazonium cation. p-Nitrophenyldiazonium cation is a stronger electrophile than p-toluene diazonium cation. Therefore, it couples preferentially with phenol.
\n\"NCERT<\/p>\n

Question 62. How will you bring out the following conversion?<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 63. How will you carry out the following conversion?<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 64. How will you carry out the following conversion?<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 65.How will you carry out the following conversion?<\/strong>
\n\"NCERT
\n Solution:<\/strong>
\n\"NCERT
\n\"NCERT<\/p>\n

Matching Column Type<\/strong><\/p>\n

Question 66. Match the reactions given in Column 1 with the statements given in Column II.<\/strong>
\n\"NCERT
\n Solution:<\/strong> (i \u2014\u00bb d), (ii \u2014\u00bbc), (iii -\u00bb a), (iv-> b)
\n\"NCERT<\/p>\n

Question 67. Match the compounds given in Column I with the items given in Column II.<\/strong>
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\n Solution:<\/strong> (i -> b), (ii \u2014\u00bb a), (iii \u2014\u00bb d), (iv -> c)
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Assertion and Reason Type Questions<\/strong><\/p>\n

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:<\/strong>
\n (a) Both Assertion and Reason are wrong.<\/strong>
\n (b) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.<\/strong>
\n (c) Assertion is correct but Reason is wrong.<\/strong>
\n (d) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.<\/strong>
\n (e) Assertion is wrong but Reason is correct.<\/strong><\/p>\n

Question 68. Assertion (A): Acylation of amines gives a monosubstituted product, whereas alkylation of amines gives polysubstituted product.<\/strong>
\n Reason (R): Acyl group sterically hinders the approach of further acyl groups.<\/strong>
\n Solution:<\/strong> (c) Amines on acetylation give monosubstituted product, while on alkylation gives polysubstitution product as well.<\/p>\n

Question 69. Assertion (A): Hofmann\u2019s bromamide reaction is given by primary amines. Reason (R): Primary amines are more basic than secondary amines.<\/strong>
\n Solution:<\/strong> (a) Hofmann\u2019s bromamide reaction is given by amides, not by amines. Moreover, primary amines are basic than secondary amines.<\/p>\n

Question 70. Assertion (A): N-Ethylbenzene sulphonamide is soluble in alkali.<\/strong>
\n Reason (R): Hydrogen attached to nitrogen in sulphonamide is strongly acidic.<\/strong>
\n Solution:<\/strong>
\n\"NCERT<\/p>\n

Question 71. Assertion (A): N, N-diethylbenzene sulphonamide is insoluble in alkalf. Reason (R): Sulphonyl group attached to nitrogen atom is strong electron withdrawing group.<\/strong>
\n Solution:<\/strong> (b) N, N-diethyl benzene sulphonamide is insoluble in alkali because it has no acidic hydrogen.
\nSulphonyl group attached to nitrogen atom is electron withdrawing group.<\/p>\n

Question 72. Assertion (A): Only a small amount of HCl is required in the reduction of nitro compounds with iron scrap and HCl in the presence of steam.<\/strong>
\n Reason (R): FeCl2<\/sub> formed gets hydrolysed to release HCl during the reaction.<\/strong>
\n Solution:<\/strong> (d) Fe + 2HCl >FeCl2<\/sub> + 2[H]
\nNascent hydrogen reduces nitro compounds.
\nFeCl2<\/sub> + H2<\/sub>O(g) > FeO + 2HCl<\/p>\n

Question 73. Assertion (A): Aromatic 1\u00b0 amines can be prepared by Gabriel phthalimide synthesis.<\/strong>
\n Reason (R): Aryl halides undergo nucleophilic substitution with anion formed by phthalimide.<\/strong>
\n Solution:<\/strong> (a) Aromatic amines are never obtained by Gabriel phthalimide synthesis.<\/p>\n

Question 74. Assertion (A): Acetanilide is less basic than aniline.<\/strong>
\n Reason (R): Acetylation of aniline results in decrease of electron density on nitrogen.<\/strong>
\n Solution:<\/strong> (d) Acetanilide is less basic than aniline because electron density of nitrogen is lowered by acetyl group.<\/p>\n

Long Answer Type Questions<\/strong><\/p>\n

Question 75. A hydrocarbon \u2018A\u2019, (C4<\/sub>H8<\/sub>) on reaction with HC1 gives a compound \u2018B'(C4<\/sub>H9<\/sub>Cl), which on reaction with 1 mol of NH3<\/sub> gives compound \u2018C\u2019, (C4<\/sub>H11<\/sub>N). On reacting with NaN02 and HC1 followed by treatment with water, compound \u2018C\u2019 yields an optically active alcohol, \u2018D\u2019. Ozonolysis of \u2018A\u2019 gives 2 moles of acetaldehyde. Identify compounds \u2018A\u2019 to \u2018D\u2019. Explain the reactions involved.<\/strong>
\n Solution:<\/strong>
\n\"NCERT
\n\"NCERT<\/p>\n

Question 76. A colourless substance \u2018A\u2019 (C6<\/sub>H7<\/sub>N) is sparingly soluble in water and gives a water soluble compound \u2018B\u2019 on treating with mineral acid. On reacting with CHCl3<\/sub> and alcoholic potash \u2018A\u2019 produces an obnoxious smell due to the formation of compound \u2018C\u2019. Reaction of \u2018A\u2019 with benzenesulphonyl chloride gives compound \u2018D\u2019 which is soluble in alkali. With NaNO2<\/sub>\u00a0and HCl, \u2018A\u2019 forms compound \u2018E\u2019 which reacts with phenol in alkaline medium to give an orange dye \u2018F’ Identify compounds \u2018A\u2019 to \u2018F’<\/strong>
\n Solution:<\/strong>
\n\"NCERT
\n\"NCERT<\/p>\n

Question 77. Predict the reagent or the product in the following reaction sequence.<\/strong>
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\n Solution:<\/strong>
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NCERT Exemplar Class 12 Chemistry Solutions<\/h2>\n