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NCERT Exemplar Problems Class 12 Chemistry

Chapter 13 Amines

Multiple Choice Questions

Single Correct Answer Type

Question 1. Which of the following is a 3° amine?

(a) 1 -Methylcyclohexylamine (c) tert-Butylamine

Solution: (b)

Question 2. The correct IUPAC name for CH2 = CHCH2NHCH3 is

(a) allylmethylamine (b) 2-amino-4-pentene

(c) 4-aminopent-l-ene . (d) N-methylprop-2-en-l-amine.

Solution:

Question 3. Amongst the following, the strongest base in aqueous medium is .

(a) CH3NH2 (b) NCCH2NH2 (C) (CH3)2NH (d) C6H5NHCH3

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Solution: (c) 2° amine is more basic than 1° amine, i.e., (CH3)2NH is more basic than

CH3NH2. Due to -I effect of CN group, NC-CH2NH2 is less basic than CH3NH2. Further

C6H5NHCH3 is less basic than both CH3NH2 and (CH3)2NH due to delocalization of lone pair

of electrons present on the nitrogen atom into benzene ring. Hence, the decreasing order of

amines is:

(CH3)2NH > CH3NH2 > C6H5NHCH3 > NC – CH2NH2

Question 4. Which of the following is the weakest Bronsted base?

Solution: (a) Due to delocalization of lone pair of electrons on the N-atom into the benzene

ring, C6H5NH2 is the weakest base.

Resonating Structure of Aniline

Question 5. Benzylamine may be alkylated as shown in the following equation:

C6H5CH2NH2 + R – X > C6H5CH2NHR

Which of the following alkyl halides is best suited for this reaction through SN1

mechanism? .

(a) CH3Br (b) C6H5Br (c) C6H5CH2Br (d) C2H5Br

Solution: (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.

Question 6. Which of the following reagents would not be a good choice for reducing

an aryl nitro compound to an amine?

(a) H2 (excess)/Pt (b) LiAlH4 in ether




(c) Fe and HCl (d) Sn and HCl

Solution: (b) Aryl nitro compound cannot be converted into amine using LiAlH4 in ether.

Question 7. In order to prepare a 10 amine‘from an alkyl halide with simultaneous

addition of one CH2 group in the carbon chain, the reagent used as source of nitrogen

is

(a) sodium amide, NaNH2

(b) sodium azide, NaN2

(c) potassium cyanide, KCN

(d) potassium phthalimide, C6H4(CO2)N K

Solution: (c) KCN is used to increase number of carbon atoms.

Question 8. The source of nitrogen in Gabriel synthesis of amines is .

(a) sodium azide, NaN3

(b) sodium nitrite, NaNO2

(c) potassium cyanide, KCN

(d) potassium phthalimide, C6H4(CO2)N’KT




Solution: (d) Potassium phthalimide is the source of nitrogen in Gabriel’s synthesis.

Question 9. Amongst the given set of reactants, the most appropriate for preparing 2°

amine is .

(a) 2° R – Br + NH3

(b) 2° R – Br + NaCN followed by H2/Pt

(c) 10 R – NH2 + RCHO followed by H2/Pt

(d) 1° R – Br + (2 mol) + potassium phthalimide followed by H3O+/heat

Solution: (c)

Question 10. The best reagent for converting 2-phenylpro-panamide into-2-phenyl-

propanamine is .

(a) excess H2

(b) Br2 in aqueous NaOH

(c) iodine in the presence of red phosphorus

(d) LiAlH4 in ether

Solution:

Question 11. The best reagent for converting-2-phenylpro-panamide into 1-

phenylethana- mine is .

(a) excess H2/Pt (b) NaOH /Br2

(c) NaBH4/methanol (d) LiAlH4/ether




Solution:

Question 12. Hoffmann bromamide degradation reaction is shown by .

(a) ArNH2 (b) ArCONH2 (c) ArNO2 (d) ArCH2NH2

Solution:

Question 13. The correct increasing order of basic strength for the following

compounds is

Solution:(d)




Electron withdrawing group decreases the basic strength while electron releasing groups

increases the basic strength of aniline.

Question 14. Methylamine reacts with HN03 to form .

(a) CH3-0-N = 0 (b) CH3-0-CH3

(c) CH3OH (d) CH3CHO

Solution:

Question 15. The gas evolved when methylamine reacts with nitrous acid is .

(a) NH3 (b) N2 (C) H2 (d) C2H6

Solution: (b) Chemical reaction takes place during reaction of methylamine with nitrous acid

is as follows

Question 16. In the nitration of benzene using a mixture of cone. H2SO4 and cone.

HNO3,

the species which initiates the reaction is .

(a) NO2 (b) NCf (c) NO2 (d) NH2

Solution: (c)




NO2 (Nitronium ion) electrophile initiates the process of nitration. It is obtained as:

Question 17. Reduction of aromatic nitro compounds using Fe and HCl gives .

(a) aromatic oxime (b) aromatic hydrocarbon

(c) aromatic primary amine (d) aromatic amide

Solution: (c)

Question 18. The most reactive amine towards dilute hydrochloric acid is

Solution: (b) The greater will be the strength of base, the greater will be its reactivity

towards dilute HCl. Hence, (CH3)2NH has the highest basic strength as it has the highest

reactivity.

Question 19. Acid anhydrides on reaction with primary amines give .

(a) amide (b) imide

(c) secondary amine (d) imine

Solution:




Question 20.

Solution:

Question 21. Best method for preparing primary amines from alkyl halides without

changing the number of carbon atoms in the chain is

(a) Hoffmann bromamide reaction (b) Gabriel phthalimide reaction (c) Sandmeyer

reaction (d) reaction with NH3

Solution: (b) Gabriel phthalimide synthesis is used to get primary amine from alkyl halide

without any change in number of carbon atoms.

Question 22. Which of the following compound will not undergo azo coupling reaction

with benzene diazonium chloride?

(a) Aniline (b) Phenol (c) Anisole (d) Nitrobenzene




Solution: (d) Diazonium cation is a weak electrophile and hence reacts with electron rich

compounds containing electron donating groups such as -OH, -NH2 and -OCH3 groups and

not with compounds containing electron withdrawing groups such as -NO2, etc.

Question 23. Which of the following compounds is the weakest Bronsted base?

Solution: (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.

Question 24. Among the following amines’ the strongest Bronsted base is

Solution: (d) Aniline is weaker base than NH3 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.

Question 25. The correct decreasing order of basic strength of the following species

is

H2O, NH3, OH , NH;

(a) NH3 > OH > NH3 > H2O (b) OH > NH2 > H2O > NH3

(c) NH3 > H2O > NH2 > OH (d) H2O > NH3 > OH > NH2

Solution: (a) NH2 > OH > NH3 > H2O. 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 and OH have negative charge due to which they are more basic than NH3

and H20.




Question 26. Which of the following should be most volatile?

Solution: (b) 1° and 2° amines have higher boiling points due to intermolecular H-bonding

(and hence less volatile than 3° amines and hydrocarbons of comparable molecular mass).

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?

(a) Reaction of nitrile with LiAlH4

(b) Reaction of amide with LiAlH4 followed by treatment with water.

(c) Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis

(d) Treatment of amide with bromine in aqueous solution of sodium hydroxide.

Solution: (d) Aliphatic and arylalkyl primary amines can be prepared by the reduction of the

corresponding nitriles with LiAlH4.

Heating alkyl halide with primary, secondary and tertiary amine can be prepared by

reduction of LiAlH4 followed by treatment with water.

Heating 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.

More than One Correct Answer Type




Question 28. Which of the following cannot be prepared by Sandmeyer’s reaction?

(a) Chlorobenzene (b) Bromobenzene

(c) Iodobenzene (d) Fluorobenzene

Solution: (c, d) Chloro and bromo arenes are easily prepared by Sandmeyer’s reaction.

Iodoarenes are prepared by simply warming the diazonium salt solution with aqueous KI

solution.

Fluoroarenes are prepared by Balz-Schiemann reaction.

All other reagents give aniline.

Question 29. Reduction of nitrobenzene by which of the following reagents give

aniline?

(a) Sn/HCl (b) Fe/HCl (c) H2-Pd (d) Sn/NH4OH

Solution: (a, b, c)

Question 30. Which of the following species are involved in the carbylamine test?

(a) R-NC (b) CHCl3 (c) COCl2 (d) NaN02 + HCl

Solution: (a, b) Carbylamine reaction: Amine on reaction with a mixture of CHCl3 and KOH

produces alkyl isocyanate.

R-NH2 + CHCl, + 3KOH -> RNC + 3KCl + 3H2O Only RNC and CHCl3 are involved in

carbylamine reaction.

Question 31. The reagents that can be used to convert benzene diazonium chloride to

benzene are

(a) SnCl2/HCl (b) CH3CH2OH

(c) H3PO2 (d) LiAlH4




Solution: (a, b)

Question 32. The Product of the following reaction is .

Solution:




Question 33. Arenium ion involved in the bromination of aniline is.

Solution: (a,b,c)

Arenium ion involved in the bromination of aniline are as follows

Question 34. Which of the following amines can be prepared by Gabriel synthesis?

(a) Isobutyl amine (b) 2-Phenylethylamine

(c) N-Methylbenzylamine (d) Aniline

Solution: (a, b)

Only primary aliphatic amines such as (a) (CH3)2CH-CH2NH2 and C6H5CH2NH2 (b) can be

prepared by Gabriel synthesis. 2° amines, i.e., C6H5CH2NHCH3 (C) and 1° amine,

C6H5NH2 (d), however, cannot be prepared.

Question 35. Which of the following reactions are correct?




Solution: CH3CH2NH2 + NH4C1

Question 36. Under which of the following reaction conditions, aniline gives p-nitro

derivative as the major product?

(a) Acetyl chloride/pyridine followed by reaction with cone. H2SO4 + cone.

(b) Acetic anhydride/pyridine followed by cone. H2SO4 + cone. HNO3

(c) Dil. HCl followed by reaction with cone. H2SO4 + cone. HNO3

(d) Reaction with cone. HNO3 + cone. H2S04

Solution: (a, b)

Aniline 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 + cone. H2SO4) produces p-nitroaniline preferentially as

shown below:

Question 37. Which of the following reactions belong to electrophilic aromatic

substitution?

(a) Bromination of acetanilide

(b) Coupling reaction of aryldiazonium salts

(c) Diazotisation of aniline

(d) Acylation of aniline

Solution: (a, b) Acylation is a nucleophilic substitution reaction in which H atom of -NH2 is

replaced by acyl group. Diazotisation is also a nucleophilic substitution reaction.

Short Answer Type Questions




Question 38. Which is the role of HNO3 in the nitrating mixture used for nitration of

benzene?

Solution: HNO3 acts as a base in the nitrating mixture (HNO3 + H2SO4) and provides the

electrophile.

Question 39. Why is NH2 group of aniline acetylated before carrying out nitration?

Solution: Direct nitration of aniline is not possible on account of oxidation of -NH2 group.

However, nitration can be carried after protecting the -NH2 group by acetylation to give

acetanilide which is then nitrated and finally hydrolysed to give o- and p-nitroanilines.

The acetyl group being electron withdrawing attracts the lone pair of electrons of the N-atom

towards carbonyl group.

As a result, the activating effect -NH2 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 —NHCOCH3 group is less than that of —NH2 group.

Question 40. What is the product when C6H5CH2NH2 reacts with HN02?

Solution:

Question 41. What is the best reagent to convert nitrile to primary amine?

Solution: Reduction of nitriles with sodium alcohol or LiAlH4 gives primary amine.




Question 42. Give the structure of ‘A’ in the following reaction.

Solution:

Question 43. What is Hinsberg reagent?

Solution: Hinsberg reagent is benzene sulphonyl chloride (C6H5SO2Cl).

Question 44. Why is benzene diazonium chloride not stored and is used immediately

after its preparation?

Solution: Benzene diazonium chloride is very unstable. Therefore, it cannot be stored and is

used immediately after its preparation.

Question 45. Why does acetylation of -NH2 group of aniline reduce its activating

effect?

Solution: Direct nitration of aniline is not possible on account of oxidation of -NH2 group.

However, nitration can be carried after protecting the -NH2 group by acetylation to give

acetanilide which is then nitrated and finally hydrolysed to give o- and p-nitroanilines.

The acetyl group being electron withdrawing attracts the lone pair of electrons of the N-atom

towards carbonyl group.

As a result, the activating effect -NH2 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 —NHCOCH3 group is less than that of —NH2 group.

Question 46. Explain why is MeNH2 stronger base than MeOH?

Solution: Nitrogen is less electronegative than oxygen, therefore, lone pair of electrons on

nitrogen is readily available for donation. Hence, MeNH2 is more basic than MeOH.

Question 47. What is the role of pyridine in the acylation reaction of amines?

Solution: Pyridine and other bases are used to remove the side product i.e., HCl formed

during reaction mixture.

Question 48. Under what reaction conditions (acidic/basic), the coupling reaction of

aryl diazonium chloride with aniline is carried out?

Solution: Coupling reaction of aryl diazonium chloride with aniline is carried out in mild

acidic condition (pH = 4-5).

Question 49. Predict the product of reaction of aniline with bromine in non-polar

solvent such as CS2.

Solution:

In non-polar solvent (such as CS2) the activating effect of -NH2 group is reduced (due to

resonance) and hence, mono substitution occurs only at o- and p-positions.

Question 50. Arrange the following compounds in increasing order of dipole moment:

CH3CH2CH3, CH3CH2NH2, CH3CH2OH.

Solution: CH3CH2CH3 < CH3CH2NH2 < CH3CH2OH

Since 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.




Question 51. What is the structure and IUPAC name of the compound, allyl amine?

Solution:

Question 52. Write down the IUPAC name of

Solution: N, N-dimethylbenzenamine.

Question 53. A compound Z with molecular formula C3H9N reacts with C6H5SO2Cl to

give a solid, insoluble in alkali, identify Z.

Solution: Compound ‘Z’ with molecular formula C3H9N is an aliphatic amine which on

treatment with C6H5SO2Cl 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 (C2H5NHCH3).

Question 54. A primary amine, RNH2 can be reacted with CH3-X to get secondary

amine, R NHCH3 but the only disadvantage is that 3° amine and quaternary

ammonium salts are also obtained as side products. Can you suggest a method

where RNH2 forms only 2° amine?




Solution:

Question 55. Complete the following reaction.

Solution:

Question 56. Why is aniline soluble in aqueous HCl?

Solution: Aniline forms anilinium chloride salt with aqueous HCl which is soluble in water.

Question 57. Suggest a route by which the following conversion can be

accomplished.




Solution:

Question 58. Identify A and B in the following reaction.

Solution:

Question 59. How will you carry out the following conversions?

(i) toluene –> p-toIuidine

(ii) p-toluidine diazonium chloride –> p-toluic acid




Solution:

Question 60. Write following conversions:

(i) nitrobenzene –> acetanilide

(ii) acetanilide –> p-nitroaniline

Solution:

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.

Solution: 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.

Question 62. How will you bring out the following conversion?

Solution:




Question 63. How will you carry out the following conversion?

Solution:

Question 64. How will you carry out the following conversion?




Solution:

Question 65.How will you carry out the following conversion?

Solution:




Matching Column Type

Question 66. Match the reactions given in Column 1 with the statements given in

Column II.




Solution: (i —» d), (ii —»c), (iii -» a), (iv-> b)

Question 67. Match the compounds given in Column I with the items given in Column

II.




Solution: (i -> b), (ii —» a), (iii —» d), (iv -> c)

Assertion and Reason Type Questions

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:

(a) Both Assertion and Reason are wrong.

(b) Both Assertion and Reason are correct but Reason is not the correct explanation

of Assertion.

(c) Assertion is correct but Reason is wrong.

(d) Both Assertion and Reason are correct and Reason is the correct explanation of

Assertion.

(e) Assertion is wrong but Reason is correct.

Question 68. Assertion (A): Acylation of amines gives a monosubstituted product,

whereas alkylation of amines gives polysubstituted product.




Reason (R): Acyl group sterically hinders the approach of further acyl groups.

Solution: (c) Amines on acetylation give monosubstituted product, while on alkylation gives

polysubstitution product as well.

Question 69. Assertion (A): Hofmann’s bromamide reaction is given by primary

amines. Reason (R): Primary amines are more basic than secondary amines.

Solution: (a) Hofmann’s bromamide reaction is given by amides, not by amines. Moreover,

primary amines are basic than secondary amines.

Question 70. Assertion (A): N-Ethylbenzene sulphonamide is soluble in alkali.

Reason (R): Hydrogen attached to nitrogen in sulphonamide is strongly acidic.

Solution:

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.

Solution: (b) N, N-diethyl benzene sulphonamide is insoluble in alkali because it has no

acidic hydrogen.

Sulphonyl group attached to nitrogen atom is electron withdrawing group.

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.

Reason (R): FeCl2 formed gets hydrolysed to release HCl during the reaction.

Solution: (d) Fe + 2HCl >FeCl2 + 2[H]

Nascent hydrogen reduces nitro compounds.

FeCl2 + H2O(g) > FeO + 2HCl

Question 73. Assertion (A): Aromatic 1° amines can be prepared by Gabriel

phthalimide synthesis.

Reason (R): Aryl halides undergo nucleophilic substitution with anion formed by




phthalimide.

Solution: (a) Aromatic amines are never obtained by Gabriel phthalimide synthesis.

Question 74. Assertion (A): Acetanilide is less basic than aniline.

Reason (R): Acetylation of aniline results in decrease of electron density on nitrogen.

Solution: (d) Acetanilide is less basic than aniline because electron density of nitrogen is

lowered by acetyl group.

Long Answer Type Questions

Question 75. A hydrocarbon ‘A’, (C4H8) on reaction with HC1 gives a compound

‘B'(C4H9Cl), which on reaction with 1 mol of NH3 gives compound ‘C’, (C4H11N). On

reacting with NaN02 and HC1 followed by treatment with water, compound ‘C’ yields

an optically active alcohol, ‘D’. Ozonolysis of ‘A’ gives 2 moles of acetaldehyde.

Identify compounds ‘A’ to ‘D’. Explain the reactions involved.

Solution:




Question 76. A colourless substance ‘A’ (C6H7N) is sparingly soluble in water and

gives a water soluble compound ‘B’ on treating with mineral acid. On reacting with

CHCl3 and alcoholic potash ‘A’ produces an obnoxious smell due to the formation of

compound ‘C’. Reaction of ‘A’ with benzenesulphonyl chloride gives compound ‘D’

which is soluble in alkali. With NaNO2 and HCl, ‘A’ forms compound ‘E’ which reacts

with phenol in alkaline medium to give an orange dye ‘F’ Identify compounds ‘A’ to ‘F’

Solution:




Question 77. Predict the reagent or the product in the following reaction sequence.

Solution:



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