NCERT Exemplar for Class 12 Chemistry Chapter 13

SimplyAcad has provided the NCERT Exemplar for Class 12 Chemistry below to help students learn about the different topics in a detailed manner. The exemplar will allow students to gain deep insights of all the sections and prepares you better for the upcoming examination. Chemistry requires attention to minor details and information, therefore, solving exemplars will be an effective way to increase your marks. The given exemplar contains MCQs of two different types, and Short Answer Type Questions, there are a total of 50 questions asked. Students can access the NCERT exemplar for class 12 chemistry, Chapter 13 Amines by scrolling below. Along with this, there are several NCERT exemplar for class 12 science of all the chapters provided on this platform.

Access the Solution of NCERT Exemplar Class 12 Chemistry Chapter 13 Amines

I. NCERT Exemplar Class 12 Chemistry Chapter 13 – Multiple Choice Questions (Type-I)

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

(i) 1-methylcyclohexylamine

(ii) Triethylamine

(iii) tert-butylamine

(iv) N-methyl aniline

Solution:

Option (ii) is the answer.

2. The correct IUPAC name for CH2==CHCH2 NHCH3 is

(i) Allylmethylamine

(ii) 2-amino-4-pentene

(iii) 4-aminopent-1-ene

(iv) N-methylprop-2-en-1-amine

Solution:

Option (iv) is the answer.

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

(i) CH3NH2

(ii) NCCH2NH2

(iii) (CH3)2 NH

(iv) C6H5NHCH3

Solution:

Option (iii) is the answer.

4. Which of the following is the weakest Brönsted base?

Solution:

Option (A) is the answer.

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?

(i) CH3Br

(ii) C6H5Br

(iii) C6H5CH2Br

(iv) C2H5 Br

Solution:

Option (iii) is the answer.

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

aryl nitro compound to an amine?

(i) H2 (excess)/Pt

(ii) LiAlH4 in ether

(iii) Fe and HCl

(iv) Sn and HCl

Solution:

Option (ii) is the answer.

7. To prepare a 1° amine from an alkyl halide with simultaneous addition

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

is ___________.

(i) Sodium amide, NaNH2

(ii) Sodium azide, NaN3

(iii) Potassium cyanide, KCN

(iv) Potassium phthalimide, C6H4(CO)2N–K+

Solution:

Option (iii) is the answer.

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

(i) Sodium azide, NaN3

(ii) Sodium nitrite, NaNO2

(iii) Potassium cyanide, KCN

(iv) Potassium phthalimide, C6H4(CO)2N–K+Solution:

Option (iv) is the answer.

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

amine is _____.

(i) 2° R—Br + NH3

(ii) 2° R—Br + NaCN followed by H2/Pt

(iii) 1° R—NH2 + RCHO followed by H2/Pt

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

Solution:

Option (iii) is the answer.

10. The best reagent for converting 2–phenylpropanamide into

2-phenylpropanolamine is _____.

(i) excess H2

(ii) Br2 in aqueous NaOH

(iii) iodine in the presence of red phosphorus

(iv) LiAlH4 in ether

Solution:

Option (iv) is the answer.

11. The best reagent for converting 2-phenylpropanamide into 1- phenylethanamine is ____.

(i) excess H2/Pt

(ii) NaOH/Br2

(iii) NaBH4/methanol

(iv) LiAlH4/ether

Solution:

Option (ii) is the answer.

12. Hoffmann Bromamide Degradation reaction is shown by __________.

(i) ArNH2

(ii) ArCONH2

(iii) ArNO2

(iv) ArCH2NH2Solution:

Option (ii) is the answer.

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

_________.

(i) II < III < I

(ii) III < I < II

(iii) III < II < I

(iv) II < I < III

Solution:

Option (iv) is the answer.

14. Methylamine reacts with HNO2

to form _________.

(i) CH3—O—N==O

(ii) CH3—O—CH3

(iii) CH3OH

(iv) CH3CHO

Solution:

Option (iii) is the answer.

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

(i) NH3

(ii) N2

(iii) H2

(iv) C2H6

Solution:

Option (ii) is the answer.

16. In the nitration of benzene using a mixture of conc. H2SO4

and conc. HNO3, the species which initiates the reaction is __________.

(i) NO2

(ii) NO+

(iii) NO2+

(iv) NO2–

Solution:

Option (iii) is the answer.

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

(i) aromatic oxime

(ii) aromatic hydrocarbon

(iii) aromatic primary amine

(iv) aromatic amide

Solution:

Option (iii) is the answer.

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

Solution:

Solution:

Option (ii) is the answer.

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

(i) amide

(ii) imide

(iii) secondary amine

(iv) imine

Solution:

Option (i) is the answer

20. The reaction Ar + N2Cl– → )Cu/HCl)– ArCl + N2 + CuCl is named as _________.

(i) Sandmeyer reaction

(ii) Gatterman reaction

(iii) Claisen reaction

(iv) Carbylamine reaction

Solution:

Option (ii) is the answer.

21. The best method for preparing primary amines from alkyl halides without

changing the number of carbon atoms in the chain is

(i) Hoffmann Bromamide reaction

(ii) Gabriel phthalimide synthesis

(iii) Sandmeyer reaction

(iv) Reaction with NH3

Solution:

Option (ii) is the answer.

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

with benzene diazonium chloride.

(i) Aniline

(ii) Phenol

(iii) Anisole

(iv) NitrobenzeneSolution:

Option (iv) is the answer.

23. Which of the following compounds is the weakest Brönsted base?

Solution:

Solution:

Option (iii) is the answer.

24. Among the following amines, the strongest Brönsted base is __________.

Solution;

Option (iv) is the answer.

25. The correct decreasing order of basic strength of the following species is _______. H2O, NH3, OH–, NH2–

(i) NH2– > OH – > NH3 > H2O

(ii) OH– > NH2– > H2O > NH3

(iii) NH3 > H2O > NH2– > OH–

(iv) H2O > NH3> OH– > NH2–

Solution:

Option (i) is the answer.

26. Which of the following should be most volatile?

(i) II

(ii) IV

(iii) I

(iv) III

Solution:

Option (ii) is the answer.

27. Which of the following methods of preparation of amines will not give the same number

of carbon atoms in the chain of amines as in the reactant?

(i) The reaction of nitrite with LiAlH4.

(ii) The reaction of the amide with LiAlH4

followed by treatment with water.

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

hydrolysis.

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

Solution:

Option (iv) is the answer.

II. NCERT Exemplar Class 12 Chemistry Chapter 13 – Multiple Choice Questions (Type-II)

Note: In the following questions, two or more options may be correct.

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

(i) Chlorobenzene

(ii) Bromobenzene

(iii) Iodobenzene

(iv) Fluorobenzene

Solution:

Option (iii) and (iv) are the answers.

29. Reduction of nitrobenzene by which of the following reagent gives aniline?

(i) Sn/HCl

(ii) Fe/HCl

(iii) H2-Pd

(iv) Sn/NH4OH

Solution;

Option (i), (ii) and (iii) are the answers.

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

(i) R—NC

(ii) CHCl3

(iii) COCl2

(iv) NaNO2 + HCl

Solution:

Option (i) and (ii) are the answers.

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

benzene are __________.

(i) SnCl2/HCl

(ii) CH3CH2OH

(iii) H3PO2

(iv) LiAlH4

Solution:

Option (ii) and (iii) are the answers.

32. The product of the following reaction is __________.

Solution:

Option (A) and (B) is the answer.

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

Solution:

Option (i), (ii) and (iii) are the answers.

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

(i) Isobutyl amine

(ii) 2-Phenylethylamine

(iii) N-methyl benzylamine

(iv) Aniline

Solution:

Option (i) and (ii) are the answers.

35. Which of the following reactions is correct?

Solution:

Option (i) and (iii) are the answers.

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

as the major product?

(i) Acetyl chloride/pyridine followed by reaction with conc. H2SO4 + conc. HNO3

(ii) Acetic anhydride/pyridine followed by conc. H2SO4 + conc. HNO3

(iii) Dil. HCl followed by reaction with conc. H2SO4 + conc. HNO3

(iv) Reaction with conc. HNO3 + conc.H2SO4

Solution:

Option (i) and (ii) are the answers.

37. Which of the following reactions belong to electrophilic aromatic substitution?

(i) Bromination of acetanilide

(ii) Coupling reaction of aryldiazonium salts

(iii) Diazotisation of aniline

(iv) Acylation of aniline

Solution:

Option (i) and (ii) are the answers.

III. NCERT Exemplar Class 12 Chemistry Chapter 13 – Short Answer Type

38. What is the role of HNO3 in the nitrating mixture used for nitration of benzene?

Solution:

The nitrating mixture is the mixture of a 1:1 solution of HNO3 and H2SO4 and is used for the nitration of organic compounds. It acts as a base and provides electrophile in the nitration process of benzene.

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

Solution:

To control the nitration reaction and tarry oxidation products and nitro derivatives products formation, the NH2 group of aniline is acetylated before carrying out nitration. Here, the major product is p-nitroaniline.

40. What is the product when C6H5CH2NH2 reacts with HNO2?

Solution:

C6H5CH2NH2 reacts with HNO2 to form unstable diazonium salt, which in turn gives alcohol.

C6H5CH2NH2 + HNO2 → C6H5CH2OH + N2 + H2O

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

Solution:

The best reagents for the conversion of nitrile to primary amine are LiAlH4 and Sodium/Alcohol. By reduction, the nitriles can be converted into a corresponding primary amine.

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

Solution:

The product formed in this chemical reaction is 3-Methylnitrobenzene.

43. What is Hinsberg reagent?

Solution:

Benzenesulphonyl chloride, or C6H5SOCl, is commonly known as Hinsberg’s reagent. Hinsberg’s reagent is be used to distinguish primary, secondary and tertiary amines.

44. Why is benzene diazonium chloride not stored and used immediately after its preparation?

Solution:

Benzene diazonium chloride is highly soluble in water at high temperature and is itself very stable at low temperature. It should be used immediately after its preparation as it is unstable.

45. Why does the acetylation of —NH2 group of aniline reduce its activating effect?

Solution:

The acetylation of —NH2 group of aniline reduces its activating effect because the lone pair of electrons on the nitrogen of acetanilide interacts with the oxygen atom due to resonance.

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

Solution:

MeNH2 is a stronger base than MeOH because of the lower electronegativity and the presence of the lone pair of electrons on the nitrogen atom in MeNH2.

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

Solution:

The activating effect of –NH2 group can be controlled by protecting the -NH2 group by acetylation with acetic anhydride in the presence of pyridine and then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine. Pyridine is a base that is used to get rid of HCl as a side product from the reaction.

48. Under what reaction conditions (acidic/basic), the coupling reaction of aryldiazonium chloride with aniline is carried out?

Solution:

This reaction is carried out in a mild basic medium. This is an electrophilic substitution reaction. Aryldiazonium chloride reacts with aniline to form a yellow dye of p-Aminoazobenzene.

49. Predict the product of the reaction of aniline with bromine in a non-polar solvent such as CS2.

Solution:

The products formed in the reaction of aniline with bromine in a non-polar solvent such as CS2 are 4-Bromoaniline and 2-Bromoaniline, where 4-Bromoaniline is the major product.

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

CH3CH2CH3, CH3CH2NH2, CH3CH2OH

Solution:

CH3CH2CH3 < CH3CH2NH2 < CH3CH2OH

The dipole moment of CH3CH2OH is greater than that of CH3CH2NH2. CH3CH2CH3 has the least dipole moment among the three given compounds because it is almost a non-polar molecule.

NCERT Exemplar For Class 12 Science

The NCERT exemplars are an effective study material for scoring higher marks in the examination paper. Students must practise these additional questions for their own benefits, as these are curated by the best subject-matter experts to boost both knowledge and confidence. Students can easily access the ncert exemplar for class 12 science by visiting our website SimplyAcad and solve all the questions listed to secure maximum marks.

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