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Important Questions for Class 12 Chemistry Chapter 13 - Amines.

Class 12 chemistry important questions with answers are provided here for chapter 13 Amines. These important questions are based on the CBSE board curriculum and correspond to the most recent Class 12 chemistry syllabus. By practising these Class 12 important questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 12 Annual examinations as well as other entrance exams such as NEET and JEE.

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Class 12 Amines Important Questions with Answers

Short Answer Type Questions

Q1. What is the role of HNO3 in the nitrating mixture used for the nitration of benzene?

Answer:

HNO3 acts as a base in the nitrating mixture and provides the electrophile NO2+ ion for the nitration of benzene.

Q2. Why is the NH2 group of aniline acetylated before carrying out nitration?

Answer:

NH2 group of aniline is acetylated first so that controlled nitration can occur at the para position. If the NH2 group of aniline is not acetylated, then a mixture of ortho, meta and para products will form.

Q3. What is the product when C6H5CH2NH2 reacts with HNO2?

Answer:

C6H5CH2OH is formed.

C6H5CH2NH2 reacts with HNO2 to form unstable diazonium salt, giving alcohol. Thus, when C6H5CH2NH2 reacts with HNO2, benzyl alcohol is formed along with N2 and H2O.

C6H5CH2NH2 + HNO2 → C6H5CH2OH + N2 + H2O

When C6H5CH2NH2 reacts with HNO2 at low temperature, stable Diazonium salt is formed, and no N2 gas is evolved.

Q4. What is the best reagent to convert nitrile to primary amine?

Answer:

LiAlH4 and Sodium/Alcohol are the best reagents for converting nitrile to a primary amine. The nitriles are transformed into a corresponding primary amine through reduction.

Q5. Give the structure of ‘A’ in the following reaction.

Answer:

m-nitrotoluene is formed.

Q6. What is Hinsberg reagent?

Answer:

Hinsberg reagent is an alternative name for benzene sulfonyl chloride (C6H5SO2Cl). It is used to detect and distinguish primary, secondary, and tertiary amines in a given sample.

Q7. Why is benzene diazonium chloride not stored and is used immediately after its preparation?

Answer:

Benzene diazonium chloride cannot be stored and is used immediately after its preparation because of its unstable nature. With a slight increase in temperature, it dissociates to give nitrogen gas.

Q8. Why does acetylation of -NH2 group of aniline reduce its activating effect?

Answer:

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

The activating effect of the –NH2 group is controlled by protecting the -NH2 group by acetylation with acetic anhydride and then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.

Q9. Explain why MeNH2 is a stronger base than MeOH?

Answer:

Nitrogen is less electronegative than oxygen; therefore lone pairs of electrons on nitrogen are readily available for donation. Hence, MeNH2 is more basic than MeOH.

Q10. What is the role of pyridine in the acylation reaction of amines?

Answer:

Pyridine acts as an acceptor for the acid by-product formed in the reaction. Thus, it removes the side product, i.e. HCl, from the reaction mixture.

Q11. Under what reaction conditions (acidic/basic) the coupling reaction of aryl diazonium chloride with aniline is carried out?

Answer:

Coupling reaction of aryl diazonium chloride with aniline is carried out in mildly acidic conditions, i.e. pH=4−5.

Q12. Predict the product of the reaction of aniline with bromine in a nonpolar solvent such as CS2.

Answer:

The products formed in the reaction of aniline with bromine in a nonpolar solvent such as CS2 are 4-Bromoaniline and 2-Bromoaniline, where 4-Bromoaniline is the major product. In a nonpolar solvent medium, the activating effect of the –NH2 group of aniline is reduced because of resonance, and thus, mono-substitution occurs only at ortho- and para-positions.

Q13. Arrange the following compounds in increasing order of dipole moment. CH3CH2CH3, CH3CH2NH2, CH3CH2OH.

Answer:

CH3CH2CH3 < CH3CH2NH2 < CH3CH2OH

Q14. What is the structure and IUPAC name of the compound, allylamine?

Answer:

The IUPAC name of allylamine is 3-Amino-1-propene.

Structure of allylamine:

Q15. A compound Z with molecular formula C3H9N reacts with C6H5SO2Cl to give a solid, insoluble in alkali. Identify Z.

Answer:

C3H9N reacts with C6H5SO2Cl or Hinsberg’s reagent to give a solid, insoluble in alkali, which means that C3H9N is a secondary amine. The product obtained in this reaction has no replaceable hydrogen attached to the nitrogen atom of the amine group.

Thus the structure of the given amine is:

The chemical reaction involved in this case is:

Q16. A primary amine, RNH2 can be reacted with CH3-X to get secondary amine, R-NHCH3 but the only disadvantage is that 3º amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where RNH2 forms only 2º amine?

Answer:

Carbylamine reaction is shown by 1° amine only, which replaces two hydrogen atoms attached to the nitrogen atom of the NH2 group with one carbon atom. The isocyanide will give a secondary amine with one methyl group on catalytic reduction.

Q17. Complete the following reaction.

Answer:

The reaction exhibits azo-coupling of phenols. In mild alkaline conditions, phenol moiety participates in the azo-coupling, and the para position of phenol is occupied.

Q18. Why is aniline soluble in aqueous HCI?

Answer:

Aniline forms the salt anilinium chloride, which is water-soluble.

C6H5NH2 + HCl → [C6H5NH3]+Cl

Q19. Suggest a route by which the following conversion can be accomplished.

Answer:

We can convert it as

Q20. Identify A and B in the following reaction.

Answer:

Here, A and B will be

Q21. How will you carry out the following conversions?

(i) toluene → p-toluidine

(ii) p-toluidine diazonium chloride → p-toluic acid

Answer:

(i) toluene → p-toluidine

(ii) p-toluidine diazonium chloride → p-toluic acid

Q22. Write following conversions:

(i) nitrobenzene → acetanilide

(ii) acetanilide → p-nitroaniline

Answer:

(i) nitrobenzene → acetanilide

img 1

(ii) acetanilide → p-nitroaniline

img 2

Q23. A solution contains 1 g mol each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. 1 g mol of an alkaline solution of phenol is added to this. Predict the major product. Explain your answer.

Answer:

This reaction is an example of electrophilic aromatic substitution. In an alkaline medium, phenol generates phenoxide ion, which is more electron-rich than phenol and hence more reactive for an electrophilic attack. The electrophile in this reaction is aryl diazonium cation. The stronger the electrophile the faster the reaction. p-Nitrophenyldiazonium cation is a stronger electrophile than p-toluene diazonium cation. Therefore, it couples preferentially with phenol.

Q24. How will you bring out the following conversion?

Answer:

We can convert p-nitro aniline to 3,4,5-tribromobenzene as-

Q25. How will you carry out the following conversion?

Answer:

We can convert benzene to p-nitroaniline as-

imgQ26. How will you carry out the following conversion?

img 3

Answer:

We can convert aniline to m-bromonitrobenzene as:

img 4

Q27. How will you carry out the following conversions?

im

Answer:

(i) We can convert aniline to 3,5-dibromonitrobenzene as

im1

(ii) We can convert aniline to 3,5-dibromo-4-iodonitrobenzene as

im 2

Long Answer Type Questions

Q1. A hydrocarbon ‘A’, (C4H8) on reaction with HCl gives a compound ‘B’, (C4H9CI), which on reaction with 1 mol of NH3 gives compound ‘C’, (C4H11N). On reacting with NaNO2 and HCI followed by treatment with water, compound ‘C’ yields optically active alcohol, ‘D’. Ozonolysis of ‘A’ gives 2 moles of acetaldehyde. Identify compounds ‘A’ to ‘D’. Explain the reactions involved.

Answer:

The addition of HCl has occurred on ‘A’. This implies ‘A’ is an alkene.

NH2 substitutes Cl in compound ‘B’ to give ‘C’.

‘C’ gives a diazonium salt with NaNO2/HCl that liberates N2 to give optically active alcohol.

This means that ‘C’ is an aliphatic amine.

The number of carbon atoms in amine is the same as in compound ‘A’.

Since products of ozonolysis of compound ‘A’ are CH3—CH = O and O = CH—CH3.

Thus, compound ‘A’ is CH3—CH= CH—CH3.

Based on the structure of ‘A’ reactions can be explained as follows:

Q2. A colourless substance ‘A’ (C6H7N) is sparingly soluble in water and gives a water-soluble compound ‘B’ on treatment with mineral acid. On reacting with CHCI3 and alcoholic potash, ‘A’ produces an obnoxious smell due to compound ‘C’ formation. The reaction of ‘A’ with benzene sulphonyl chloride gives compound ‘D’, which is soluble in alkali. With NaNO2 and HCI, ‘A’ forms compound ‘E’, which reacts with phenol in an alkaline medium to give’ F’ orange dye. Identify compounds ‘A’ to ‘F’.

Answer:

Here,

A = Aniline

B = Anilinium chloride

C = Benzeneisonitrile

D = N-phenylbenzenesulphonamide

E = Benzenediazonium chloride

F = 4-hydroxyazobenzene


Q3. Predict the reagent or the product in the following reaction sequence.

Answer:

The reagent or the product in the following reaction sequence are:

1 = Sn/HCl

2 = 4-methyl-2-nitroacetanilide

3 = H2/OH

4 = Toluenediazonium chloride

5 = H3PO2/H2O and Cu+

Test your Knowledge on Important questions class 12 chemistry chapter 13 organic compounds containing nitrogen!

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