Important Questions Class 12 Chemistry Chapter 13

Important Questions Class 12 Chemistry Chapter 13

Important Questions for CBSE Class 12 Chemistry Chapter 13 – Amines

Chemistry is one of the most rewarding subjects in CBSE, and the key to get excellent scores  is to understand the concepts, formulas, equations, and how to apply them.

Extramarks offers Class 12 Chapter 13 Chemistry Important Questions which are reviewed by subject matter experts and prepared according to the latest CBSE curriculum. These answers are written in a clear concise manner and are taken from NCERT books. These questions are easily accessible on  the Extramarks’ website as per the student’s convenience.

CBSE Class 12 Chemistry Chapter-13 Important Questions

Study Important Question for Class 12 Chemistry Chapter 13 – Amines

Amines are organic compounds derived from ammonia that have one or more alkyl or aryl groups attached to the nitrogen atom. Amines have many functions in living things, including bioregulation, neurotransmission, and predator protection. Because of their high biological activity, many amines are used as pharmaceuticals and treatments.

Chapter 13 Class 12 Chemistry Important Questions provided by Extramarks will help students with understanding the types of questions expected  in board exams. Given below are  short answer questions along with concise   answers of 2 marks each.. Students can access the link provided to go through a  set of questionnaires Class 12 Chemistry Chapter 13 Important Questions.

Very Short Answer Questions – 1 Mark

Q1. In their aqueous solutions, arrange the following compounds in increasing order of basic strength: (NH3)3NH2, (CH3)2NH, (CH3)3N

A1. The order of basicity.

(CH3)

(CH3)3 N > 2 NH > CH3NH2 > (CH3)3 N > NH3

Q2. Give the IUPAC name for H2N — CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2

A2. But-3-ene-1-amine (IUPAC name)

Q3. Write the n-methylethanamine structure. 

A3. H3C—H2C—NH—CH2 is the structure of n-methylethanamine.

Short Answer Questions – 2 Marks

Q1. Explain Hinsberg’s test for identifying primary, secondary, and tertiary amines. Write down the chemical equations for the reactions as well.

Ans. Hinsberg test: When benzenesulphonyl chloride (C6H5SO2Cl), also known as Hinsberg’s reagent, reacts with primary and secondary amines, it produces sulphonamides.

N-ethylbenzenesulphonyl amide is created by reacting benzenesulphonyl chloride with primary amine. The hydrogen linked to nitrogen in sulphonamide is highly acidic due to the strong electron withdrawing sulphonyl group. As a result, it dissolves in alkali.

Secondary Amines: In the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed. Because N, N-diethylbenzene sulphonamide lacks a hydrogen atom linked to the nitrogen atom, it is not acidic and thus insoluble in alkali.

Tertiary Amines: Benzenesulphonyl chloride does not react with them. This property of amines reacting differently with benzenesulphonyl chloride is used to differentiate primary, secondary, and tertiary amines, as well as to separate an amine mixture.

Q2. Give one chemical test to differentiate between the following compounds:

(a) Dimethylamine and thylamine

Ans. The carbylamine test can distinguish between methylamine and dimethylamine:

When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, they produce foul-smelling isocyanides or carbylamines. The Carbylamine test detects methylamine (an aliphatic primary amine) but not dimethylamine.

CH3 – NH2 +  CHCl3 +  3KOH−→ΔCH3 – NC  +  3KCl  +  3H2

(CH3)2NH  +  CHCl3 +  3KOH−→Δ No reaction

(b)Secondary and tertiary amines 

Ans. Allowing secondary and tertiary amines to react with Hinsberg’s reagent (Benzenesulphonyl chloride, C6H5SO2Cl) distinguishes them. When secondary amines react with Hinsberg’s reagent, an alkali-insoluble product is formed. For example, N, Ndiethylamine reacts with Hinsberg’s reagent to form N, Ndiethylbenzenesulphonamide, which is insoluble in alkali. In contrast, Hinsberg’s reagent does not react with tertiary amines.

(c)  Aniline and ethylamine

Ans. The azo-dye test can distinguish between ethylamine and aniline. At 0-5°C, aromatic amines react with HNO2 (NaNO2 + dil.HCl) to produce a dye, which is then reacted with an alkaline solution of 2-naphthol. The dye is typically yellow, red, or orange in colour. Aliphatic amines exhibit rapid effervescence under comparable conditions (due to the development of N2 gas).

(d) Methanol and methylamine

Ans. Methylamine performs the carbylamine test, which involves heating a primary amine with chloroform and potassium hydroxide to produce foul-smelling isocyanides. In contrast, ethanol does not produce a positive carbylamine test result.

(e)  N, N-dimethylamine and methylamine

Ans. N,N-dimethylamine is a secondary amine, whereas methylamine is a primary amine, and the only test used to distinguish between the two is the carbylamines test.

When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, they produce foul-smelling isocyanides or carbylamines. The Carbylamine test detects methylamine (an aliphatic primary amine) but not N,N-dimethylamine.

Q3.  Give Reasons

(i) Optical activity is found in quaternary ammonium salts with four different alkyl groups.

Ans. Because nitrogen atoms lack a lone pair of electrons, there is no rapid inversion in quaternary ammonium compounds. The nitrogen atom’s optical activity is due to its sigma bonding to four alkyl groups.

(ii) Alkylamines have a higher basicity than ammonia.

Ans. Because of its electron-releasing nature, the alkyl group (R) pulls electrons towards nitrogen in alkyl amine, making the unshared electron pair more available for sharing with the acid’s proton. As a result, alkyl amines have a more basic nature than ammonia.

(iii) Gabriel phthalimide synthesis cannot produce aniline.

Ans. Because aryl halides cannot undergo nucleophilic substitution with the anion produced by phthalimide, the Gabriel phthalimide synthesis technique cannot yield phthalimide.

(iv) Ethylamine is water soluble, but aniline is not.

Ans. Intermolecular hydrogen bonds are formed when ethylamine is mixed with water. However, because of the presence of a large hydrophobic C6H5 group, aniline does not form many hydrogen bonds with water. As a result, aniline is insoluble in water.

(v) Amines dissolve in diluted HCl.

Ans. When the amines receive a H + ion from the acid, they become charged, allowing them to form (strong) ion-dipole interactions with water molecules. It basically turns into salt and dissolves in a glass of water in the same way that table salt does.

(vi) Amines have a lower boiling point than comparable molecular mass alcohols.

Ans. Amines have lower boiling points than alcohols because nitrogen is less electronegative than oxygen. Because of this, the NH bond is less polar than the OH bond, and the hydrogen bond between amines and alcohols is weaker.

(vii) Friedel-Crafts reaction does not occur in aniline.

Ans. Because AlCl3 (the Lewis acid used as a catalyst in Friedel crafts reactions) lacks electrons, it acts as a Lewis base and does not undergo Friedel craft reactions. It attacks aniline’s lone pair of nitrogens, forming an insoluble complex that precipitates  and halts the process. As a result, aniline does not undergo the Friedel-Crafts reaction.

(viii) When aniline reacts with bromine water, it readily forms 2, 4, 6-tribromoaniline.

Ans. Aniline is a highly activating group due to the nitrogen atom and the low + I impact of hydrogen. This produces a dense electron cloud in benzene, resulting in a strong reaction with Br water and the formation of 2, 4, 6-Tribromoaniline.

(ix) Aromatic amine diazonium salts are more stable than aliphatic amine diazonium salts.

Ans. Because of resonance, the positive charge on the benzene ring disperses. This resonance is responsible for the diazonium ion’s stability. Aromatic amine diazonium salts are therefore more stable than aliphatic amine diazonium salts.

(x) Despite the fact that the amino group is o, p-directing in aromatic electrophilic substitution reactions, aniline on nitration produces a significant amount of m-nitroaniline.

Ans. Nitration occurs in an acidic environment. In an acidic medium, aniline is protonated to produce anilinium ion (which is meta-directing).

Short Answer Questions – 3 Marks

Q1. Explain your reasoning for the following:

(i) The pKb value of aniline is greater than that of methylamine.

(ii) Ethylamine dissolves in water, whereas aniline does not.

(iii) The boiling points of primary amines are higher than those of tertiary amines.

A1.

(i) Because the lone pair of electrons on the nitrogen atom become delocalized over the benzene ring and are unavailable for protonation due to resonance in aniline, which is absent in alkylamine, aniline is less basic than methylamine.

(ii) Ethylamine is soluble in water due to its ability to form H-bonds with water, whereas aniline is insoluble in water due to its larger hydrocarbon component, which tends to retard H-bond formation.

(iii) Primary amines undergo extensive intermolecular H-bonding due to the presence of two H-atoms on the N-atom, whereas tertiary amines do not undergo H-bonding due to the absence of an H-atom on the N-atom. As a result, the boiling points of primary amines are higher than those of 3° amines.

Q2. Give reasons for the following:

(i) Acetylation reduces aniline’s activation effect.

(ii) CH3NH2 is more basic than C6H5NH2.

(iii) Despite the fact that —NH2 is an o/p directing group, aniline on nitration yields a significant amount of m-nitroaniline.

A2.

(i) Acetylation of aniline reduces its activation effect because the acetyl group, as an electron withdrawing group, attracts the N-lone atom’s pair of electrons towards the carboxyl group, making the N-lone atom’s pair of electrons less available for donation to the benzene ring via resonance.

(ii) CH3NH2 is more basic than aniline because the lone pair of electrons on the nitrogen atom are available for donation, whereas in aniline the lone pair of electrons on the nitrogen atom is delocalised over the benzene ring and thus unavailable for donation.

(iii) Nitration in an acidic medium causes aniline to be protonated, yielding anilinium ion, which is indirecting.

Long Answer Questions – 5 Marks

Q1. A colourless substance ‘A’ (C6H7N) is sparingly soluble in water and, when treated with mineral acid, yields a water-soluble compound ‘B.’ When ‘A’ reacts with CHCI3 and alcoholic potash, it emits an unpleasant odour due to the formation of compound ‘C.’ The reaction of ‘A’ with benzene sulphonyl chloride produces compound ‘D,’ which is alkali soluble. ‘A’ forms compound ‘E’ with NaNO2 and HCI, which reacts with phenol in an alkaline medium to produce ‘F’ orange dye. Compounds ‘A’ through ‘F’ must be identified.

A1. Here,

A = Aniline

B = Anilinium chloride

C = Benzeneisonitrile

D = N-phenylbenzenesulphonamide

E = Benzenediazonium chloride

F = 4-हयड्रोक्सयजोबेन्ज़ेने