NCERT Questions for Class 12 Chemistry Chapter 9 – Amines

Important Questions with Solutions of Class 12 Chemistry Chapter 9 – Amines

1) Describe one chemical observation for the segregation between the below pairs of compounds:

(a) Methylamine & dimethylamine

Ans – Methylamine & dimethylamine can be classified by carbylamines test:

Carbylamine Test: Aliphatic & aromatic primary amines give foul-smelling isocyanides or carbylamines while heating with chloroform & ethanolic potassium hydroxide. The Carbylamine test remains positive for methylamine (primary amine) but looks different for dimethylamine.

(b) Methylamine & methanol

Ans – Methylamine offers the carbylamine test where a primary amine after heating with chloroform & potassium hydroxide creates a bad-smelling isocyanide. Meanwhile, ethanol won’t provide a positive carbylamine test.

(c) Ethanol & ethanaamine

Ans – Ethanol test: Ethanol on heating becomes turbid with Lucas reagent (anhy. ZnCl & conc. HCI).

Ethanamine test: Ethanamine combined with HNO₂, NaNO₂ mixture & dilute HCI offers powerful effervescence called nitrogen gas is created.

(d) Methylamine & N, N-dimethylamine

Ans – N, N-dimethylamine is a secondary amine whereas methylamine is a primary amine. Hence, the carbylamines test is the one that can classify between a primary & a secondary amine.

Aliphatic & aromatic primary amines give bad-smelling isocyanides/carbylamines during the heating process with chloroform & ethanolic potassium hydroxide. The Carbylamine test remains positive for methylamine which is an aliphatic primary amine, but reacts differently for N, N-dimethylamine.

(e) Secondary & tertiary amines

Ans – The reaction of secondary and tertiary amines with Hinsberg’s reagent classifies them as Benzenesulphonyl chloride, C₆H₅SO₂CI. Secondary amines collate with Hinsberg’s reagent to give an alkali-insoluble substance. N, Ndiethylamine reacts with Hinsberg’s reagent to give N, Ndiethylbenzenesulphonamide & is insoluble in alkali. However, Hinsberg’s reagent does not have any reaction with tertiary amines.

(f) Ethylamine & aniline

Ans – The azo-dye test can classify between ethylamine & aniline. A reaction of aromatic amines with HNO₂ (NaNO₂ + dilute HCI) at 0-5°C gives a dye, the formed dye later reacts with an alkaline solution of 2-naphthol. The color of the dye is usually orange, yellow, or red. Based on similar conditions, aliphatic amines give a brisk effervescence (owing to the formation of N₂ gas).

2) Explain the Hinsberg test for detecting primary, secondary, and tertiary amines. Give the chemical equations of the reaction steps.

Ans: Hinsberg Test: Benzenesulphonyl chloride (C₆H₅SO₂CI), also called Hinsberg’s reagent, undergoes a reaction with primary and secondary amines to give sulphonamides.

(i)Primary amines: Benzenesulphonyl chloride reacts with primary amines to give N-ethylbenzenesulphonyl amide. The sulphonyl group is a powerful electron withdrawing group, the hydrogen attached to nitrogen of sulphonamide is very acidic and thus alkali-soluble.

(ii) Secondary amines: N, N-diethylbenzenesulphonamide is formed by reacting with primary amines. It has no acid character, so it does not dissolve in alkali, as N, Diethyl Benzene sulphonamide does not contain a hydrogen atom bonded with the nitrogen atom.

(iii) Tertiary amines: Benzenesulphonyl chloride reacts not at all with tertiary amines. The difference in the reaction of amines with benzenesulphonyl chloride has been used as a distinction between primary, secondary, and tertiary amines. It also helps to differentiate the amines mixture.

3) Give the proper reason & justify them!

(a) The quaternary ammonium salts have all 4 different optically active alkyl groups.

Ans – Quaternary ammonium compounds don’t show quick inversion since the nitrogen atoms lack a single pair of electrons. Nitrogen is a chiral center because its sigma-bonding with 4 alkyl groups causes optical activity.

(b) The creation of aniline is impossible using Gabriel phthalimide synthesis.

Ans – Aryl halides cannot compete for nucleophilic substitution where the anion is produced during the phthalimide. Hence, the Gabriel phthalimide synthesis is unable to yield aniline.

(c) The C-N-C bond in trimethyl amine is defined as 108^∘

Ans – The angle of the bond [C-N-C] remains 108^∘ because of the repulsive force between the massive methyl groups on either side of the nitrogen atoms.

(d) Methylamine in water combines with ferric chloride to produce hydrated ferric oxide as a precipitate.

Ans – Methylamine reacts with ferric chloride in the presence of water and forms ferric hydroxide. Methylamine has a stronger base compared to water because of the +1 action formation of the -CH3 group. Thus, water absorbs H+ ions from its environment leaving behind methylamine that creates OH- ions. Methyl amine is a base-forming hydroxide ion after getting dissolved in water.

(e) Diazonium salts of aromatic amines remain more stable than aliphatic amines. Why?

Ans – The positive charge present in the benzene ring disperses because of the resonance factor. The stability of the diazonium ion is the direct outcome of resonance. Therefore, the diazonium salt of aromatic amine is largely more stable than the aliphatic amine one.

(f) Alkylamines have a stronger base than ammonia.

Ans – The electron-releasing alkyl group (R) attracts the electrons over nitrogen in alkyl amine by making the lone electron pair easily available to link with the proton from acid. Thus, alkyl amines portray a more basic character than ammonia.

(g) Ethylamine can be soluble in water whereas aniline cannot. Why?

Ans – In the case of ethylamine, intermolecular H₂ bonds exist after combining with water. Since aniline possesses a massive hydrophobic group -C₆H₅⁺ aniline doesn’t make hydrogen bonds with water extensively. Hence, aniline remains insoluble in water.

(h) 1° amines boil at temperatures higher compared to 2° amines and can boil at higher temperatures than 3º amines.

Ans – Primary amines are more in boiling points than tertiary amines since they possess replacement hydrogen atoms to which hydrogen can be bonded to tertiary amines. The tertiary amine’s boiling point increased based on hydrogen bonding as extensive heat is needed to break down these hydrogen bonds.

(i) Even if the amino group is o, p-directing over an aromatic electrophilic substitution process, aniline on nitration creates a considerable range of m-nitroaniline.

Ans – Nitration occurs in an acidic medium. Aniline gets protonated to give anilinium ions in an acidic medium (often called meta-directing).

(j) The pkb point of benzeneamine remains 6.33 whereas the ammonia has 4.75.

Ans – The benzylamine’s pK_b point is higher than ammonia since ammonia has a stronger base than benzylamine. A higher pK_b value shows lower basicity.

(k) Sulphanilic acid remains insoluble in water.

Ans – Sulphanilic acid is insoluble in water & acids as it cannot form hydrogen bonds using water molecules. Whereas, the compound can be soluble in aqueous mineral acids like HF, HCI, and HNO3 since these mineral acids can form a hydrogen bond with them.

(l) Why Gabriel phthalimide synthesis is incorporated to build primary amines.

Ans – Primary alkyl halides are converted into primary amines using the above synthesis. As no secondary and tertiary amines are formed in this synthesis, pure primary amines are produced due to this reaction. Hence, this reaction is used for primary amines and N-alkyl phthalimide is created as an outcome.

(m) Amines can be soluble in dilute HCI

Ans – The amines get charged after accepting an H+ ion from the acid, hence it can perform (robust) ion-dipole connection with water molecules by converting salt & getting dissolved as though that might happen with the table salt and glass of water.

(n) Amines contain fewer boiling points than alcohols of similar molecular masses.

Ans – Since nitrogen is less electronegative than oxygen, amines boil at lower temperatures when compared to alcohol. Therefore, the N-H bond is relatively less polar than the O-H bond and has a weaker hydrogen bonding over amines & alcohols.

(o) Aniline won’t process Friedel-Crafts reaction.

Ans – Since the reagent AlCl₃ is an electron-deficient compound, it operates as a Lewis base, and won’t process Friedel crafts reaction but acts as a Lewis base. It targets the sole pair of nitrogen in aniline and forms an insoluble complex that precipitates to terminate the reaction. Hence, aniline won’t process the Friedal-Crafts reaction.

(p) Aniline gives 2, 4, and 6-tribromoaniline while reacting with bromine water. Is it true?

Ans – Due to the presence of nitrogen atoms and the reduced +1 effect of hydrogen, aniline acts as an activating group. It causes an extremely dense electron cloud in benzene with a strong reaction over bromine water to create 2, 4, 6-tribromoaniline.