Chapter 9: Amines Case Study Questions | chemistry Class 12 CBSE

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A pharmaceutical chemist is working on synthesizing a new antihistaminic drug that contains a tertiary amino group. She starts with nitrobenzene and needs to prepare N,N-dimethylaniline. In the first step, she reduces nitrobenzene to aniline using iron scrap and hydrochloric acid. She prefers this reagent over Sn/HCl because FeCl2 formed gets hydrolysed to release HCl, so only a small amount of HCl is needed to initiate the reaction. She then plans to alkylate aniline with excess methyl iodide in the presence of sodium carbonate. Understanding the structure and reactivity of amines is crucial to designing efficient drug synthesis pathways.

Question 1: Why is iron scrap and HCl preferred over Sn/HCl for the reduction of nitrobenzene to aniline?

FeCl2 formed during the reaction gets hydrolysed, releasing HCl back into the reaction mixture.
Therefore, only a small (catalytic) amount of HCl is required to initiate the reaction, making it more economical.

Question 2: Write the reaction of aniline with excess methyl iodide in the presence of sodium carbonate and name the final product.

Aniline reacts with excess CH3I in the presence of Na2CO3 to give N,N,N-trimethylanilinium iodide (a quaternary ammonium salt).
The reaction proceeds stepwise: C6H5NH2 → C6H5NHCH3 → C6H5N(CH3)2 → [C6H5N(CH3)3]+ I−, the final product being trimethylphenylammonium iodide.

Question 3: N,N-Dimethylaniline is a tertiary amine. Explain why tertiary aromatic amines are weaker bases than their aliphatic counterparts, and also weaker than secondary aromatic amines like N-methylaniline, using pKb values and structural reasoning.

In aromatic amines, the lone pair on nitrogen is in conjugation with the benzene ring, making it less available for protonation, hence weaker basicity than aliphatic amines.
N,N-Dimethylaniline has pKb = 8.92, while N-methylaniline has pKb = 9.30; lower pKb means stronger base, so N,N-dimethylaniline is a slightly stronger base than N-methylaniline due to +I effect of two methyl groups.
However, both are much weaker bases than aliphatic amines (pKb ~3–4) because the aryl group withdraws the lone pair through resonance, reducing its availability for accepting a proton from an acid.

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Case Set 2

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During a forensic chemistry investigation, a technician needs to identify whether an unknown organic compound containing nitrogen is a primary, secondary, or tertiary amine. The compound has molecular formula C3H9N. The technician applies three chemical tests: the carbylamine test (heating with CHCl3 and alc. KOH), the Hinsberg test (reaction with benzenesulphonyl chloride and NaOH), and the nitrous acid test (reaction with NaNO2 + HCl at 0°C). The results are recorded and analyzed to identify all possible isomers and classify them. This systematic approach is commonly used in organic qualitative analysis laboratories.

Question 1: Which isomers corresponding to molecular formula C3H9N are primary amines? Write their IUPAC names.

Propan-1-amine (CH3CH2CH2NH2) and propan-2-amine (CH3CHNH2CH3) are primary amines with formula C3H9N.
N-methylethanamine (CH3NHCH2CH3) is a secondary amine, and N,N-dimethylmethanamine [(CH3)3N] is a tertiary amine with this formula.

Question 2: Describe how the Hinsberg test distinguishes between primary, secondary, and tertiary amines.

Primary amine reacts with benzenesulphonyl chloride to form N-alkylbenzenesulphonamide, which is soluble in alkali (due to acidic N–H).
Secondary amine forms N,N-dialkylbenzenesulphonamide, which is insoluble in alkali (no N–H present). Tertiary amine does not react with benzenesulphonyl chloride at all.

Question 3: The carbylamine reaction is used as a confirmatory test for primary amines. Explain the reaction with propan-1-amine, state why secondary and tertiary amines do not give this test, and mention the observable change.

Propan-1-amine reacts with CHCl3 and alc. KOH on heating: CH3CH2CH2NH2 + CHCl3 + 3KOH → CH3CH2CH2NC + 3KCl + 3H2O, forming propyl isocyanide (carbylamine), which has a foul/offensive smell.
Secondary and tertiary amines cannot give carbylamine reaction because they lack a free –NH2 group; the reaction requires two hydrogen atoms on nitrogen for the intermediate to form isocyanide.
The observable change is the production of an intensely foul-smelling substance (isocyanide/carbylamine), which serves as a positive test for the presence of a primary amine.

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Case Set 3

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A student is studying the synthesis of azo dyes in a chemistry laboratory. She prepares benzenediazonium chloride by treating aniline with NaNO2 and HCl at 273–278 K. She notes that the diazonium salt must be prepared fresh and used immediately. She then uses coupling reactions to synthesize two azo dyes: one by coupling with phenol and another by coupling with aniline. The azo dyes produced show vivid orange and yellow colours. The student also notes that arenediazonium salts are more stable than alkyldiazonium salts. This practical demonstrates the commercial importance of diazonium salts in the textile and food industry.

Question 1: Write the chemical equation for the preparation of benzenediazonium chloride from aniline.

C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl− + NaCl + 2H2O
This reaction is called diazotisation and must be carried out at 273–278 K to prevent decomposition of the diazonium salt.

Question 2: Write the coupling reaction of benzenediazonium chloride with phenol and name the product. Why is the coupling reaction an electrophilic substitution?

C6H5N2+Cl− + HO-C6H5 → C6H5–N=N–C6H4–OH (p-hydroxyazobenzene, an orange dye) + HCl
It is electrophilic substitution because the diazonium ion (Ar–N2+) acts as a weak electrophile and attacks the electron-rich phenol ring at the para position.

Question 3: Explain why arenediazonium salts are more stable than alkyldiazonium salts. Also explain the role of resonance in stabilising the arenediazonium ion.

Alkyldiazonium salts are highly unstable and decompose rapidly to give alcohols and N2 gas even at low temperatures.
Arenediazonium salts are stabilised by resonance: the positive charge on nitrogen is delocalised into the aromatic ring through resonance, spreading the charge over multiple atoms and lowering the energy of the ion.
This extended resonance/conjugation involving the benzene ring gives four resonance structures to the arenediazonium ion, making it stable enough to exist in solution at 273–278 K for a short time, unlike alkyldiazonium salts which have no such resonance stabilisation.

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Case Set 4

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A chemistry teacher is demonstrating the physical properties of amines to her Class 12 students. She presents samples of methylamine (gas), butylamine (liquid), and aniline (colourless liquid that turns brown on storage). She explains how hydrogen bonding affects boiling points: primary amines have higher boiling points than tertiary amines of similar molecular mass because of intermolecular H-bonding. She also compares the solubility of lower and higher amines in water and organic solvents, and discusses why aniline turns coloured on storage. The students record observations in a table for comparison with alcohols and alkanes of similar molar masses.

Question 1: Why do lower aliphatic amines dissolve in water but higher amines do not?

Lower aliphatic amines (up to 3 carbons) are soluble in water because they can form hydrogen bonds with water molecules through their N–H bonds.
As molar mass increases, the size of the hydrophobic alkyl chain increases, which outweighs the hydrogen bonding ability, making higher amines insoluble in water.

Question 2: Arrange the following in increasing order of boiling points and justify: n-butylamine, diethylamine, triethylamine.

Increasing order of boiling points: triethylamine < diethylamine < n-butylamine.
Triethylamine (3°) has no N–H for H-bonding so lowest bp; diethylamine (2°) has one N–H so intermediate bp; n-butylamine (1°) has two N–H atoms available for H-bonding giving maximum intermolecular association and highest bp among these isomers.

Question 3: Compare the boiling points of amines with alcohols and alkanes of similar molar mass, and explain why aniline turns coloured on storage. Also state why amines have lower boiling points than alcohols of comparable mass.

Alcohols have higher boiling points than amines of comparable molar mass because O is more electronegative (3.5) than N (3.0), forming stronger hydrogen bonds; e.g., n-butanol (bp 390.3 K) > n-butylamine (bp 350.8 K) > isobutane (bp 300.8 K).
Aniline and other arylamines are initially colourless but turn brown/dark on storage due to atmospheric oxidation of the aromatic amine by oxygen in air.
Amines have lower boiling points than alcohols because N–H···N hydrogen bonds are weaker than O–H···O hydrogen bonds due to lower electronegativity of nitrogen compared to oxygen, leading to less intermolecular association.

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Case Set 5

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In an organic synthesis lab, a researcher is using the Gabriel phthalimide synthesis to prepare pure primary amines. The method involves treating phthalimide with ethanolic KOH to form potassium phthalimide, followed by reaction with an alkyl halide, and then alkaline hydrolysis to obtain the primary amine. Meanwhile, a colleague is using the Hoffmann bromamide degradation reaction, treating an amide with Br2 in aqueous NaOH solution to get a primary amine containing one fewer carbon than the amide. Both methods are compared for their efficiency, selectivity, and limitations in preparing primary aliphatic and aromatic amines.

Question 1: State one major advantage of Gabriel phthalimide synthesis over ammonolysis of alkyl halides for preparing primary amines.

Gabriel phthalimide synthesis gives pure primary amine as the only product, unlike ammonolysis which gives a mixture of primary, secondary, tertiary amines and quaternary ammonium salts.
The method is selective because phthalimide anion can react only once with the alkyl halide, preventing over-alkylation.

Question 2: Write the structural formula and IUPAC name of the amide that would give propan-1-amine by Hoffmann bromamide degradation reaction.

The amine produced (propan-1-amine) has 3 carbon atoms, so the amide must have 4 carbon atoms.
The amide is CH3CH2CH2CONH2, IUPAC name: butanamide.

Question 3: Explain why aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis. Also explain the mechanism of Hoffmann bromamide degradation indicating the carbon count change.

Aromatic primary amines cannot be prepared by Gabriel synthesis because aryl halides (ArX) do not readily undergo nucleophilic substitution reactions with the phthalimide anion; the benzene ring deactivates the carbon towards nucleophilic attack.
In Hoffmann bromamide degradation: R–CO–NH2 + Br2 + 4NaOH → R–NH2 + Na2CO3 + 2NaBr + 2H2O; during this reaction, the alkyl/aryl group migrates from the carbonyl carbon to the nitrogen atom (rearrangement step).
As a result, the amine formed has one carbon atom fewer than the starting amide — e.g., butanamide (4C) gives propan-1-amine (3C) — making this reaction useful for 'descending' the amine series.

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Chapter 9: Amines | Case Study Questions

Passage

Question 1: Why is iron scrap and HCl preferred over Sn/HCl for the reduction of nitrobenzene to aniline?

Question 2: Write the reaction of aniline with excess methyl iodide in the presence of sodium carbonate and name the final product.

Question 3: N,N-Dimethylaniline is a tertiary amine. Explain why tertiary aromatic amines are weaker bases than their aliphatic counterparts, and also weaker than secondary aromatic amines like N-methylaniline, using pKb values and structural reasoning.

Passage

Question 1: Which isomers corresponding to molecular formula C3H9N are primary amines? Write their IUPAC names.

Question 2: Describe how the Hinsberg test distinguishes between primary, secondary, and tertiary amines.

Question 3: The carbylamine reaction is used as a confirmatory test for primary amines. Explain the reaction with propan-1-amine, state why secondary and tertiary amines do not give this test, and mention the observable change.

Passage

Question 1: Write the chemical equation for the preparation of benzenediazonium chloride from aniline.

Question 2: Write the coupling reaction of benzenediazonium chloride with phenol and name the product. Why is the coupling reaction an electrophilic substitution?

Question 3: Explain why arenediazonium salts are more stable than alkyldiazonium salts. Also explain the role of resonance in stabilising the arenediazonium ion.

Passage

Question 1: Why do lower aliphatic amines dissolve in water but higher amines do not?

Question 2: Arrange the following in increasing order of boiling points and justify: n-butylamine, diethylamine, triethylamine.

Question 3: Compare the boiling points of amines with alcohols and alkanes of similar molar mass, and explain why aniline turns coloured on storage. Also state why amines have lower boiling points than alcohols of comparable mass.

Passage

Question 1: State one major advantage of Gabriel phthalimide synthesis over ammonolysis of alkyl halides for preparing primary amines.

Question 2: Write the structural formula and IUPAC name of the amide that would give propan-1-amine by Hoffmann bromamide degradation reaction.

Question 3: Explain why aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis. Also explain the mechanism of Hoffmann bromamide degradation indicating the carbon count change.

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