CBSE Class 12 Chemistry Amines MCQs Set 07

Multiple Choice Questions

Question 1. Which of the following is a 3° amine?
(a) 1-methylcyclohexylamine
(b) Triethylamine
(c) tert-butylamine
(d) N-methylaniline
Answer: (b) Triethylamine
Triethylamine represents a tertiary amine where three ethyl groups are linked to the central nitrogen atom.
In simple words: A tertiary amine is a compound where the nitrogen atom is bonded to three carbon chains and has no hydrogen atoms attached directly to it.

Exam Tip: To identify the degree of an amine, count only the carbon atoms directly bonded to the nitrogen atom.

Question 2. Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?
(a) H2(excess)/Pt
(b) LiAlH4 in ether
(c) Fe and HCl
(d) Sn and HCl
Or
Which of the following would not be a good choice for reducing nitrobenzene to aniline?
(a) H2/Ni
(b) LiAlH4
(c) Fe and HCl
(d) Sn and HCl
Answer: (b) LiAlH4 in ether
LiAlH4 is a highly powerful reducing agent but is not a good choice to turn nitrobenzene into aniline. It actually converts nitrobenzene into azobenzene instead.
In simple words: Lithium aluminium hydride does not produce aniline from nitrobenzene. Instead, it links two benzene rings together to make azobenzene.

Exam Tip: Scrap iron with hydrochloric acid is the preferred choice for this reduction because the iron chloride byproduct gets hydrolyzed to generate its own acid catalyst.

Question 3. The best reagent for converting propanamide into propanamine is
(a) excess H2
(b) Br2 in aqueous NaOH
(c) iodine in the presence of red phosphorus
(d) LiAlH4 in ether
Answer: (d) LiAlH4 in ether
To transform propanamide into propanamine while keeping the same carbon count, we must use LiAlH4 in ether. The other option, Br2/NaOH, decreases the carbon chain length.
In simple words: Lithium aluminium hydride reduces the carbonyl group of the amide directly into a methylene group without losing any carbon atoms.

Exam Tip: Check the number of carbon atoms in both reactant and product: use LiAlH4 if the carbon count remains the same, and Hofmann degradation if the product has one less carbon.

Question 4. Ammonolysis of ethyl chloride followed by reaction of the amine so formed with 1 mole of methyl chloride gives an amine that
(a) reacts with Hinsberg reagent to form a product soluble in an alkali
(b) on reaction with nitrous acid, produced nitrogen gas
(c) reacts with benzene sulphonyl chloride to form a product that is insoluble in alkali
(d) does not react with Hinsberg reagent
Answer: (c) reacts with benzene sulphonyl chloride to form a product that is insoluble in alkali
To begin, ammonolysis of ethyl chloride produces ethylamine, which is a primary amine. Reacting this with one mole of methyl chloride generates N-ethylmethylamine, a secondary amine. Secondary amines combine with Hinsberg's reagent to produce a compound that cannot dissolve in basic solutions since it has no acidic hydrogen.
In simple words: The reaction steps form a secondary amine. It reacts with Hinsberg's reagent, but the resulting product cannot dissolve in alkali because it has no acidic hydrogen atom left on the nitrogen.

Exam Tip: Remember that secondary sulfonamides are insoluble in alkali due to the absence of any acidic hydrogen on the nitrogen atom.

Question 5. In order to prepare a 1° amine from an alkyl halide with simultaneous addition of one —CH2 group in the carbon chain, the reagent used as source of nitrogen is
(a) Sodium amide, NaNH2
(b) Sodium azide, NaN3
(c) Potassium cyanide, KCN
(d) Potassium phthalimide, C6H4(CO)2N- K+
Answer: (c) Potassium cyanide, KCN
Treating an alkyl halide with potassium cyanide adds a cyano group, which increases the carbon count by one. Later, reducing this nitrile group produces a primary amine containing an additional —CH2 group.
In simple words: Potassium cyanide introduces one more carbon into the chain along with a nitrogen atom. Reducing this nitrile later yields a primary amine with an extra carbon.

Exam Tip: Look for changes in carbon chain length; KCN followed by reduction is a classic way to extend the carbon chain during amine synthesis.

Question 7. The solubility in water of \( \text{C}_6\text{H}_5\text{NH}_2 \) (I), \( (\text{C}_2\text{H}_5)_2\text{NH} \) (II) and \( \text{C}_2\text{H}_5\text{NH}_2 \) (III) increase in the order of:
(a) II < III < I
(b) I < II < III
(c) III < II < I
(d) II < I < III
Answer: (b) I < II < III
Aromatic amines such as aniline (I) show the lowest solubility in water because of their large hydrophobic phenyl ring. Secondary amines (II) exhibit lower solubility than primary amines (III) since primary amines possess more hydrogen atoms to establish hydrogen bonds with water molecules.
In simple words: Aniline has a large, water-hating ring which makes it least soluble. Diethylamine is secondary and has only one hydrogen for water-bonding, while ethylamine is primary and has two, making it the most soluble.

Exam Tip: Solubility decreases as the size of the hydrophobic alkyl or aryl group increases, and as the number of hydrogen atoms available for H-bonding decreases.

Question 8. Two isomers, \( \text{n-C}_4\text{H}_9\text{NH}_2 \) and \( (\text{C}_2\text{H}_5)_2\text{NH} \) have molar mass of 73 each. Which of the following statement is correct about their boiling points?
(a) The boiling point of \( \text{n-C}_4\text{H}_9\text{NH}_2 \) is higher than that of \( (\text{C}_2\text{H}_5)_2\text{NH} \).
(b) The boiling point of \( (\text{C}_2\text{H}_5)_2\text{NH} \) is higher than that of \( \text{n-C}_4\text{H}_9\text{NH}_2 \).
(c) Both the amines will have the same boiling point.
(d) The boiling point of both the amines will be lower than that of water.
Answer: (a) The boiling point of \( \text{n-C}_4\text{H}_9\text{NH}_2 \) is higher than that of \( (\text{C}_2\text{H}_5)_2\text{NH} \).
The primary amine contains two hydrogen atoms linked to nitrogen, allowing it to create stronger intermolecular hydrogen bonds than the secondary amine. Therefore, the primary isomer possesses a higher boiling point.
In simple words: Because primary amines have more hydrogen atoms on their nitrogen, they stick together more tightly. It takes more energy (heat) to break these connections, giving them higher boiling points.

Exam Tip: For isomeric amines of comparable molecular mass, the boiling point trend always follows the order: primary > secondary > tertiary.

Question 9. Which statement is wrong about amine?
(a) Aniline and other arylamines are usually colourless but get coloured on storage due to atmospheric oxidation.
(b) Amines are less polar than alcohol.
(c) The order of boiling point of isomeric amines is: Primary < secondary < tertiary.
(d) Higher amines are essentially insoluble in water.
Answer: (c) The order of boiling point of isomeric amines is: Primary < secondary < tertiary.
The actual boiling point sequence for isomeric amines is primary > secondary > tertiary owing to the reduction in the number of hydrogen atoms accessible for intermolecular hydrogen bonding.
In simple words: Primary amines have the highest boiling points among isomers, not the lowest, because they can form more extensive hydrogen bonds.

Exam Tip: Read option labels very carefully. Ensure you do not confuse the less-than symbol (<) with the greater-than symbol (>) when evaluating trends.

Question 10. Amongst the following, the strongest base in aqueous medium is
(a) \( \text{CH}_3\text{NH}_2 \)
(b) \( (\text{CH}_3)_2\text{NH} \)
(c) \( (\text{CH}_3)_3\text{N} \)
(d) \( \text{C}_6\text{H}_5\text{NH}_2 \)
Or
Out of the following, the strongest base in aqueous solution is
(a) methylamine
(b) dimethylamine
(c) trimethylamine
(d) aniline
Answer: (b) \( (\text{CH}_3)_2\text{NH} \) (or dimethylamine)
In an aqueous environment, the basicity of methyl-substituted amines is governed by a blend of the inductive effect, solvation, and steric hindrance. This mixture causes the secondary amine, dimethylamine, to be the strongest base.
In simple words: In water, dimethylamine has the best combination of electron-donating groups and stabilization by water molecules, making it a stronger base than the others.

Exam Tip: Remember the basic strength order in water for methyl-substituted amines: secondary > primary > tertiary > ammonia (213 order).

Question 12. IUPAC name of product formed by reaction of methylamine with two moles of ethyl chloride is
(a) N,N-dimethylethanamine
(b) N,N-diethylmethanamine
(c) N-methylethanamine
(d) N-ethyl-N-methylethanamine
Answer: (d) N-ethyl-N-methylethanamine
Reacting methylamine with two moles of ethyl chloride produces diethylmethylamine. Following IUPAC rules, the longest carbon chain (ethyl group) acts as the parent chain, resulting in N-ethyl-N-methylethanamine.
In simple words: Two ethyl groups replace the hydrogens on methylamine. The longest continuous carbon chain is two carbons (ethanamine), so the other groups are named as substituents.

Exam Tip: For IUPAC naming of tertiary amines, identify the longest carbon chain as the parent amine, and write the other alkyl groups alphabetically with the prefix 'N-'.

Question 13. Match the reactions given in column I with the statements given in column II and choose the correct option from the codes given below.

Codes:
(a) A-(i), B-(ii), C-(iii)
(b) A-(ii), B-(iii), C-(i)
(c) A-(i), B-(iii), C-(ii)
(d) A-(iii), B-(i), C-(ii)
Answer: (b) A-(ii), B-(iii), C-(i)
The carbylamine reaction serves as a diagnostic test to identify primary amines. The Hofmann bromamide degradation process forms an amine with one fewer carbon atom than the parent amide. Ammonolysis refers to the nucleophilic substitution of alkyl halides with ammonia.
In simple words: Carbylamine reaction detects primary amines, Hofmann degradation steps down the carbon chain, and ammonolysis is the reaction of alkyl halides with ammonia.

Exam Tip: Quickly associate Hofmann bromamide with a decrease in carbon count, and Carbylamine with the detection of primary amines by their foul smell.

Question 14. An organic compound \( (\text{C}_3\text{H}_9\text{N}) \) (A), when treated with nitrous acid, gave an alcohol and \( \text{N}_2 \) gas was evolved. (A) on warming with \( \text{CHCl}_3 \) and caustic soda gave (C), which on reduction gave isopropylmethylamine. Predict the compound (A).
(a) \( (\text{CH}_3)_2\text{CH—NH}_2 \)
(b) \( \text{CH}_3\text{—CH}_2\text{—NH—CH}_3 \)
(c) \( (\text{CH}_3)_2\text{N—CH}_3 \)
(d) \( \text{CH}_3\text{—CH}_2\text{—CH}_2\text{—NH}_2 \)
Answer: (a) \( (\text{CH}_3)_2\text{CH—NH}_2 \)
Compound (A) represents isopropylamine. Because (A) reacts with nitrous acid to liberate nitrogen gas and produce an alcohol, it has to be a primary aliphatic amine. Its treatment with chloroform and caustic soda (the carbylamine test) forms isopropyl isocyanide, which on reduction generates isopropylmethylamine.
In simple words: (A) must be a primary amine since it releases nitrogen gas when treated with nitrous acid. Reducing its carbylamine product gives isopropylmethylamine, confirming the starting skeleton is isopropyl.

Exam Tip: Note that the reduction of an alkyl isocyanide (R-NC) always yields a secondary methylamine (R-NH-CH3).

Question 15. —NH2 group in the aniline being an activating group proceeds reaction at
(a) only para-position
(b) only ortho-position
(c) meta-position
(d) Both (a) and (b)
Answer: (d) Both (a) and (b)
The amino substituent possesses a lone electron pair on nitrogen that enhances the electron density at the ortho and para sites of the ring via resonance. This renders both locations highly reactive to electrophilic substitution.
In simple words: The amino group pushes electrons into the benzene ring, making both the ortho and para positions highly reactive toward electrophiles.

Exam Tip: The —NH2 group is extremely activating. Unless protected, reactions like bromination will lead to trisubstitution at both ortho and para positions.

Question 16. The stability of arenediazonium ion is explained on the basis of resonance. Which of the following resonating structure is incorrect?
(a) Structure (a)
(b) Structure (b)
(c) Structure (c)
(d) Structure (d)
Answer: (d) Structure (d)
Resonating structures of the arenediazonium ion delocalize the positive charge from the diazonium group to the ortho and para positions of the aromatic ring. Option (d) is incorrect since it falsely depicts a negative charge on the ring.
In simple words: The positive charge is distributed around the benzene ring's ortho and para positions. Structure (d) is wrong because it incorrectly shows a negative charge on the ring.

Exam Tip: Make sure you draw the positive charges on the carbon atoms of the ring (at ortho and para positions) during resonance representation.

Question 17. The correct name of the given reaction is
\( \text{Ar—N}_2^+\text{X}^- \xrightarrow[\text{Cu powder}]{\text{HBr}} \text{Ar—Br} + \text{N}_2 \)
(a) Hofmann bromamide degradation reaction
(b) Gabriel phthalimide synthesis
(c) Carbylamine reaction
(d) Gattermann reaction
Answer: (d) Gattermann reaction
The variation of the Sandmeyer reaction in which halogen acids are utilized with copper powder rather than cuprous salts is known as the Gattermann reaction.
In simple words: When copper powder is used as a catalyst instead of a copper halide salt to introduce halogen into the ring, the reaction is named after Gattermann.

Exam Tip: Identify Gattermann reaction by looking for metallic copper powder (Cu) instead of cuprous salts (Cu2X2).

Question 18. The correct name of the given reaction is
\( \text{Ar—N}_2^+\text{X}^- \xrightarrow{\text{CuCN/KCN}} \text{Ar—CN} + \text{N}_2 \)
(a) Sandmeyer's reaction
(b) Gabriel phthalimide synthesis
(c) Carbylamine reaction
(d) Hofmann bromamide degradation reaction
Answer: (a) Sandmeyer's reaction
This transformation is referred to as Sandmeyer's reaction, in which cuprous cyanide is employed to replace the diazonium group with a cyano group.
In simple words: The reaction that uses cuprous salts to substitute a diazonium group with a chlorine, bromine, or cyano group is called Sandmeyer's reaction.

Exam Tip: Remember that cuprous salts are the signature reagent for Sandmeyer reactions, providing a highly efficient way to synthesize substituted benzenes.

Question 19. Benzene diazonium chloride on hydrolysis gives
(a) phenol
(b) chlorobenzene
(c) benzene
(d) aniline
Answer: (a) phenol
Heating an aqueous mixture of benzene diazonium chloride triggers its hydrolysis, producing phenol and releasing nitrogen gas besides hydrochloric acid.
In simple words: When warmed in water, the diazonium salt decomposes and reacts to produce phenol, releasing nitrogen gas.

Exam Tip: Keep diazonium salts cold (0–5°C) during prep; otherwise, they react with water to yield phenol.

Question 20. A student was preparing aniline in the lab. She took a compound "X" and reduced it in the presence of Ni as a catalyst. What could be the compound "X"?
(a) Nitrobenzene
(b) 1-nitrohexane
(c) Benzonitrile
(d) 1-hexanenitrile
Answer: (a) Nitrobenzene
Catalytic reduction of nitrobenzene using nickel as the catalyst generates aniline, which represents an aromatic primary amine.
In simple words: Reducing the nitro group on a benzene ring using nickel and hydrogen produces aniline.

Exam Tip: Ensure that the carbon skeleton in your chosen option matches the product (aromatic ring for aniline, aliphatic chain for hexylamines).

Question 22. Consider the following table:

X, Y and Z are
(a) X = Ethylnitrile, Y = \( \text{H}_2\text{O} \), Z = Phenyl isocyanide
(b) X = Acetamide, Y = \( \text{HBF}_4 \), Z = Phenyl cyanide
(c) X = Acetamide, Y = \( \text{CH}_3\text{CH}_2\text{OH} \), Z = Phenyl isocyanide
(d) X = Ethylnitrile, Y = \( \text{H}_3\text{PO}_2 + \text{H}_2\text{O} \), Z = Phenyl cyanide
Answer: (c) X = Acetamide, Y = \( \text{CH}_3\text{CH}_2\text{OH} \), Z = Phenyl isocyanide
Reduction of acetamide (X) with lithium aluminium hydride produces ethylamine. Ethanol (Y) reduces benzene diazonium chloride to benzene. Heating aniline with chloroform and alcoholic KOH (Z) generates phenyl isocyanide.
In simple words: Acetamide reduces to ethylamine; ethanol acts as a mild reducing agent to turn diazonium chloride into benzene; and aniline undergoes the carbylamine reaction to make phenyl isocyanide.
Exam Tip: Know your mild reducing agents for diazonium salts: both ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)) and hypophosphorous acid (\( \text{H}_3\text{PO}_2 \)) can convert diazonium salts to benzene.

Question 23. Which of the following reaction(s) is/are electrophilic substitution?
I. Bromination of acetanilide
II. Coupling reaction of aryl diazonium salts
III. Diazotisation of aniline
IV. Acylation of aniline
Choose the correct answer from the options given below:
(a) I and II
(b) I, II and III
(c) III and IV
(d) I and IV
Answer: (a) I and II
Bromination of acetanilide and diazo coupling of aryl diazonium salts represent electrophilic aromatic substitution reactions. On the other hand, diazotisation and acylation are nucleophilic processes occurring at the nitrogen atom.
In simple words: Reactions that add groups onto the benzene ring are electrophilic substitutions. Reactions occurring at the nitrogen atom itself are nucleophilic.

Exam Tip: Keep in mind that benzene rings act as nucleophiles, so substitutions occurring on the ring are electrophilic substitutions.

Question 25. Which of the following compound will not undergo azo coupling reaction with benzene diazonium chloride?
(a) Aniline
(b) Phenol
(c) Anisole
(d) Nitrobenzene
Answer: (d) Nitrobenzene
Diazo coupling necessitates a highly activated benzene ring bearing strong electron-releasing substituents such as —NH2 or —OH. Nitrobenzene features a powerful electron-withdrawing nitro group, which deactivates the ring and stops the reaction.
In simple words: Diazonium ions are weak electrophiles and only attack benzene rings that are highly rich in electrons. Nitrobenzene has a group that pulls electrons away, leaving it too unreactive to couple.

Exam Tip: Azo coupling requires strongly activating electron-donating groups like —OH, —NH2, or —OCH3 because the diazonium ion is a weak electrophile.

Assertion-Reason

Directions In the following questions. An Assertion (A) is followed by a corresponding Reason (R). Use the following keys to choose the appropriate answer.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A),
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A),
(c) (A) is true, but (R) is false,
(d) (A) is false, but (R) is true.

Question 26. Assertion (A): Only a small amount of HCl is required in the reduction of nitro compounds with iron scrap and HCl in the presence of steam.
Reason (R): FeCl2 formed gets hydrolysed to release HCl during the reaction.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Just a small amount of hydrochloric acid is required to initiate the reduction since the ferrous chloride formed during the process undergoes hydrolysis to regenerate hydrochloric acid.
In simple words: We only need a tiny amount of HCl to start the reaction because the iron chloride produced gets hydrolyzed, constantly regenerating more HCl.

Exam Tip: This auto-regeneration of HCl is why Fe/HCl is the preferred chemical method for industrial aniline preparation over Sn/HCl.

Question 27. Assertion (A): Hofmann’s bromamide reaction is given by primary amides.
Reason (R): Primary amines are less basic than secondary amines.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
It is correct that only primary amides participate in Hofmann's bromamide degradation. It is likewise true that primary amines are less basic than secondary amines in water, yet this fact does not explain why the Hofmann reaction is unique to primary amides.
In simple words: Both statements are scientifically true, but the basicity of amines is unrelated to the mechanism of amide degradation.

Exam Tip: Primary amides are required because the mechanism of Hofmann degradation involves removal of an N-H proton to form a bromoamide intermediate.

Question 28. Assertion (A): Aniline cannot be prepared by Gabriel phthalimide synthesis.
Reason (R): Aryl halides do not undergo nucleophilic substitution.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
Aniline cannot be prepared through Gabriel phthalimide synthesis. Nevertheless, the reason is incorrect since aryl halides are capable of undergoing nucleophilic substitution under extreme conditions, even though they are generally highly unreactive toward standard nucleophilic attack.
In simple words: Phthalimide cannot substitute on chlorobenzene because aryl halides are extremely unreactive under normal conditions. However, saying they 'do not undergo' nucleophilic substitution at all is incorrect, as it can occur under harsh conditions.

Exam Tip: Gabriel phthalimide synthesis is limited strictly to primary aliphatic amines because the phthalimide nucleophile cannot displace aromatic halogens.

Question 29. Assertion (A): Tertiary butyl amine can be prepared by the action of —NH3 on tert-butyl bromide.
Reason (R): Tertiary butyl bromide being 3° alkyl halide prefers to undergo elimination on the treatment with a base.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
Combining ammonia with tertiary butyl bromide produces isobutylene as the major product rather than tertiary butyl amine. This happens because tertiary butyl bromide is a tertiary alkyl halide that undergoes elimination instead of substitution when treated with a basic reagent like ammonia.
In simple words: Tertiary alkyl halides are bulky, so they react with bases (like ammonia) to lose HBr and form alkenes rather than undergoing substitution to form amines.

Exam Tip: To prepare a tertiary alkylamine, use alternative methods since nucleophilic substitution on a tertiary halide predominantly yields alkenes.

Question 30. Assertion (A): Acylation of amines gives a monosubstituted product, whereas alkylation of amines gives polysubstituted product.
Reason (R): Acyl group sterically hinders the approach of further acyl groups.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
Acylation produces a monosubstituted compound because the electron-withdrawing carbonyl group of the acyl moiety diminishes the nucleophilicity of the nitrogen atom, halting subsequent reactions. It is not steric hindrance that prevents further reaction.
In simple words: The added acyl group pulls electrons away from nitrogen, making it a very weak nucleophile that will not react further.

Exam Tip: Alkylation activates the amine further due to the +I effect of alkyl groups, leading to over-alkylation, while acylation deactivates it.

Question 31. Assertion (A): In acylation reaction of amines, equilibrium shifts to the right hand side in the presence of pyridine.
Reason (R): In the presence of strong base, HCl is removed and reaction shifts toward the right hand side.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Pyridine, acting as a stronger base than the starting amine, combines with the hydrochloric acid byproduct to eliminate it. Based on Le Chatelier’s principle, extracting a product drives the chemical equilibrium to the right side.
In simple words: Pyridine neutralizes the HCl acid produced in the reaction. Removing the product pulls the reaction forward to make more products.

Exam Tip: Pyridine is standard in acylation reactions to consume the acid byproduct and shift equilibrium forward.

Question 32. Assertion (A): N, N-diethylbenzene sulphonamide is insoluble in alkali.
Reason (R): Sulphonyl group attached to nitrogen atom is strong electron withdrawing group.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
N,N-diethylbenzene sulphonamide is unable to dissolve in basic solutions because it lacks an acidic hydrogen atom bonded to the nitrogen. While the electron-withdrawing power of the sulphonyl group is real, it only increases the acidity of available N-H hydrogens, which does not apply here as the molecule has no such hydrogen.
In simple words: Both statements are correct, but the reason it does not dissolve in alkali is the absence of any hydrogen atom on the nitrogen, rather than the withdrawing nature of the sulphonyl group itself.

Exam Tip: For primary amines, the sulfonamide has a remaining acidic N-H and is soluble in base. For secondary amines, it lacks an N-H and is insoluble.

Question 33. Assertion (A): In strongly acidic solutions, aniline becomes more reactive towards electrophilic reagents.
Reason (R): The amino group being completely protonated in strongly acidic solution, the lone pair of electrons on the nitrogen is no longer available for resonance.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In a strongly acidic environment, aniline undergoes protonation to yield the anilinium ion. Since the nitrogen atom bears a positive charge, it acts as a powerful deactivating group, rendering the benzene ring far less reactive toward electrophilic substitution.
In simple words: Strong acids turn aniline into the positive anilinium ion. This positive charge pulls electron density out of the ring, making it much harder for electrophiles to react.

Exam Tip: Protonation of the amino group deactivates the aromatic ring and directs incoming electrophiles to the meta position.

Question 34. Assertion (A): Acetanilide is less basic than aniline.
Reason (R): Acetylation of aniline results in decrease of electron density on nitrogen.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
In acetanilide, the carbonyl group withdraws the nitrogen lone pair via resonance, which reduces its availability for protonation. Thus, acetanilide is a weaker base than aniline.
In simple words: The carbonyl group in acetanilide acts like an electron sponge, pulling nitrogen's electrons away so they cannot easily bond with a proton.

Exam Tip: Acetylation is a very useful way to reduce aniline's high reactivity and basicity during organic synthesis.

Question 35. Assertion (A): Besides ortho- and para-nitroaniline, nitration of aniline in an acidic medium also gives the meta derivative.
Reason (R): In acidic medium aniline gets protonated forming anilinium ion.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Nitration in strongly acidic mixtures protonates aniline to form the anilinium cation. This species is highly deactivating and meta-directing, which accounts for the synthesis of a substantial portion of the meta isomer.
In simple words: The acid in the nitrating mixture converts aniline into the anilinium ion. This ion directs the nitro group to the meta position, producing 47% of m-nitroaniline.

Exam Tip: Note the yields of aniline nitration: 51% para, 47% meta, and 2% ortho.

Question 36. Assertion (A): Benzene diazonium salts are soluble in water.
Reason (R): They are covalent in nature, so they are soluble in water.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
Benzene diazonium salts dissolve easily in water since they are ionic substances, rather than covalent. Because water is highly polar, it easily solvates these charged species.
In simple words: Diazonium salts are ionic, like table salt, which is why they dissolve so well in water. Thus, statement (R) is completely false.

Exam Tip: Diazonium salts are ionic solids that dissolve in polar solvents (water) but decompose easily if warmed.