60 free MCQs on Amines with worked answers and explanations. Nitrogen-containing organic compounds. Covers classification (primary, secondary, tertiary), basicity comparison, preparation methods, and reactions including diazotization and coupling, key for understanding dyes and pharmaceuticals.
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In aniline, the nitrogen's lone pair is pulled into the benzene ring through resonance, leaving less electron density available to accept a proton - which is why aniline is a much weaker base than methylamine, where the lone pair is fully available.
Q1.
Which functional group is present in amines?
Answer: A. -NH<sub>2</sub>
Why: Amines contain the amino (-NH<sub>2</sub>) functional group derived from ammonia.
Q2.
What is the IUPAC name of CH<sub>3</sub>NH<sub>2</sub>?
Answer: A. Methanamine
Why: CH<sub>3</sub>NH<sub>2</sub> is methanamine (commonly called methylamine).
Q3.
Primary amines have how many alkyl groups attached to nitrogen?
Answer: A. One
Why: Primary amines (R-NH<sub>2</sub>) have one alkyl or aryl group on the nitrogen.
Q4.
Which amine has the formula (CH<sub>3</sub>)<sub>2</sub>NH?
Answer: B. Secondary
Why: Dimethylamine (CH<sub>3</sub>)<sub>2</sub>NH has two alkyl groups on nitrogen, making it a secondary amine.
Q5.
Which amine has three alkyl groups attached to nitrogen?
Answer: C. Tertiary
Why: Tertiary amines have three alkyl/aryl groups on nitrogen.
Q6.
Aniline is an example of which type of amine?
Answer: B. Aromatic primary
Why: Aniline (C<sub>6</sub>H<sub>5</sub>NH<sub>2</sub>) is an aromatic primary amine.
Q7.
Which gas is produced when amines react with water?
Answer: A. They form basic solutions
Why: Amines are basic and accept protons from water to form alkaline solutions.
Q8.
Amines are basic because nitrogen has:
Answer: A. A lone pair of electrons
Why: The lone pair on nitrogen enables amines to accept protons (Bronsted base) or donate to Lewis acids.
Q9.
Which compound is used to test for primary amines using the carbylamine reaction?
Answer: A. Chloroform and KOH
Why: The carbylamine reaction uses chloroform and alcoholic KOH; primary amines give foul-smelling isocyanides.
Q10.
What is the product of the carbylamine reaction with a primary amine?
Answer: A. Isocyanide
Why: Primary amines react with CHCl<sub>3</sub>/KOH to form isocyanides (carbylamines), which have a very offensive smell.
Q11.
Which compound gives a positive carbylamine test?
Answer: A. Primary amine
Why: Only primary amines give the characteristic foul-smelling isocyanide in the carbylamine test.
Q12.
The basicity order of amines in gas phase is:
Answer: A. 3° > 2° > 1° > NH<sub>3</sub>
Why: In the gas phase, more alkyl groups increase electron density on N, so 3° > 2° > 1° > NH<sub>3</sub>.
Q13.
Aniline reacts with acetic anhydride to form:
Answer: A. Acetanilide
Why: Aniline reacts with acetic anhydride to form acetanilide (N-phenylethanamide).
Q14.
Which reagent is used to distinguish between primary, secondary, and tertiary amines?
Answer: A. Hinsberg's reagent (benzenesulfonyl chloride)
Why: Hinsberg's reagent (C<sub>6</sub>H<sub>5</sub>SO<sub>2</sub>Cl) reacts differently with primary, secondary, and tertiary amines.
Q15.
What type of hybridisation does the nitrogen in amines have?
Answer: A. sp<sup>3</sup>
Why: Nitrogen in amines is sp<sup>3</sup> hybridised with a lone pair, giving a pyramidal geometry.
Q16.
Which of the following is an aromatic amine?
Answer: A. Aniline
Why: Aniline (aminobenzene) is an aromatic amine where NH<sub>2</sub> is attached to a benzene ring.
Q17.
Amines are formed by reduction of:
Answer: A. Nitro compounds
Why: Reduction of nitro compounds (e.g., nitrobenzene) gives amines.
Q18.
Which is used to reduce nitrobenzene to aniline in the laboratory?
Answer: A. Fe and HCl
Why: Iron (or tin) with hydrochloric acid reduces nitrobenzene to aniline.
Q19.
Diazotisation converts a primary aromatic amine to:
Answer: A. Diazonium salt
Why: Diazotisation with NaNO<sub>2</sub>/HCl at 0-5°C converts ArNH<sub>2</sub> to ArN2+Cl- (diazonium salt).
Q20.
Which of the following is a secondary amine?
Answer: A. Dimethylamine
Why: Dimethylamine (CH<sub>3</sub>)<sub>2</sub>NH has two methyl groups on nitrogen, making it secondary.
Q21.
Why is aniline less basic than methylamine?
Answer: A. Lone pair is delocalised into benzene ring
Why: In aniline, the N lone pair is delocalised into the aromatic ring, reducing its availability for protonation.
Q22.
The basicity order in aqueous solution for aliphatic amines is:
Answer: A. 2° > 3° > 1° > NH<sub>3</sub>
Why: In aqueous solution, solvation effects make secondary amines most basic due to optimum balance of inductive effect and solvation.
Q23.
What is formed when aniline reacts with Hinsberg's reagent?
Answer: A. Soluble sulfonamide salt
Why: Primary aromatic amines give a sulfonamide that dissolves in NaOH (has one acidic H on N).
Q24.
What happens when a secondary amine reacts with Hinsberg's reagent?
Answer: A. Forms insoluble sulfonamide (no N-H, insoluble in NaOH)
Why: Secondary amines give an N-disubstituted sulfonamide which is insoluble in NaOH.
Q25.
Tertiary amines with Hinsberg's reagent:
Answer: A. Do not react (no N-H bond)
Why: Tertiary amines have no N-H bond and so do not react with Hinsberg's reagent.
Q26.
What is the Sandmeyer reaction?
Answer: A. Replacement of diazonium group by CN, Cl, or Br using CuCN or CuX
Why: Sandmeyer reaction replaces the diazonium group with CN, Cl, or Br using corresponding cuprous salts.
Q27.
Gattermann reaction converts diazonium salt to:
Answer: A. ArCl or ArBr using Cu and HX
Why: Gattermann reaction uses Cu powder and HCl/HBr (instead of CuCl/CuBr) to replace diazonium with halide.
Q28.
Coupling of diazonium salt with phenol in alkaline conditions gives:
Answer: A. Azo dye
Why: Azo coupling forms a highly coloured azo compound (-N=N-) used in dyes.
Q29.
Gabriel phthalimide synthesis is used to prepare:
Answer: A. Primary aliphatic amines
Why: Gabriel synthesis converts phthalimide to primary aliphatic amines via alkylation and hydrolysis.
Q30.
Hofmann bromamide reaction converts an amide to:
Answer: A. Primary amine with one less carbon
Why: Hofmann degradation (amide + Br<sub>2</sub>/NaOH) gives a primary amine with one fewer carbon than the original amide.
Q31.
Which compound is formed when aniline is treated with bromine water?
Answer: A. 2,4,6-Tribromoaniline
Why: The NH<sub>2</sub> group strongly activates the ring; bromine water produces 2,4,6-tribromoaniline as a white precipitate.
Q32.
What is the product of reaction between aniline and excess methyl iodide followed by AgOH treatment?
Answer: A. Trimethylphenylammonium hydroxide
Why: Exhaustive methylation gives the quaternary ammonium hydroxide (trimethylphenylammonium hydroxide).
Q33.
Which reaction produces an amine from an amide using Br<sub>2</sub> and NaOH?
Answer: A. Hofmann degradation
Why: Hofmann degradation or bromamide reaction converts RCONH<sub>2</sub> to RNH<sub>2</sub> (one less carbon).
Q34.
The nitrous acid test distinguishes primary, secondary, and tertiary amines. Primary aliphatic amines give:
Answer: A. N<sub>2</sub> gas evolution
Why: Primary aliphatic amines react with HNO<sub>2</sub> to form unstable diazonium salts that immediately release N<sub>2</sub>.
Q35.
Secondary aliphatic amines with nitrous acid give:
Answer: A. N-Nitrosamine (yellow oily liquid)
Why: Secondary amines react with HNO<sub>2</sub> to form N-nitroso compounds (yellow oily liquids).
Q36.
Acetylation of aniline protects the amino group because:
Answer: A. Acetamide group is less activating and less susceptible to oxidation
Why: N-acetylation converts the strongly activating -NH<sub>2</sub> to -NHCOCH<sub>3</sub>, making the ring less reactive and protecting the group.
Q37.
Which reagent converts nitrile (RCN) to primary amine?
Answer: A. LiAlH<sub>4</sub>
Why: LiAlH<sub>4</sub> reduces nitriles to primary amines (RCN → RCH<sub>2</sub>NH<sub>2</sub>).
Q38.
What is the major product when aniline is acetylated and then nitrated?
Answer: A. p-Nitroacetanilide (para-isomer predominates)
Why: The acetamido group is an ortho/para director; para-product predominates due to steric reasons.
Q39.
Which type of reaction is diazotisation?
Answer: A. Primary aromatic amine reacts with NaNO<sub>2</sub>/HCl at 0-5°C
Why: Diazotisation must be performed at 0-5°C; higher temperatures cause decomposition of the diazonium salt.
Q40.
The Reimer-Tiemann reaction introduces a formyl group into:
Answer: A. Phenol
Why: The Reimer-Tiemann reaction is specific to phenol; it introduces a -CHO group ortho to -OH.
Q41.
Arrange in order of increasing basicity: aniline, diphenylamine, ammonia, cyclohexylamine.
Answer: A. Diphenylamine < aniline < ammonia < cyclohexylamine
Why: Aryl groups decrease basicity through conjugation. Cyclohexylamine (aliphatic) > NH<sub>3</sub> > aniline > diphenylamine.
Q42.
Why does Hofmann degradation give a primary amine with one less carbon than the amide?
Answer: A. The carbonyl carbon is lost as CO<sub>2</sub> during rearrangement
Why: During Hofmann rearrangement, the acyl nitrene rearranges; the C=O is lost as CO<sub>2</sub>, giving RNH<sub>2</sub> with one fewer carbon.
Q43.
Diazonium salts are stable only at:
Answer: A. 0-5°C
Why: Above ~5°C, diazonium salts decompose rapidly; they must be used immediately after preparation at 0-5°C.
Q44.
In the Balz-Schiemann reaction, diazonium salt is converted to ArF using:
Answer: A. BF<sub>4</sub><sup>-</sup> (fluoroborate anion)
Why: Balz-Schiemann reaction: ArN2+BF<sub>4</sub><sup>-</sup> is thermally decomposed to give the aryl fluoride (ArF).
Q45.
Which of the following does NOT undergo diazotisation?
Answer: A. Dimethylamine (secondary amine)
Why: Diazotisation requires a primary amine; secondary amines react with HNO<sub>2</sub> to form N-nitroso compounds, not diazonium salts.
Q46.
Exhaustive methylation of a tertiary amine followed by AgOH treatment gives a quaternary ammonium hydroxide; heating this causes:
Answer: A. Hofmann elimination to give alkene
Why: Heating quaternary ammonium hydroxide gives Hofmann elimination: alkene + tertiary amine + water.
Q47.
Which of the following is used in azo dye synthesis?
Answer: A. Diazonium salt coupling with activated aromatic compound
Why: Azo dyes are made by coupling an electrophilic diazonium ion with a nucleophilic (electron-rich) aromatic ring.
Q48.
Why is the basicity of aniline much lower than that of aliphatic amines?
Answer: A. Lone pair on N is delocalised into the pi system of benzene ring, making it less available
Why: Resonance between N lone pair and the aromatic pi system in aniline reduces the electron density on N significantly.
Q49.
Trimethylamine is less basic than dimethylamine in aqueous solution because:
Answer: A. It is more sterically hindered for solvation by water despite three +I groups
Why: In water, trimethylamine is less solvated than dimethylamine, so the latter is effectively more basic in aqueous solution.
Q50.
What is the product of reacting aniline with p-toluenesulfonyl chloride (Hinsberg's reagent)?
Answer: A. 4-Methylbenzenesulfonanilide (soluble in NaOH)
Why: Primary amine gives monosubstituted sulfonamide with one N-H still present, making it soluble in NaOH.
Q51.
Which statement about the Curtius rearrangement is correct?
Answer: A. Acyl azide rearranges to isocyanate, which hydrolyses to primary amine
Why: Curtius rearrangement: RCON<sub>3</sub> (acyl azide) → R-N=C=O (isocyanate) → RNH<sub>2</sub> upon hydrolysis.
Q52.
In the Leuckart reaction, formaldehyde and formic acid convert an amine to:
Answer: A. N-Methylated amine
Why: Leuckart reaction uses HCHO and HCOOH to N-methylate amines via reductive amination.
Q53.
Which compound gives N<sub>2</sub> gas, N-nitrosamine, and no reaction respectively with HNO<sub>2</sub>?
Answer: A. 1°, 2°, 3° aliphatic amines
Why: 1° aliphatic → N<sub>2</sub>; 2° → N-nitrosamine; 3° → no reaction with HNO<sub>2</sub>.
Q54.
Why do electron-withdrawing groups on the benzene ring decrease the basicity of arylamines?
Answer: A. They further reduce electron density on nitrogen via induction and resonance
Why: EWGs like -NO<sub>2</sub>, -Cl withdraw electron density from the ring and from nitrogen, reducing its ability to donate lone pair.
Q55.
Which of the following will form an insoluble product with Hinsberg's reagent that is also insoluble in NaOH?
Answer: A. Diethylamine (secondary amine)
Why: Secondary amines give N,N-disubstituted sulfonamides with no N-H; this is insoluble in NaOH.
Q56.
The order of reactivity of aliphatic amines towards acylation (electrophilic at carbonyl) is:
Answer: A. 3° < 2° < 1°
Why: More alkyl groups on N increase steric hindrance, making tertiary amines least reactive toward acylation (3° do not acylate easily).
Q57.
In Gabriel synthesis, which nitrogen compound is first alkylated and then hydrolysed?
Answer: A. Potassium phthalimide
Why: Gabriel synthesis uses potassium phthalimide as the nitrogen source; after alkylation, hydrazinolysis or acid hydrolysis gives the primary amine.
Q58.
Para-nitroaniline is less basic than aniline because:
Answer: A. -NO<sub>2</sub> group withdraws electrons from nitrogen through conjugation
Why: The strong electron-withdrawing -NO<sub>2</sub> group at para position withdraws electron density from N through resonance, drastically reducing basicity.
Q59.
Which product results from Hofmann elimination of (CH3CH<sub>2</sub>)3N+CH<sub>2</sub>CH<sub>3</sub> OH-?
Answer: A. Ethene + triethylamine
Why: Hofmann elimination of tetraethylammonium hydroxide gives ethene (beta elimination) and triethylamine.
Q60.
Which amine is prepared by the Schmidt reaction from a carboxylic acid and hydrazoic acid?
Answer: A. Primary amine with one less carbon
Why: Schmidt reaction: RCOOH + HN<sub>3</sub> → RNH<sub>2</sub> (primary amine with one fewer carbon, via isocyanate intermediate).