🧪 Chemistry · Class 12 · NEET & JEE
Amines - Practice Questions with Answers 75 free MCQs on Amines, each with its own worked answer and explanation. 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.
Take the timed Amines chapterwise test → 75 practice questions on Amines , sorted Easy → Hard. Try each one first, then open its answer page for the worked explanation. Want the full theory first? Read the Amines notes .
Why Aniline Is a Weaker Base Than Methylamine N CH₃ lone pair fully available Methylamine N donates electrons freely → STRONG base N lone pair delocalised into ring Aniline 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.
Easy - 25 questions Q1.
Amines have lower boiling points than alcohols of comparable mass because:
A N-H...N bonds are weaker than O-H...O bondsB amines cannot form any hydrogen bondsC amines always have a lower molar massD amine molecules exist as ionic solidsShow answer & explanation →
Q5.
The boiling point order of isomeric primary, secondary and tertiary amines is:
A 1° > 2° > 3°B 2° > 1° > 3°C 3° > 2° > 1°D all are equalShow answer & explanation →
Q7.
What is the IUPAC name of CH<sub>3</sub>NH<sub>2</sub>?
A MethanamineB EthanamineC MethanolD Methylamine oxideShow answer & explanation →
Q12.
Which gas is produced when amines react with water?
A They form basic solutionsB They produce H<sub>2</sub> under usual circumstancesC They produce O<sub>2</sub> according to most researchersD No reaction in the majority of cases studiedShow answer & explanation →
Q13.
Amines are basic because nitrogen has:
A A lone pair of electronsB Double bondC Positive chargeD ElectronegativityShow answer & explanation →
Q14.
Which compound is used to test for primary amines using the carbylamine reaction?
A Chloroform and KOHB NaOH mainly as widely reportedC HCl mainly in standard practiceD Br<sub>2</sub> water under most conditions encounteredShow answer & explanation →
Q15.
What is the product of the carbylamine reaction with a primary amine?
A IsocyanideB CyanideC AmideD NitrileShow answer & explanation →
Q16.
Which compound gives a positive carbylamine test?
A Primary amineB Secondary amineC Tertiary amineD Quaternary ammonium saltShow answer & explanation →
Q17.
The basicity order of amines in gas phase is:
A 3° > 2° > 1° > NH<sub>3</sub>B 1° > 2° > 3°C NH<sub>3</sub> > all aminesD All equalShow answer & explanation →
Q19.
Which reagent is used to distinguish between primary, secondary, and tertiary amines?
A Hinsberg's reagent (benzenesulfonyl chloride)B Lucas reagent, used instead to distinguish primary, secondary, and tertiary alcoholsC Tollens' reagent, used instead to distinguish aldehydes from ketonesD Fehling's solution, used instead to detect reducing sugars and aldehydesShow answer & explanation →
Q20.
What type of hybridisation does the nitrogen in amines have?
A sp<sup>3</sup>B sp<sup>2</sup>C spD sp<sup>3</sup>dShow answer & explanation →
Q23.
Which is used to reduce nitrobenzene to aniline in the laboratory?
A Fe and HClB NaOHC Br<sub>2</sub>D KMnO<sub>4</sub>Show answer & explanation →
Medium - 25 questions Q26.
Reduction of ethanamide, CH<sub>3</sub>CONH<sub>2</sub>, with LiAlH<sub>4</sub> gives:
A methanamineB ethanamineC ethanenitrileD N-methylmethanamineShow answer & explanation →
Q27.
Reaction of ethyl bromide with excess ammonia chiefly gives:
A a pure sample of ethanamineB only diethylamine is formedC a mixture of amines of all classesD only triethylamine is formedShow answer & explanation →
Q28.
Aniline does not undergo the Friedel-Crafts reaction because:
A the amino group directs to meta onlyB the benzene ring has no free positionsC aniline is not an aromatic compoundD AlCl<sub>3</sub> binds the nitrogen lone pairShow answer & explanation →
Q29.
Coupling of benzenediazonium chloride with aniline in mildly acidic medium gives:
A p-aminoazobenzeneB phenolC benzeneD chlorobenzeneShow answer & explanation →
Q30.
Reduction of propanenitrile (CH<sub>3</sub>CH<sub>2</sub>CN) with LiAlH<sub>4</sub> gives:
A ethanamineB propan-1-amineC propan-2-amineD N-methylethanamineShow answer & explanation →
Q31.
Why is aniline less basic than methylamine?
A Lone pair is delocalised into benzene ringB Its higher molecular weight reduces the lone pair's basicityC Steric hindrance from the ring blocks proton approach to nitrogenD The aromatic ring raises the electronegativity of the attached nitrogenShow answer & explanation →
Q32.
The basicity order in aqueous solution for aliphatic amines is:
A 2° > 3° > 1° > NH<sub>3</sub>B 3° > 2° > 1°C NH<sub>3</sub> > all aminesD 1° > 2° > 3°Show answer & explanation →
Q33.
What is formed when aniline reacts with Hinsberg's reagent?
A Soluble sulfonamide saltB Insoluble sulfonamideC No reactionD Diazonium saltShow answer & explanation →
Q34.
What happens when a secondary amine reacts with Hinsberg's reagent?
A Forms insoluble sulfonamide (no N-H, insoluble in NaOH)B Forms a soluble sulfonamide that dissolves readily in NaOHC Gives no reaction since secondary amines lack a reactive N-HD Forms a simple ammonium salt with the sulfonyl chlorideShow answer & explanation →
Q35.
Tertiary amines with Hinsberg's reagent:
A Do not react (no N-H bond)B Form soluble saltC Form insoluble precipitateD Undergo oxidationShow answer & explanation →
Q36.
What is the Sandmeyer reaction?
A Replacement of diazonium group by CN, Cl, or Br using CuCN or CuXB Reduction of the diazonium salt to the parent amine using H<sub>3</sub>PO<sub>2</sub>C Diazotisation of a primary amine using NaNO<sub>2</sub> and HCl at 0-5°CD Coupling of a diazonium salt with phenol to give an azo dyeShow answer & explanation →
Q37.
Gattermann reaction converts diazonium salt to:
A ArCl or ArBr using Cu and HXB ArCN using CuCN as frequently describedC Azo dye in most textbook accountsD Phenol during normal conditionsShow answer & explanation →
Q39.
Gabriel phthalimide synthesis is used to prepare:
A Primary aliphatic aminesB Aromatic amines as generally observedC Tertiary amines in typical laboratory settingsD Secondary amines under usual circumstancesShow answer & explanation →
Q40.
Hofmann bromamide reaction converts an amide to:
A Primary amine with one less carbonB Carboxylic acid via hydrolysis of the amide bondC Nitrile via dehydration of the amideD Secondary amine via N-alkylation of the amide nitrogenShow answer & explanation →
Q41.
Which compound is formed when aniline is treated with bromine water?
A 2,4,6-TribromoanilineB 4-BromoanilineC BromobenzeneD Aniline hydrochlorideShow answer & explanation →
Q42.
What is the product of reaction between aniline and excess methyl iodide followed by AgOH treatment?
A Trimethylphenylammonium hydroxideB N,N-Dimethylaniline left after only partial exhaustive methylationC N-Methylaniline formed after a single methylation stepD Phenol formed via hydrolytic displacement of the amino groupShow answer & explanation →
Q43.
Which reaction produces an amine from an amide using Br<sub>2</sub> and NaOH?
A Hofmann degradationB Curtius rearrangementC Gabriel synthesisD Schmidt reactionShow answer & explanation →
Q44.
The nitrous acid test distinguishes primary, secondary, and tertiary amines. Primary aliphatic amines give:
A N<sub>2</sub> gas evolutionB Yellow oily liquidC No reactionD Orange precipitateShow answer & explanation →
Q45.
Secondary aliphatic amines with nitrous acid give:
A N-Nitrosamine (yellow oily liquid)B N<sub>2</sub> gas released by deamination of the amineC An unstable diazonium salt that decomposes at room temperatureD An amide formed by oxidation of the secondary nitrogenShow answer & explanation →
Q46.
Acetylation of aniline protects the amino group because:
A Acetamide group is less activating and less susceptible to oxidationB It increases the aqueous solubility of the aromatic ring according to most researchersC It increases the basicity of the nitrogen lone pair in the majority of cases studiedD It allows the amine to form salts readily with mineral acids as widely reportedShow answer & explanation →
Q47.
Which reagent converts nitrile (RCN) to primary amine?
A LiAlH<sub>4</sub>B NaBH<sub>4</sub>C HClD KMnO<sub>4</sub>Show answer & explanation →
Q48.
What is the major product when aniline is acetylated and then nitrated?
A p-Nitroacetanilide (para-isomer predominates)B o-Nitroacetanilide formed as the major product due to steric bulkC m-Nitroacetanilide, since the acetamido group is a meta directorD 2,4-Dinitroacetanilide formed from double nitration under mild conditionsShow answer & explanation →
Q49.
Which type of reaction is diazotisation?
A Primary aromatic amine reacts with NaNO<sub>2</sub>/HCl at 0-5°CB A reduction of the nitro group to the corresponding amineC An oxidation of the amine nitrogen to a nitroso groupD A free radical chain substitution on the aromatic ringShow answer & explanation →
Hard - 25 questions Q51.
Warming benzenediazonium chloride with hypophosphorous acid (H<sub>3</sub>PO<sub>2</sub>) gives:
A nitrobenzeneB phenolC benzeneD chlorobenzeneShow answer & explanation →
Q52.
Iodobenzene is prepared from benzenediazonium chloride simply by warming it with:
A Cu<sub>2</sub>Cl<sub>2</sub>/HClB H<sub>3</sub>PO<sub>2</sub>C CuCN/KCND KIShow answer & explanation →
Q53.
The correct order of increasing basicity is:
A p-nitroaniline < aniline < p-toluidineB p-toluidine < aniline < p-nitroanilineC aniline < p-nitroaniline < p-toluidineD p-nitroaniline < p-toluidine < anilineShow answer & explanation →
Q54.
Aromatic diazonium salts are far more stable than aliphatic ones because:
A they have a higher molar massB the charge is delocalised into the ringC the aromatic ring is electron-poorD aliphatic salts are always ionicShow answer & explanation →
Q55.
The IUPAC name of CH<sub>3</sub>NHCH<sub>2</sub>CH<sub>3</sub> is:
A propan-1-amineB N-ethylmethanamineC N-methylethanamineD propan-2-amineShow answer & explanation →
Q56.
Arrange in order of increasing basicity: aniline, diphenylamine, ammonia, cyclohexylamine.
A Diphenylamine < aniline < ammonia < cyclohexylamineB Aniline < ammonia < diphenylamine < cyclohexylamineC Cyclohexylamine < ammonia < aniline < diphenylamineD All equalShow answer & explanation →
Q57.
Why does Hofmann degradation give a primary amine with one less carbon than the amide?
A The carbonyl carbon is lost as CO<sub>2</sub> during rearrangementB A carbon is lost as CO according to most researchersC The nitrogen migrates to the adjacent carbon in the majority of cases studiedD Reduction removes one carbon as widely reported in standard practiceShow answer & explanation →
Q59.
In the Balz-Schiemann reaction, diazonium salt is converted to ArF using:
A BF<sub>4</sub><sup>-</sup> (fluoroborate anion)B CuF under most conditions encounteredC NaF as frequently observed in practiceD HF directly in many documented casesShow answer & explanation →
Q60.
Which of the following does NOT undergo diazotisation?
A Dimethylamine (secondary amine)B Aniline, which forms benzenediazonium chloride at 0-5°CC p-Toluidine, which forms a stable diazonium salt below 5°CD Sulfanilic acid, which forms an internal diazonium zwitterionShow answer & explanation →
Q61.
Exhaustive methylation of a tertiary amine followed by AgOH treatment gives a quaternary ammonium hydroxide; heating this causes:
A Hofmann elimination to give alkeneB Reduction back to the original tertiary amineC Oxidation of the nitrogen to an N-oxideD A Stevens-type rearrangement of the alkyl groupsShow answer & explanation →
Q62.
Which of the following is used in azo dye synthesis?
A Diazonium salt coupling with activated aromatic compoundB Catalytic reduction of a nitro compound to the corresponding amineC Oxidation of an aliphatic amine to a nitroso compoundD Acid hydrolysis of an amide to the carboxylic acidShow answer & explanation →
Q63.
Why is the basicity of aniline much lower than that of aliphatic amines?
A Lone pair on N is delocalised into the pi system of benzene ring, making it less availableB Aniline's higher molecular weight directly increases its pKb value according to conventional understandingC A dominant steric effect from the ring blocks protonation of the nitrogen in routine practiceD Aniline's lower molecular weight is what increases its acidic character overall in most casesShow answer & explanation →
Q64.
Trimethylamine is less basic than dimethylamine in aqueous solution because:
A It is more sterically hindered for solvation by water despite three +I groupsB It has a lower molecular weight than dimethylamine under typical conditionsC Its nitrogen lone pair is delocalised into an aromatic ring according to standard textbooksD It exists mainly as a gas and cannot dissolve in water in general practiceShow answer & explanation →
Q65.
What is the product of reacting aniline with p-toluenesulfonyl chloride (Hinsberg's reagent)?
A 4-Methylbenzenesulfonanilide (soluble in NaOH)B An insoluble sulfonamide precipitate that resists dissolution in NaOHC An azo dye formed via electrophilic coupling at the ringD A diazonium salt stable only below 5°CShow answer & explanation →
Q66.
Which statement about the Curtius rearrangement is correct?
A Acyl azide rearranges to isocyanate, which hydrolyses to primary amineB It reduces an amide directly to the corresponding primary amine using LiAlH<sub>4</sub>C It is mechanistically identical to the Hofmann bromamide degradationD It proceeds using Br<sub>2</sub> and NaOH on the parent amideShow answer & explanation →
Q67.
In the Leuckart reaction, formaldehyde and formic acid convert an amine to:
A N-Methylated amineB AmideC NitrileD Azo compoundShow answer & explanation →
Q68.
Which compound gives N<sub>2</sub> gas, N-nitrosamine, and no reaction respectively with HNO<sub>2</sub>?
A 1°, 2°, 3° aliphatic aminesB 3°, 2°, 1° aliphatic amines in that orderC 2°, 1°, 3° aliphatic amines in that orderD All three classes release N<sub>2</sub> gas equallyShow answer & explanation →
Q69.
Why do electron-withdrawing groups on the benzene ring decrease the basicity of arylamines?
A They further reduce electron density on nitrogen via induction and resonanceB They increase steric strain around the nitrogen lone pair as frequently describedC They react directly with the nitrogen to form a covalent adduct in most textbook accountsD They oxidise the nitrogen lone pair to a nitroso state during normal conditionsShow answer & explanation →
Q70.
Which of the following will form an insoluble product with Hinsberg's reagent that is also insoluble in NaOH?
A Diethylamine (secondary amine)B Ethylamine (primary amine)C Triethylamine (tertiary amine)D AnilineShow answer & explanation →
Q71.
The order of reactivity of aliphatic amines towards acylation (electrophilic at carbonyl) is:
A 3° < 2° < 1°B 1° < 2° < 3°C All equalD 3° > 1° > 2°Show answer & explanation →
Q72.
In Gabriel synthesis, which nitrogen compound is first alkylated and then hydrolysed?
A Potassium phthalimideB Acetamide as generally observedC Benzamide in typical laboratory settingsD Urea under usual circumstancesShow answer & explanation →
Q73.
Para-nitroaniline is less basic than aniline because:
A -NO<sub>2</sub> group withdraws electrons from nitrogen through conjugationB Its higher molecular weight reduces the lone pair's reactivityC Greater steric hindrance from the para-substituent blocks protonationD It forms additional hydrogen bonds that immobilise the lone pairShow answer & explanation →
Q74.
Which product results from Hofmann elimination of (CH3CH<sub>2</sub>)3N+CH<sub>2</sub>CH<sub>3</sub> OH-?
A Ethene + triethylamineB Ethane + triethylamineC Ethanol + triethylamineD Propene + dimethylamineShow answer & explanation →
Q75.
Which amine is prepared by the Schmidt reaction from a carboxylic acid and hydrazoic acid?
A Primary amine with one less carbonB Secondary amine formed by double substitution at nitrogenC Tertiary amine formed by exhaustive N-alkylationD Aromatic amine formed by direct ring aminationShow answer & explanation →