Organic Compounds Containing Nitrogen — NEET Chemistry MCQs with answers

Organic Compounds Containing Nitrogen is a Class 12 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 559 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Amines · Diazonium salts · Nitro compounds and cyanides/isocyanides · Amines & Diazonium Salts · Cyanides and isocyanides · Amines · Diazonium salts · Amines · Amino acids · Proteins · Carbohydrates · Nucleic acids · Vitamins · Hormones/Proteins · Synthetic polymers · Enzymes · Basicity of amines · Preparation · Distinguishing tests · Basicity comparison · Carbylamine reaction · Nitro and cyanide compounds · Diazonium coupling · Amine nucleophilicity vs basicity · Diazonium salt applications · Amine basicity · Diazonium salt structure · Isocyanide hydrogenation · Diazonium reactions summary · Amine preparation summary · Nitro/nitrite/isocyanide isomerism · Hinsberg test statements · EAS on aniline summary · Reduction methods comparison · Sandmeyer vs Gattermann comparison · Nitrile vs isocyanide reduction comparison · Carbylamine test scope comparison · Diazonium coupling partners comparison · Diazonium coupling misconception · Amine basicity misconception

8 free Organic Compounds Containing Nitrogen practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Amines

   A mixture containing a primary, a secondary, and a tertiary amine is treated with benzenesulfonyl chloride and excess KOH. Which sequence correctly describes how the three amines can be physically separated from the resulting mixture?

   • A.Simply filter the mixture; all three amines are automatically separated into three distinct filtered layers
   • B.Extract the whole basic reaction mixture directly with ether; the tertiary amine alone dissolves in ether while both sulfonamides stay in the aqueous layer, achieving full separation in one step
   • C.Filter off the insoluble N,N-dialkyl sulfonamide (from the secondary amine); acidify the remaining alkaline filtrate with dilute HCl to reprecipitate the N-alkyl sulfonamide (from the primary amine, now converted back from its KOH-soluble salt) while the tertiary amine remains dissolved as its water-soluble hydrochloride salt; basify this filtrate and extract with ether to recover the free tertiary amine; finally, hydrolyse each isolated sulfonamide separately with strong acid to liberate the pure primary and secondary amines
   • D.Simple fractional distillation of the crude reaction mixture separates the three amines directly based on differing boiling points, without any chemical treatment
   Show answer and explanation

   Answer: C. Filter off the insoluble N,N-dialkyl sulfonamide (from the secondary amine); acidify the remaining alkaline filtrate with dilute HCl to reprecipitate the N-alkyl sulfonamide (from the primary amine, now converted back from its KOH-soluble salt) while the tertiary amine remains dissolved as its water-soluble hydrochloride salt; basify this filtrate and extract with ether to recover the free tertiary amine; finally, hydrolyse each isolated sulfonamide separately with strong acid to liberate the pure primary and secondary amines

   Hinsberg's reagent converts the primary amine to a KOH-soluble sulfonamide, the secondary amine to a KOH-insoluble sulfonamide, and leaves the tertiary amine unreacted (soluble in acid); filtering, then acidifying the filtrate, then basifying and ether-extracting, separates all three, after which acid hydrolysis of each isolated sulfonamide regenerates the pure free amines.

   After treatment with benzenesulfonyl chloride/excess KOH: the secondary amine's sulfonamide (no N-H left) is insoluble in KOH and can be filtered off directly as a solid. The primary amine's sulfonamide (one acidic N-H) is dissolved in the alkaline filtrate as its potassium salt, and the tertiary amine (unreacted, since it has no N-H) is also present in this filtrate as a basic species (partly as its salt in the aqueous alkaline mixture, or as a separate layer, depending on exact conditions) — acidifying this filtrate with dilute HCl reprotonates the sulfonamide anion, causing the neutral primary-amine sulfonamide to reprecipitate (now separable by filtration), while the tertiary amine is protonated to its water-soluble ammonium chloride salt and stays in solution. Basifying this remaining aqueous solution with NaOH frees the tertiary amine, which (being water-insoluble as the free base) can be extracted into ether and isolated. Finally, each isolated sulfonamide (from the primary and secondary amines) is separately hydrolysed under strong acid conditions to cleave the S-N bond and liberate the pure free primary and secondary amines respectively. This full, correctly ordered sequence achieves clean three-way separation, which is precisely why Hinsberg's reagent is valued analytically beyond simple qualitative identification.

   Common mistake: Assuming a single filtration or extraction step can separate all three amine classes at once

2. Question 2 · hard · Amines

   Gabriel synthesis is attempted using potassium phthalimide and 2-bromobutane (a secondary alkyl halide) in place of a primary alkyl halide. Compared to using a primary halide, the major complication expected is:

   • A.The reaction proceeds with an even better SN2 yield than with primary halides, since secondary carbons are more electrophilic
   • B.Potassium phthalimide reacts exclusively at the aromatic ring of the phthalimide instead of at nitrogen when a secondary halide is used
   • C.The reaction instead proceeds by Hoffmann bromamide degradation, converting the phthalimide into an amine with one less carbon
   • D.A significant proportion of the secondary substrate undergoes E2 elimination (competing with the desired SN2 substitution) because the phthalimide anion, while a good nucleophile, is also a moderately strong base, and secondary substrates are inherently more prone to elimination, lowering the yield of the desired N-alkylphthalimide
   Show answer and explanation

   Answer: D. A significant proportion of the secondary substrate undergoes E2 elimination (competing with the desired SN2 substitution) because the phthalimide anion, while a good nucleophile, is also a moderately strong base, and secondary substrates are inherently more prone to elimination, lowering the yield of the desired N-alkylphthalimide

   The phthalimide anion is both a nucleophile and a base; with a secondary alkyl halide, steric hindrance around the reacting carbon slows SN2 substitution while making the competing E2 elimination pathway (base removing a β-hydrogen) more favourable, so the yield of the desired N-alkylphthalimide substitution product drops relative to primary substrates.

   For primary alkyl halides, the phthalimide anion's attack at the relatively unhindered primary carbon proceeds efficiently via SN2, and elimination is a minor pathway since primary substrates have limited elimination tendency and typically require stronger, more hindered bases to eliminate significantly. With a secondary alkyl halide such as 2-bromobutane, however, two things change: (1) steric hindrance around the secondary carbon slows the rate of backside SN2 attack, and (2) the phthalimide anion, although chosen as a nucleophile, retains base character, and secondary substrates are considerably more prone to E2 elimination (loss of a β-hydrogen and the leaving group to form an alkene) than primary substrates are. The two effects compound: substitution becomes slower while elimination becomes more competitive, so a genuinely significant fraction of 2-bromobutane is converted to butene(s) instead of the desired N-(sec-butyl)phthalimide, lowering the overall yield and making Gabriel synthesis considerably less clean and less commonly used for secondary substrates compared to primary ones.

   Common mistake: Assuming Gabriel synthesis works equally well and cleanly with any class of alkyl halide, ignoring substrate-dependent SN2/E2 competition

3. Question 3 · medium · Amines

   Among aniline, p-nitroaniline, p-chloroaniline, and p-toluidine, which compound is the strongest base in aqueous solution?

   • A.p-Nitroaniline
   • B.p-Chloroaniline
   • C.p-Toluidine (p-methylaniline)
   • D.Aniline (unsubstituted)
   Show answer and explanation

   Answer: C. p-Toluidine (p-methylaniline)

   The -CH3 group in p-toluidine is a weak electron donor (+I, and hyperconjugation into the ring), which pushes additional electron density onto the amino nitrogen through the conjugated ring system, making p-toluidine slightly more basic than aniline; both -Cl and -NO2, being net electron-withdrawing, make the corresponding anilines less basic than aniline itself, with -NO2 the strongest deactivator of the group.

   For para-substituted anilines, basicity is controlled by whether the substituent donates or withdraws electron density from the ring (and, by conjugation, from the amino nitrogen). Ranking these four: p-toluidine (-CH3, a weak +I/hyperconjugative donor) is the most basic, slightly exceeding aniline itself, because extra electron density is pushed toward the ring and, via conjugation, helps stabilise the resulting anilinium cation's charge distribution and slightly increases nitrogen's electron density. Aniline is the unsubstituted baseline. p-Chloroaniline is slightly less basic than aniline because chlorine's inductive electron-withdrawal (-I) outweighs its weak resonance donation (+M) at this position. p-Nitroaniline is by far the weakest base of the four because -NO2 is powerfully electron-withdrawing both by induction and by resonance (its own conjugation with the ring competes directly with and diminishes the amino lone pair's availability), giving the overall order p-toluidine > aniline > p-chloroaniline > p-nitroaniline.

   Common mistake: Assuming any ring substituent uniformly increases basicity, or reversing the direction of -NO2's effect

4. Question 4 · medium · Amines

   Rank the following in order of DECREASING base strength (in aqueous solution): aniline, p-toluidine (p-methylaniline), p-nitroaniline.

   • A.p-nitroaniline > aniline > p-toluidine
   • B.aniline > p-nitroaniline > p-toluidine
   • C.p-nitroaniline > p-toluidine > aniline
   • D.p-toluidine > aniline > p-nitroaniline
   Show answer and explanation

   Answer: D. p-toluidine > aniline > p-nitroaniline

   The electron-donating methyl group in p-toluidine increases electron density on nitrogen (raising basicity above aniline), while the strongly electron-withdrawing nitro group in p-nitroaniline pulls electron density away from nitrogen through both resonance and induction (drastically lowering basicity), giving the order p-toluidine > aniline > p-nitroaniline.

   All three are aromatic primary amines, so their basicity is governed by how substituents on the ring affect availability of the nitrogen lone pair, in addition to the baseline resonance delocalization already present in aniline. In p-toluidine, the para-methyl group is weakly electron-donating (+I, hyperconjugation) and pushes additional electron density into the ring and onto nitrogen, partially compensating for the resonance loss and making it a slightly stronger base than plain aniline. In p-nitroaniline, the para-nitro group is strongly electron-withdrawing by both resonance (extending conjugation to pull the amine lone pair deeply into the nitro group's resonance system, forming a quinonoid-like structure) and induction, drastically reducing electron density on nitrogen and making it a much weaker base than aniline — nitroaniline's basicity is so reduced that it is barely basic in comparison. Hence the overall order of decreasing base strength is p-toluidine > aniline > p-nitroaniline.

   Common mistake: Forgetting that electron-withdrawing para substituents can conjugate with the amino lone pair, dramatically reducing basicity

5. Question 5 · medium · Amines

   The C-N-C bond angle in aniline is slightly different from the ideal tetrahedral angle (109.5 degrees) expected for a purely sp3 nitrogen. What does this indicate about aniline's nitrogen hybridization?

   • A.The bond angle difference is due to steric repulsion from ortho hydrogens only, with no electronic contribution
   • B.Aniline's nitrogen is purely sp3 with no deviation from a typical alkylamine's geometry
   • C.Aniline's nitrogen is purely sp, giving a linear arrangement
   • D.Aniline's nitrogen shows partial sp2 character, intermediate between sp3 (pyramidal) and sp2 (planar), due to resonance delocalization of the lone pair into the ring, which flattens the geometry somewhat compared to a simple alkylamine
   Show answer and explanation

   Answer: D. Aniline's nitrogen shows partial sp2 character, intermediate between sp3 (pyramidal) and sp2 (planar), due to resonance delocalization of the lone pair into the ring, which flattens the geometry somewhat compared to a simple alkylamine

   Because the nitrogen lone pair in aniline conjugates with the aromatic ring's pi system, the nitrogen adopts partial sp2 character, flattening its geometry somewhat compared to a purely pyramidal sp3 alkylamine, which explains the deviation from the ideal tetrahedral bond angle.

   In a simple alkylamine like methylamine, nitrogen is essentially sp3 hybridized with a pyramidal geometry, similar to ammonia. In aniline, however, the nitrogen lone pair is not purely localized — it participates in resonance conjugation with the adjacent aromatic ring's pi-electron system (this conjugation is what makes the ring activated and ortho/para-directing). For this conjugation to be effective, the lone pair's p-orbital character must align with the ring's pi system, which favors a more planar (sp2-like) arrangement around nitrogen rather than a fully pyramidal sp3 one. As a result, aniline's nitrogen exhibits partial sp2 character — a geometry intermediate between the fully pyramidal sp3 nitrogen of an alkylamine and the fully planar sp2 nitrogen of an amide (where conjugation with a carbonyl is even more complete). This is reflected in a slightly different C-N-C(H) bond angle and a lower inversion barrier at nitrogen compared to purely aliphatic amines, and it is a direct structural consequence of the resonance that also explains aniline's reduced basicity.

   Common mistake: Assuming aniline's nitrogen has identical hybridization/geometry to a simple aliphatic amine, ignoring resonance flattening

6. Question 6 · medium · Amines

   A mixture of n-propylamine (1 degree), N-methylethanamine (2 degree), and N,N-dimethylethanamine (3 degree) needs to be separated into pure components. Outline the correct use of the Hinsberg method for this separation.

   • A.Treat the mixture with benzenesulfonyl chloride and excess KOH; the tertiary amine remains as an unreacted separate layer (removed by separation), the primary amine's sulfonamide dissolves as a soluble salt (separated from insoluble solid), and mild acid hydrolysis of the isolated soluble salt regenerates pure primary amine, while the insoluble secondary sulfonamide precipitate is filtered off and hydrolyzed separately to regenerate pure secondary amine
   • B.Simple fractional distillation alone is sufficient and the Hinsberg reagent is unnecessary for this separation
   • C.All three amines form identical sulfonamide precipitates that cannot be distinguished by KOH solubility
   • D.The tertiary amine reacts fastest with benzenesulfonyl chloride, so it should be removed first by filtration of its sulfonamide precipitate
   Show answer and explanation

   Answer: A. Treat the mixture with benzenesulfonyl chloride and excess KOH; the tertiary amine remains as an unreacted separate layer (removed by separation), the primary amine's sulfonamide dissolves as a soluble salt (separated from insoluble solid), and mild acid hydrolysis of the isolated soluble salt regenerates pure primary amine, while the insoluble secondary sulfonamide precipitate is filtered off and hydrolyzed separately to regenerate pure secondary amine

   Hinsberg's reagent followed by KOH separates the mixture because each amine class behaves distinctly: the tertiary amine is unreactive (separable as a distinct layer), the primary amine's sulfonamide dissolves in KOH (separable by filtration from the insoluble secondary sulfonamide), and each fraction can then be hydrolyzed back to the pure free amine.

   This is the classic practical application of the Hinsberg test for preparative separation, not just qualitative identification. Step 1: treat the mixture with benzenesulfonyl chloride; the tertiary amine (N,N-dimethylethanamine) has no N-H and does not react, remaining as a separate, immiscible unreacted amine layer that can be physically separated (e.g., by extraction/distillation). Step 2: the primary amine (n-propylamine) forms a sulfonamide with one remaining acidic N-H, and upon adding excess KOH, this is deprotonated to a water-soluble salt. Step 3: the secondary amine (N-methylethanamine) forms a sulfonamide with no remaining N-H, which stays as an insoluble solid precipitate even in excess KOH. Step 4: filter off the insoluble secondary sulfonamide precipitate from the aqueous solution containing the dissolved primary amine salt. Step 5: acidify and then hydrolyze (or simply acid-hydrolyze) the filtrate to regenerate pure primary amine, and separately hydrolyze the isolated solid sulfonamide (typically with strong acid or base under more forcing conditions) to regenerate pure secondary amine. This sequential exploitation of differing N-H content and resulting KOH solubility is exactly how Hinsberg's method achieves practical, complete separation of a three-component amine mixture.

   Common mistake: Believing physical methods like distillation alone can cleanly separate amines of similar boiling points without chemical differentiation

7. Question 7 · medium · Amines

   A qualitative organic analysis lab has an unlabeled liquid known to be one of: n-propylamine, N-methylethanamine, or N,N-dimethylethanamine. The student runs the carbylamine test first (negative: no foul odor), then the Hinsberg test on a fresh sample, observing an insoluble precipitate that persists in excess KOH. What is the identity of the unknown?

   • A.n-Propylamine
   • B.N-methylethanamine (a secondary amine)
   • C.N,N-dimethylethanamine (a tertiary amine)
   • D.The data is contradictory and no single compound fits both observations
   Show answer and explanation

   Answer: B. N-methylethanamine (a secondary amine)

   A negative carbylamine test rules out the primary amine (which requires two N-H bonds and would give a positive test), while an insoluble Hinsberg precipitate that persists in excess KOH is diagnostic of a secondary amine (its sulfonamide has no remaining N-H to be deprotonated); together these two observations uniquely identify N-methylethanamine.

   This is a combined diagnostic reasoning problem using two class-selective tests together. The carbylamine test requires two N-H bonds on nitrogen (found only in primary amines) to proceed through the double dehydrohalogenation pathway with dichlorocarbene; a negative result (no foul odor) therefore rules out n-propylamine (the primary amine option) since it would have given a strongly positive result. This leaves either the secondary amine (N-methylethanamine) or the tertiary amine (N,N-dimethylethanamine) as candidates. The Hinsberg test then distinguishes between these two: a tertiary amine has no N-H bond at all and does NOT react with benzenesulfonyl chloride, so it would produce NO precipitate whatsoever (the unreacted amine would simply remain as a separate immiscible layer) — but the student did observe an insoluble precipitate, which rules out the tertiary amine option. A secondary amine, in contrast, DOES react with benzenesulfonyl chloride to form a sulfonamide that has no remaining N-H bond (both original positions being now occupied by the two carbon substituents and the sulfonyl group), making the precipitate insoluble even in excess KOH — this exactly matches both observations. Therefore, the unknown must be N-methylethanamine, the secondary amine.

   Common mistake: Using only one diagnostic test result in isolation instead of combining both tests to narrow down to a unique amine class

8. Question 8 · easy · Diazonium salts

   Which of the following best describes the physical state/handling of benzenediazonium chloride in practice?

   • A.It is always isolated as a stable dry crystalline solid stored at room temperature
   • B.It is generally used immediately as a cold aqueous solution rather than isolated and stored
   • C.It exists only as a gas above 0°C and cannot be handled in solution
   • D.It is insoluble in water and must be handled as a suspension
   Show answer and explanation

   Answer: B. It is generally used immediately as a cold aqueous solution rather than isolated and stored

   Aryldiazonium salts are thermally unstable and (especially in dry form) potentially explosive, so they are conventionally generated and used immediately as cold aqueous solutions rather than isolated as dry solids.

   Because diazonium salts decompose on warming (and dry diazonium salts can be shock-sensitive/explosive), standard laboratory and industrial practice is to generate the diazonium salt in ice-cold aqueous solution and use it immediately in the next step (Sandmeyer, Gattermann, coupling, hydrolysis) without isolating it as a solid.

   Common mistake: Assuming diazonium salts can be stored like ordinary stable salts

Questions about Organic Compounds Containing Nitrogen for NEET

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