Q.Classify the following amines as primary, secondary or tertiary:

Showing the 12 most recent of 14 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Choose the incorrect statement. (A) Propan-2-amine can be obtained by reacting acetoxime with $\mathrm{Na} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ (B) Aniline cannot be prepared by Phthalimide reaction (C) The decreasing order of basic strength of amines in aqueous solution is Ethanamine > $\mathrm{N}, \mathrm{N}$-Dimethylaniline > Benzenamine (D) Fluorobenzene cannot be prepared from Benzenediazonium chloride by Sandmeyer's reaction because Fluorination of the Diazonium salt is highly endothermic in nature

   ›Reveal solutionSolution

   Checking each statement, A, B and C are correct; D is the incorrect statement — fluorobenzene is not made by Sandmeyer's reaction (it is made by the Balz–Schiemann route), but the reason given ("fluorination of the diazonium salt is highly endothermic") is not the valid explanation, so the statement is wrong.

   (A) Correct. Acetoxime $(CH_3)_2C=NOH$ on reduction with Na/$C_2H_5OH$ gives $(CH_3)_2CH\text{-}NH_2$, propan-2-amine. Oximes reduce to primary amines. ✓

   (B) Correct. The Gabriel (phthalimide) synthesis needs an SN2 displacement on the alkyl halide; aryl halides do not undergo this, so aniline cannot be prepared by the phthalimide reaction. The statement is true. ✓

   (C) Correct. Basicity in water: ethanamine (aliphatic 1° amine) is the strongest; among the aromatics, $N,N$-dimethylaniline is more basic than aniline (two $+I$ methyl groups on N). Order ethanamine $>$ $N,N$-dimethylaniline $>$ benzenamine. ✓

   (D) Incorrect. It is true that fluorobenzene is not obtained by Sandmeyer's reaction (Sandmeyer uses Cu(I) salts for Cl, Br, CN). But aryl fluorides are made by the Balz–Schiemann reaction (heating the diazonium tetrafluoroborate), not blocked by any "highly endothermic fluorination" of the diazonium salt — indeed C–F bond formation is strongly exothermic. The reasoning is false, making this the incorrect statement.

   ✓Final answer

   The correct option is (D) — Fluorobenzene cannot be prepared from Benzenediazonium chloride by Sandmeyer's reaction because Fluorination of the Diazonium salt is highly endothermic in nature.

2. KCET 2025Set D-41 markMCQ
   Q.Match the following with their pKa values

   Acid	pKa
   (I) Phenol

   (a) 16 | | (II) p-Nitrophenol |

   (b) 0.78 | | (III) Ethyl alcohol |

   (c) 10 | | (IV) Picric acid |

   (d) 7.1 | (A) I – c, II – d, III – a, IV – b (B) I – a, II – d, III – c, IV – b (C) I – a, II – b, III – c, IV – d (D) I – b, II – a, III – d, IV – c

   ›Reveal solutionSolution

   Rank the four compounds by acid strength using resonance and the $-\mathrm{NO_2}$ electron-withdrawing effect, then assign the $\mathrm{p}K_a$ values in the reverse order (stronger acid $\Rightarrow$ smaller $\mathrm{p}K_a$).

   Step 1 — The governing principle.

   $\mathrm{p}K_a = -\log K_a$

   So a stronger acid has a larger $K_a$ and therefore a SMALLER $\mathrm{p}K_a$. An acid is strong when its conjugate base is stable — i.e. when the negative charge left behind after losing $\mathrm{H^+}$ can be spread out.

   Step 2 — Rank the four species by conjugate-base stability.

   (III) Ethyl alcohol, $\mathrm{C_2H_5OH}$ — the ethoxide ion $\mathrm{C_2H_5O^-}$ has its negative charge localised entirely on oxygen, with no resonance at all. Worse, the ethyl group is electron-releasing (+I), which pushes electron density onto the already-negative oxygen and destabilises it further. ⇒ weakest acid ⇒ highest $\mathrm{p}K_a$ = 16 (a).

   (I) Phenol, $\mathrm{C_6H_5OH}$ — the phenoxide ion delocalises its negative charge into the benzene ring by resonance (onto the ortho and para carbons). This resonance stabilisation makes phenol far more acidic than an alcohol — about $10^6$ times so. ⇒ $\mathrm{p}K_a$ = 10 (c).

   (II) p-Nitrophenol — a $-\mathrm{NO_2}$ group at the para position is strongly electron-withdrawing by both $-I$ and $-R$ effects. Crucially, at the para position the nitro group can accept the negative charge by resonance directly onto its own oxygen atoms, giving the phenoxide extra stabilisation on top of the ring delocalisation. ⇒ markedly more acidic than phenol ⇒ $\mathrm{p}K_a$ = 7.1 (d).

   (IV) Picric acid (2,4,6-trinitrophenol) — three nitro groups (two ortho, one para), all withdrawing electrons and all able to delocalise the negative charge of the phenoxide. Their effects add up, making the conjugate base extremely stable. Picric acid is so acidic it rivals a mineral acid. ⇒ strongest acid ⇒ lowest $\mathrm{p}K_a$ = 0.78 (b).

   Step 3 — Assemble the acidity order and the matching.

   $\underbrace{\text{Picric acid}}_{0.78} \;>\; \underbrace{\text{p-nitrophenol}}_{7.1} \;>\; \underbrace{\text{Phenol}}_{10} \;>\; \underbrace{\text{Ethyl alcohol}}_{16} \qquad (\text{decreasing acid strength})$

   	Acid	$\mathrm{p}K_a$	Label
   I	Phenol	10	c
   II	p-Nitrophenol	7.1	d
   III	Ethyl alcohol	16	a
   IV	Picric acid	0.78	b

   So the matching is I – c, II – d, III – a, IV – b, which is option (A).

   Quick elimination check: every wrong option assigns ethyl alcohol (III) something other than 16, or gives phenol the picric-acid value — both chemically impossible.

   ✓Final answer

   The correct option is (A) — I – c, II – d, III – a, IV – b.

   ANSWER: A

3. KCET 2025Set D-41 markMCQ
   Q.Arrange the following compounds in their decreasing order of reactivity towards Nucleop addition reaction. $\mathrm{CH_3COCH_3, CH_3COC_2H_5, CH_3CHO}$ (A) $\mathrm{CH_3CHO > CH_3COCH_3 > CH_3COC_2H_5}$ (B) $\mathrm{CH_3COCH_3 > CH_3CHO > CH_3COC_2H_5}$ (C) $\mathrm{CH_3COC_2H_5 > CH_3COCH_3 > CH_3CHO}$ (D) $\mathrm{CH_3CHO > CH_3COC_2H_5 > CH_3COCH_3}$

   ›Reveal solutionSolution

   Rank by the two effects that control nucleophilic addition — the $+I$ (electron-releasing) effect of alkyl groups and their steric bulk. Both make more/larger alkyl groups less reactive, so aldehyde > methyl ketone > ethyl ketone.

   Step 1 — What makes a carbonyl reactive towards a nucleophile.

   The $\mathrm{C{=}O}$ bond is polarised because oxygen is far more electronegative than carbon:

   $\mathrm{{}^{\delta+}C{=}O^{\delta-}}$

   A nucleophile attacks the electron-deficient carbonyl carbon, and in doing so the carbon rehybridises from planar $sp^2$ to tetrahedral $sp^3$. Two factors therefore govern the rate:

   1. Electronic factor: the greater the positive charge ($\delta^+$) on the carbonyl carbon, the more strongly it attracts the nucleophile → faster.
   2. Steric factor: the more crowded the carbonyl carbon, the harder it is for the nucleophile to approach, and the more strained the resulting crowded $sp^3$ (tetrahedral) product → slower.

   Step 2 — Compare the three compounds by their substituents.

   Compound	Groups on the carbonyl C
   $\mathrm{CH_3CHO}$ (ethanal)	one $\mathrm{CH_3}$ + one H
   $\mathrm{CH_3COCH_3}$ (propanone)	two $\mathrm{CH_3}$
   $\mathrm{CH_3COC_2H_5}$ (butan-2-one)	one $\mathrm{CH_3}$ + one $\mathrm{C_2H_5}$

   Step 3 — Apply the electronic ($+I$) factor.

   Alkyl groups are electron-releasing ($+I$ effect). Pushing electron density towards the carbonyl carbon reduces its $\delta^+$, making it less attractive to a nucleophile.

   • $\mathrm{CH_3CHO}$ has only one alkyl group (the H contributes no $+I$) → largest $\delta^+$ → most reactive.
   • $\mathrm{CH_3COCH_3}$ has two alkyl groups → $\delta^+$ reduced further.
   • $\mathrm{CH_3COC_2H_5}$ has two alkyl groups, and ethyl has a stronger $+I$ effect than methyl → $\delta^+$ reduced the most → least reactive.

   Step 4 — Apply the steric factor.

   The same ordering emerges independently:

   • Ethanal's carbonyl carbon bears a tiny H — nearly unhindered.
   • Propanone bears two methyls — moderately hindered.
   • Butan-2-one bears a methyl and a bulkier ethyl — the most hindered.

   Both effects reinforce each other (which is why the trend is so reliable), giving:

   $\boxed{\mathrm{CH_3CHO} > \mathrm{CH_3COCH_3} > \mathrm{CH_3COC_2H_5}}$

   Step 5 — The general rule this illustrates.

   $\mathrm{HCHO} > \mathrm{RCHO} > \mathrm{RCOR'}$

   i.e. formaldehyde > other aldehydes > ketones towards nucleophilic addition — and within each class, reactivity falls as the alkyl groups get larger and more numerous.

   Step 6 — Rejecting the distractors.

   • (B) puts a ketone above the aldehyde — contradicts both the $+I$ and steric arguments.
   • (C) is the complete reverse of the correct order.
   • (D) correctly places the aldehyde first but then ranks the bulkier, more electron-rich ethyl methyl ketone above propanone, which is backwards.

   ✓Final answer

   The correct option is (A) — $\mathrm{CH_3CHO > CH_3COCH_3 > CH_3COC_2H_5}$.

   ANSWER: A

4. KCET 2025Set D-41 markMCQ
   Q.Which of the following reaction/s does not yield an amine? I. $\mathrm{R-X + NH_3 \xrightarrow{\Delta} (alc)}$ II. $\mathrm{R-C \equiv N \xrightarrow{H_2/Ni, Na(Hg)/C_2H_5OH}}$ III. $\mathrm{R-C \equiv N + H_2O \xrightarrow{H^+}}$ IV. $\mathrm{R-C=NH_2 + 4[H] \xrightarrow{i) LiAlH_4, ii) H_2O}}$ (A) Both I and III (B) Only II (C) Only III (D) Both II and IV

   ›Reveal solutionSolution

   Check each route: three are amine-forming reductions/substitutions; nitrile hydrolysis (III) gives a carboxylic acid, so it is the only one that fails.

   Step 1 — Reaction I: $\mathrm{R-X + NH_3 \xrightarrow{\Delta,\ alc.}}$

   This is ammonolysis of an alkyl halide — a nucleophilic substitution in which ammonia attacks the carbon bearing the halogen:

   $\mathrm{R-X + NH_3 \longrightarrow R-NH_2 + HX}$

   It does give an amine (in practice a mixture of $1^\circ$, $2^\circ$, $3^\circ$ amines and the quaternary salt, because the product amine is itself nucleophilic). Since the question only asks whether an amine is obtained, I yields an amine.

   Step 2 — Reaction II: $\mathrm{R-C\equiv N \xrightarrow{H_2/Ni \ \text{or}\ Na(Hg)/C_2H_5OH}}$

   This is the reduction (Mendius reaction) of a nitrile. Catalytic hydrogenation over Ni, or nascent hydrogen from sodium amalgam in ethanol, adds hydrogen across the $\mathrm{C\equiv N}$ triple bond:

   $\mathrm{R-C\equiv N \xrightarrow{\ [H]\ } R-CH_2-NH_2}$

   This is a standard preparation of a primary amine with one carbon more than the parent halide. II yields an amine.

   Step 3 — Reaction III: $\mathrm{R-C\equiv N + H_2O \xrightarrow{H^+}}$

   Here water, not hydrogen, is the reagent, and the conditions are acidic hydrolysis. The nitrogen leaves as ammonia/ammonium and the carbon ends up as a carboxyl group:

   $\mathrm{R-C\equiv N \xrightarrow[\ H_2O\ ]{H^+} R-CONH_2 \xrightarrow[\ H_2O\ ]{H^+} R-COOH + NH_4^+}$

   The product is a carboxylic acid. No amine is formed — the nitrogen is expelled as ammonium salt. III does NOT yield an amine.

   Step 4 — Reaction IV: $\mathrm{R-C=NH \ (\text{imine/amide}) + 4[H] \xrightarrow{i)\ LiAlH_4,\ ii)\ H_2O}}$

   $\mathrm{LiAlH_4}$ is a powerful hydride reducing agent; the aqueous work-up then liberates the free base. Reduction of a $\mathrm{C=N}$ (imine) — or of an amide, which is what this route amounts to — delivers the corresponding amine:

   $\mathrm{R-CONH_2 \xrightarrow[ii)\ H_2O]{i)\ LiAlH_4} R-CH_2-NH_2}$

   IV yields an amine.

   Step 5 — Collect.

   Amine formed: I ✓, II ✓, IV ✓. Amine not formed: III only.

   Option (A) wrongly includes I, (B) wrongly names II, and (D) wrongly names II and IV — all of which do produce amines.

   ✓Final answer

   The correct option is (C) — Only III (acidic hydrolysis of a nitrile gives a carboxylic acid, not an amine).

   ANSWER: C

5. COMEDK 2025Set 2025-A1 markMCQ
   Q.Choose the correct statement: (A) Aliphatic amines are weaker bases than $\mathrm{NH}_3$ while aromatic amines are stronger bases than $\mathrm{NH}_3$. (B) Gabriel phthalimide synthesis is used for preparing both Ethyl amine and Aniline. (C) Anilinium ion is less resonance stabilised than Aniline. (D) Sec-butylamine is optically inactive because Nitrogen atom of the $-\mathrm{NH}_2$ group is achiral.

   ›Reveal solutionSolution

   The key idea is to evaluate each statement about amine basicity, synthesis, resonance, and chirality. Only statement (C) is correct: anilinium ion is less resonance-stabilized than aniline.

   Concept & Intuition

   Amines are organic derivatives of ammonia. Their basicity depends on how well the lone pair on nitrogen is available for protonation. Resonance, inductive effects, and hybridization all matter. Gabriel phthalimide synthesis is a classic method for making primary amines, but it fails for aromatic amines like aniline. Chirality at nitrogen is tricky because nitrogen inverts rapidly, so sec-butylamine is not optically active due to that inversion, not because the nitrogen is achiral. Let’s check each option.

   1. Option (A): Aliphatic amines are stronger bases than NH₃ because alkyl groups donate electron density (inductive effect), making the lone pair more available. Aromatic amines are weaker bases than NH₃ because the lone pair is delocalized into the benzene ring (resonance), reducing availability. So the statement says the opposite — false.

   2. Option (B): Gabriel phthalimide synthesis uses phthalimide and an alkyl halide to make primary amines. It works for alkyl halides (e.g., ethyl bromide → ethylamine). But aniline cannot be made this way because aryl halides (like chlorobenzene) do not undergo nucleophilic substitution easily under these conditions. So false.

   3. Option (C): Aniline has resonance between the nitrogen lone pair and the benzene ring, stabilizing the molecule. When aniline is protonated to form anilinium ion, the lone pair is used to bind H⁺, so resonance is lost. The anilinium ion is therefore less resonance-stabilized than aniline. This is correct.

   4. Option (D): Sec-butylamine has a chiral carbon (the carbon attached to NH₂ has four different groups), so the molecule is optically active. The nitrogen atom itself is not a chiral center because the lone pair inverts rapidly (like an umbrella flipping), so optical activity is not due to nitrogen chirality. But the statement says sec-butylamine is optically inactive — false, because the carbon is chiral.

   Watch out

   A common mistake is thinking that nitrogen inversion makes a molecule optically inactive even when a chiral carbon is present. In sec-butylamine, the carbon is the chiral center, not the nitrogen.

   Tip

   For basicity comparisons: alkyl groups push electrons → stronger base; resonance with an aromatic ring pulls electron density → weaker base.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

6. COMEDK 2025Set 2025-E1 markMCQ
   Q.Choose the correct statement from the options given (A) The decreasing order of basic nature of the following amines is: Methylamine $>$ Dimethylamine $>$ Trimethylamine $>$ Aniline. (B) The intermolecular bonding in primary amines is stronger than in secondary amines. (C) Benzene diazonium chloride when reacted with Aniline in presence of dil. HCl at 273 K yields $\mathrm{C}_6 \mathrm{H}_5-\mathrm{N}=\mathrm{N}-\mathrm{NH}-\mathrm{C}_6 \mathrm{H}_5$ (D) On heating an aliphatic primary amine with $\mathrm{CHCl}_3$ in presence of Ethanolic KOH , a Nitrile is formed

   ›Reveal solutionSolution

   Primary amines (two N–H bonds) hydrogen-bond more strongly than secondary amines (one N–H), so statement (B) is the correct one.

   Assess each option:

   • (A) In aqueous solution the basicity order of methylamines is (CH$_3$)$_2$NH > CH$_3$NH$_2$ > (CH$_3$)$_3$N > aniline (a balance of +I, solvation and steric effects). The stated order MeNH$_2$ > Me$_2$NH is wrong.
   • (B) A primary amine R–NH$_2$ has two N–H bonds and can form more hydrogen bonds than a secondary amine R$_2$NH (only one N–H). Hence primary amines have stronger intermolecular H-bonding (and higher boiling points than secondary amines of comparable mass) — correct.
   • (C) Coupling of benzene diazonium chloride with aniline gives, under the usual conditions, an azo dye (p-aminoazobenzene, C-coupling), not simply the stated diazoamino product — statement not correct.
   • (D) The carbylamine (isocyanide) reaction of a primary amine with CHCl$_3$ + ethanolic KOH gives an isocyanide (R–NC), not a nitrile (R–CN) — incorrect.

   The correct statement is (B).

   ✓Final answer

   The correct option is (B) — The intermolecular bonding in primary amines is stronger than in secondary amines.

7. COMEDK 2024Set 2024-M1 markMCQ
   Q.Which one of the following shows the correct increasing order of basic nature of the given compounds? A: Phenylmethanamine B: N-Ethylethanamine C:N, N-Dimethylaniline D : N, N-Dimethylmethanamine (A) $\mathrm{B}<\mathrm{C}<\mathrm{A}<\mathrm{D}$ (B) A < D < B < C (C) D < B < A < C (D) C < A < D < B

   ›Reveal solutionSolution

   Ranking the four amines by the availability of the nitrogen lone pair gives $\text{C} < \text{A} < \text{D} < \text{B}$ (aromatic amine weakest, secondary aliphatic strongest) — option (D).

   Identify the compounds

   • A — Phenylmethanamine (benzylamine), $\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$: a primary aliphatic amine; the ring is one carbon away, so the lone pair is not delocalised.
   • B — N-Ethylethanamine (diethylamine), $(\text{C}_2\text{H}_5)_2\text{NH}$: a secondary aliphatic amine.
   • C — N,N-Dimethylaniline, $\text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2$: an aromatic amine; the N lone pair is delocalised into the ring, making it much less available.
   • D — N,N-Dimethylmethanamine (trimethylamine), $(\text{CH}_3)_3\text{N}$: a tertiary aliphatic amine.

   Reasoning

   The more available the nitrogen lone pair, the stronger the base.

   1. C is weakest. In N,N-dimethylaniline the lone pair is conjugated into the benzene ring, so it is least available — aromatic amines are far weaker bases than aliphatic amines ($\text{p}K_{aH} \approx 5.1$).
   2. A next. Benzylamine is a primary aliphatic amine; the ring is insulated by the $\text{CH}_2$, so it behaves as an ordinary primary amine ($\text{p}K_{aH} \approx 9.3$).
   3. D next. Trimethylamine (tertiary) is a stronger base than a primary amine but suffers reduced solvation of its conjugate acid ($\text{p}K_{aH} \approx 9.8$).
   4. B strongest. Diethylamine (secondary) has the best balance of inductive donation and cation solvation, giving the highest basicity ($\text{p}K_{aH} \approx 11$).

   Increasing basic strength: $\text{C} < \text{A} < \text{D} < \text{B}$.

   ✓Final answer

   Increasing order of basic nature is $\mathbf{C < A < D < B}$ — option (D).

8. COMEDK 2023Set 2023-E1 markMCQ
   Q.Select the strongest base from the given compounds: [A] p- $\mathrm{NO}_2-\mathrm{C}_6 \mathrm{H}_4 \mathrm{NH}_2$ [B] $\mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}_2-\mathrm{NH}_2$ [C] $\mathrm{m}-\mathrm{NO}_2-\mathrm{C}_6 \mathrm{H}_4 \mathrm{NH}_2$ [D] $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$ (A) [A] (B) [C] (C) [B] (D) [D]

   ›Reveal solutionSolution

   Strongest base = [B] = benzylamine, which is listed as option (C).

   Concept: basicity of amines depends on the availability of the lone pair on nitrogen.

   [D] Aniline, C6H5-NH2: the N lone pair is delocalised into the benzene ring -> weak base.

   [A] p-Nitroaniline: the -NO2 group withdraws electrons by both -I and -R (and para -R is strongly deactivating) -> even weaker base (weakest).

   [C] m-Nitroaniline: -NO2 withdraws by -I only from the meta position -> weaker than aniline but stronger than the para isomer.

   [B] Benzylamine, C6H5-CH2-NH2: the nitrogen is attached to an sp3 CH2, NOT directly to the ring, so its lone pair is NOT in conjugation with the ring and remains fully available. It behaves essentially like an aliphatic amine -> STRONGEST base of the set.

   Strongest base = [B] = benzylamine, which is listed as option (C).

   ✓Final answer

   The correct option is (C) — [B]

   ANSWER: C

9. COMEDK 2023Set 2023-M1 markMCQ
   Q.Rank the following compounds in order of increasing basicity. (A) 4 < 2 < 1 < 3 (B) 4 < 1 < 3 < 2 (C) 4 < 3 < 1 < 2 (D) 2 < 1 < 3 < 4

   ›Reveal solutionSolution

   Basicity depends on the availability of the nitrogen lone pair for protonation. The order of increasing basicity is benzamide (4) < o‑nitroaniline (3) < aniline (1) < benzylamine (2), so the correct ranking is 4 < 3 < 1 < 2, which corresponds to option (C).

   The key concept is lone‑pair availability. A base is strong when its lone pair is “free” to accept a proton. Anything that stabilizes the lone pair (by delocalization or electron withdrawal) makes the compound less basic; anything that pushes electron density toward nitrogen makes it more basic. Here, all four compounds have a nitrogen that can be protonated, but the groups attached to the benzene ring dramatically affect how much that lone pair is tied up in resonance or pulled away by inductive effects.

   Let’s work through each compound step by step.

   1. Compound 4 – Benzamide (C₆H₅CONH₂) The nitrogen is part of an amide group. The lone pair on nitrogen is strongly delocalized into the adjacent carbonyl (C=O) via resonance:

   $\text{R–C(=O)–NH}_2 \longleftrightarrow \text{R–C(O⁻)–NH}_2^+$

   This resonance makes the lone pair much less available for protonation. Additionally, the carbonyl oxygen is more electronegative and pulls electron density inductively. Result: benzamide is the least basic of the four.

   2. Compound 3 – o‑Nitroaniline (2‑nitroaniline)

      Here the NH₂ is directly on the ring, but an ortho nitro group (NO₂) is present. The nitro group is strongly electron‑withdrawing both inductively (through σ‑bonds) and by resonance (it can accept electron density from the ring). This withdrawal reduces electron density on the NH₂ nitrogen. Moreover, the ortho position allows a direct resonance interaction: the lone pair on NH₂ can be delocalized into the nitro group, further stabilizing the neutral amine and making it harder to protonate. So o‑nitroaniline is less basic than aniline (compound 1), but still more basic than benzamide because the amide resonance is even more effective at tying up the lone pair.

   3. Compound 1 – Aniline (C₆H₅NH₂)

      The NH₂ is directly attached to the benzene ring. The lone pair on nitrogen can be delocalized into the aromatic ring (resonance with the π‑system). This delocalization makes aniline a weaker base than aliphatic amines (like benzylamine). However, there is no strong electron‑withdrawing group like NO₂ or C=O to further reduce basicity. Aniline is therefore more basic than compounds 3 and 4, but less basic than benzylamine.

   4. Compound 2 – Benzylamine (C₆H₅CH₂NH₂)

      Here the nitrogen is separated from the benzene ring by a CH₂ group. This insulating methylene group prevents direct resonance between the nitrogen lone pair and the aromatic ring. The only effect of the benzene ring is a weak inductive withdrawal through the CH₂, which is very small. Benzylamine behaves essentially like a primary aliphatic amine (e.g., methylamine) and is the most basic of the four.

   Watch out

   A common mistake is to think that because aniline’s lone pair is delocalized into the ring, it is “very weak.” But compared to amides and nitroanilines, aniline is actually moderately basic. The amide resonance is far more effective at stabilizing the neutral form than simple aromatic delocalization.

   Tip

   A quick mental shortcut: Amide < Nitroaniline < Aniline < Benzylamine — the farther the nitrogen is from the ring (and from electron‑withdrawing groups), the stronger the base.

   Now assemble the order from least basic to most basic:

   4 (benzamide) < 3 (o‑nitroaniline) < 1 (aniline) < 2 (benzylamine).

   This matches option (C).

   ✓Final answer

   The correct option is (C).

   ANSWER: C

10. COMEDK 2022Set 20221 markMCQ
    Q.Which of the following is highly basic? (A) Diphenylamine (B) Benzylamine (C) Aniline (D) Triphenylamine

    ›Reveal solutionSolution

    Most basic = benzylamine.

    Concept: basicity of amines depends on the availability of the nitrogen lone pair.

    • Benzylamine, C6H5-CH2-NH2: the ring is insulated from N by an sp3 CH2, so the lone pair is NOT delocalised into the ring. It behaves like an aliphatic amine and is strongly basic (pKb ~ 4.7).
    • Aniline, C6H5-NH2: the lone pair is delocalised into the ring -> weakly basic (pKb ~ 9.4).
    • Diphenylamine: two rings pull the lone pair -> far weaker base.
    • Triphenylamine: three rings + steric crowding -> essentially non-basic.

    Most basic = benzylamine.

    ✓Final answer

    The correct option is (B) — Benzylamine

    ANSWER: B

11. KCET 2021Set B-21 markMCQ
    Q.Which of the following compound on heating given $N_2O$? (A) $\text{Pb(NO}_3)_2$ (B) $\text{NH}_4\text{NO}_3$ (C) $\text{NH}_4\text{NO}_2$ (D) $\text{NaNO}_3$

    ›Reveal solutionSolution

    Gentle thermal decomposition of ammonium nitrate is the lab preparation of nitrous oxide: $\mathrm{NH_4NO_3 \to N_2O + 2H_2O}$.

    1. The concept — internal redox in an ammonium salt.

    In $\mathrm{NH_4NO_3}$ the same compound contains nitrogen in two very different oxidation states: $-3$ in the $\mathrm{NH_4^+}$ cation and $+5$ in the $\mathrm{NO_3^-}$ anion. On heating they undergo an intramolecular redox reaction, meeting at the intermediate state $+1$ — which is exactly the oxidation state of N in $\mathrm{N_2O}$:

    $\mathrm{NH_4NO_3} \xrightarrow{\;\Delta\;(\sim 250^\circ\text{C})\;} \mathrm{N_2O} + 2\mathrm{H_2O}$

    This is the standard laboratory preparation of nitrous oxide ("laughing gas").

    2. Why the other three do not give $\mathrm{N_2O}$.

    • (A) $\mathrm{Pb(NO_3)_2}$ — a heavy-metal nitrate; it decomposes to the oxide, giving brown $\mathrm{NO_2}$ and $\mathrm{O_2}$:

    $2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}$

    • (C) $\mathrm{NH_4NO_2}$ — here the nitrogen states are $-3$ and $+3$; they meet at $0$, giving dinitrogen, not $\mathrm{N_2O}$ (this is the lab preparation of pure $\mathrm{N_2}$):

    $\mathrm{NH_4NO_2} \xrightarrow{\Delta} \mathrm{N_2} + 2\mathrm{H_2O}$

    • (D) $\mathrm{NaNO_3}$ — an alkali-metal nitrate; it merely loses oxygen to become the nitrite:

    $2\mathrm{NaNO_3} \xrightarrow{\Delta} 2\mathrm{NaNO_2} + \mathrm{O_2}$

    3. The distinction to remember.

    $\mathrm{NH_4NO_2} \to \mathrm{N_2}$ (nitrite $\to$ nitrogen); $\mathrm{NH_4NO_3} \to \mathrm{N_2O}$ (nitrate $\to$ nitrous oxide). The extra oxygen in the nitrate is what raises the product's nitrogen from $0$ to $+1$.

    ✓Final answer

    The correct option is (B) — $\mathrm{NH_4NO_3}$.

    ANSWER: B

12. KCET 2021Set B-21 markMCQ
    Q.$K_a$ values for acids $\mathrm{H_2SO_3}$, $\mathrm{HNO_2}$, $\mathrm{CH_3COOH}$ and HCN are respectively $1.3 \times 10^{-2}$, $4 \times 10^{-4}$, $1.8 \times 10^{-5}$ and $4 \times 10^{-10}$, which of the above acids produces stronger conjugate base in aqueous solution? (A) $\mathrm{H_2SO_3}$ (B) $\mathrm{HNO_2}$ (C) $\mathrm{CH_3COOH}$ (D) HCN

    ›Reveal solutionSolution

    The strength of a conjugate base is inversely related to the acid’s $K_a$ — the weakest acid gives the strongest conjugate base. HCN has the smallest $K_a$ ($4 \times 10^{-10}$), so its conjugate base ($\mathrm{CN^-}$) is the strongest. The correct option is (D).

    The key idea is the conjugate acid–base relationship: for any acid HA, its conjugate base A⁻ is what remains after the acid donates a proton. A strong acid readily gives up its proton, leaving behind a weak, stable conjugate base that has little tendency to re-accept a proton. Conversely, a weak acid holds its proton tightly, so its conjugate base is much more eager to grab a proton — that is, it is a stronger base.

    Quantitatively, for a conjugate pair in water, the product of the acid dissociation constant $K_a$ and the base dissociation constant $K_b$ of the conjugate base equals $K_w$ ($1.0 \times 10^{-14}$ at 25°C):

    $K_a \times K_b = K_w$

    So $K_b = K_w / K_a$. A smaller $K_a$ means a larger $K_b$ — a stronger conjugate base. Therefore, to find which acid produces the strongest conjugate base, we simply look for the acid with the smallest $K_a$.

    1. List the given $K_a$ values clearly:

       • $\mathrm{H_2SO_3}$: $1.3 \times 10^{-2}$
       • $\mathrm{HNO_2}$: $4 \times 10^{-4}$
       • $\mathrm{CH_3COOH}$: $1.8 \times 10^{-5}$
       • HCN: $4 \times 10^{-10}$

    2. Compare the magnitudes. The smallest $K_a$ is $4 \times 10^{-10}$, belonging to HCN. It is many orders of magnitude smaller than the next smallest ($1.8 \times 10^{-5}$). This means HCN is by far the weakest acid in the list.

    3. Apply the inverse relationship. Since $K_b \propto 1/K_a$, the conjugate base of HCN ($\mathrm{CN^-}$) has the largest $K_b$ and is therefore the strongest base among the four conjugate bases.

    Watch out

    A common mistake is to pick the acid with the largest $K_a$ (strongest acid) thinking it produces a strong conjugate base. The opposite is true: the stronger the acid, the weaker its conjugate base. Here, $\mathrm{H_2SO_3}$ has the largest $K_a$, so its conjugate base ($\mathrm{HSO_3^-}$) is the weakest base — not what the question asks.

    Tip

    You don’t need to calculate $K_b$ values at all. Just rank the acids by $K_a$: the smallest $K_a$ wins. If the numbers are given in scientific notation, compare the exponents first — here $10^{-10}$ is clearly smaller than $10^{-2}$, $10^{-4}$, or $10^{-5}$.

    ✓Final answer

    The acid that produces the strongest conjugate base is HCN, so the correct option is (D).