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

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Consider the following $C_6H_5NH_2$ (I), $C_2H_5NH_2$ (II), $NH_3$ (III), $(C_2H_5)_2NH$ (IV) The correct pair with lowest $pK_b$ and highest $pK_b$ values respectively is (A) IV, I (B) III, II (C) II, IV (D) I, IV

   ›Reveal solutionSolution

   Among the four amines, diethylamine is the strongest base (lowest $pK_b$) and aniline is the weakest (highest $pK_b$), because aniline's lone pair is tied up in ring resonance while diethylamine benefits from two electron-donating ethyl groups.

   Concept and Intuition

   $pK_b$ measures how weak a base is — lower $pK_b$ means a stronger base (more available lone pair on nitrogen to accept a proton), higher $pK_b$ means a weaker base. For aromatic amines like aniline, the nitrogen lone pair is delocalised into the benzene ring (resonance), making it far less available and hence aniline is a much weaker base than any simple alkyl amine. Among aliphatic amines in aqueous solution, secondary amines are generally more basic than primary amines (more alkyl groups donating electron density through +I effect, and $NH_3$ is even less basic still since it has none), though steric hindrance and solvation partially offset this — this net ordering, confirmed by standard $pK_b$ tables, is: diethylamine > ethylamine > ammonia > aniline (in basicity).

   Step-by-Step Solution

   1. $C_6H_5NH_2$ (I): aniline — the lone pair on N delocalises into the aromatic ring, making it a very weak base. Highest $pK_b$.
   2. $C_2H_5NH_2$ (II): a simple primary alkyl amine, moderately basic — one ethyl group's +I effect.
   3. $NH_3$ (III): no alkyl groups at all, less basic than either alkyl amine but more basic than aniline (no resonance loss).
   4. $(C_2H_5)_2NH$ (IV): a secondary alkyl amine with two ethyl groups donating electron density (+I effect) to nitrogen, and in aqueous solution this out-weighs the modest steric/solvation penalty, making it the strongest base of the four, i.e. lowest $pK_b$.
   5. So the ordering of $pK_b$ (weakest to strongest base) is I > III > II > IV, giving lowest $pK_b$ = IV and highest $pK_b$ = I.

   Common Mistakes

   • Assuming primary amines are always more basic than secondary ones (true in the gas phase, but the question is implicitly about aqueous basicity, the standard exam convention, where secondary alkyl amines usually edge out primary).
   • Forgetting aniline's resonance delocalisation and treating it like an ordinary amine.

   ✓Final answer

   The correct option is (A) — IV, I.

   ANSWER: A

2. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Consider the given sequence of reactions. $C_6H_5MgBr \xrightarrow[(ii)\ H^+]{(i)\ CO_2} A \xrightarrow[(ii)\ \Delta]{(i)\ NH_3} B \xrightarrow{Br_2\ |\ KOH} C \xrightarrow[Py]{CH_3I} D$ The incorrect statement about B, C and D from the following is (A) In the conversion of B to C, one carbon decreases (B) D has lower $pK_b$ than C (C) Both C and D respond to carbylamine reaction (D) C undergoes diazotization but not D

   ›Reveal solutionSolution

   Tracing the sequence (Grignard→acid→amide→Hofmann degradation→N-methylation) identifies B = benzamide, C = aniline, D = N-methylaniline; the false statement is that "both C and D respond to carbylamine reaction" — the test is primary-amine-specific, and D is secondary.

   Concept and Intuition

   The carbylamine (isocyanide) test — heating an amine with $CHCl_3$/alcoholic KOH to give a foul-smelling isocyanide — is a diagnostic test SPECIFIC to primary amines (both aliphatic and aromatic), because it requires an N–H2 group to form the intermediate dichlorocarbene adduct and lose 2 HCl. Secondary and tertiary amines give no reaction. Likewise, diazotisation (with $HNO_2/HCl$, cold) to form a diazonium salt is also primary-amine-specific; secondary amines instead form N-nitrosamines.

   Step-by-Step Solution

   1. $C_6H_5MgBr + CO_2$ then $H^+$: the Grignard adds across $CO_2$'s carbonyl, protonation on workup gives benzoic acid, $A = C_6H_5COOH$.
   2. $A + NH_3$ forms ammonium benzoate; heating dehydrates it to the amide: $B = C_6H_5CONH_2$ (benzamide).
   3. $B \xrightarrow{Br_2/KOH}$ Hofmann bromamide degradation converts the amide to a primary amine with ONE FEWER carbon (the carbonyl carbon leaves as $CO_3^{2-}$/isocyanate intermediate): $C = C_6H_5NH_2$ (aniline).
   4. $C \xrightarrow[Py]{CH_3I}$ methylates the amine nitrogen once (pyridine mops up the HI byproduct, limiting over-alkylation here): $D = C_6H_5NHCH_3$ (N-methylaniline), a secondary amine.
   5. Check (A): B (7 carbons incl. carbonyl C) → C (6 ring carbons only) — yes, one carbon is lost. TRUE.
   6. Check (B): alkyl/methyl substitution on the N of aniline increases electron density on N (inductive donation), making D a slightly stronger base (lower $pK_b$) than C in water. TRUE.
   7. Check (C): carbylamine test requires a primary amine (–NH₂). C is primary → responds. D is secondary (–NH–CH₃, no free NH₂) → does NOT respond. So "both C and D respond" is FALSE.
   8. Check (D): diazotisation with cold $HNO_2$ needs a primary aromatic amine to form $Ar{-}N_2^+$; C (aniline) does this, while D (secondary) instead gives an N-nitroso amine, not a diazonium salt — so "C undergoes diazotisation but not D" is TRUE.
   9. The one incorrect statement, therefore, is (C).

   Common Mistakes

   • Forgetting that N-alkylation converts a primary amine into a secondary one, which then fails both the carbylamine test and diazotisation.
   • Assuming basicity always increases monotonically with methyl substitution regardless of solvent/sterics — while true here in water for mono-methylation, it's worth remembering the trend can reverse with further alkylation (di-methyl is often less basic than mono in water) — not needed for this question but a common follow-up trap.

   ✓Final answer

   The correct option is (C) — Both C and D respond to carbylamine reaction.

   ANSWER: C

3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Consider the following set of reactions and statements given about X and Y $Y \xleftarrow[(ii)\ H_2O]{(i)\ LiAlH_4}$ [FIGURE] (a benzene ring with a $-CONH_2$ substituent, i.e. benzamide) $\xrightarrow{Br_2/OH^-} X$ I. p$K_b$ of X is greater than Y II. Both on reaction with $NaNO_2$ and HCl at 273-298K form stable diazonium salts III. Both can be prepared by ammonolysis of corresponding chlorides The incorrect statements are (A) I, II only (B) II, III only (C) I only (D) I, III only

   ›Reveal solutionSolution

   X = aniline (via Hofmann bromamide degradation) and Y = benzylamine (via LiAlH₄ reduction) from benzamide. Statement I (about basicity) is actually true; statements II (diazonium stability) and III (ammonolysis route) are the incorrect ones.

   Concept and Intuition

   Benzamide, $C_6H_5CONH_2$, can be converted two different ways:

   • $LiAlH_4$ reduces the amide carbonyl all the way to a $CH_2$ group, giving benzylamine, Y = $C_6H_5CH_2NH_2$ (an amine attached to the ring through a $CH_2$ spacer — behaves like an aliphatic primary amine).
   • $Br_2/OH^-$ triggers the Hofmann bromamide degradation, which removes the carbonyl carbon entirely and gives the amine with one less carbon directly on the ring: X = aniline, $C_6H_5NH_2$ (a genuinely aromatic amine, its N lone pair conjugated with the ring).

   Step-by-Step Solution

   1. Identify X = aniline, Y = benzylamine, as above.
   2. Statement I — $pK_b(X) > pK_b(Y)$: aniline's nitrogen lone pair delocalises into the benzene ring, making it far less available for protonation, i.e. aniline is a much weaker base than benzylamine (whose $CH_2$ spacer isolates the lone pair from the ring). A weaker base has a higher $pK_b$. So aniline's $pK_b$ (~9.4) is indeed greater than benzylamine's $pK_b$ (~4.7) — statement I is true (not one of the incorrect ones).
   3. Statement II — both form stable diazonium salts with $NaNO_2/HCl$ at 273–298 K: only aromatic primary amines (aniline) give diazonium salts stable enough to isolate/use in that temperature range. Benzylamine, though attached to a ring, has its amino nitrogen on an $sp^3$ carbon (behaves as a primary aliphatic amine for this reaction) — its diazonium salt is unstable and decomposes immediately (releasing $N_2$, forming alcohols/alkenes). So statement II is false — it is one of the incorrect statements.
   4. Statement III — both can be prepared by ammonolysis of the corresponding chloride: benzylamine can indeed be made by ammonolysis of benzyl chloride ($C_6H_5CH_2Cl + NH_3$), which proceeds readily since it's a benzylic (activated) halide. But aniline is not obtained this way from chlorobenzene — aryl C–Cl bonds resist nucleophilic substitution under ordinary ammonolysis conditions due to resonance strengthening of the C–Cl bond, so this route does not work for X. So statement III is false — it is also one of the incorrect statements.
   5. Incorrect statements = II, III only.

   Common Mistakes

   • Assuming both amines behave identically towards diazotisation just because both ultimately derive from the same benzamide precursor — the location of the amino group (on the ring vs on a $CH_2$) is what matters.
   • Reading statement I's claim at face value as "obviously wrong" without checking it — it is in fact chemically correct here, which is the trap of this question.

   ✓Final answer

   The correct option is (B) — II, III only.

   ANSWER: B

4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.Consider the following Statement-I: $\mathrm{CH_3NH_2}$ is more basic than $\mathrm{NH_3}$ but $\mathrm{C_6H_5NH_2}$ is less basic than $\mathrm{NH_3}$. Statement-II: The order of basic strength of amines in aqueous phase follows the order $\mathrm{(C_2H_5)_3N > (C_2H_5)_2NH > C_2H_5NH_2}$ The correct answer is (A) Both statement-I and statement-II are correct (B) Both statement-I and statement-II are not correct (C) Statement-I is correct, but statement-II is not correct (D) Statement-I is not correct, but statement-II is correct

   ›Reveal solutionSolution

   Statement-I (relative basicity of methylamine/ammonia/aniline) is a correct, standard fact. Statement-II's claimed order $3° > 2° > 1°$ for ethylamines in water is wrong — aqueous basicity is not simply +I count, because steric hindrance to solvation of the bulky ammonium ion matters too.

   Concept and Intuition

   Basicity of amines depends on how available the nitrogen lone pair is to accept a proton, AND (in aqueous solution specifically) on how well the resulting ammonium cation is stabilised by hydrogen-bonding solvation. Alkyl groups donate electron density (+I), which should increase basicity with more alkyl substitution, but each alkyl group also sterically blocks water molecules from approaching and solvating the $-\mathrm{NH}^+$ centre. In water, this steric/solvation effect competes with (and for bulky groups, dominates) the electronic effect, producing the experimentally observed non-monotonic order for simple trialkylamines.

   Step-by-Step Solution

   1. Statement-I: $\mathrm{CH_3NH_2}$ vs $\mathrm{NH_3}$ — the methyl group donates electron density inductively onto N, making methylamine a stronger base than ammonia. $\mathrm{C_6H_5NH_2}$ (aniline) vs $\mathrm{NH_3}$ — the nitrogen lone pair conjugates into the benzene ring (resonance delocalisation), making it much less available to bind $\mathrm{H^+}$, so aniline is a weaker base than ammonia. Both parts are textbook-correct, so Statement-I is TRUE.
   2. Statement-II: claims $(\mathrm{C_2H_5})_3\mathrm{N} > (\mathrm{C_2H_5})_2\mathrm{NH} > \mathrm{C_2H_5NH_2}$ in aqueous phase.
   3. Experimentally (and as taught in NCERT), the correct aqueous order for these ethylamines is $(\mathrm{C_2H_5})_2\mathrm{NH} > \mathrm{C_2H_5NH_2} > (\mathrm{C_2H_5})_3\mathrm{N} > \mathrm{NH_3}$ — the tertiary amine, despite having the most +I donation, drops BELOW the secondary and primary amines because its three bulky ethyl groups sterically hinder solvation of its ammonium ion, reducing the stabilisation of the protonated form.
   4. So the specific monotonic order given in Statement-II ($3° > 2° > 1°$) is not correct as stated (real order places $2° > 1° > 3°$) — Statement-II is FALSE.
   5. Hence: Statement-I correct, Statement-II incorrect.

   Common Mistakes

   • Applying a simple "more alkyl groups = more basic" rule uniformly to aqueous-phase amine basicity — this ignores the solvation/steric factor that reverses the naive order for bulkier amines in water.
   • Not distinguishing gas-phase basicity trends (where the naive $3°>2°>1°$ order for amines often does hold, since there's no solvent to hinder) from aqueous-phase trends (where it doesn't) — Statement-II is specifically about the aqueous phase.

   ✓Final answer

   The correct option is (C) — Statement-I is correct, but statement-II is not correct.

   ANSWER: C

5. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.Which of the following has the lowest $pK_b$ value? (A) $\mathrm{CH_3-NH_2}$ (B) $\mathrm{(CH_3CH_2)_2NH}$ (C) aniline (benzene ring with an $\mathrm{NH_2}$ substituent) (D) N-methylaniline (benzene ring with an $\mathrm{NH-CH_3}$ substituent)

   ›Reveal solutionSolution

   Basicity order here is: aliphatic secondary amine > aliphatic primary amine >> aromatic secondary amine > aromatic primary amine, so diethylamine has the lowest $pK_b$ (strongest base).

   Concept and Intuition

   Basicity of an amine depends on how available the nitrogen lone pair is to accept a proton.

   • In aromatic amines (aniline, N-methylaniline), the nitrogen lone pair is delocalised into the benzene ring by resonance, making it far less available to bind $H^+$. This makes aniline and N-methylaniline much weaker bases (higher $pK_b$, around 9–9.5) than any simple aliphatic amine.
   • In aliphatic amines, alkyl groups push electron density onto nitrogen via the $+I$ (inductive) effect, which should make more-substituted amines more basic. In aqueous solution, this trend is moderated by solvation (a more crowded/hydrophobic nitrogen is harder to hydrate and stabilise as $R_3NH^+$), so the typically observed order for basicity in water is secondary > primary > tertiary > $NH_3$ for simple alkylamines. Diethylamine (a secondary amine) is therefore a slightly stronger base than methylamine (a primary amine), giving it the lower (more negative-leaning, i.e., smaller) $pK_b$ value.

   Step-by-Step Solution

   1. Compare aliphatic vs aromatic amines first: aniline and N-methylaniline both have $pK_b \approx 9$–$9.5$ because of resonance delocalisation of the N lone pair into the ring — these are ruled out as "lowest $pK_b$" candidates.
   2. Between the two aliphatic amines, methylamine ($CH_3NH_2$) has $pK_b \approx 3.4$, while diethylamine ($(CH_3CH_2)_2NH$) has $pK_b \approx 3.0$ — lower than methylamine.
   3. So diethylamine has the lowest $pK_b$ among the four, i.e., it is the strongest base.

   Common Mistakes

   • Assuming aromatic amines are always more basic because of the "bigger" molecule — resonance delocalisation actually makes them weaker bases, not stronger.
   • Forgetting that "lowest $pK_b$" means "strongest base," not "weakest."

   ✓Final answer

   The correct option is (B) — $(CH_3CH_2)_2NH$.

   ANSWER: B

6. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.Arrange the following in increasing order of their $pK_b$ values. A. Aniline B. Benzylamine C. Ethylamine (A) A < B < C (B) C < B < A (C) B < C < A (D) A < C < B

   ›Reveal solutionSolution

   Basicity (and hence pKb) among these three amines is governed by whether the nitrogen lone pair is conjugated with the benzene ring: aniline (conjugated, weak base, high pKb) < benzylamine (insulated by CH₂, moderately strong base) < ethylamine (purely aliphatic, strongest base, lowest pKb) — so increasing pKb is ethylamine < benzylamine < aniline.

   Concept and Intuition

   A more basic amine has lower pKb (it is a stronger base, so equilibrium favours accepting a proton more, giving a smaller pKb). Aromatic amines like aniline are weak bases because the nitrogen lone pair delocalizes into the ring (resonance with the ring), making it far less available to accept a proton. Purely aliphatic amines like ethylamine have no such delocalization and are strongly basic (also aided by +I effect and better solvation of the resulting ammonium ion). Benzylamine's nitrogen is separated from the ring by a $-\mathrm{CH_2}-$ group, so there is no resonance delocalization into the ring — its basicity is close to (but very slightly less than) a simple aliphatic amine, because of a small inductive electron-withdrawing effect transmitted through the CH₂ from the phenyl ring.

   Step-by-Step Solution

   1. Aniline ($\mathrm{C_6H_5NH_2}$): lone pair conjugated with the aromatic ring $\Rightarrow$ weakest base $\Rightarrow$ highest $\mathrm{pK_b}$ ($\approx 9.4$).
   2. Benzylamine ($\mathrm{C_6H_5CH_2NH_2}$): no conjugation (CH₂ insulates the nitrogen from the ring) $\Rightarrow$ moderately strong base, close to an aliphatic amine but slightly weaker due to inductive effect ($\mathrm{pK_b} \approx 4.7$).
   3. Ethylamine ($\mathrm{C_2H_5NH_2}$): purely aliphatic, +I effect, no ring at all $\Rightarrow$ strongest base $\Rightarrow$ lowest $\mathrm{pK_b}$ ($\approx 3.3$).
   4. Increasing order of $\mathrm{pK_b}$: ethylamine (C) < benzylamine (B) < aniline (A), i.e. C < B < A.

   Common Mistakes

   • Assuming benzylamine is as weak a base as aniline just because it "has a benzene ring" — the key is whether the lone pair is directly conjugated with the ring, which it is not in benzylamine.
   • Mixing up pKb (higher = weaker base) with pKa/basicity strength (higher = stronger base) and inverting the order.

   ✓Final answer

   The correct option is (B) — C < B < A.

   ANSWER: B

7. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.Arrange the following in decreasing order of their basicity A. 4-methoxyaniline: $\mathrm{4\text{-}CH_3O\text{-}C_6H_4\text{-}NH_2}$ B. 4-methylbenzylamine: $\mathrm{4\text{-}CH_3\text{-}C_6H_4\text{-}CH_2NH_2}$ C. 4-nitroaniline: $\mathrm{4\text{-}O_2N\text{-}C_6H_4\text{-}NH_2}$ (A) B > C > A (B) B > A > C (C) A > B > C (D) A > C > B

   ›Reveal solutionSolution

   Whether the amine nitrogen is conjugated with the aromatic ring is the dominant factor: a benzylamine (not conjugated) beats any aniline (conjugated); among anilines, an electron-donating para substituent boosts basicity and an electron-withdrawing one kills it.

   Concept and Intuition

   Basicity of an amine depends on how available its nitrogen lone pair is to accept a proton.

   • In aniline-type amines, the $-NH_2$ is attached directly to the ring, so its lone pair is pulled into resonance with the aromatic $\pi$ system. This delocalisation makes the lone pair much less available, so anilines are always much weaker bases than comparable aliphatic amines.
   • In 4-methylbenzylamine, the $-NH_2$ is on a $CH_2$ group, one carbon removed from the ring. This $sp^3$ methylene completely breaks conjugation between the nitrogen lone pair and the ring — so this amine behaves essentially like a plain primary aliphatic amine (strongly basic, lone pair fully available), and is therefore much more basic than either aniline derivative.
   • Among the two anilines: a para-methoxy group is electron-donating (+M, releases electron density into the ring and onto nitrogen), increasing basicity relative to plain aniline; a para-nitro group is strongly electron-withdrawing (-M, actively pulls the lone pair towards itself through the ring), decreasing basicity drastically.

   Step-by-Step Solution

   1. Recognise B (4-methylbenzylamine) is NOT an aniline — its N is isolated from ring conjugation by a $CH_2$ spacer, so it is the strongest base of the three by a wide margin.
   2. Compare A (4-methoxyaniline) and C (4-nitroaniline), both true anilines whose lone pairs are ring-conjugated.
   3. $-OCH_3$ at the para position donates electron density by resonance into the ring, partially compensating for the conjugative loss and making A more basic than unsubstituted aniline.
   4. $-NO_2$ at the para position withdraws electron density strongly by resonance, making the lone pair on N in C even less available than in plain aniline — the weakest base of the three.
   5. Overall order: B (most basic) > A > C (least basic).

   Common Mistakes

   • Treating all three as simple anilines and ranking purely by substituent effect, forgetting that B's nitrogen isn't even conjugated to the ring — this is the deciding factor, not a minor correction.
   • Reversing the substituent effect: thinking $-NO_2$ increases basicity because it's "electron-withdrawing = more acidic character helps stabilize the free base" — it's the opposite; withdrawing groups always reduce amine basicity by pulling the lone pair away.

   ✓Final answer

   The correct option is (B) — B > A > C.

   ANSWER: B

8. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.The decreasing order of $pK_a$ of the following aromatic amines

   (a) $\mathrm{C_6H_5NH_2}$ (aniline)

   (b) $\mathrm{C_6H_5CH_2NH_2}$ (benzylamine)

   (c) a benzene ring bonded to $\mathrm{N(H)-CH_3}$ ($N$-methylaniline)

   (d) a benzene ring bonded to $\mathrm{N(CH_3)_2}$ ($N,N$-dimethylaniline) (A)

   (a) >

   (b) >

   (d) >

   (c) (B)

   (a) >

   (c) >

   (d) >

   (b) (C)

   (b) >

   (d) >

   (c) >

   (a) (D)

   (b) >

   (c) >

   (d) > (a)

   ›Reveal solutionSolution

   Benzylamine (aliphatic-like, lone pair not conjugated) is the most basic; among the anilines, N,N-dimethylaniline > N-methylaniline > aniline, since alkyl +I effects partly offset the lone-pair delocalization into the ring, giving decreasing $pK_a$ order (b) > (d) > (c) > (a).

   Concept and Intuition

   Basicity of an amine depends on how available its nitrogen lone pair is for protonation. In aniline, the lone pair on N conjugates (delocalizes) into the aromatic ring, making it much less available and hence aniline is a weak base. Benzylamine has a $\mathrm{CH_2}$ group between the ring and the nitrogen, which breaks this conjugation — the lone pair behaves like that of a normal aliphatic amine, making benzylamine considerably more basic than any of the anilines. Among the anilines themselves, replacing N–H with N–CH$_3$ groups adds electron density to nitrogen through the inductive (+I) effect of the alkyl groups, which increases basicity somewhat (aniline < N-methylaniline < N,N-dimethylaniline), even though the lone pair is still partly delocalized into the ring in all three.

   Step-by-Step Solution

   1. Benzylamine (b): $\mathrm{CH_2}$ spacer prevents conjugation $\Rightarrow$ most basic, highest $pK_{aH}$ (~9.3).
   2. N,N-dimethylaniline (d): two methyl (+I) groups increase electron density at N compared to aniline $\Rightarrow$ $pK_{aH} \approx 5.1$.
   3. N-methylaniline (c): one methyl (+I) group $\Rightarrow$ $pK_{aH} \approx 4.85$, less basic than the dimethyl version but more than plain aniline.
   4. Aniline (a): no alkyl substitution, lone pair most delocalized into ring $\Rightarrow$ least basic, $pK_{aH} \approx 4.6$.
   5. Decreasing $pK_a$ order: (b) > (d) > (c) > (a).

   Common Mistakes

   • Assuming more methyl substitution always means more basic without accounting for conjugation with the ring still being present in (c) and (d).
   • Forgetting that benzylamine is essentially an aliphatic amine in behavior, not an aromatic amine, and so is far more basic than all the anilines.

   ✓Final answer

   The correct option is (C) — (b) > (d) > (c) > (a).

   ANSWER: C

9. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.Given below are two statements Assertion (A): Ethylamine is stronger base than ammonia Reason (R): It is due to -I effect of ethyl group The correct answer is (A) Both A and R are correct and R is correct explanation of A. (B) Both A and R are correct but R is not correct explanation of A. (C) A is correct but R is incorrect. (D) A is incorrect but R is correct.

   ›Reveal solutionSolution

   Ethylamine is a stronger base than ammonia because the ethyl group is electron-donating (+I effect), not electron-withdrawing (-I effect). So Assertion (A) is correct, but Reason (R) is incorrect. The correct option is (C).

   Concept & Intuition: Basicity of Amines

   Basicity in amines is all about the availability of the lone pair on nitrogen to accept a proton. The more electron-rich the nitrogen, the stronger the base. Alkyl groups like ethyl are electron-donating through the +I effect (inductive effect). They push electron density toward the nitrogen, making the lone pair more available. In contrast, the -I effect withdraws electrons, which would decrease basicity. The question tests whether you know the correct direction of the inductive effect for an alkyl group.

   ---

   Step-by-step reasoning:

   1. Identify the Assertion (A): "Ethylamine is a stronger base than ammonia."

      • Ammonia (NH₃) has a nitrogen with a lone pair.
      • Ethylamine (CH₃CH₂NH₂) has an ethyl group attached to nitrogen.
      • Alkyl groups are electron-releasing (+I effect). This increases electron density on nitrogen, making it more willing to accept a proton.
      • Therefore, ethylamine is indeed a stronger base than ammonia. Assertion (A) is correct.

   2. Identify the Reason (R): "It is due to -I effect of ethyl group."

      • The -I effect means electron-withdrawing (e.g., from groups like -NO₂, -CN, -F).
      • The ethyl group is actually electron-donating (+I effect), not withdrawing.
      • So the reason given is factually wrong. Reason (R) is incorrect.

   3. Evaluate the relationship:

      • Since (R) is incorrect, it cannot be a correct explanation of (A), even though (A) is true.
      • This matches option (C): "A is correct but R is incorrect."

   Watch out

   A common mistake is to think alkyl groups are electron-withdrawing because they are "bigger" or because of some confusion with resonance effects. Remember: alkyl groups always donate electrons via the inductive effect.

   Tip

   A quick memory aid: Alkyl = electron donor (think of it as "pushing" electrons toward nitrogen). The more alkyl groups attached (up to a point), the stronger the base — except in solution where steric hindrance and solvation effects can reverse the order.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

10. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
    Q.Basicity order of the following amines in aqueous medium i) N-Methylaniline ii) N, N-Dimethylaniline iii) Aniline (A) i > ii > iii (B) i > iii > ii (C) iii > ii > i (D) ii > i > iii

    ›Reveal solutionSolution

    In aqueous medium, the basicity of aniline derivatives is governed by a balance between the +I effect of methyl groups and the loss of resonance stabilisation due to steric hindrance. The order is: N,N-dimethylaniline > N-methylaniline > aniline, so the correct option is (D).

    Why This Approach Works

    Basicity of amines in water depends on the availability of the lone pair on nitrogen to accept a proton. For aniline and its N-methyl derivatives, two opposing factors are at play:

    • Inductive effect (+I): Methyl groups are electron-donating, which increases electron density on nitrogen, making it more basic.
    • Resonance effect: In aniline, the lone pair is delocalised into the benzene ring, reducing basicity. N-methylation can hinder this resonance due to steric clash between the methyl groups and the ortho-hydrogens of the ring, forcing the lone pair to be more localised on nitrogen.

    In aqueous solution, the solvation of the protonated ammonium ion also matters, but here the key is the balance of these two effects.

    Step-by-Step Reasoning

    1. Baseline: Aniline (iii)

       Aniline’s lone pair is strongly delocalised into the aromatic ring via resonance. This makes it the least basic among the three because the lone pair is less available for protonation. Its conjugate acid is stabilised by resonance, but the free base is even more stabilised, so the equilibrium favours the free base.

    2. Effect of one methyl group: N-Methylaniline (i)

       Adding one methyl group to nitrogen introduces a +I effect, which pushes electron density toward nitrogen, increasing basicity. However, the methyl group is small enough that the lone pair can still participate in resonance with the ring, though slightly less effectively than in aniline. The net result is a modest increase in basicity over aniline.

    3. Effect of two methyl groups: N,N-Dimethylaniline (ii)

       Two methyl groups provide a stronger +I effect, which would suggest even higher basicity. But more importantly, the two methyl groups create significant steric hindrance with the ortho-hydrogens of the benzene ring. This forces the N(CH₃)₂ group to twist out of the plane of the ring, breaking the resonance between the lone pair and the aromatic π-system. The lone pair becomes almost entirely localised on nitrogen, making it much more available for protonation. This steric inhibition of resonance outweighs the inductive effect, making N,N-dimethylaniline the most basic of the three.

    4. Order in aqueous medium

       Combining these:

       • N,N-Dimethylaniline (ii) is the most basic.
       • N-Methylaniline (i) is intermediate.
       • Aniline (iii) is the least basic. Hence, the order is ii > i > iii.

    Watch out

    A common mistake is to think that more methyl groups always mean higher basicity due to the +I effect alone. In aniline derivatives, steric hindrance can dominate, reversing the expected trend. For example, in the gas phase, the order is different because solvation effects are absent.

    Tip

    A quick way to remember: If the nitrogen’s lone pair can be planar with the ring, resonance reduces basicity. Bulky substituents force non-planarity, “freeing” the lone pair and boosting basicity.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

11. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.Which of the following produces nitrogen gas after the reaction with nitrous acid? (A) $\mathrm{(CH_3)_3N}$ (B) $\mathrm{C_2H_5NHC_2H_5}$ (C) $\mathrm{(C_2H_5)_3N}$ (D) $\mathrm{C_2H_5NH_2}$

    ›Reveal solutionSolution

    Nitrous acid's reaction with amines depends on how many N–H bonds are available: only primary amines form an unstable diazonium intermediate that decomposes with N₂ evolution, so $C_2H_5NH_2$ (ethylamine) is the one that liberates nitrogen gas.

    Concept and Intuition

    The reaction of amines with nitrous acid ($HNO_2$) is a standard qualitative test to distinguish 1°/2°/3° amines:

    • Primary amines (one N–H replaced twice, i.e. two N–H bonds present) form a diazonium salt via nitrosation, which for aliphatic amines is thermally unstable even at 0–5°C and decomposes instantly, releasing nitrogen gas and forming an alcohol (plus other minor products) — vigorous effervescence is the visible clue.
    • Secondary amines (only one N–H bond) form a stable, often oily, yellow N-nitrosamine ($R_2N-NO$) — no gas evolved.
    • Tertiary amines (no N–H bonds) cannot be nitrosated at nitrogen in the same way; they simply form an unstable nitrite salt at low temperature with no clean gas-evolving reaction.

    Step-by-Step Solution

    1. $(CH_3)_3N$: a tertiary amine (no N–H) → no diazonium/nitrosamine chemistry possible → no $N_2$.
    2. $C_2H_5NHC_2H_5$ (diethylamine): a secondary amine (one N–H) → forms a stable N-nitrosamine, not a gas-releasing diazonium salt → no $N_2$.
    3. $(C_2H_5)_3N$: a tertiary amine → same reasoning as (1) → no $N_2$.
    4. $C_2H_5NH_2$ (ethylamine): a primary amine (two N–H bonds) → reacts with $HNO_2$ to form the diazonium salt $C_2H_5N_2^+$, which is unstable and decomposes immediately at ordinary temperatures, releasing $N_2$ gas and giving ethanol.
    5. So the amine that produces nitrogen gas is ethylamine, option (D).

    Common Mistakes

    • Forgetting that aliphatic (unlike aromatic) diazonium salts are too unstable to isolate — they decompose instantly with gas evolution, which is exactly why this is used as a distinguishing test rather than a synthetic route to isolable diazonium salts.
    • Confusing secondary amines' nitrosamine formation (no gas) with primary amines' gas-releasing decomposition — both involve "N-nitroso" chemistry but with very different outcomes.

    ✓Final answer

    The correct option is (D) — $C_2H_5NH_2$.

    ANSWER: D

12. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.Arrange the following in decreasing order of their $\mathrm{pK_b}$ values:

    (a) $\mathrm{CH_3NH_2}$;

    (b) $\mathrm{(CH_3)_3N}$;

    (c) Benzylamine (a benzene ring with a $\mathrm{CH_2NH_2}$ substituent);

    (d) N-Methylbenzylamine (a benzene ring with a $\mathrm{CH_2NHCH_3}$ substituent) (A) d > a > c > b (B) a > b > d > c (C) d > c > b > a (D) a > c > d > b

    ›Reveal solutionSolution

    Basicity in water depends on three competing effects (inductive donation, steric hindrance to solvation, and how well the resulting cation is hydrogen-bonded), and the one rock-solid, textbook fact here — that $\mathrm{CH_3NH_2}$ is a stronger base than $\mathrm{(CH_3)_3N}$ in water — is enough to pick out option (C) uniquely.

    Concept and Intuition

    Gas-phase basicity of amines increases with more alkyl substitution (more +I groups push electron density onto N). But in water, the picture changes: the protonated ammonium cation must be stabilised by hydrogen-bonding with solvent, and bulky alkyl groups around N get in the way of this solvation. For the methylamine series this gives the well-known aqueous order $\mathrm{(CH_3)_2NH} > \mathrm{CH_3NH_2} > \mathrm{(CH_3)_3N} > \mathrm{NH_3}$ — i.e. even though trimethylamine has three +I methyls, steric hindrance to solvation makes it a weaker base than methylamine. The same steric penalty applies even more strongly once a bulky benzyl group is involved, so a benzylic secondary amine can end up weaker than a benzylic primary amine.

    Step-by-Step Solution

    1. $pK_b$ is large for a weak base and small for a strong base, so "decreasing $pK_b$" means listing weakest base first.
    2. Fact: in water, $\mathrm{CH_3NH_2}$ (a) is a stronger base than $\mathrm{(CH_3)_3N}$ (b) — so $pK_b(a) < pK_b(b)$, meaning b must be listed before a in a decreasing-$pK_b$ list.
    3. Test each option for the relative position of a and b:
       • (A) d > a > c > b — a before b: implies $pK_b(a)>pK_b(b)$. Contradicts step 2.
       • (B) a > b > d > c — a before b: same contradiction.
       • (C) d > c > b > a — b before a: consistent with $pK_b(b) > pK_b(a)$. ✓
       • (D) a > c > d > b — a before b: contradiction.
    4. Only (C) survives. (It is also chemically sensible for the rest of the order: benzylamine's lone $-\mathrm{CH_2Ph}$ group is bulkier than methyl and gives less solvation-friendly stabilisation than $\mathrm{CH_3NH_2}$, so benzylamine (c) is weaker than a; and going to the secondary N-methylbenzylamine (d) adds even more steric bulk around N, making it the weakest of all four — matching d > c > b > a.)

    Common Mistakes

    • Assuming "more alkyl/aryl substitution always means more basic" — this is true in the gas phase but can reverse in water due to solvation/steric effects.
    • Forgetting that $pK_b$ and base strength are inversely related (a larger $pK_b$ means a weaker base).

    ✓Final answer

    The correct option is (C) — d > c > b > a.

    ANSWER: C