Question 1:

   Question: Given below are two statements:

       Statement-I: Oxidation of p-nitrotoluene with acidic $\mathrm{KMnO}_4$ gives an acid that is stronger than benzoic acid.

       Statement-II: Reduction of p-nitrotoluene with $\mathrm{Sn}/\mathrm{HCl}$ followed by neutralization gives an amine that is more basic than aniline.

    In light of the above statements, choose the most appropriate answer from the options given below.

   Options: 

       A) Statement-I is incorrect but Statement-II is correct

       B) Both Statement-I and Statement-II are correct

       C) Both Statement-I and Statement-II are incorrect

       D) Statement-I is correct but Statement-II is incorrect

   Correct Answer: B

   Year: RE-NEET 2026

   Solution: Oxidation of p-nitrotoluene with acidic $\mathrm{KMnO}_4$ converts it into p-nitrobenzoic acid. Because of the -I and -M effects of the nitro group, p-nitrobenzoic acid is a stronger acid than benzoic acid. Reduction of p-nitrotoluene with $\mathrm{Sn}/\mathrm{HCl}$ followed by neutralization reduces the $-\mathrm{NO}_2$ group to an amino group ($-\mathrm{NH}_2$), forming p-toluidine (p-methylaniline). This is more basic than aniline due to the +I effect and hyperconjugation of the methyl group.

   Step Solution:

    1.  Identify Oxidation Product: p-nitrotoluene + acidic $\mathrm{KMnO}_4$ $\rightarrow$ p-nitrobenzoic acid.

    2.  Evaluate Acidity: The $-\mathrm{NO}_2$ group is an Electron Withdrawing Group (EWG) exhibiting -I and -M effects, which stabilize the carboxylate ion more than in benzoic acid, making it a stronger acid.

    3.  Identify Reduction Product: p-nitrotoluene + $\mathrm{Sn}/\mathrm{HCl}$ $\rightarrow$ p-toluidine (after neutralization).

    4.  Evaluate Basicity: The $-\mathrm{CH}_3$ group is an Electron Donating Group (EDG) exhibiting +I and hyperconjugation effects, which increase electron density on the nitrogen lone pair compared to aniline, making it more basic.

    5.  Conclusion: Both statements regarding increased acidity and increased basicity are true.

   Difficulty level: Hard

   Concept Name: Electronic Effects on Acidity and Basicity (-I, -M, +I, Hyperconjugation)

   Short cut solution: EWG (like $-\mathrm{NO}_2$) increases acidity; EDG (like $-\mathrm{CH}_3$) increases basicity of aromatic systems.

Question 2

   Question: The correct order of decreasing basic strength of the given amines is: Answer (1)Sol. Lower is the value of $pK_b$ , higher is the basicity Also aliphatic amines are stronger bases than aromatic amines. $pK_b$  : Benzenamine > N-Methylaniline > Ethanamine > N Ethylethanamine Basic strength : N-Ethylethanamine > Ethanamine > N-Methylaniline > Benzenamine

   Options:

       A. N-ethylethanamine > ethanamine > N-methylaniline > benzenamine

       B. benzenamine > ethanamine > N-methylaniline > N-ethylethanamine

       C. N-methylaniline > benzenamine > ethanamine > N-ethylethanamine

       D. N-ethylethanamine > ethanamine > benzenamine > N-methylaniline

   Correct Answer: A

   Year: NEET 2025

   Solution: Lower is the value of $\mathrm{pK_b}$, higher is the basicity. Also, aliphatic amines are stronger bases than aromatic amines. $\mathrm{pK_b}$ values: Benzenamine > N-Methylaniline > Ethanamine > N-Ethylethanamine. Basic strength: N-Ethylethanamine > Ethanamine > N-Methylaniline > Benzenamine.

   Step Solution:

    1.  Identify types: Categorize as aliphatic ($2^\circ$ N-ethylethanamine, $1^\circ$ ethanamine) and aromatic ($2^\circ$ N-methylaniline, $1^\circ$ benzenamine).

    2.  Primary Rule: Aliphatic amines are more basic than aromatic amines because their lone pairs are not delocalized into a benzene ring.

    3.  Compare Aliphatic: N-ethylethanamine ($2^\circ$) is more basic than ethanamine ($1^\circ$) due to the inductive $+I$ effect of two ethyl groups.

    4.  Compare Aromatic: N-methylaniline ($2^\circ$) is more basic than benzenamine ($1^\circ$) because the methyl group provides a $+I$ effect to the nitrogen.

    5.  Final Order: $2^\circ$ Aliphatic > $1^\circ$ Aliphatic > $2^\circ$ Aromatic > $1^\circ$ Aromatic.

   Difficulty Level: Medium

   Concept Name: $\mathrm{pK_b}$ and Inductive Effects

   Short cut solution: Aliphatic is always stronger than aromatic; for simple alkyl chains, $2^\circ > 1^\circ$.

Question 12

   Question: The correct order of the basic strength of methyl substituted amines in aqueous solution is

   Options:

       A. $\mathrm{CH_3NH_2 > (CH_3)_2NH > (CH_3)_3N}$

       B. $\mathrm{(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N}$

       C. $\mathrm{(CH_3)_3N > CH_3NH_2 > (CH_3)_2NH}$

       D. $\mathrm{(CH_3)_3N > (CH_3)_2NH > CH_3NH_2}$

   Correct Answer: B

   Year: NEET 2019

   Solution: Basicity in aqueous solution depends on the $+I$ effect, hydrogen bonding (solvation), and steric factors. For the small $\mathrm{CH_3}$ group, $2^\circ$ amines are the strongest as they balance these factors. Stability due to H-bonding predominates over the $+I$ effect in $1^\circ$ vs $3^\circ$ amines.

   Step Solution:

    1.  Aqueous Factor: Recognize that in water, solvation (H-bonding) is as important as the inductive effect.

    2.  $2^\circ$ Supremacy: In methyl amines, $\mathrm{(CH_3)_2NH}$ is always the strongest base.

    3.  Solvation vs Sterics: $\mathrm{(CH_3)_3N}$ ($3^\circ$) is the weakest because three methyl groups create steric hindrance that prevents water from stabilizing the cation via H-bonding.

    4.  Comparison: $\mathrm{CH_3NH_2}$ ($1^\circ$) is stronger than $3^\circ$ due to better solvation.

    5.  Result: $2^\circ > 1^\circ > 3^\circ$ ($\mathrm{(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N}$).

   Difficulty Level: Medium

   Concept Name: Solvation and Steric Hindrance

   Short cut solution: Remember the code 213 for methyl amines in water ($2^\circ > 1^\circ > 3^\circ$).

Question 15

   Question: The correct increasing order of basic strength for the following compounds is

   Options:

       A. III < I < II

       B. III < II < I

       C. II < I < III

       D. II < III < I

   Correct Answer: C

   Year: NEET 2017

   Solution: Increasing order: p-nitroaniline < aniline < p-toluidine.

   Step Solution:

    1.  Analyze Substituents: $-\mathrm{CH_3}$ (in Toluidine) is an electron-donating group (EDG); $-\mathrm{NO_2}$ (in Nitroaniline) is an electron-withdrawing group (EWG).

    2.  EDG Effect: The $+I$ and hyperconjugation effects of the methyl group increase electron density on Nitrogen, making III the most basic.

    3.  EWG Effect: The strong $-R$ and $-I$ effects of the nitro group pull electrons away from Nitrogen, making II the least basic.

    4.  Reference: Aniline (I) has no substitution and falls in the middle.

    5.  Final Sequence: II (Nitro) < I (Aniline) < III (Toluidine).

   Difficulty Level: Easy

   Concept Name: Substituent Effects (EDG/EWG)

   Short cut solution: EDG (Toluidine) increases basicity; EWG (Nitro) decreases it.

Question 19

   Question: The correct statement regarding the basicity of arylamines is

   Options:

       A. arylamines are more basic than alkylamines because of aryl group

       B. arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp - hybridised

       C. arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalised by interaction with the aromatic ring $\pi$-electron system

       D. arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalised by interaction with the aromatic ring $\pi$ -electron system.

   Correct Answer: C

   Year: NEET-I 2016

   Solution: In arylamines, the lone pair of electrons on the nitrogen atom is delocalised over the benzene ring, making it unavailable for donation. Thus, arylamines are less basic than alkylamines.

   Step Solution:

    1.  Define Arylamine: An amine where Nitrogen is directly attached to an aromatic ring.

    2.  Mechanism: The lone pair on Nitrogen participates in resonance with the $\pi$ electrons of the ring.

    3.  Availability: Because of this delocalization, the electrons are less likely to be donated to a proton ($H^+$).

    4.  Contrast: In alkylamines, the $+I$ effect of alkyl groups actually pushes more electron density toward the Nitrogen.

    5.  Conclusion: Alkylamines > Arylamines in basicity.

   Difficulty Level: Easy

   Concept Name: Resonance Effect / Delocalization

   Short cut solution: Resonance = less available electrons = lower basicity.

Question 31

   Question: Which of the following compounds is most basic?

   Options:

       A.

       B.

       C.

       D.

   Correct Answer: B

   Year: 2011 Mains

   Solution: In benzylamine, the electron pair present on the nitrogen is not delocalised with the benzene ring.

   Step Solution:

    1.  Check Attachment: In benzylamine ($C_6H_5CH_2NH_2$), Nitrogen is attached to an $sp^3$ hybridized carbon, not the ring directly.

    2.  Analyze Resonance: The $-\mathrm{CH_2}-$ group acts as a spacer, preventing the lone pair from entering the benzene ring's $\pi$ system.

    3.  Compare: In aniline and its derivatives, the lone pair is directly on the ring and is delocalized.

    4.  Basicity: A localized lone pair is more available for donation than a delocalized one.

    5.  Result: Benzylamine is significantly more basic than aniline or nitroaniline.

   Difficulty Level: Easy

   Concept Name: Localization of Lone Pair

   Short cut solution: Benzylamine is an aliphatic amine; it doesn't "know" the ring is there for resonance purposes.

Question 37

   Question: Which of the following is more basic than aniline?

   Options:

       A. Benzylamine

       B. Diphenylamine

       C. Triphenylamine

       D. p-Nitroaniline

   Correct Answer: A

   Year: 2006

   Solution: Electron withdrawing groups decrease basicity. Lone pair electrons are more delocalised in diphenylamine and triphenylamine. In benzylamine, the electron pair is not delocalised, hence it is more basic than aniline.

   Step Solution:

    1.  Reference: Aniline has one ring delocalizing its lone pair.

    2.  Multi-ring: Diphenylamine (2 rings) and Triphenylamine (3 rings) have even more delocalization, making them less basic than aniline.

    3.  Nitro effect: $-\mathrm{NO_2}$ is an EWG that further drains electron density.

    4.  Local Nitrogen: In benzylamine, the lone pair is localized on the nitrogen atom.

    5.  Conclusion: Localized electrons (Benzylamine) > Delocalized electrons (Aniline).

   Difficulty Level: Easy

   Concept Name: Electronic Effects on Basicity

   Short cut solution: Localized lone pair always beats delocalized lone pair in basicity.

Question 51

   Question: Which one of the following order is wrong, with respect to the property indicated?

   Options:

       A. Benzoic acid > phenol > cyclohexanol (acid strength)

       B. Aniline > cyclohexylamine > benzamide (basic strength)

       C. Formic acid > acetic acid > propanoic acid (acid strength)

       D. Fluoroacetic acid > chloroacetic acid > bromoacetic acid (acid strength)

   Correct Answer: B

   Year: 1994

   Solution: Basic strength decreases as: cyclohexylamine > aniline > benzamide. Lesser basicity in aniline and benzamide is due to participation of the lone pair of electrons in resonance.

   Step Solution:

    1.  Verify Option B: It lists Aniline > Cyclohexylamine.

    2.  Compare Types: Cyclohexylamine is aliphatic (localized lone pair + ring $+I$ effect). Aniline is aromatic (delocalized lone pair).

    3.  Evaluate Benzamide: It is an amide where the lone pair is in resonance with a carbonyl oxygen, making it the weakest.

    4.  Correct Order: Cyclohexylamine (strongest) > Aniline > Benzamide (weakest).

    5.  Check Others: A, C, and D are all correct based on resonance and inductive effects. Thus, B is the wrong order.

   Difficulty Level: Medium

   Concept Name: Comparative Acid-Base Strength

   Short cut solution: Aliphatic amines are always stronger than aromatic amines; Option B says the opposite.