Question

Given below are two statements:
Statement-I : Oxidation of \(p\)-nitrotoluene with acidic \(\text{KMnO}_4\) gives an acid that is stronger than benzoic acid.
Statement-II : Reduction of \(p\)-nitrotoluene with Sn/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.

• 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.

Hint:

Electronic effects rule organic reactivity: - Electron Withdrawing Groups (\(-\text{NO}_2, -\text{CN}\)) \(\rightarrow\) Increase acidity, decrease basicity. - Electron Donating Groups (\(-\text{CH}_3, -\text{OCH}_3\)) \(\rightarrow\) Decrease acidity, increase basicity.

Answer 1

The Correct Option is D

Concept: Chemical properties of aromatic compounds are heavily governed by substituent electronic effects.

• Acidity of Benzoic Acids: Electron-withdrawing groups (EWGs) stabilize the carboxylate conjugate base via inductive (\(-I\)) and resonance (\(-M\)) effects, which significantly enhances the parent acid's strength.

• Basicity of Anilines: Electron-withdrawing groups diminish the availability of the nitrogen lone pair for donation, lowering the basicity, whereas electron-donating groups (EDGs) elevate it.

Step 1: Analyzing Statement-I
The chemical transformation of \(p\)-nitrotoluene with strong acidic \(\text{KMnO}_4\) leads to the exhaustive oxidation of the benzylic methyl (\(-\text{CH}_3\)) group down to a carboxylic acid group (\(-\text{COOH}\)): \[ p\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_3 \xrightarrow{\text{KMnO}_4, \text{ H}^+} p\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{COOH} \quad \text{(\(p\)-nitrobenzoic acid)} \] The nitro group (\(-\text{NO}_2\)) situated at the para-position acts as a strong electron-withdrawing group via both powerful resonance (\(-M\)) and inductive (\(-I\)) pathways. It pulls electron density away from the aromatic ring, dispersing the negative charge on the carboxylate anion formed upon deprotonation: \[ p\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{COO}^- \] Because of this stabilization, \(p\)-nitrobenzoic acid is a notably stronger acid than unsubstituted benzoic acid (\(\text{C}_6\text{H}_5\text{COOH}\)). Thus, Statement-I is completely correct.

Step 2: Analyzing Statement-II
Treating \(p\)-nitrotoluene with metallic tin in hydrochloric acid (Sn/HCl) followed by an alkaline neutralization steps selectively reduces the nitro (\(-\text{NO}_2\)) group down to an amino group (\(-\text{NH}_2\)), leaving the methyl group unaffected: \[ p\text{-O}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_3 \xrightarrow{\text{1. Sn/HCl, 2. OH}^-} p\text{-H}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_3 \quad \text{(\(p\)-toluidine)} \] The resulting compound is \(p\)-toluidine. The methyl group (\(-\text{CH}_3\)) attached at the para-position behaves as an electron-donating group through inductive (\(+I\)) and hyperconjugative mechanisms. It pushes electron density into the aromatic ring, making the lone pair on the amino nitrogen more accessible for protonation. Therefore, \(p\)-toluidine is a

stronger base than aniline (\(\text{C}_6\text{H}_5\text{NH}_2\)). Looking closely at Statement-II, it states that the reduction yields an amine that is

more basic than aniline. Since \(p\)-toluidine is indeed more basic than aniline, Statement-II is scientifically correct as well. Note on potential official key variance: Let us carefully re-verify standard textbook evaluations. If an option states (2) Both are correct, let us verify if there is any nuance. The reduction of \(p\)-nitrotoluene yields \(p\)-toluidine. Since \(p\)-toluidine contains a \(\text{-CH}_3\) group which is an electron-donating group (\(+I\) and hyperconjugation), it increases the electron density on nitrogen, making it a stronger base than aniline. Hence Statement-II is absolutely correct. Both statement-I and statement-II are true. Thus, option (2) is the accurate choice.