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Question #44032

Difficulty: 8/10 | 🟢 Free

The Question

An equimolar mixture of 4-methylaniline and (S)-2-amino-3-phenylpropanoic acid is treated with one equivalent of benzenesulfonyl chloride ( $\ce{PhSO2Cl}$ ) in an aqueous buffer solution at $\text{pH} = 7.0$ and $298\text{ K}$ . Which of the following represents the major organic product(s) isolated from this reaction?

Options

A.
$N$ -(4-methylphenyl)benzenesulfonamide and unreacted (S)-2-amino-3-phenylpropanoic acid
B.
Unreacted 4-methylaniline and $N$ -(1-carboxylato-2-phenylethyl)benzenesulfonamide
C.
$N$ -(4-methylphenyl)benzenesulfonamide and $N$ -(1-carboxylato-2-phenylethyl)benzenesulfonamide
D.
Unreacted 4-methylaniline and unreacted (S)-2-amino-3-phenylpropanoic acid

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Aniline Resonance Pitfalls|Hinsberg Reagent Selectivity|Amino Acid Zwitterion pH Sensitivity

Why This Question Is Tricky

Common Mistakes

The problem tests the competition between inherent nucleophilicity and pH-dependent speciation (zwitterion formation).

Students often fall into the trap of assuming the aliphatic amino group of the amino acid will react faster with the Hinsberg reagent than the resonance-stabilized aromatic amine.

However, they fail to account for the acid-base equilibrium at physiological pH (pH = 7.0), where the amino acid exists as a zwitterion with a protonated, non-nucleophilic ammonium group ( $\ce{-NH3^+}$ ).

Meanwhile, the weaker but unprotonated aromatic amine retains its lone pair and successfully undergoes sulfonylation.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Amines - Class 12
Concept	Aniline Resonance Pitfalls
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free