Question:medium

Given below are two statements:
Statement (I) : All the following compounds react with p-toluenesulfonyl chloride.
$C_6H_5NH_2 \quad      (C_6H_5)_2NH                \quad (C_6H_5)_3N$
Statement (II) : Their products in the above reaction are soluble in aqueous NaOH.
In the light of the above statements, choose the correct answer from the options given below.

Updated On: Feb 3, 2026
  • Both Statement I and Statement II is false
  • Statement I is true but Statement II is false
  • Statement I is false but Statement II is true
  • Both Statement I and Statement II is true
Show Solution

The Correct Option is A

Solution and Explanation

To address the stated problem, an examination of the two provided statements is required, focusing on the chemical reactivity of the specified compounds with p-toluenesulfonyl chloride and the solubility of the resulting products in aqueous NaOH.

Statement I Analysis:

Statement I asserts that: "All the following compounds react with p-toluenesulfonyl chloride."

The compounds under consideration are: \(C_6H_5NH_2\) (aniline), \((C_6H_5)_2NH\) (diphenylamine), and \((C_6H_5)_3N\) (triphenylamine).

Rationale:

  • Aniline (\(C_6H_5NH_2\)): As a primary amine, it reacts with p-toluenesulfonyl chloride, yielding a tosylamide.
  • Diphenylamine (\((C_6H_5)_2NH\)): A secondary amine, capable of forming a tosyl derivative, though its reactivity is typically lower than primary amines.
  • Triphenylamine (\((C_6H_5)_3N\)): A tertiary amine, which generally does not react with p-toluenesulfonyl chloride under standard conditions.

Consequently, Statement I is invalidated by the non-reactivity of triphenylamine with p-toluenesulfonyl chloride.

Statement II Analysis:

Statement II claims that: "Their products in the above reaction are soluble in aqueous NaOH."

Rationale:

  • The reaction products of aniline and diphenylamine with p-toluenesulfonyl chloride are tosylamides.
  • Tosylamides are typically insoluble in aqueous NaOH, as they lack acidic hydrogens necessary for salt formation.

Therefore, Statement II is false due to the insolubility of the products in aqueous NaOH.

Concluding Determination:

The accurate conclusion is: "Both Statement I and Statement II is false".

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