Given below are two statements :
Statement (I) : All the following compounds react with p-toluenesulfonyl chloride.
C₆H₅NH₂ (C₆H₅)₂NH (C₆H₅)₃N
Statement (II) : Their products in the above reaction are soluble is aqueous NaOH.
In the light of the above statements, choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Count the N–H hydrogens: TsCl needs one to react, and NaOH needs one left behind to dissolve the product.

1. Reaction needs N–H. p-Toluenesulfonyl chloride (TsCl) is a Hinsberg-type reagent. The amine nitrogen attacks sulfur, and an N–H hydrogen leaves with Cl as HCl. With no N–H, there is no stable sulfonamide.
2. Solubility needs N–H. A sulfonamide dissolves in aqueous NaOH only if it still has an N–H. That hydrogen is acidic because the strongly electron-withdrawing SO₂ group sits next to it, so NaOH removes it and forms a soluble sodium salt.

   Amine	N–H count	Reacts with TsCl?	Product soluble in NaOH?
   C₆H₅NH₂ (1°)	2	Yes, gives ArSO₂NHC₆H₅	Yes, one acidic N–H left
   (C₆H₅)₂NH (2°)	1	Yes, gives ArSO₂N(C₆H₅)₂	No, no N–H left
   (C₆H₅)₃N (3°)	0	No	No product

3. So. Statement I is false because triphenylamine does not react. Statement II is false because the product from diphenylamine is insoluble in NaOH. Both are false, which is option (c).

The idea to keep. In Hinsberg-type tests, track the N–H hydrogens. Primary amines give NaOH-soluble sulfonamides, secondary amines give insoluble ones, and tertiary amines give no sulfonamide at all.

Answer from the NTA final answer key. Solution by PrepWiser.