What happens when Hinsberg derivative is added to acidic solution? The Hinsberg test of $\ce{C5H14N2}$ compound produces a solid that is insoluble in 10% aq. $\ce{NaOH}.$ This solid derivative dissolves in 10% aq. $\ce{H2SO4}.$ Which of the following would best fit these facts?


I think b or d is the answer.
The reason being it's insoluble in NaOH. That implies it should not contain 1 degree amine. Can anyone clarify any further?
" compound produces a solid" indicates that a reaction has occured.(hence no 3° amine group(according to this)). Further this is insoluble in base so there is a 2° amine group.

"This solid dissolves in acid" directly implies that there is a 3° amine group.

Hence there is both 2° amine group as well as 3° amine group in the compound.

Hence option b is the correct answer.

References:
1. Solomon and Fryle Hinsberg Test