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?

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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?

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    $\begingroup$ Edit your question to explain why you think that and you will get answers. $\endgroup$ – Waylander Jul 20 '19 at 8:18
  • $\begingroup$ Think about what base and acid solubility mean in the context of the question. $\endgroup$ – user55119 Jul 20 '19 at 19:45

" 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

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    $\begingroup$ You might edit this answer to give detail on what the Hinsburg test is - I had never heard of it and had to look it up. $\endgroup$ – Waylander Jul 21 '19 at 17:16

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