Question

An organic compound $C_{5}H_{14}N_{2}$ on reaction with $PhSO_{2}Cl$ produces a solid, which is insoluble in $10\%$ aqueous $NaOH$ . This solid dissolves in $10\%$ aqueous sulphuric acid. Then the possible structure of the compound may be

• A. $\left(NH\right)_{2}\left(CH\right)_{2}\left(CH\right)_{2}\left(CH\right)_{2}N\left(\left(CH\right)_{3}\right)_{2}$
• B. $\left(\left(CH\right)_{3}\right)_{2}\left(NCH\right)_{2}\left(CH\right)_{2}\left(NHCH\right)_{3}$
• C. $\left(NH\right)_{2}\left(CH\right)_{2}C\left(\left(CH\right)_{3}\right)_{2}\left(CH\right)_{2}\left(NH\right)_{2}$
• D. $\left(\left(CH\right)_{3}\right)_{2}\left(NCH\right)_{2}N\left(\left(CH\right)_{3}\right)_{2}$

Answer 1

Answer: (B)

$\left(C H_{3}\right)_{2}NCH_{2}CH_{2}NHCH_{3}$ is a $C_{5}H_{14}N_{2}$ compound which has a secondary amino group. It produces a solid that is insoluble in 10% aq. NaOH . This solid derivative dissolves in 10% aq. Sulfuric acid.