Question

Consider the following sequence of reactions :

[A]$\left(C_3{H}_{9}N\right)\stackrel{NaN{O}_2/HCl}{⟶}\left[B\right]\left(C_3{H}_{8}O\right)$

$\underset{K_{r}C{r}_2{O}_7}{\overset{H_{2}S{O}_4}{\to }}\left(C_3{H}_6{O}_2\right)\left[C\right]$

The compound [A] is :

• A. $C{H}_{3}C{H}_{2}NHC{H}_3$
• B. $C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2$
• C. $(C{H}_3$)${}_2$$CHN{H}_2$
• D. $\left(C{H}_3\right)$${}_{3}N$

Answer 1

The correct option is B $C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2$

We all know that primary amine($C{H}_{3}C{H}_{2}C{H}_{2}N{H}_2$) reacts with the nitrous acid$\left(NaN{O}_2/HClorHN{O}_2\right)$ to form diazonium salt($C{H}_{3}C{H}_{2}C{H}_2-N_2^{+}$). Further addition of water leads to the formation of primary alcohol ($C{H}_{3}C{H}_{2}C{H}_{2}OH$). This primary alcohol getsoxidised in presence of potassium dichromate to form carboxylic acid($C{H}_{3}C{H}_{2}COOH$) So the correct option is B.