Question

In the series of reaction

$C_{6}H_{5}NH_{2}\underset{0 - 5{^\circ}C}{\overset{\quad NaNO_{2}/HCl\quad}{\rightarrow}}X\underset{CH_{2}O}{\overset{\quad HNO_{2}\quad}{\rightarrow}}Y + N_{2} + HCl$ X and Y are respectively

• A. $C_{6}H_{5} - N = N - C_{6}H_{5},C_{6}H_{5}N_{2}^{\oplus}Cl^{\Theta}$
• B. $C_{6}H_{5}N_{2}^{\oplus}Cl^{\Theta},C_{6}H_{5} - N = N - C_{6}H_{5}$
• C. $C_{6}H_{5}N_{2}^{\oplus}Cl^{\Theta},C_{6}H_{5}NO_{2}$
• D. $C_{6}H_{5}NO_{2},C_{6}H_{6}$

Answer 1

Answer: (C)

$C_{6}H_{5}N_{2}^{\oplus}Cl^{\Theta},C_{6}H_{5}NO_{2}$

Answer 2

$\underset{(Y)}{C_{6}H_{5}NO_{2}} + N_{2} + HCl$