In the series of reactions: C6H5NH2 xrightarrow[0.5 oC]NaNO2/HClX xrightarrow[CH2]HNO2Y+N2+HCl X and Y are, respectively:

Question

In the series of reactions: C6H5NH2 xrightarrow[0.5 oC]NaNO2/HClX xrightarrow[CH2]HNO2Y+N2+HCl X and Y are, respectively: (A)  C6H5-N=N-C6H5,C6H5N2⊕ Cl-  (B)  C6H5N2⊕ Cl-,C6H5-N=N-C6H5  (C)  C6H5N2⊕ Cl-,C6H5NO2  (D)  C6H5NO2,C6H6

Tardigrade Admin · Accepted Answer

Primary amine groups when reacts with textNaN textO text2 text/HCl or textHN textO2 at text0 texto textC, it undergoes diazotisation and forms diazonium salt (-N=N-). So, the compound X will be textC text6 textH text5 textN= textN- textCl Benzene diazonium chloride) When benzene diazonium chloride reacts with textHN textO text2 in presence of HCHO nitrobenzene is obtained. The reactions are as under: C6H5NH2 xrightarrow[0-5 oC]NaNO2/HCl undersetBenzene diazonium chloride mathopC6H5N=NCl xrightarrow[CH2O]HNO2C6H5NO2+N2+HCl

Q. In the series of reactions: $C_{6} H_{5} N H_{2} N a N O_{2} / H Cl 0.5^{o} C X H N O_{2} C H_{2} Y + N_{2} + H Cl$ X and Y are, respectively:

$C_{6} H_{5} - N = N - C_{6} H_{5}, C_{6} H_{5} N_{2}^{\oplus} C l^{-}$

$C_{6} H_{5} N_{2}^{\oplus} C l^{-}, C_{6} H_{5} - N = N - C_{6} H_{5}$

$C_{6} H_{5} N_{2}^{\oplus} C l^{-}, C_{6} H_{5} N O_{2}$

$C_{6} H_{5} N O_{2}, C_{6} H_{6}$

Q. In the series of reactions: C6​H5​NH2​NaNO2​/HCl0.5oC​XHNO2​CH2​​Y+N2​+HCl X and Y are, respectively:

C6​H5​−N=N−C6​H5​,C6​H5​N2⊕​Cl−

C6​H5​N2⊕​Cl−,C6​H5​−N=N−C6​H5​

C6​H5​N2⊕​Cl−,C6​H5​NO2​

C6​H5​NO2​,C6​H6​

Q. In the series of reactions: $C_{6} H_{5} N H_{2} N a N O_{2} / H Cl 0.5^{o} C X H N O_{2} C H_{2} Y + N_{2} + H Cl$ X and Y are, respectively:

$C_{6} H_{5} - N = N - C_{6} H_{5}, C_{6} H_{5} N_{2}^{\oplus} C l^{-}$

$C_{6} H_{5} N_{2}^{\oplus} C l^{-}, C_{6} H_{5} - N = N - C_{6} H_{5}$

$C_{6} H_{5} N_{2}^{\oplus} C l^{-}, C_{6} H_{5} N O_{2}$

$C_{6} H_{5} N O_{2}, C_{6} H_{6}$