Nitrobenzene on treatment with zinc dust and aqueous ammonium chloride gives

Question

Nitrobenzene on treatment with zinc dust and aqueous ammonium chloride gives (A) C 6 H 5 N = N - C 6 H 5 (B) C 6 H 5 NH 2 (C) C 6 H 5 NO (D) C 6 H 5 NHOH

Tardigrade Admin · Accepted Answer

When nitrobenzene is reduced with a neutral reducing agent

(Z n

dust

+ N H_{4} Cl, A l - H g

couple and water), pheylhydroxylamine is formed.

Nitrobenzene C_{6} H_{5} N O_{2} + 2 [H] Z n dust + N H_{4} Cl - H_{2} O Nitrosobenzene C_{6} H_{5} NO < b r / > 2 [H] Phenyl hydroxylamine C_{6} H_{5} N H O H

Q. Nitrobenzene on treatment with zinc dust and aqueous ammonium chloride gives

$C_{6} H_{5} N = N - C_{6} H_{5}$

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

$C_{6} H_{5} NO$

$C_{6} H_{5} N H O H$

Q. Nitrobenzene on treatment with zinc dust and aqueous ammonium chloride gives

C6​H5​N=N−C6​H5​

C6​H5​NH2​

C6​H5​NO

C6​H5​NHOH

When nitrobenzene is reduced with a neutral reducing agent (Zn dust +NH4​Cl,Al−Hg couple and water), pheylhydroxylamine is formed. Nitrobenzene C6​H5​NO2​​+2[H]Zn dust +NH4​Cl−H2​O​ Nitrosobenzene C6​H5​NO​<br/>2[H]​Phenyl hydroxylamineC6​H5​NHOH​

$C_{6} H_{5} N = N - C_{6} H_{5}$

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

$C_{6} H_{5} NO$

$C_{6} H_{5} N H O H$