Question

Write chemical equation for the formation of chlorobenzene by Sandmeyer reaction. How can the following compounds be obtained from chlorobenzene? Clarify giving chemical equations.
(i) Phenol
(ii) Toluene
(iii) Diphenyl

Hint:

Sandmeyer: Ar–N$_2^+$ \(\xrightarrow{\mathrm{CuX}}\) Ar–X.
From chlorobenzene: Dow (NaOH, 623 K, high pressure) \(\to\) phenol.
Wurtz–Fittig with CH$_3$Cl \(\to\) toluene.
Fittig (2 ArX + 2 Na) \(\to\) biaryl (diphenyl).

Answer 1

Formation of chlorobenzene (Sandmeyer reaction):
Step 1: Diazotization of aniline
\[ \mathrm{C_6H_5NH_2 + NaNO_2 + 2HCl \ (0{-}5^\circ C) \;\rightarrow\; C_6H_5N_2^+Cl^- + 2H_2O + NaCl} \]
Step 2: Replacement by Cl (CuCl)
\[ \mathrm{C_6H_5N_2^+Cl^- \xrightarrow{CuCl} C_6H_5Cl + N_2 \uparrow} \]
Conversions starting from chlorobenzene:
(i) Chlorobenzene \(\rightarrow\) Phenol (Dow process):
\[ \mathrm{C_6H_5Cl \xrightarrow[\ 300\,\mathrm{atm}\ ]{NaOH\ (aq),\,623\,K} \ C_6H_5ONa \xrightarrow{H^+} C_6H_5OH + Na^+} \]
(ii) Chlorobenzene \(\rightarrow\) Toluene (Wurtz–Fittig):
\[ \mathrm{C_6H_5Cl + CH_3Cl + 2Na \xrightarrow{dry\ ether} C_6H_5CH_3 + 2NaCl} \]
(iii) Chlorobenzene \(\rightarrow\) Diphenyl (Fittig):
\[ \mathrm{2\,C_6H_5Cl + 2Na \xrightarrow{dry\ ether} C_6H_5{-}C_6H_5 + 2NaCl} \]