Question

Write short notes on the following: (i) Coupling reaction (ii) Sandmeyer reaction (iii) Friedel-Craft reaction. (2+1½+1½=5)
OR
Write chemical equations for obtaining the following from benzene diazonium chloride: (i) Phenyl cyanide (ii) Bromobenzene (iii) Nitrobenzene (iv) p-hydroxyazobenzene (v) p-Aminoazobenzene. (1+1+1+1+1=5)

Hint:

Diazonium chemistry: with cuprous salts the \(-N_2^+\) group leaves as \(N_2\) (Sandmeyer), while with phenol or aniline it stays and couples to form a coloured azo dye. Friedel-Crafts needs anhydrous \(AlCl_3\).

Answer 1

Option 1: Short notes

Step 1: Coupling reaction. Benzene diazonium chloride \(C_6H_5N_2^+Cl^-\) contains a diazonium cation that behaves as a weak electrophile. It attacks electron-rich aromatic rings such as phenol or aniline at the para position (electrophilic aromatic substitution) to give brightly coloured azo dyes carrying the \(-N=N-\) linkage.
With phenol (mildly alkaline medium): \(C_6H_5N_2Cl + C_6H_5OH \rightarrow p\text{-}HOC_6H_4N=NC_6H_5 + HCl\) (p-hydroxyazobenzene, orange).
With aniline (mildly acidic medium): \(C_6H_5N_2Cl + C_6H_5NH_2 \rightarrow p\text{-}H_2NC_6H_4N=NC_6H_5 + HCl\) (p-aminoazobenzene, yellow).

Step 2: Sandmeyer reaction. The diazonium group \(-N_2^+\) is replaced by \(-Cl\), \(-Br\) or \(-CN\) when the diazonium salt is warmed with the corresponding cuprous halide or cuprous cyanide; nitrogen gas is evolved.
\(C_6H_5N_2Cl \xrightarrow{CuCl/HCl} C_6H_5Cl + N_2\)
\(C_6H_5N_2Cl \xrightarrow{CuBr/HBr} C_6H_5Br + N_2\)
\(C_6H_5N_2Cl \xrightarrow{CuCN/KCN} C_6H_5CN + N_2\)

Step 3: Friedel-Crafts reaction. Aromatic hydrocarbons undergo alkylation or acylation in the presence of anhydrous \(AlCl_3\) (a Lewis acid), which generates the attacking electrophile (\(R^+\) or acylium \(RCO^+\)).
Alkylation: \(C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 + HCl\) (toluene).
Acylation: \(C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl\) (acetophenone).

Option 2: Equations from benzene diazonium chloride

Step 1: (i) Phenyl cyanide (benzonitrile) by Sandmeyer reaction using cuprous cyanide:
\(C_6H_5N_2Cl \xrightarrow{CuCN/KCN} C_6H_5CN + N_2\)

Step 2: (ii) Bromobenzene by Sandmeyer reaction using cuprous bromide:
\(C_6H_5N_2Cl \xrightarrow{CuBr/HBr} C_6H_5Br + N_2\)

Step 3: (iii) Nitrobenzene by warming with sodium nitrite in presence of copper powder (via the fluoroborate):
\(C_6H_5N_2Cl + NaNO_2 \xrightarrow{Cu} C_6H_5NO_2 + N_2 + NaCl\)

Step 4: (iv) p-hydroxyazobenzene by coupling with phenol in mildly alkaline medium:
\(C_6H_5N_2Cl + C_6H_5OH \rightarrow p\text{-}HOC_6H_4N=NC_6H_5 + HCl\)

Step 5: (v) p-Aminoazobenzene by coupling with aniline in mildly acidic medium:
\(C_6H_5N_2Cl + C_6H_5NH_2 \rightarrow p\text{-}H_2NC_6H_4N=NC_6H_5 + HCl\)

\[\boxed{\text{Both alternatives use the reactions of benzene diazonium chloride.}}\]