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 (concept-and-role approach)

Step 1: Coupling reaction (dye formation). Here the diazonium ion acts as the electrophile rather than losing nitrogen. Because the \(-N_2^+\) group is only weakly electrophilic, it can substitute only into strongly activated rings (phenols, aromatic amines), joining two rings through an azo \(-N=N-\) bridge. The extended conjugation of the product makes it coloured, so these are used as azo dyes.
Phenol coupling: \(C_6H_5N_2Cl + C_6H_5OH \rightarrow p\text{-}HOC_6H_4N=NC_6H_5 + HCl\)
Aniline coupling: \(C_6H_5N_2Cl + C_6H_5NH_2 \rightarrow p\text{-}H_2NC_6H_4N=NC_6H_5 + HCl\)

Step 2: Sandmeyer reaction (nitrogen displacement). Unlike coupling, here the whole \(-N_2^+\) leaves as \(N_2\) and a cuprous salt supplies the new group through a radical pathway. It is a clean route to aryl halides and aryl nitriles that are hard to make by direct substitution.
\(C_6H_5N_2Cl \xrightarrow{CuCl/HCl} C_6H_5Cl + N_2\); \(C_6H_5N_2Cl \xrightarrow{CuBr/HBr} C_6H_5Br + N_2\); \(C_6H_5N_2Cl \xrightarrow{CuCN/KCN} C_6H_5CN + N_2\).

Step 3: Friedel-Crafts reaction (ring functionalisation). Anhydrous \(AlCl_3\) polarises the alkyl or acyl halide to release a carbocation or acylium ion, which substitutes a ring hydrogen. Alkylation attaches an alkyl group; acylation attaches a \(-COR\) group.
\(C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 + HCl\); \(C_6H_6 + CH_3COCl \xrightarrow{AlCl_3} C_6H_5COCH_3 + HCl\).

Option 2: Equations from benzene diazonium chloride (grouped by mechanism)

Step 1: Sandmeyer set (i and ii). Cuprous salts replace \(-N_2^+\) with loss of \(N_2\).
(i) \(C_6H_5N_2Cl \xrightarrow{CuCN/KCN} C_6H_5CN + N_2\) (phenyl cyanide).
(ii) \(C_6H_5N_2Cl \xrightarrow{CuBr/HBr} C_6H_5Br + N_2\) (bromobenzene).

Step 2: Nitro group introduction (iii). \(C_6H_5N_2Cl + NaNO_2 \xrightarrow{Cu} C_6H_5NO_2 + N_2 + NaCl\) (nitrobenzene).

Step 3: Coupling set (iv and v). The intact diazonium ion couples at the para position of an activated ring.
(iv) With phenol: \(C_6H_5N_2Cl + C_6H_5OH \rightarrow p\text{-}HOC_6H_4N=NC_6H_5 + HCl\) (p-hydroxyazobenzene).
(v) With aniline: \(C_6H_5N_2Cl + C_6H_5NH_2 \rightarrow p\text{-}H_2NC_6H_4N=NC_6H_5 + HCl\) (p-aminoazobenzene).

\(\boxed{\text{Sandmeyer and nitration lose } N_2; \text{ coupling keeps the } -N=N- \text{ bridge.}}\)