Sandmeyer reaction — conversion of diazonium salt to various products

medium CBSE NEET JEE-MAIN NCERT Class 12 3 min read

Question

What is the Sandmeyer reaction? Show how a benzene diazonium salt can be converted to chlorobenzene, bromobenzene, and cyanobenzene. Also, how is iodobenzene prepared from diazonium salt?

(NCERT Class 12, Chapter 13 — Amines)


Solution — Step by Step

Aniline reacts with nitrous acid (NaNO₂ + HCl) at 0-5°C to form benzene diazonium chloride:

C6H5NH2+NaNO2+2HCl0-5°CC6H5N2+Cl+NaCl+2H2O\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0\text{-}5°C} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}

The low temperature is critical — above 5°C, the diazonium salt decomposes.

The Sandmeyer reaction uses Cu₂X₂/HX (cuprous halide in the corresponding acid):

Chlorobenzene:

C6H5N2+ClCuCl/HClC6H5Cl+N2\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCl/HCl}} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2

Bromobenzene:

C6H5N2+ClCuBr/HBrC6H5Br+N2\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuBr/HBr}} \text{C}_6\text{H}_5\text{Br} + \text{N}_2

Cyanobenzene:

C6H5N2+ClCuCN/KCNC6H5CN+N2\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCN/KCN}} \text{C}_6\text{H}_5\text{CN} + \text{N}_2

In each case, N₂ gas is released and the N2+\text{N}_2^+ group is replaced by the nucleophile.

For iodobenzene, we use KI directly (no copper catalyst needed):

C6H5N2+Cl+KIC6H5I+N2+KCl\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{KI} \rightarrow \text{C}_6\text{H}_5\text{I} + \text{N}_2 + \text{KCl}

This is NOT called the Sandmeyer reaction — it’s simply a replacement reaction. The Sandmeyer reaction specifically involves Cu(I) salts as catalysts.

For fluorobenzene, we use the Balz-Schiemann reaction:

C6H5N2+Cl+HBF4C6H5N2+BF4ΔC6H5F+BF3+N2\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{HBF}_4 \rightarrow \text{C}_6\text{H}_5\text{N}_2^+\text{BF}_4^- \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{F} + \text{BF}_3 + \text{N}_2

The diazonium fluoroborate is thermally decomposed to give fluorobenzene.


Why This Works

Diazonium salts are so reactive because the N2+\text{N}_2^+ group is an excellent leaving group — it leaves as stable N₂ gas. This makes the C-N bond easy to break, allowing replacement by various nucleophiles. The Cu(I) catalyst in the Sandmeyer reaction facilitates electron transfer, making the replacement of Cl, Br, and CN efficient.

This reaction is synthetically very important because it allows us to introduce groups onto the benzene ring that are otherwise difficult to attach directly. The sequence amine → diazonium → product gives us access to aryl halides and aryl cyanides.


Alternative Method — Summary of Diazonium Reactions

ProductReagentReaction Name
ArClCuCl/HClSandmeyer
ArBrCuBr/HBrSandmeyer
ArCNCuCN/KCNSandmeyer
ArIKI(Direct replacement)
ArFHBF₄, then heatBalz-Schiemann
ArOHH₂O, warmHydrolysis
ArHH₃PO₂Deamination

For NEET, remember: Sandmeyer = Cu(I) catalyst for Cl, Br, CN. For iodo, no copper needed (just KI). For fluoro, it’s Balz-Schiemann (HBF₄). This distinction between reaction names is tested frequently.


Common Mistake

Students call every reaction of diazonium salts a “Sandmeyer reaction.” This is wrong. The Sandmeyer reaction ONLY involves Cu(I) salts (CuCl, CuBr, CuCN). The conversion to ArI using KI, to ArF using HBF₄, or to ArOH using water are all different named reactions. NEET questions specifically test whether you know which replacement uses which reagent and which name.

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