Sandmeyer, Gattermann & Diazonium Replacement Reactions

CHEDiazonium Salts & Reactions

Replacement of the diazonium group $\text{-}N_2^+$ by halogens is the standard way to put $Cl$ , $Br$ , $I$ or $F$ onto a benzene ring in good yield and at the correct position. ISC expects the right reagent for each halogen and the distinction between Sandmeyer and Gattermann.

Sandmeyer (cuprous halide / cyanide)

C_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl + N_2 \qquad C_6H_5N_2^+Cl^- \xrightarrow{CuBr/HBr} C_6H_5Br + N_2

CuCN

similarly gives benzonitrile

C_6H_5CN

Gattermann (copper powder + HX)

C_6H_5N_2^+Cl^- \xrightarrow{Cu/HCl} C_6H_5Cl + N_2

same product as Sandmeyer but generally lower yield

Iodo and fluoro (Balz-Schiemann)

C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + N_2 + KCl \qquad C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + N_2 + BF_3

iodide needs no copper; fluoride uses the diazonium fluoroborate

Sandmeyer reaction: warm the diazonium salt with a cuprous halide; $C_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl$ and with $CuBr/HBr \rightarrow C_6H_5Br$ (plus $N_2$ ).
Gattermann reaction: uses copper powder with $HCl$ / $HBr$ instead of cuprous halide; same products but generally lower yields.
Iodobenzene needs no copper catalyst: simply warm the diazonium salt with potassium iodide, $C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + N_2$ .
Fluorobenzene by Balz-Schiemann: treat the diazonium salt with $HBF_4$ to get the diazonium fluoroborate, which on heating gives $C_6H_5F + N_2 + BF_3$ .
Sandmeyer also installs $\text{-CN}$ ( $CuCN$ ) to give benzonitrile, extending it beyond halogens.
Choice rule: use Sandmeyer ( $Cu^+$ salt) for $Cl$ / $Br$ / $CN$ when you want better yields; Gattermann (Cu/HX) is the simpler-reagent alternative for $Cl$ / $Br$ .
These diazonium replacements give a single, correctly positioned aryl halide, whereas direct halogenation of an arene can give isomer mixtures.
The cuprous halide acts as a catalyst that promotes the radical replacement of $\text{-}N_2^+$ by the halogen.
All these replacements expel $N_2$ gas, which is the driving force for displacing the unstable $\text{-}N_2^+$ group.

Where the marks go

Using a cuprous halide for iodobenzene — $I$ needs only $KI$ , no copper catalyst.
Quoting Sandmeyer for fluorobenzene — $F$ is installed by Balz-Schiemann ( $HBF_4$ , then heat the fluoroborate), not by a cuprous salt.
Confusing Sandmeyer (cuprous halide $CuX$ ) with Gattermann (copper powder $Cu$ + $HX$ ); both must be paired with the matching acid $HX$ .
Writing cupric ( $Cu^{2+}$ ) instead of cuprous ( $Cu^+$ ) salts for the Sandmeyer reaction.
Forgetting that $N_2$ is evolved in every replacement — omitting it leaves the equation unbalanced.

How the board asks it

Conversionthe reagent-for-each-halogen rules
How will you convert benzene diazonium chloride into (i) chlorobenzene, (ii) bromobenzene, (iii) iodobenzene and (iv) fluorobenzene? Give the reagents and conditions in each case.
Distinguishsandmeyer ( $CuX$ ) vs gattermann ( $Cu/HX$ )
Distinguish between the Sandmeyer reaction and the Gattermann reaction with respect to the reagents used and the yield of the aryl halide obtained.
Predict the productsandmeyer extended to $CuCN$ giving benzonitrile
What is the organic product when benzene diazonium chloride is warmed with cuprous cyanide ( $CuCN$ )? Name the product, draw its structure and name the gas evolved.
Give reasonsdiazonium route gives a single correctly positioned aryl halide
Account for the fact that aryl iodides are best prepared from a diazonium salt and $KI$ rather than by direct iodination of benzene.
Identify / classifybalz-schiemann route to fluorobenzene
An aromatic amine $A$ ( $C_6H_7N$ ) on treatment with $NaNO_2/HCl$ at $0$ – $5^\circ C$ gives $B$ , which on warming with $HBF_4$ followed by heating gives $C$ ( $C_6H_5F$ ). Identify $A$ , $B$ and $C$ and write the equations involved.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.