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MHTCET 13 Apr 2026 Shift 2 Paper

Question : 85 of 150

Marks: +1, -0

What is the importance of Sandmeyer's reaction?

To obtain R-I from R-Cl
To obtain Ar-X by replacing diazo group of AR-N ${ }_{2}^{+}-\mathrm{X}$ - using $\mathrm{CuCl} / \mathrm{HCl}$
To obtain R-F from R-Cl using AgF
To obtain $\mathrm{R}-\mathrm{CH}_{2}-\mathrm{R}$ from $\mathrm{R}-\mathrm{CO}-\mathrm{R}$ using $\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$

Solution:

Concept:

Sandmeyer's reaction is used to convert aryl diazonium salts into aryl halides by replacing the diazo group with a halogen or cyanide using cuprous salts.

Explanation:

In this reaction, the diazonium group

(-\mathrm{N}_2^+)

of an aryl diazonium salt is replaced by chlorine, bromine, or cyanide in the presence of cuprous salts.

For example, benzene diazonium chloride reacts with cuprous chloride in hydrochloric acid to give chlorobenzene:

\mathrm{ArN}_2^+\mathrm{Cl}^- \xrightarrow{\mathrm{CuCl}/\mathrm{HCl}} \mathrm{ArCl} + \mathrm{N}_2

Here, the diazo group is replaced by a chlorine atom, forming an aryl chloride.

Thus, Sandmeyer's reaction is important for preparing aryl halides from aryl diazonium salts.

The other options represent different reactions.

Option A is the Finkelstein reaction, which gives alkyl iodides from alkyl chlorides.

Option C is the Swarts reaction, which gives alkyl fluorides using

\mathrm{AgF}

Option D is the Clemmensen reduction, which reduces carbonyl compounds using

\mathrm{Zn-Hg}/\mathrm{HCl}

Answer:

Option B: To obtain

\mathrm{ArX}

by replacing the diazo group of

\mathrm{ArN}_2^+\mathrm{X}^-

using

\mathrm{CuCl}/\mathrm{HCl}

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