Question:
What is the correct sequence of reagents used for converting nitrobenzene into m-dibromobenzene ?
What is the correct sequence of reagents used for converting nitrobenzene into m-dibromobenzene ?
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Always check the directing influence of groups. Since we want a {meta} product, we must do the bromination while the meta-directing $-NO_2$ group is still present.
Updated On: Feb 2, 2026
- Sn/HCl / KBr / Br2 / H+
- NaNO2 / HCl / KBr / H+
- Sn/HCl / NaNO2/HCl / CuBr/HBr
- Br2/Fe / Sn/HCl / NaNO2/HCl / CuBr/HBr
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The Correct Option is D
Solution and Explanation
Step 1: Nitrobenzene has an $-NO_2$ group which is meta-directing. So, first brominate with $Br_2/Fe$ to get m-bromonitrobenzene.
Step 2: Reduce the $-NO_2$ to $-NH_2$ using $Sn/HCl$ to get m-bromoaniline.
Step 3: Convert the $-NH_2$ to a diazonium salt using $NaNO_2/HCl$.
Step 4: Replace the diazonium group with $Br$ using the Sandmeyer Reaction ($CuBr/HBr$) to get m-dibromobenzene.
Step 2: Reduce the $-NO_2$ to $-NH_2$ using $Sn/HCl$ to get m-bromoaniline.
Step 3: Convert the $-NH_2$ to a diazonium salt using $NaNO_2/HCl$.
Step 4: Replace the diazonium group with $Br$ using the Sandmeyer Reaction ($CuBr/HBr$) to get m-dibromobenzene.
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