The product formed, when benzene diazonium fluoroborate is heated with aqueous sodium nitrite solution in the presence of copper is
- fluorobenzene
- benzene
- phenol
- p-nitrophenol
- nitrobenzene
The Correct Option is
Approach Solution - 1
The reaction of benzene diazonium fluoroborate with aqueous sodium nitrite in the presence of copper leads to the formation of nitrobenzene. This reaction is a typical example of the Sandmeyer reaction, where a diazonium salt undergoes a substitution reaction in the presence of copper and sodium nitrite, forming a nitro group on the benzene ring.
The correct option is (E) : nitrobenzene
Approach Solution -2
The reaction involves benzene diazonium fluoroborate (\(C_6H_5N_2BF_4\)) reacting with aqueous sodium nitrite (NaNO2) solution in the presence of copper.
This is a variation of the Sandmeyer reaction. The diazonium salt is converted to a nitro group (\(-NO_2\)). The reaction proceeds as follows:
\(C_6H_5N_2BF_4 + NaNO_2 \xrightarrow{Cu} C_6H_5NO_2 + N_2 + NaBF_4\)
Benzene diazonium fluoroborate reacts with sodium nitrite in the presence of copper to form nitrobenzene.
Therefore, the answer is nitrobenzene.
Top Questions on Amines
- The best reagent for converting propanamide into propanamine is:
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(b) Aniline does not undergo Friedel–Crafts reaction.
(c) Aniline on nitration gives a substantial amount of meta product. - Assertion (A): Acetamide (structure: \(\text{C}_6\text{H}_5\text{NHCOCH}_3\)) is less basic than aniline.
Reason (R): Acetylation of aniline results in decrease of electron density on nitrogen. - The chemical test which can be used to distinguish between ethanamine and aniline is:
- If amines are arranged in increasing order of their basic strength in gaseous phase, the correct order will be:
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