In the reaction,P, Q and R respectively are: | vedclass

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(A) The reaction sequence is as follows:$1$. The conversion of aniline to benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$ is achieved using $NaNO_2 + \

Vedclass · Accepted Answer

$NaNO_2 + 	ext{dil. } HCl, HBF_4, Cu + NaNO_2$

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(A) The reaction sequence is as follows:$1$. The conversion of aniline to benzene diazonium chloride $(C_6H_5N_2^+Cl^-)$ is achieved using $NaNO_2 + 	ext{dil. } HCl$ at $0-5 \ ^\circ C$. Thus,$P = NaNO_2 + 	ext{dil. } HCl$.$2$. The conversion of benzene diazonium chloride to benzene diazonium fluoroborate $(C_6H_5N_2^+BF_4^-)$ is achieved by adding fluoroboric acid $(HBF_4)$. Thus,$Q = HBF_4$.$3$. The conversion of benzene diazonium fluoroborate to nitrobenzene $(C_6H_5NO_2)$ is achieved by heating with sodium nitrite in the presence of copper powder. Thus,$R = Cu + NaNO_2$.Therefore,$P, Q$ and $R$ are $NaNO_2 + 	ext{dil. } HCl, HBF_4, Cu + NaNO_2$ respectively.

In the reaction,$P, Q$ and $R$ respectively are:

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