How will you obtain monobromobenzene from aniline?
(N/A) To obtain monobromobenzene from aniline,the following two-step process is used:
$1$. Diazotization: Aniline is treated with $NaNO_2$ and $HCl$ at a low temperature of $273-278 \ K$ to form benzene diazonium chloride.
$2$. Sandmeyer's Reaction: The benzene diazonium chloride is then treated with cuprous bromide $(Cu_2Br_2)$ in the presence of hydrobromic acid $(HBr)$ to yield bromobenzene.
The reaction sequence is:
$C_6H_5NH_2$ $\xrightarrow{NaNO_2 + HCl, 273-278 \ K} C_6H_5N_2^+Cl^-$ $\xrightarrow{Cu_2Br_2/HBr} C_6H_5Br$
$1$. Diazotization: Aniline is treated with $NaNO_2$ and $HCl$ at a low temperature of $273-278 \ K$ to form benzene diazonium chloride.
$2$. Sandmeyer's Reaction: The benzene diazonium chloride is then treated with cuprous bromide $(Cu_2Br_2)$ in the presence of hydrobromic acid $(HBr)$ to yield bromobenzene.
The reaction sequence is:
$C_6H_5NH_2$ $\xrightarrow{NaNO_2 + HCl, 273-278 \ K} C_6H_5N_2^+Cl^-$ $\xrightarrow{Cu_2Br_2/HBr} C_6H_5Br$
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Similar Questions
Which of the following are intermediates in the Sandmeyer reaction?
$(i)$ $C_6H_5N^{+} \equiv NCl^{-}$
$(ii)$ $C_6H_5N^{+} \equiv N$
$(iii)$ $\overset{\centerdot }{C}_6H_5$
$(iv)$ $C_6H_5Cl$
$(i)$ $C_6H_5N^{+} \equiv NCl^{-}$
$(ii)$ $C_6H_5N^{+} \equiv N$
$(iii)$ $\overset{\centerdot }{C}_6H_5$
$(iv)$ $C_6H_5Cl$
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