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Question

(B) $1$. The reaction of nitrobenzene with $Sn/HCl$ is a reduction reaction that converts the nitro group $(-NO_2)$ into an amino group $(-NH_2)$,form

Vedclass · Accepted Answer

$X$: Aniline,$Y$: Bromobenzene

Vedclass · Answer

(B) $1$. The reaction of nitrobenzene with $Sn/HCl$ is a reduction reaction that converts the nitro group $(-NO_2)$ into an amino group $(-NH_2)$,forming aniline $(X)$.$2$. Aniline reacts with $NaNO_2/HCl$ at low temperatures $(0-5 \ ^\circ C)$ to undergo diazotization,forming benzenediazonium chloride.$3$. The subsequent reaction with $CuBr$ (Sandmeyer reaction) replaces the diazonium group with a bromine atom,resulting in the formation of bromobenzene $(Y)$.

The products $X$ and $Y$ in the following reaction sequence are:

Similar Questions

What is the correct order of basicity for the following compounds?
$1$. $C_6H_5NH_2$ (Aniline)
$2$. $CH_3CH_2NH_2$ (Ethylamine)
$3$. $(CH_3)_2NH$ (Dimethylamine)
$4$. $CH_3CONH_2$ (Acetamide)

Product $(A)$ is

Identify the products $y$ and $x$ in the following reaction sequence:
$C_6H_5CONH_2 \xrightarrow{NaOBr} y$
$C_6H_5CONH_2 \xrightarrow[(ii) H_3O^+]{(i) C_6H_5SO_2Cl / py, \Delta} x$

Which among the following compounds has the lowest boiling point?

Consider the following reactions:
$[P] \xrightarrow[(i) \ NaNO_2/HCl, \ 0 - 5^oC]{(ii) \ \beta-naphthol/NaOH} \text{Colored Solid}$
$[P] \xrightarrow{Br_2/H_2O} C_7H_6NBr_3$
The compound $[P]$ is:

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