In the above conversion,the correct sequence of reagents to be added is:

  • A
    $(i) Fe / H^{+}, (ii) HONO, (iii) CuCl, (iv) KMnO_4, (v) Br_2$
  • B
    $(i) KMnO_4, (ii) Br_2 / Fe, (iii) Fe / H^{+}, (iv) Cl_2$
  • C
    $(i) Br_2 / Fe, (ii) Fe / H^{+}, (iii) HONO, (iv) CuCl, (v) KMnO_4$
  • D
    $(i) Br_2 / Fe, (ii) Fe / H^{+}, (iii) KMnO_4, (iv) Cl_2$

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Which of the following compounds does not produce an amine upon hydrolysis?

Three isomers $A$,$B$,and $C$ (molecular formula $C_8H_{11}N$) give the following results:
$A$ and $C$ $\xrightarrow{\text{Diazotization}}$ $P + Q$ $\xrightarrow[(ii) \text{ oxidation } (KMnO_4 + H^{+})]{(i) \text{ Hydrolysis}}$ $R$ (product of $A$) and $S$ (product of $C$)
Identify $A$,$B$,and $C$ from the given options.

$CH_3CH_2CONH_2$ $\xrightarrow{Br_2 / KOH_{(aq)}} X$ $\xrightarrow{C_6H_5COCl} Y$. Which statement is true?

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Aniline on treatment with $NaNO_2 / HCl$ at about $0^\circ C$ gives a compound $A$, which on further treatment with $CuCN / KCN$ gives a compound $B$, which when reacts with $H_2 / Ni$ gives a compound $C$. The compound $C$ on reaction with $HNO_2$ gives a compound $D$. The structure of $D$ is:

Match List-$I$ with List-$II$ and select the correct answer.
List-$I$ List-$II$
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$2.$ Nitrobenzene $b.$ Sulpha drug
$3.$ Sulphanilamide $c.$ Solvent in Friedel-Crafts reaction
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