Aniline,when diazotized in cold and then treated with $N,N$-dimethylaniline,gives a coloured product. Its structure would be:

  • A
    $(CH_3)_2N-C_6H_4-N=N-C_6H_5$
  • B
    $(CH_3)_2N-C_6H_4-NH-C_6H_5$
  • C
    $CH_3NH-C_6H_4-N=N-C_6H_4-NHCH_3$
  • D
    $CH_3-C_6H_4-N=N-C_6H_4-NH_2$

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Among the following,the strongest nucleophile is

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

Reaction of $R-CONH_2$ with a mixture of $Br_2$ and $KOH$ gives $R-NH_2$ as the main product. The intermediates involved in this reaction are :-
$(A)$ $R-CONHBr$
$(B)$ $R-NH-Br$
$(C)$ $R-N=C=O$
$(D)$ $R-CONBr_2$

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Decarboxylation of all six possible isomers of diaminobenzoic acids $C_6H_3(NH_2)_2COOH$ yields three products $A, B$ and $C$. Three acids give product $A$,two acids give product $B$ and one acid gives product $C$. The melting point of product $C$ is .....$^{\circ} C$.

The product$(s)$ of the following reaction sequence is(are):
Aniline $\xrightarrow[\text{ii) } KBrO_3/HBr]{\text{i) Acetic anhydride/pyridine}}$ $\xrightarrow[\text{iv) } NaNO_2/HCl, 273-278 \ K]{\text{iii) } H_3O^+, \text{heat}}$ $\xrightarrow{\text{v) } Cu/HBr} \text{Product}$

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