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

  • A
    Formation of $X$ is Hofmann degradation
  • B
    Formation of $Y$ is Schotten-Baumann reaction
  • C
    $X$ is more basic than $Y$
  • D
    All

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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.

Suppose the following reaction:
The product $D$ and $F$ are related as:

$NBS \xrightarrow{KOBr} (A)$. The product $(A)$ is:

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Amongst the compounds given,the one that would form a brilliant colored dye on treatment with $NaNO_2$ in dil. $HCl$ followed by addition to an alkaline solution of $\beta$-naphthol is

Write $IUPAC$ names of the following compounds and classify them into primary,secondary,and tertiary amines.
$(i)$ $(CH_3)_2CHNH_2$
$(ii)$ $CH_3(CH_2)_2NH_2$
$(iii)$ $CH_3NHCH(CH_3)_2$
$(iv)$ $(CH_3)_3CNH_2$
$(v)$ $C_6H_5NHCH_3$
$(vi)$ $(CH_3CH_2)_2NCH_3$
$(vii)$ $m-BrC_6H_4NH_2$

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