$A$ colourless substance $'A'$ $(C_6H_7N)$ is sparingly soluble in water and gives a water-soluble compound $'B'$ on treating with mineral acid. On reacting with $CHCl_3$ and alcoholic potash,$'A'$ produces an obnoxious smell due to the formation of compound $'C'$. Reaction of $'A'$ with benzenesulphonyl chloride gives compound $'D'$ which is soluble in alkali. With $NaNO_2$ and $HCl$,$'A'$ forms compound $'E'$ which reacts with phenol in an alkaline medium to give an orange dye $'F'$. Identify compounds $'A'$ to $'F'$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) The substance $'A'$ is $C_6H_5NH_2$ (Aniline).
$1$. $C_6H_5NH_2 + HCl \rightarrow C_6H_5NH_3^+Cl^-$ (Anilinium chloride,$'B'$).
$2$. $C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$ (Phenyl isocyanide,$'C'$).
$3$. $C_6H_5NH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5NHSO_2C_6H_5 + HCl$ ($N$-phenylbenzenesulphonamide,$'D'$).
$4$. $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278 \ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$ (Benzenediazonium chloride,$'E'$).
$5$. $C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{OH^-} C_6H_5-N=N-C_6H_4OH$ ($p$-hydroxyazobenzene,orange dye,$'F'$).

Explore More

Similar Questions

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

Give the structures of $A$,$B$ and $C$ in the following reactions:
$(i)$ $CH_3CH_2I$ $\xrightarrow{NaCN} A$ $\xrightarrow[Partial\,hydrolysis]{OH^{-}} B$ $\xrightarrow{NaOH+Br_2} C$
$(ii)$ $C_6H_5N_2Cl$ $\xrightarrow{CuCN} A$ $\xrightarrow{H_2O/H^{+}} B$ $\xrightarrow[\Delta]{NH_3} C$
$(iii)$ $CH_3CH_2Br$ $\xrightarrow{KCN} A$ $\xrightarrow{LiAlH_4} B$ $\xrightarrow[0\ ^oC]{HNO_2} C$
$(iv)$ $C_6H_5NO_2$ $\xrightarrow{Fe/HCl} A$ $\xrightarrow[273\ K]{NaNO_2+HCl} B$ $\xrightarrow[\Delta]{H_2O/H^{+}} C$
$(v)$ $CH_3COOH$ $\xrightarrow[\Delta]{NH_3} A$ $\xrightarrow{NaOBr} B$ $\xrightarrow{NaNO_2/HCl} C$
$(vi)$ $C_6H_5NO_2$ $\xrightarrow{Fe/HCl} A$ $\xrightarrow[273\ K]{HNO_2} B$ $\xrightarrow{C_6H_5OH} C$

$A$ given nitrogen-containing aromatic compound $A$ reacts with $Sn/HCl,$ followed by $HNO_2$ to give an unstable compound $B$. $B,$ on treatment with phenol,forms a beautiful coloured compound $C$ with the molecular formula $C_{12}H_{10}N_2O.$ The structure of compound $A$ is

$(A)$ $\xrightarrow{KOBr} (B)$ $\xrightarrow[KOH]{CHCl_3} (C)$ $\xrightarrow{LiAlH_4} \text{Structure of } (C) \text{ is}$

Difficult
View Solution

The correct option$(s)$ for the following sequence of reactions is(are):
$(A)$ $Q = KNO_2, W = LiAlH_4$
$(B)$ $R =$ benzenamine,$V = KCN$
$(C)$ $Q = AgNO_2, R =$ phenylmethanamine
$(D)$ $W = LiAlH_4, V = AgCN$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo