In the reaction $C_6H_5CHO + C_6H_5NH_2 \to C_6H_5N=CHC_6H_5 + H_2O$,the compound $C_6H_5N=CHC_6H_5$ is known as

In the sequence of the following reactions,$X$ and $Y$ are respectively: $CH_3OH$ $\xrightarrow{HI} CH_3I$ $\xrightarrow{KCN} CH_3CN$ $\xrightarrow{\text{reduction}} X$ $\xrightarrow{HNO_2} Y$

The correct order of base strength in benzene as a solvent for the given amines is:
$I: CH_3CH_2NH_2$
$II: (CH_3CH_2)_2NH$
$III: (CH_3CH_2)_3N$

Select the percentage in which these amine derivatives are formed in the reaction mixture along with the reason for the nitration of aniline:

$PhNH_2 + Ph_3COH \xrightarrow{H^{+}} P$ (major) (not an $N$-derivative). The product $(P)$ is:

Compound $III$ is
$\xrightarrow[{ii)\,CuCl/HCl}]{{i)\,HNO_2\,(0 - 5\,^oC)}}I$ $\xrightarrow{{Sn/HCl}}II$ $\xrightarrow[{ii)\,H_3PO_2 + H_2O}]{{i)HNO_2\,(0 - 5\,^oC)\,}}III$