How will you convert aniline into benzyl alcohol.

What is the major product "P" of the following reaction ?

$\mathrm{A}$ and $\mathrm{B}$ in the following reactions are:

In the reaction 

the product $\left[\mathrm{Y}\right]$ is

Give reasons : Aromatic diazonium salts are more stable than aliphatic diazonium salts.

In the above chemical reaction, intermediate "$\mathrm{X}$" and reagent/condition "$\mathrm{A}$" are :

The correct sequence of correct reagents for the following transformation is :-

In the following reaction sequence, $\left[\mathrm{C}\right]$ is :

Give the structures of A and B in the following sequence of reactions:

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}{\mathrm{Cl}}^{-} \underset{∆}{\overset{\mathrm{CuCN}}{\to }} \mathrm{A} \stackrel{{\mathrm{H}}_2\mathrm{O}/{\mathrm{H}}^{+}}{\to } \mathrm{B}$

How will you bring about the following transformations?

Aniline to Benzene diazonium chloride

Consider the above reaction, the Product "P" is :

 Three isomers A, B and C (mol. formula $C_8{H}_{11}N$  ) give the following results:

$\mathrm{A}$ and $\mathrm{C}$$$\begin{gathered} \stackrel{\mathrm{Dianotization}}{\to }\mathrm{P}+\mathrm{Q}\underset{\left(\mathrm{ii}\right) \mathrm{acidation} \\ \left({\mathrm{KMnO}}_4+{\mathrm{H}}^{+}\right)}{\overset{\left(\mathrm{i}\right) \mathrm{Hydrolyds}}{\to }}\mathrm{R} \left(\mathrm{product} \mathrm{of} \mathrm{A}\right)+\mathrm{S} \left(\mathrm{product} \mathrm{of} \mathrm{C}\right) \end{gathered}$$

R has lower boiling point than S

$\mathrm{B}\stackrel{{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{SO}}_2\mathrm{Cl}}{\to }$ alkali-insoluble product

$\mathrm{A}, \mathrm{B}$ and $\mathrm{C},$ respectively are: