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:

• A

  
• Correct answer
  B

  
• C

  
• D

  

Hence, the correct answer is option: 'B'.