The product of the reaction of ${\left(\mathrm{C}{\mathrm{H}}_3\right)}_3\mathrm{C}\mathrm{C}\mathrm{H}\left(\mathrm{O}\mathrm{H}\right)\mathrm{C}{\mathrm{H}}_3$ with conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ is

• A

  ${\left(\mathrm{C}{\mathrm{H}}_3\right)}_3\mathrm{C}\mathrm{H}=\mathrm{C}{\mathrm{H}}_2$
• Correct answer
  B

  ${\left(\mathrm{C}{\mathrm{H}}_3\right)}_2\mathrm{C}=\mathrm{C}{\left(\mathrm{C}{\mathrm{H}}_3\right)}_2$
• C

  $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}\mathrm{H}=\mathrm{C}{\mathrm{H}}_2$
• D

  $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}\mathrm{H}=\mathrm{C}\mathrm{H}\mathrm{C}{\mathrm{H}}_3$

Solution

The product of the reaction of $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}\mathrm{C}\mathrm{H}\left(\mathrm{O}\mathrm{H}\right)\mathrm{C}{\mathrm{H}}_3$ with conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ is ${\left(\mathrm{C}{\mathrm{H}}_3\right)}_2\mathrm{C}=\mathrm{C}{\left(\mathrm{C}{\mathrm{H}}_3\right)}_2$

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