A is an optically inactive alkyl chloride which on reaction with aqueous $\mathrm{KOH}$ gives $\mathrm{B}$. $\mathrm{B}$ on heating with $\mathrm{Cu}$ at $300^{\circ} \mathrm{C}$ gives an alkene $\mathrm{C}$, what are $\mathrm{A}$ and $\mathrm{C}$

• A

  $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}, \mathrm{CH}_{2}=\mathrm{CH}_{2}$
• B

  $\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{MeCH}=\mathrm{CH} \cdot \mathrm{Me}$
• Correct answer
  C

  $\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$
• D

  $\mathrm{Me}_{2} \mathrm{CH} . \mathrm{CH}_{2} \mathrm{Cl} \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$

Solution

OR

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