Compound $\mathbf{X}\left(\mathrm{C}_4 \mathrm{H}_6 \mathrm{OCl}_2\right)$ reacts with cold water to form $\mathbf{Y}\left(\mathrm{C}_4 \mathrm{H}_7 \mathrm{O}_2 \mathrm{Cl}\right)$. $\mathbf{X}$ on heating with aq. KOH , followed by acidification, gives $\mathbf{Z}\left(\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_3\right)$, which can be resolved into enantiomers. $\mathbf{X}$ is-

• A

  $\mathrm{CH}_3-\mathrm{CCl}_2-\mathrm{CH}_2-\mathrm{CHO}$
• B

  $\mathrm{ClCH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{COCl}$
• C

  
  
• Correct answer
  D

  $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CHCl}-\mathrm{COCl}$

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