One per cent composition of an organic compound $A$ is, carbon : $85.71 \%$ and hydrogen $14.29 \%$. Its vapour density is 14 . Consider the following reaction sequence
Identify $\underline{C}$,

• A

  
  
• Correct answer
  B

  $\mathrm{HO}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CO}_2 \mathrm{H}$
• C

  $\mathrm{HO}-\mathrm{CH}_2-\mathrm{CO}_2 \mathrm{H}$
• D

  $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CO}_2 \mathrm{H}$

Solution

$\begin{aligned} & \mathrm{C}=85.71 \% \quad=\frac{85.71}{12}=7.14 ; \quad \frac{7.14}{7.14}=1 \\ & \mathrm{H}=14.29 \% \quad=\frac{14.29}{1}=14.29 ; \quad \frac{14.29}{7.14}=2 \end{aligned}$ $\therefore \quad$ Empirical formula $=\mathrm{CH}_2$ and, empirical formula weight $=12+2=14$ Again, molecular formula weight $\begin{aligned} & =2 \times \text { vapour density } \\ & =2 \times 14=28 \\ \therefore \quad n & =\frac{28}{14}=2 \end{aligned}$ $\therefore$ Molecular formula $=\left(\mathrm{CH}_2\right)_2=\mathrm{C}_2 \mathrm{H}_4$

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