Question

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 $A {\overset{Cl_2/H_2O}{\longrightarrow}} B \underset{(2)H_3O^+}{\overset{(1)KCN/EtOH}{\longrightarrow}}C$

• A.
• B. $HO - CH _{2}- CH _{2}- CO _{2} H$
• C. $HO - CH _{2}- CO _{2} H$
• D. $CH _{3}- CH _{2}- CO _{2} H$

Answer 1

Answer: (B)

$HO - CH _{2}- CH _{2}- CO _{2} H$

Answer 2

$C=85.71 \%=\frac{85.71}{12}=7.14 ; \,\,\,\frac{7.14}{7.14}=1$
$H=14.29 \%=\frac{14.29}{1}=14.29 ; \,\,\,\,\frac{14.29}{7.14}=2$
$\therefore$ Empirical formula $= CH _{2}$
and, empirical formula weight $=12+2=14$
Again, molecular formula weight
$=2 \times$ vapour density
$=2 \times 14=28$
$\therefore n=\frac{28}{14}=2$
$\therefore$ Molecular formula $=\left( CH _{2}\right)_{2}= C _{2} H _{4}$