Q. An organic compound $X$ on analysis gives $24.24$ per cent carbon and $4.04$ per cent hydrogen. Further, sodium extract of $1.0\, g$ of $X$ gives $2.90\, g$ of silver chloride with acidified silver nitrate solution. The compound $X$ may be represented by two isomeric structures $Y$ and $Z$. $Y$ on treatment with aqueous potassium hydroxide solution gives a dihydroxy compound while $Z$ on similar treatment gives ethanal. Find out the molecular formula of $X$ and gives the structure of $Y$ and $Z$
IIT JEEIIT JEE 1989Some Basic Concepts of Chemistry
Solution:
Mass of chlorine in $1.0 \,g\, X =\frac{35.5}{143.5} \times 2.9 =0.717\, g$
Now, the empirical formula can be derived as :
C
H
Cl
% wt :
24.24
4.04
71.72
Mole :
2
4
2
Simple ratio :
1
2
1
$\Rightarrow$ Empirical formula $= CH_2Cl$.
Because $X$ can be represented by two formula of which one gives a dihydroxy compound with $KOH$ indicates that $X$ has two chlorine atoms per molecule.
$\Rightarrow X = C_2H_4Cl$, with two of its structural isomers.
$ \underset{I}{Cl - CH_2 - CH_2 - Cl}$ and $ \underset{II}{CH_3 - CHCl_2}$
On treatment with $KOH$, $1$ will give ethane-$1,2$-diol, hence it is
$Y. Z$ on treatment with $KOH $ will give ethanal as
| C | H | Cl | |
|---|---|---|---|
| % wt : | 24.24 | 4.04 | 71.72 |
| Mole : | 2 | 4 | 2 |
| Simple ratio : | 1 | 2 | 1 |
