Some Basic Concepts of Chemistry 2 Question 44
44. An organic compound $X$ on analysis gives 24.24 per cent carbon and 4.04 per cent hydrogen. Further, sodium extract of $1.0 \mathrm{~g}$ of $X$ gives $2.90 \mathrm{~g}$ 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 $Z$.
(1989, 5M)
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Solution:
- Mass of chlorine in $1.0 \mathrm{~g} X=\frac{35.5}{143.5} \times 2.9=0.717 \mathrm{~g}$
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 $=\mathrm{CH}_{2} \mathrm{Cl}$.
Because $X$ can be represented by two formula of which one gives a dihydroxy compound with $\mathrm{KOH}$ indicates that $X$ has two chlorine atoms per molecule.
$\Rightarrow X=\mathrm{C}{2} \mathrm{H}{4} \mathrm{Cl}_{2}$ with two of its structural isomers.
$$ \mathrm{Cl}-\mathrm{CH}{2}-\mathrm{CH}{2}-\mathrm{Cl} \text { and } \mathrm{CH}{3}-\mathrm{CHCl}{2} $$
On treatment with $\mathrm{KOH}$, I will give ethane-1, 2-diol, hence it is $Y$. $Z$ on treatment with $\mathrm{KOH}$ will give ethanal as
$$ (Y) $$
