An organic compound X on analysis gives 2.90 g silver choride with acidified silver nitrate solution. The compound X may be represented by two isomeric structures Y and Z. Y on treatment withaqueous potassium hydroxide solution gives a dihydroxy compound while Z on similar treatment gives ethanal. Find out the molecular formula of Z and gives the structure of Y and Z.

An organic compound X on analysis gives 2.90 g silver choride with aci

1.	An organic compound X on analysis gives 2.90 g silver choride with acidified silver nitrate solution. The compound X may be represented by two isomeric structures Y and Z. Y on treatment withaqueous potassium hydroxide solution gives a dihydroxy compound while Z on similar treatment gives ethanal. Find out the molecular formula of Z and gives the structure of Y and Z.
Answer» SOLUTION :Mass of chlorin in `1.0 g X = (35.5)/(143.5) xx 2.9 = 0.717 g` Now, the empirical formula can be derived as : `rArr` Empirical formula `= CH_(2)Cl`. Because X canbe represented by two formula of which one GIVES a dihydroxy compound with `KOH` indicates that x has two chlorine atoms per molecules. `rArr X = C_(2)H_(4)Cl_(2)` 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, I` will GIVE ethane -1,2-diol, hence it is Y, Z one treatment with `KOH` will give ETHANAL as `ClCH_(2)CH_(2)Cl+OH^(-)rarr underset((Y))(underset(OH)underset(\|)(CH_(2))-underset(OH)underset(\|)(CH_(2)))` `CH_(3)CHCl_(2)+KOH rarr underset("Unstable")(CH_(3)CH(OH)_(2))OVERSET(-H_(2)O)rarrunderset((Z))(CH_(3)CHO)`