An aromatic hydrocarbobn (A) containingC = 91.3% and H = 8.7% on treatmentwith chlorine gave there isomeric monochloro compound (X),(Y) and (Z), eachhaving28% chlorine . On oxidation withKMnO_(4) all the threegavemononbasicacids. The acid form (X) on distillation with soda line gave benzenewhile thoseform (Y) and (Z) gavemonochiorobenzene. What formulawould you assign to the variouscomnpounds?

An aromatic hydrocarbobn (A) containingC = 91.3% and H = 8.7% on treat

1.	An aromatic hydrocarbobn (A) containingC = 91.3% and H = 8.7% on treatmentwith chlorine gave there isomeric monochloro compound (X),(Y) and (Z), eachhaving28% chlorine . On oxidation withKMnO_(4) all the threegavemononbasicacids. The acid form (X) on distillation with soda line gave benzenewhile thoseform (Y) and (Z) gavemonochiorobenzene. What formulawould you assign to the variouscomnpounds?
Answer» Solution : Molecular formula of `(A) = C_(7) H_(8)`. `4^(@) D.U`. And `C:H = 1:1` suggest benzene ring with with one `(-Me)` group. So `(A)` is touene. REACTIONS: Molecular of formula of `(X,Y,Z) = C_(7) H_(7) Cl` Molecular mass of `(X,Y,Z) = 12xx7+7+35.5 = 126.5` Percentage of `Cl` in `(X,Y,Z) = (35.5xx100)/(126.5) = 28%`