An aromatic hydrocarbon (A) (mol. Wt.= 78) contains 92.3% of carbon (A), on treatment with bromine in the dark, produced (B) which contains 48.85% of carbon, 3.19% of hydrogen and 50.96% of bromine. (B), on heating with CH_(3)Br and Na in etherical solution, gave (C ) containing 91.3% of C and 8.7%of H. (C ) on oxidation, produced a monobasic acid (D). The sodium salt of (D), on distillation with soda lime, gave (A). Determine the structures of (A), (B),(C ) and (D).

An aromatic hydrocarbon (A) (mol. Wt.= 78) contains 92.3% of carbon (A

1.	An aromatic hydrocarbon (A) (mol. Wt.= 78) contains 92.3% of carbon (A), on treatment with bromine in the dark, produced (B) which contains 48.85% of carbon, 3.19% of hydrogen and 50.96% of bromine. (B), on heating with CH_(3)Br and Na in etherical solution, gave (C ) containing 91.3% of C and 8.7%of H. (C ) on oxidation, produced a monobasic acid (D). The sodium salt of (D), on distillation with soda lime, gave (A). Determine the structures of (A), (B),(C ) and (D).
Answer» Solution :Moles of `C:H= (92.3)/(12): (7.7)/(1)` `=7.7:7.7` `=1: 1` Empirical formula of (A) is CH and empirical formula weight is 13.As the molecular weight of (A) is 78, `n= (78)/(13)=6` Molecular formula of (A) is `C_(6)H_(6)` In COMPOUND (B): moles of `C:H:Br= (45.85)/(12): (3.19)/(1): (50.96)/(80)` `=3.82: 3.19: 0.637` `=6:5:1` As (B) must be `C_(6)H_(5)Br`, it has to be a substitution product and not the addition product. Further, (C ) is produced by HEATING `C_(6)H_(5)Br` with `CH_(3)Br` and Na (Fitting reaction), (C ) must be `C_(6)H_(5)CH_(3)`. This is ALSO supported by the FOLLOWING data, i.e., for (C ) being `C_(7)H_(8)` Moles of `C: H= (91.3)/(12): (8.7)/(1)` `=1: 1.143` `=7:8` (C ) on oxidation, will give `C_(6)H_(5)COOH` (D), the sodium salt of which on distillation with soda lime will give `C_(6)H_(6)`(A).