896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28 gat STP. Hydrogenation of 'A' gives 2-methyl pentane. Also 'A' on hydration in the presence of H_(2)SO_(4) and HgSO_(4) gives a ketone 'B' moelcular formula C_(6)H_(12)O. The ketone 'B' gives a positive iodoformtest. Find the structure of 'A' and give the reactions involved.

896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 1

1.	896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28 gat STP. Hydrogenation of 'A' gives 2-methyl pentane. Also 'A' on hydration in the presence of H_(2)SO_(4) and HgSO_(4) gives a ketone 'B' moelcular formula C_(6)H_(12)O. The ketone 'B' gives a positive iodoformtest. Find the structure of 'A' and give the reactions involved.
Answer» SOLUTION :To calculate molar MASS of hydrocarbon .A. it is given that 896 ml of Hydrocarbon .A. `C_(X)H_(y)` Weighs 3.28 gat STP. 22400 ml of has `AC_(x)H_(y)` mass `= 3.28 xx 22400` ml/896ml = 831 g/MOL. Hence, the molar mas of A = 831 g mol. Thus, empirical formula `=C_(3)H_(5)` Empirical formula weight `= 3 xx 12= 36 + 5 = 41`. Molecular formula = (Empirical formula)n n = Mol. mass/Empirical mass = 831/41 = 2.02. Molecular formula `= [C_(3)H_(5)]_(2) = C_(6)H_(10)` To detremine the structure of (A) and (B) : `C_(6)H_(10)` on hydrogenation with 2 moles of `H_(2)` gives `C_(6)H_(12)` and structure is 2-methylpentane. (A) on hydration in presence of dil. `H = H_(2)SO_(4)` and `HgSO_(4)` gives `C_(6)H_(12)O` which gives Iodoformtest positive. Hence, structure of `A = (CH_(3))_(2)CH-CH-C=CH` (4-methylpentlyne) and (B) is 4-methylpentanone-2.