Question

An oxygen containing organic compound was found to contain 53% carbon and 13% hydrogen. Its vapour density is 23. The compound reacts with sodium metal to liberate hydrogen. Identify the functional isomer of this compound.

Answer 1

% O = 100 - (53 + 13) = 34
`:.` Ratio of atoms, `C : H : O = (53)/(12) : (13)/(1) : (34)/(16) = 4.4 : 13 : 2.125 = 2 : 6 : 1`
`:. E.F. = C_(2)H_(6)O` and E.F. wt `= 2 xx 12 + 6 xx 1 + 1 xx 16 = 46`
Mol. wt. `= 2 xx V.D. = 2 xx 23 = 46`
`:.` n = Mol. wt/E.F. wt. `= (46)/(46) = 1`
Thus, M.F. `= 1 xx E.F. = C_(2)H_(6)O`
Since the given organic compound reacts with sodium o liberate `H_(2)` gas, therefore, it must be ethanol.
`underset("Ethanol")(2CH_(3)CH_(2)OH)+2Na rarr 2CH_(3)CH_(2)ONa+H_(2)`
The functional isomer of ethanol is methoxymethane, i.e., `CH_(3)OCH_(3)`.