i. (A) reacts with `Na` to give `H_(2)` and thus it contains `(-OH)` group.
ii. The molecular weight of `(A) (C_(4)H_(10)O_(2))` is `90`. If one `(OH)` group, then
`90 gm(A)` with `Na` gives `11200 ml H_(2)` (i.e, half mole `H_(2))`
`0.90 gm` o f `(A)` gives `(11200xx0.90)/(90)=112ml` of `H_(2)` at `STP`
Since `0.90 Na` gives `224 ml H_(2)` at `STP`, it contians two `(-OH)` groups.
iii. Keeping in view the above facts `(A)` is ,
iv. `(A)` shows optical isomerism of which `(B)` form is optically active having two isomers `(D)` and `(E )`, `(C )` being the inactive form, therefore `(A), (B), (C )`, and `(D)` are: