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0.90 gm of an organic compound C(4)H(10)...

`0.90 gm` of an organic compound `C_(4)H_(10)O_(2) (A)` when treated with sodium gives `224 ml` of hydrogen at `NTP`. Compound (A) can be separated into fraction (B) and
(C ), by crystallisation of which the fraction (B) is resolved into isomers (D) and (E ). Write down the structural formula of (A) to (E ) with proper reasoning.

Text Solution

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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:
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