A certain hydrocarbon `(A)` was found to contain `85.7 %` carbon and `14.3 %` hydrogen. This compound consumes `1` molar equivalent of hydrogen to give a saturated hydrocarbon `(B)`. One gram of hydrocarbon `(A)` just decolourised `38.05 gm` of a `5 %` solution (by weight) of `Br_2` in `C Cl_4`. Compound `(A)` on oxidation with concentrated `KMn_4` gave compound `(C)` (molecular formula, `C_(4) H_(8) O)` and acetic acid. Compound `(C)` could easily be prepared by the action of acidic aqueous mercuric sulphate on 2-butyne. Determine the molecular formula of `(A)` and deduce the structures of `(A), (B)`, and `(C)`.

Explanation

Moles of `C: H` in (A) `=(85.7)/(12): (14.7)/(1)= 7.14: 14.3= 1:2`
`therefore` empirical formula of (A) is `CH_(2)` and empirical formula weight is 14. As (A) consumes I molar equivalent of hydrogen to give a saturated hydrocarbon (B), (A) should have one double bond. Reaction of (A) with `Br_(2)` will also be an addition reaction in which 1 mole of (A) will combine with 1 mole of `Br_(2)`. Thus, Moles of (A)= moles of `Br_(2)` or `(1)/(M)= ((5 xx 38.05)//100)/(160) " " {{:("M = mol . wt of (A)"),(Br_(2)=160):}}`
`therefore M=84`
`therefore n=("molecular formula wt.")/("empirical formula wt")=(84)/(14)=6`
`therefore` molecular formula of (A) is `C_(6)H_(12)`
Now from the given reaction sequece, we have,
`{:(C_(6)H_(12)underset(KMnO_(4))overset((O))rarrC_(4)H_(8)O+ CH_(3)COOH),("(A) I (C)" ),(" "IIuarr H^(+)//HgSO_(1)),(" "CH_(3).C -=C.CH_(3)),(" 2-butyne"):}`
We conclude that `C_(6)H_(12)` (A) must be represented as `C_(4)H_(8)=CH.CH_(3)` (step I) and `C_(4)H_(8)O` (C ) must be a ketone
`CH_(3).overset(overset(C )(||))(C )CH_(2).CH_(3)` (step II)
further, addition of `H_(2)` to compound (A) will produce `C_(6)H_(14)` (B) (an alkane). The structure of (A), (B) and (C ) may be represented as