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

(i) For empirical formula :
`|{:("Element","Percentage","Relative no. of atoms","Silplest ratio"),(,,=("percentage")/("at mass"),),(C,85.7,(85.7)/(12)=7.14,1),(H,14.3,(14.3)/(1)=14.3,2):}|`
Therefore, empirical formula `= CH_2`.
(ii) One gram of hydrocabon uses `38.05 gm` of `5 %` solution of `Br_2` in `C Cl_4`, since it is alkene, `1 mol (A)` uses `1 mol` of `Br_2`.
`:. (38.05 xx 5)/(100) gm Br_2` is used by `1 gm` of `(A)`
`:. 160 gm` of `Br_2` is used by `(160 xx 100)/(38.05 xx 5 gm)` of `(A)`
`84 gm of (A)`
`:. Mw of (A) = 84`
Empirical formula w.t. of `(A)`, i.e., `CH_2 = 14`
Now, `n = (Mw)/("Empirical formula w.t") = (84)/(14) = 6`
`:.` Molecular formula `= ("Empirical formula")_n`
=`(CH_2)_6 = C_6 H_12`.
(iii)
(iv) Proceed reverse to obtain the structure of `(A)`
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