A compound `(A)` is a sodium salt of dibasic acid and on electrolysis gives a gas `(B),` gas `CO_(2)` at anode, and `H_(2)(g)` at cathode.
Gas `(B)` does not give any test for unsaturation, `i.e.,` it does not decolourise `Br_(2)` water solution and aqueous `KMnO_(4)` solution.
Gas `(B)` on reaction with `(H_(2)+Pt)` at room temperature does not react but at `120^(@)C`, gas `(B)` gives butane. Identfy `(A)` and `(B)`

Explanation

`(A)` should be a `6-C-`atom because `(B)` on reduction at `120^(@)C` gives butane `[ 4 C` atoms of butane `+2(-COOH)` group `].`
`(A)` can be `:`

`(A)` cannot be `(II)` because the product obtained by this will be alkene that gives test for unsaturation, but according to the problem, `(B)` does not give this test. So `(A)` is `(I)`

The product can be `(C)` also, due to the stabilisation of `1^(@)` radical to `2^(@)` radical, but one has to proceed according to the problem given .