An optically active compound A `(C_(10)H_(14))` gets oxidised to benzoic acid `(C_6H_5COOH)` by alkaline `KMnO_4`. However, compound B, which is an optically inactive isomer of A does not get oxidised by alkaline `KMnO_4`. Identify A and B .

Explanation

As A is oxidised to `C_(6)H_(5)COOH`, A is a substituted benzene which has only one side chain consisting of four carbon atoms. Again, as A is optically active, there must be an unsymmetric carbon atom present in the slide chain. So, the side chain is `-CH(CH_(3))CH_(2)CH_(3)` and A is sec-butylbenzene, `C_(6)H_(5)CH (CH_(3))CH_(2)CH_(3)`. B, an isomer of A does not get oxidised by alkaline `KMnO_(4)`. Thus, there is no benzylic hydrogen in the compound. So, the side chain is `-C(CH_(3))_(3)`. The compound B is tert-butylbenzene, `C_(6)H_(5)C(CH_(3))_(3)`