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I know that in side chain oxidation with the given reagent follows oxidation of benzylic carbon (if atleast one benzylic hydrogen is present). So my answer would be
But in my answer key only it is correct for nitro substituted anthracene. But why? What is the reason? Thanks for clearing my doubt!
In each case the permanganate oxidises the most electron rich ring.
In the nitro-substituted case, that is the unsubstituted ring, probably first at the position alpha to the nitro group, forming a phenol, then para to that to a quinone but permanganate keeps on chewing until the phthalic acid is formed.
In the aniline case the most electron rich ring is the one with the NH2 attached. Oxidation of anilines by acid permanganate give a polymer called "aniline black" Wiki link. Though I can't find a reaction specific to 1-Naphthylamine other than one that documents the destruction of it by sulfuric/permanaganate without identifying the products here. The site of first oxidation is p to the aniline. This creates a reactive intermediate which is attacked by a second aniline as the first step of polymer formation.
