# Basic chemistry of the Shinoda test for flavonoids?

In this test, four pieces of magnesium fillings (ribbon) are added to the ethanolic extract followed by a few drops of concentrated hydrochloric acid. A reddish colour indicates the presence of flavonoid.

What is the basic chemistry of the test? What are the chemical reactions taking place in the test that lead to the formation of red precipitate, and also, why have we used magnesium and concentrated HCl in the test?

I found one source indicating that the reaction involves the conversion of the flavonoid to the corresponding anthocyanidin. The structure of that resulting anthocyanidin contains an extended system of conjugated $\pi$-electrons, which constitute a chromophore. As it happens, the electrons of molecules containing conjugated $\pi$ systems have energy level transitions corresponding to absorbance in the visible (and also ultraviolet) region of the electromagnetic spectrum. This accounts for their perceived color.
As for the precise mechanism involved, I was not able to find any sources whatsoever. It's obvious that the flavonoid is reduced, with the magnesium metal being the reducing agent. It could be the process occurs by single electron transfer (SET) with radical and anionic intermediates, or via a cyclic transition state and carbenoid intermediates with coordination of the magnesium, or something else entirely. In any case, the purpose of the acid is likely to activate the carbonyl of the flavonoid to reduction by protonating it, and providing a source of $\ce{H+}$ ions for the subsequent formation of water.