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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?

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2 Answers 2

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Here is a diagram of the Clemmensen reduction. The metal supplies the electrons and the hydrochloric acid supplies the protons to achieve the carbonyl reduction. The mechanism of reactions that occur on metal surfaces is poorly understood, but here is a link to the current thinking on the mechanism for the Clemmensen reduction.

enter image description here

The Shinoda test is similar to the Clemmensen reduction in that a carbonyl group is reduced. However, in this case, a reductive-elimination occurs. Due to the stability (extended conjugation) of the anthocyanidin product, instead of the reduction going all the way the to the corresponding methylene compound, the hydroxyl-like intermediate undergoes elimination to create the stable anthocyanidin. In the Shinoda test, magnesium is used in place of the zinc-mercury amalgam. The Shinoda test is a test for the presence of flavones. If they are present in the test sample then they are reduced to anthocyanidins under the Shinoda reaction conditions as shown below. The conjugation in flavinoid compounds produces a yellow color, while the extended conjugation in the resultant anthocyanidin shifts the color further out to the red region of the visible spectrum. The dramatic change in color makes this a simple visual test for the presence of flavones.

enter image description here

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  • $\begingroup$ Note also that the Shinoda test can provide an identification of the Flavonoids type. In instance, Flavones give orange anthocyanidin and Flavonols give red color, whereas flavanones give purplish color. $\endgroup$ Commented Dec 29, 2021 at 13:23
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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.

(I find it unlikely that a definitive mechanism for this reduction has been determined, if only because the mechanisms of many similar reductions involving metals and mineral acids widely used in synthetic organic chemistry have yet to be entirely resolved.)

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