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علم الكيمياء : الكيمياء العضوية : مواضيع عامة في الكيمياء العضوية :

Oxidation of alcohols

المؤلف:  Jonathan Clayden , Nick Greeves , Stuart Warren

المصدر:  ORGANIC CHEMISTRY

الجزء والصفحة:  ص194-196

2025-05-08

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So far, the metals we have used have had one oxidation state other than zero: Li(I), Mg (II), and Zn (II). If we want to oxidize organic compounds, we need metals that have at least two higher oxidation states and that means transition metals. The most important by far is chromium, with Cr (III) and Cr (VI) as the useful oxidation states. Orange Cr (VI) compounds are good oxidizing agents: they remove hydrogen from organic compounds and are themselves reduced to green Cr (III). There are many Cr (VI) reagents used in organic chemistry, some of the more important ones are related to the polymeric oxide CrO3. This is the anhydride of chromic acid and water breaks up the polymer to give a solution of chromic acid. Pyridine also breaks up the polymer to give a complex. This (Collins’ reagent) was used to oxidize organic compounds but it is rather unstable and pyridinium dichromate (PDC) and pyridinium chlorochromate (PCC) are usually now preferred, especially as they are soluble in organic solvents such as CH2Cl2.

Oxidation by these reagents of the various primary and secondary alcohols we have been making in this chapter takes us to a higher oxidation level. Oxidation of primary alcohols gives aldehydes and then carboxylic acids, while oxidation of secondary alcohols gives ketones. Note that you can’t oxidize tertiary alcohols (without breaking a C–C bond).

You will notice that the oxidation steps involve the removal of two hydrogen atoms and/or the addition of one oxygen atom. In Chapter 6 you saw that reduction meant the addition of hydrogen (and can also mean the removal of oxygen). Hiding behind these observations is the more fundamental idea that reduction requires the addition of electrons while oxidation requires the removal of electrons. If we used basic reagents, we could remove the OH proton from a primary alcohol, but to get the aldehyde we should have to remove a C–H proton as well with a pair of electrons. We should have to expel a hydride ion H− and this doesn’t happen. So, we need some reagent that can remove a hydrogen atom and a pair of electrons. That defines an oxidizing agent.

Here Cr (VI) can remove electrons to make Cr (III). It does so by a cyclic mechanism on a Cr (VI) ester. One hydrogen atom is removed (from the OH group) to make the ester and the second is removed (from carbon) in the cyclic mechanism. Notice how the arrows stop on the Cr atom and start again on the Cr=O bond, so two electrons are added to the chromium. This actually makes Cr (IV), an unstable oxidation state, but this gives green Cr (III) by further reactions.

Two examples of the use of PCC in these oxidations come from Vogel. Hexanol is oxidized to hexanal in dichloromethane solution and commercial carveol (an impure natural product) to pure carvone with PCC supported on alumina in hexane solution. In both cases the pure aldehyde or ketone was isolated by distillation.

But a word of warning: stronger oxidizing agents like calcium hypochlorite or sodium hypochlorite (bleach) may oxidize primary alcohols all the way to carboxylic acids, especially in water. This is the case with p-chloro benzyl alcohol and the solid acid is easily isolated by the type of acid/base extraction we met in the previous chapter.

You will find further discussion of oxidizing agents in later chapters of the book. We have introduced them here so that you can see how primary and secondary alcohols, made by addition of organometallic reagents, can be oxidized to aldehydes or ketones so that the process can be repeated. A secondary alcohol, which could be made in two ways, can be oxidized with the pyridine–CrO3 complex to the ketone and reacted with any Grignard or organolithium compound to give a range of tertiary alcohols.

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