Reactivity in Chemistry

Reduction & Oxidation Reactions

RO12.  Organic Redox

   The idea of oxidation states is not normally applied to organic compounds, but it can be useful to do so.  When we do, we can gain some insight into certain reactions of organic molecules.

   For example,carbon dioxide, CO2, can be thought of as having carbon in an oxidized state.   If we apply the usual oxidation state rule, carbon dioxide is overall neutral and contains two oxygens, each with 2- charge.  To counter that charge, carbon must be in oxidation state 4+.

   On the other hand, methane, CH4, can be thought of as having carbon in a reduced state. If we apply the usual oxidation state rule here,  methane is overall neutral but contains four protons.  That means the carbon must be in a 4- oxidation state.  Of course, the carbon does not behave as if it has a minus four charge.  But we will see that this sort of exercise can be useful for book-keeping  purposes.

 

Problem RO12.1.

Assign the formal oxidation state to carbon in the following molecules.

a)  methanol, CH3OH        b)  formaldehyde or methanal, CH2O    c)  carbonate, CO32-    d) hydrogen cyanide, HCN 

e) ethane, CH3CH3   f) ethene, CH2CH2    g) ethyne, CHCH 

 

 The general trend here is that the more bonds there are to oxygen, the more oxidized is carbon.  The more bonds there are to hydrogen, the more reduced is carbon. 

  Adding a hydrogen nucleophile to  a carbonyl electrophile is routinely referred to as a reduction.  For example, adding sodium borohydride to methanal would result in reduction to form methanol.  Of course, a hydride is really a proton plus two electrons.  We could write an equation for the reduction of methanal that looks a lot like the redox reactions we see in a table of standard reduction potentials.

CH2O   +   H-   +   H+    →   CH3OH 

or

CH2O   +   2 e-   +   2 H+    →   CH3OH 

  It stands to reason that the opposite reaction, the conversion of methanol to methanal, is a two electron oxidation.

CH3OH   →   CH2O   +   2 e-   +   2 H+  

  We know how to accomplish the reduction of methanal, at least on paper.  We just add a complex metal hydride, such as lithium aluminum hydride or sodium borohydride, to the carbonyl compound.  After an acidic aqueous workup to remove all the lithium and aluminum compounds, we get methanol.  For practical reasons, methanol may be difficult to isolate this way, but that's the general idea of the reaction.

  How do we accomplish the reverse reaction?

  One way would be to provide a hydride acceptor in the reaction, so that we could catch hydride as it comes off the methanol.  The most well-known such entity is NAD+, of course.  There are biological oxidations that employ NAD+ for this reason.

  More generally, the reaction can be accomplished in a number of ways, on paper, by separating out the two tasks involved.  We need something to accept the two protons: that's a base.  We need something to accept the two electrons: that's an oxidizing agent.

   For the latter task, there are a number of high oxidation state transition metal compounds that are quite willing to accept two electrons.  One of the most widely employed is Cr(VI), which accepts two electrons to become Cr(IV).  A number of other methods are available, having been developed partly to avoid the toxicity of chromium salts, but let's look at the chromium case as an example. 

   A simple chromium(VI) compound is chromium trioxide.  A simple base is pyridine.  If we took these two reagents together with benzyl alcohol in a solvent such as dichloromethane, what would happen?  OK, you might not want to try this, because chromium trioxide has an alarming capacity to cause spontaneous combustion in organic compounds, but we can do it on paper.

   Is chromium trioxide a nucleophile or an electrophile?  That Cr(VI) is pretty electrophilic, surely.  So what part of the benzyl alcohol is nucleophilic?  The oxygen atom.  When we mix these things, the oxygen atom would likely coordinate to the chromium. 

   When the oxygen atom coordinates to the chromium, the oxygen gets a positive formal charge.  It is now motivated to lose a proton.  That's what the pyridine is for.

  Now we have accomplished one of the goals of the reaction.  We have removed a proton from benzyl alcohol.  We have one more proton and two electrons left. The second proton will have to come from the carbon attached to the oxygen; that's the place where we need to form a carbonyl.

   But wait a minute.  You can't take two protons off the same molecule, can you?  And certainly not from two atoms that are right next to each other.  Doesn't that generate an unstable dianion? 

   Not this time.  The chromium is there to accept two electrons.  We won't generate an anion at all, as far as the benzyl alcohol is concerned. It is oxidized to benzaldehyde.

 

Problem RO12.2.

A completely different outcome to this reaction would be obtained in aqueous solution because of the equilibrium that exists between a carbonyl and the geminal diol (or hydrate) in water.  Instead of obtaining an aldehyde, a carboxylic acid would be obtained via a second reduction.  Provide a mechanism for this reaction.

 

Oxidation of alcohols is strongly dependent on conditions.  In general, there needs to be a hydrogen on the alcoholic carbon (H-C-O-H).  If there is no hydrogen on that carbon, the alcohol is pretty difficult to oxidise to a ketone.  If there is one hydrogen on that carbon -- that is, if the alcohol is secondary -- then the alcohol becomes a ketone.

If there are two hydrogens on that alcoholic carbon (H2C-OH), i.e. if the alcohol is a primary one, then two different products may result.  Removal of just one hydrogen from the alcoholic carbon, and replacement with an additional bond to oxygen, results in formation of an aldehyde.  On the other hand, replacement of the second hydrogen from the alcoholic carbon, and replacement with another oxygen, would lead to formation of a carboxylic acid.

That second case -- replacement of the second hydrogen with an oxygen -- only happens in aqueous media.  The aldehyde that forms after the first oxidation (H-C=O) must become hydrated (H-C(OH)2) in order for the second oxidation to occur, making the carboxylic acid (HO-C=O).  As a result, if a primary alcohol is oxidised via a reagent that requires water as a solvent, the carboxylic acid results.  If an organic-soluble reagent is used, the reaction stops at the aldehyde.

The most common methods for mildly converting primary alcohols to aldehydes are Swern oxidations and Dess-Martin oxidations.  Dess-Martin oxidations employ a high-oxidation state iodine compound -- that's I(V), compared to the more commonly encountered I- ion. The reduction product is an I(III) compound.  Swern oxidations employ sulfur in a moderately high oxidation state of zero; the sulfur is reduced to S2- in the Me2S side product.

In both cases, the oxidation mechanism is similar to the one illustrated with chromium oxide and pyridine.  The oxygen donates to the oxidizing atom (the chromium, the sulfur, or the iodine). Deprotonation of the carbon leads to formation of the C=O bond and reduction of the oxidising agent by two electrons. 

With Swern oxidations, the mechanism has an added "priming" step, because the thionyl oxidant (S=O) must first be activated; the thionyl oxygen donates to the carbonyl of the oxalyl chloride, (COCl)2.  The sulfur is then ready to accept an alcohol donor.  Once the alcohol undergoes oxidation, the oxygen from the thionyl group is completley transferred to the oxalyl group, forming both carbon dioxide and carbon monoxide in a subsequent disproportionation reaction.

Oddly enough, although PCC looks like a good candidate for aqueous oxidation (it is ionic, after all), it is frequently absorbed onto a solid surface (usually alumina, Al2O3) and used as a heterogeneous catalyst in organic solvent.  Like most heterogeneous catalysts, it works rather slowly, but it is pretty selective for aldehyde formation.

Most "ionic" oxidants really are used in the presence of water, and they do convert primary alcohols into carboxylic acids.  Examples include sodium chromate, Na2CrO4, dichromate, Na2Cr2O7, and potassium permanganate, KMnO4.  These reagents are much harsher than Swern conditions or DMP, however, and they can lead to extensive decomposition of the reactant.  Chromates are sometimes prepared as a solution called Jones reagent, in which the oxidant is pre-mixed with sulfuric and and water; the reagent also contains some acetone, to help solubilise the organic compound to be oxidised.

 

Problem RO12.3.

Fill in the missing products.

 

Problem RO12.4.

Fill in the blanks in the following synthesis.

 

 

 

 

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use.

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

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This material is based upon work supported by the National Science Foundation under Grant No. 1043566.

Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

 

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