What is the meaning of reduction and oxidation in organic chemistry? In general chemistry I learned an OIL RIG mnemonic, i.e. oxidation is loss and reduction is gain (of electrons).

But what do we say about their usage in organic chemistry? I have seen their usage a lot of times in the form of reducing agents and oxidizing agents but never really understood it. What really makes a substance a good oxidizing or a good reducing agent?


In organic chemistry reduction is usually the gain of hydrogen or the loss of oxygen and oxidation is usually the gain of oxygen or the loss of hydrogen.

Reduction of alkenes (gain of hydrogen):

Reduction of alkenes

Oxidation of secondary alcohols (loss of hydrogen):

Oxidation of secondary alcohols

The strength of an oxidising or reducing agent can be evaluated by measuring its reduction potential. Good oxidising agents have highly positive $E^\circ$ values, meaning they are 'more keen' to gain electrons, whereas good reducing agents have highly negative $E^\circ$ values. For example:

$$ \begin{align} \ce{MnO4- + 8 H+ + 5 e- &<=> Mn^2+ + 4 H2O} &\quad E^\circ &= \pu{+1.51 V} \\ \ce{Cr2O7^2- + 14 H+ + 6 e- &<=> 2Cr^3+ + 7 H2O} &\quad E^\circ &= \pu{+1.33 V} \end{align} $$

Since the redox potential for the $\ce{MnO4-}$ system is more positive, $\ce{MnO4-}$ is a stronger oxidising agent than $\ce{Cr2O7^2-}$ under acidic conditions.

The reduction potential is affected by the $\mathrm{pH}$ (along with many other factors — temperature, concentration of reactants — see the Wikipedia article for more discussion). Under neutral or alkaline conditions, $\ce{MnO4-}$ undergoes a different reaction with a less positive reduction potential, meaning it is a weaker oxidising agent under alkaline conditions that under acidic conditions.

$$\ce{MnO4- + 2 H2O + 3 e- <=> MnO2 + 4 OH-} \quad E^\circ = \pu{+0.59 V}$$

Reactive metals, such as zinc, tend to be good reducing agents. Zinc amalgam and hydrochloric acid are used in the Clemmensen reduction, to reduce ketones to alkanes.

$$\ce{Zn^2+ + 2e- <=> Zn(Hg)} \quad E^\circ = \pu{−0.7628 V}$$

Many commonly used reducing agents in organic chemistry, such as $\ce{LiAlH4}$ and $\ce{NaBH4}$, serve as sources of the hydride ion $\ce{H-}$. As an example, in the reduction of ketones using $\ce{NaBH4},$ hydrogen is added across the $\ce{C=O}$ double bond though a tetrahedral intermediate.

Reduction of ketones using NaBH4


In organic chemistry reduction is usually the gain of hydrogen or the loss of oxygen and oxidation is usually the gain of oxygen or the loss of hydrogen.

Reduction/oxidation have the same meaning in organic chemistry as in general chemistry: gain or loss of electrons. However, instead of being restricted to integer multiples of electrons, we usually deal with electron density.

Consider a homonuclear diatomic bond like $\ce{C-C}$ which equally shares the electron density between the two carbon atoms. If I replace one of the carbon atoms (EN 2.5) with a hydrogen (EN 2.2), then more electron density is donated to the carbon atom, thus "reducing" it.

Conversely, if I replace one of the carbons with fluorine to get a $\ce{C-F}$ bond, the electron density is pulled away from the carbon atom and it is oxidised. Oxygen is more electronegative than many elements, including carbon, so one can oxidise the $\ce{C-H}$ or $\ce{C-C}$ bonds of alkanes and alkenes to $\ce{C-O}$ in alcohols, and then further oxidise the carbon to $\ce{C=O}$ in aldehydes/ketones and on to $\ce{COOH}$ in carboxylic acids. Similarly, adding hydrogen across the triple bond of $\ce{N2}$ to give $\ce{NH=NH}$ (a useful reducing agent) is reduction.

Considering reduction as hydrogen donation in organic chemistry can be dangerous because many reagents we use are metal hydrides, like $\ce{NaH}$, where the hydrogen has oxidised (is pulling electron density away from) the sodium atom. The addition of $\ce{H-H}$ to Pd or Pt metal is oxidation, as is the addition of R-groups such as $\ce{(Ph)3P}$ to Pd (Suzuki, Heck reactions etc). In a Grignard reagent, an organic halide, like $\ce{PhBr}$, is reduced by the addition of metallic magnesium to form $\ce{PhMgBr}$, where the Mg is oxidised as its electron density is pulled away and deposited on the phenyl group and the bromine, leaving you with something just short of the carbanion $\ce{Ph-}$ and $\ce{Br-}$.

Hope this clarifies the issue.

  • $\begingroup$ A very interesting perspective! Thank you for your answer and welcome to Chemistry.SE. $\endgroup$ – William R. Ebenezer Jun 29 at 3:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.