Reduction and oxidation in organic chemistry

Question

What is the meaning of reduction and oxidation in organic chemistry? From what I learned in General chemistry - OIL RIG 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/a good reducing agent?

Answer 1

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):

Oxidation of secondary alcohols (loss of hydrogen):

The strength of an oxidising or reducing agent can be evaluated by measuring its reduction potential. Good oxidising agents have highly positive $\mathrm{E^\circ}$ values, meaning they are 'more keen' to gain electrons, whereas good reducing agents have highly negative $\mathrm{E^\circ}$ values. For example: $$\ce{MnO4- + 8H+ +5e- <=> Mn^{2+} + 4H2O}~~~~~~~~~~\mathrm{E^\circ = +1.51V}$$ $$\ce{Cr2O7^{2−} + 14H+ + 6e- <=> 2Cr3+ + 7H2O}~~~~~~~~~~\mathrm{E^\circ = +1.33V}$$

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

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

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.