When considering redox reactions involving organic compounds it is often confusing assigning oxidation numbers to carbon. For example consider the combustion reaction below.

CH₃CH₂CH_3(g) + O_2(g) => CO_2(g) + H₂O_(g)

Is this a redox reaction? Well it appears to be. Obviously propane is oxidised because oxygen is added and oxygen is reduced from 0 to -2. But it is difficult to see which element in propane is oxidised.

What is the oxidation state of carbon in propane?

When assigning oxidation states to carbon in organic compounds such as ethanoic acid, pictured on the right, we treat each carbon separately.

For example the CH₃ group is treated separately to the COOH group.

We ignore the c-c bond and treat the CH₃ individually.
C + 3 X 1 = 0
=> C = -3

In a similar way we treat the COOH group.
C + 2X -2 + 1 = 0
=> C = +3

Let's take propane for example.

We have two CH₃ groups and one CH₂ group.

Carbon has two oxidation states in propane.

- +3 in CH₃ and -2 in CH₂.

Calulate the oxidation state/s of carbon in the following compounds.
a) CH₃OH 0 1 2 3 4 5 6 -1 -2 -3 -4

b) HCHO 0 1 2 3 4 5 6 -1 -2 -3 -4

c) CCl₄ 0 1 2 3 4 5 6 -1 -2 -3 -4

d) CH₃COO^- 0 1 2 3 4 5 6 -1 -2 -3 -4 0 1 2 3 4 5 6 -1 -2 -3 -4

e) CH₃CH₂COOH 0 1 2 3 4 5 6 -1 -2 -3 -4 0 1 2 3 4 5 6 -1 -2 -3 -4 0 1 2 3 4 5 6 -1 -2 -3 -4

Solution