Which group is more oxidized, #-CHO# or #-CH_2OH#, and why?

There are three methods we could use to determine the relative levels of oxidation.

1. By using the oxidation number of the carbon atom

One definition of oxidation is: an increase in the oxidation number.

Let's calculate the oxidation number of #"C-1"# in ethanal.

According to the rules for calculating oxidation numbers, #"C-1"# "owns" one of the electrons in the #"C-C"# bond, both of the electrons in the #"C-H"# bond, and none of the electrons in the #"C=O"# bond.

Since #"C-1"# "owns" only three valence electrons, it has effectively "lost" an electron, so it has an oxidation number of +1.

Now, let's repeat the process for #"C-1"# in ethanol.

Here, #"C-1"# "owns" one of the electrons in the #"C-C"# bond, both of the electrons in the #"C-H"# bonds, and none of the electrons in the #"C=O"# bond.

Since #"C-1"# now "owns" five valence electrons, it has effectively "gained" an electron, so it has an oxidation number of -1.

The aldehyde carbon has a higher oxidation number than the alcohol carbon, so a #"CHO"# group is more highly oxidized than a #"CH"_2"OH"# group.

2. By counting the number of oxygen atoms

A second definition of oxidation is: an increase in the number of oxygen atoms.

Both groups contain one O atom, but the O in the aldehyde is double-bonded, so we can count it twice (as we do when determining #R,S# configurations).

Thus, the #"CHO"# group is more highly oxidized than the #"CH"_2"OH"# group.

3. By counting the number of hydrogen atoms

A third definition of oxidation is: a decrease in the number of hydrogen atoms.

#"C-1"# in the alcohol group has two H atoms attached, while #"C-1"# in the aldehyde group has one H atom attached.

Therefore, the #"CHO"# group is more highly oxidized than the #"CH"_2"OH"# group.