# When a molecule loses hydrogen atoms, does it become oxidised?

I had the following question on a bio test covering cell respiration and glycolysis, but I'm not sure if I agree with the answer.

When a molecule loses hydrogen atoms (as opposed to hydrogen ions), it becomes
B) oxidised

According to my textbook (Principles of Life, 2nd Edition), oxidation is defined as

Relative loss of electrons in a chemical reaction; either outright removal to form an ion, or the sharing of electrons with substances having a greater affinity for them, such as oxygen. Most oxidations, including biologial ones, are associated with the liberation of energy.

Hydrogen atoms have one proton and one electron in their typical state. If a molecule is losing hydrogen atoms, then its net charge wouldn't change. For example, if a (hypothetical) molecule had 7 protons and 6 electrons, it would have a charge of $+1$. If it then lost a hydrogen atom, it would now have 6 protons and 5 electrons, which still yields a net charge of $+1$. Am I missing some critical piece of information, or is this still considered oxidation?

This phrasing is only true for organic molecules. If for example, sodium hydride loses hydrogen, the sodium ion will get reduced. But since you seem to come from a biochemical background, this simplification is okay since you will be dealing with organic molecules primarily.

The idea behind that statement is that hydrogen atoms in biomolecules are typically only bound to carbon, nitrogen, oxygen and sulfur. Hydrogen is less electronegative than all these elements and therefore any $\ce{X-H}$ bond will be polarised towards $\ce{X}$; the non-hydrogen atom.

When determining oxidation states, bonds are formally cleaved heterolytically in such a way that the electronegative partner gets both electrons. Then, the electrons on the formal atomic ions created this way are counted and subtracted from the number the compound should have. Bonds between the same element are cleaved homolytically and the same procedure applied. Thus, if we take ethene ($\ce{C2H4}$, structure see below) the $\ce{C-H}$ bonding electrons are formally attributed to carbon entirely while the $\ce{C=C}$ double bond is split equally. Doing this, each carbon atom ends up with six electrons, formally, and therefore a formal charge of $2-$ or an oxidation state of $\mathrm{-II}$. If we did the same for ethyne ($\ce{C2H2}$) — a compound with two hydrogen atoms removed — we arrive at five formal electrons and thus an oxidation state of $\mathrm{-I}$. Therefore, going from ethene to ethyne is an oxidation.

\begin{align}\ce{H2C=&CH2} & \ce{H-C&#C-H}\\ \text{ethe}&\text{ne} &\text{eth}&\text{yne}\end{align}

Having gotten the theory out of the way, what does this mean for the simplified rule? Well, as I mentioned any $\ce{X-H}$ bond is polarised away from the hydrogen and thus effectively adds an additional electron to $\ce{X}$. Carbon, which is the most common bonding partner of hydrogen, is less electronegative than all the other typical elements of biomolecules, and thus will formally lose electrons to them. When removing hydrogen from a compound chemically, we will always remove at least one $\ce{C-H}$ bond in favour of a $\ce{C-C, C-O, C-N}$ or $\ce{C-S}$ bond. Thus, the carbon atom will formally lose electrons and we will have an oxidation.

This can also be rationalised in another way to determine oxidation states: the additive method. Starting from a neutral compound, assign each hydrogen atom a $\mathrm{+I}$, each oxygen $\mathrm{-II}$. Add that up and invert the sign; this is the oxidation state of all the carbons (assuming a molecule of the formula $\ce{C_mH_nO_x}$). Divide by the number of carbons to get an average oxidation state of carbon.

$$\chi(\ce{C}) = -\frac{n - 2x}{m} = \frac{2x - n}{m}$$

As you see, each hydrogen atom in a compound enters into the formula with a negative sign. Thus, removing a hydrogen must increase the average oxidation state of the carbons. Unfortunately, there is no trivial way to extend this formula to compounds containing oxygen and nitrogen or oxygen and sulfur.

If a molecule gains or looses electrons, it is reduced or oxidized, respectively. If a hydrogen atom leaves a molecule, taking it's single, formerly shared electron with it, then the molecule has lost an electron and is thus oxidized. Your argument sort of assumes that formal charge alone determines a molecule's oxidation state. So, sorry, the test answer:

B) oxidized

is correct ;)