# Why is it a reduction from a ketone to an alcohol? [closed]

Why is it a reduction from a ketone to an alcohol (using $\ce{LiAlH4}$)? I'm not able to figure out the oxidation state of carbon.

• – Ivan Neretin Jun 13 '18 at 8:45

As you can read from the link mentioned in the comment to the question, oxidation number is not a useful concept when dealing with organic compounds.

So we won't be determining oxidation number of atoms to know whether oxidation or reduction has happened.

We can follow the basic definition of oxidation and reduction to determine which one of the two is taking olace:

1. Oxidation is gain of oxygen or loss of hydrogen.

2. Reduction is loss of oxygen or gain of hydrogen.

So in case of conversion of a ketone into a 2° alcohol, two atoms of hydrogen is add for one carbonyl group.

And since hydrogen atoms are being added to the molecule (here, ketone), we can conclude that the molecule has been reduced.

Like Ivan already states in Oxidation State Of Propene, there is limited use of oxidation states in organic chemistry. However, you can still apply the schemes that I have written about in Electronegativity Considerations in Assigning Oxidation States.

In the following reaction I use $\ce{R_{1,2}}$ in place of alkyl moieties.

A case can be made for the convention that alkyl moieties have an overall oxidation state of zero.
After that you'll assign $-2$ to the oxygen in the ketone; and since the molecule is neutral the carbon has to have the oxidation state of $+2$.
In the alcohol you'll also assign $-2$ to the oxygen atom. Hydrogen is the least electronegative element and it can give away one electron, leaving it with an oxidation state of $+1$. The hydroxyl moiety therefore has an overall oxidation state of $-1$ affecting the carbon. Since there is another hydrogen the overall oxidation state of carbon is $0$.

The carbon formally goes from $+2$ to $0$, i.e. it gains electrons, therefore it is reduced.

Note that oxidation states are not actual (partial) charges. Below you can see the partial charges of acetone and 2-propanol optimised on the DF-B97D3/def2-SVP level of theory. You might note that they are quite different from the oxidation states.