Can we synthesize alcohols by the photolysis of hydrogen peroxide?

Question

As I was going through my revision of organic chemistry, I came across free radical substitution of alkanes with halogens. In a similar process where we use hydrogen peroxide, instead of chlorine, so that photolysis would yield a hydroxide radical $\ce{OH.}$ instead of chlorine radical $\ce{Cl.}$. The rest of the steps would be the regular mechanism of This way we can get an alcohol instead of alkyl chloride. Here is what I mean: \begin{align} \ce{H-O-O-H &->[h\nu]H-O. + .O-H}\\ \ce{CH3-CH3 + .O-H &-> CH3-CH2. + H2O}\\ \ce{CH3-CH. + .O-H &-> CH3-CH2-OH}\\ \end{align}

Does my hypothesis work in reality?

I am aware of the fact that dialkyl peroxides cannot be used since they are highly explosive. However, hydrogen peroxide isn't that explosive, so I think that its safe to use it. Since $\ce{H2O2}$ is a liquid, while alkanes and $\ce{Cl2}$ are gases, the reaction with $\ce{H2O2}$ could be done at low pressure. We could photolyse the $\ce{HO-OH}$ bond using light of a sufficient frequency, which probably should initiate a free radical reaction.

I did a few calculations and got the maximum wavelength of light needed to photolyse the $\ce{O-O}$ bond is $\pu{0.0084m}$, which is pretty low. So I think it's possible to photolyse $\ce{H2O2}$ and carry on the reaction.

Answer 1

I do not believe so.

The mechanism of decomposition of $\ce{H2O2}$ is:

$\ce{H2O2(g) -> 2HO.(g)}$

$\ce{H2O2(g) + .OH(g) -> H2O(g) + HOO.(g)}$

$\ce{HOO.(g) + .OH(g) -> H2O(g) + O2(g)}$

$\ce{.OH(g), HOO.(g)}$ are both radical compounds.

If you take a look at the first step in the mechanism, it is the rate limiting step and it is the same as your first step in the mechanism, the photolysis step. Therefore if $\ce{H2O2}$ undergoes partial photolysis all that will happen is that the sample will undergo decomposition quicker.

Additionally, your mechanism of oxidation has been proposed before. You can take a look at this article, in the section Mechanism of Oxidation by Peroxides: Radical or not Radical, That is the Question, in the end of subsection 4.1 and 4.2 they refer to the mechanism you proposed. They end up proving that it proceeds through a concerted mechanism that requires the presence of a suitable metal catalyst. It involves oxidative insertion or some type of $\ce{C-H}$ insertion by the metal followed by oxidation of the product of that reaction.

But say the generation of hydroxyl radicals does work out, the corresponding alcohol may not be the only product that forms. Hydroperoxides have a tendency of forming as well. This is because hydrogen peroxide tends to decompose into oxygen and water, the oxygen can interact with the alkyl radical to form hydroperoxides. Along with the fact that the alcohol formed will tend to undergo oxidation in the presence of hydrogen peroxide and hydroxyl radicals.

Also if you are still going to try the reaction, the bond dissociation energy of the $\ce{O-O}$ single bond is around 146 kJ per mole. So the energy per molecule is $\ce{2.433\times10^{-19} J}$. So the maximum wavelength of light to ionize this is $\ce{8.21\times10^{-7} m}$. Your calculation is a little off, it is in the near infrared range (you may have forgot to convert from kilojoules to joules or something like that). This is why hydrogen peroxide is stored in dark bottles and told to not be put in the presence of light.