KMnO4 + cellulose = ?

The filter paper is made of cellulose.

Use a filter funnel and add the paper,

and use the beaker to collect residues.

We pour aqueous KMnO4 on the funnel.

What will you observe?

KMnO4 is purple.

Upon pouring, the residue is green solution.

Then we confirm the product is K2MnO4.

Now we set up an ionic equation:

MnO4[-] + e[-] --> MnO4[2-]

Clearly KMnO4 get reduced.

Now we write the full equation:

2KMnO4 + 2e[-] --> K2MnO4 + MnO4[2-]

Since the reaction is under aqueous solvent,

now we add 2 protons from the products:

2KMnO4 + 2e[-] + 2H[+]

--> K2MnO4 + H2MnO4

We confirm that cellulose donate protons.

So?

Now we get some clues

Look at x, a proton will be lost,

then the -C-O-C- bond breaks downs,

and x will become a ketone.

y will also lose a proton and form a ketone.

The change is under aqueous condition,

so we have:

R2H-O-HR'2 + H2O

--> R2=O + O=R'2 + 4H[+] + 2e[-]

R2H-O-HR'2 + H2O

--> R2=O + O=R'2 + 4H[+] + 4e[-]

Now balancing the ionic equations,

we have the following full equation:

4KMnO4 + H2O + R2H-O-HR'2

---> R2=O + O=R'2 + 2K2MnO4 + 2H2MnO4