How to obtain phthalic acid from aspirin?
This is a tough synthesis, actually. The major complications are:
- The acetate group is bonded via its oxygen, not its carbon.
- The carboxylic acid group is a meta director (a moderately-deactivating substituent for electrophilic aromatic substitution, since it is an electron-withdrawing group). If you ignore it and try to get rid of the acetate group, you'll add meta, not ortho.
So, the steps we're trying are:
- Get rid of the acetate group by turning this back into 2-hydroxybenzoic acid (salicylic acid).
- Remove the
- Turn the
#"COOH"#into an ortho director (a strongly-activating group with little steric hindrance).
- Add a
- Return the first
#"COOH"#group to how it was. We'll be doing steps 4 and 5 at once.
Since there are carbonyl oxygens on both groups, we had to avoid acidic conditions in order to only affect the acetate group. So, we used
#\mathbf("OH"^(-))#on high heat (dissolved in water).
That way, the leaving group in the carboxylic acid group is identical in the forward and reverse directions, and that doesn't yield side products.
Also, the 2-hydroxybenzoic acid anion becomes the leaving group.
Work it up using water to protonate the oxide (just make the
#"pH"#higher than #14#when you use #"OH"^(-)#).
#\mathbf("Zn")#powder would remove the #"OH"#group from a benzene ring.
Now we need to turn the meta director into an ortho/para director. One way to do it is to reduce it down to a primary alcohol, since
#"OH"#is an electron-donating group (a strongly-activating group) and alkyl groups are as well. I show two options.
Friedel-Crafts Alkylation adds a
#"CH"_3#onto the benzene ring, 1,2 to the primary alcohol group as desired. You may get the 1,4 isomer, but the steric hindrance is not significant so it should be a minor product.
#"KMnO"_4#(potassium permanganate) performs a dual role here. It can oxidize primary alcohols to carboxylic acids, but it can also oxidize aromatic alkanes into carboxylic acids. Nice!