Piperic acid Oxn with alkaline CuO?

[mellow]

Antoncho previously wrote-up a Syringaldehyde prepn. via the oxidization of 3,5-dimethoxy-4-hydroxypropenylbenzene, using alkaline CuO: R-CH=CH-CH₃ + 4CuO ^__> RCHO + CHO-CH3 + 2Cu₂O

Post 293365

(Antoncho: "Double bond ox. cleavage (to aldehyde) with CuO", Chemistry Discourse) taken from the

US Patent US2516412

(http://l2.espacenet.com/dips/viewer?PN=US2516412&CY=gb&LG=en&DB=EPD)

3,5-dimethoxy-4-hydroxypropenylbenzene ...  added to a cupric oxide mixture freshly prepared from ... copper sulfate, ... sodium hydroxide, and ... water and the resulting mixture is heated to boiling under reflux for eight hours

He asked: "why then no one ever thought of possibility of applying this to piperic acid?"

Maybe they have but it didn't work so they never bothered to report their results. It seems likely that people wouldn't report failures.

I would be worried about a Cannizzaro side-reaction under the alkaline conditions with piperonal reacting with the glyoxal (aka: ethanedial, CHO-CHO) which we should also expect to get from piperic acid.

So Piperic acid Oxn: R-CH=CH-CH=CH-COO₂H + 8CuO ^__> R-CHO + CHO-CHO + CHO-COOH, (R- = 3,4-MD-phenyl-)

Possible Cannizzaro: R'-CHO + R"-CHO + H₂O ^__> R'-CH₂OH + R"-COOH  (usual conditions 50% alkali).

eg. CHO-CHO + H₂O ^__> HOCH₂-COOH

or (for piperonal): R-CHO + R-CHO + H₂O ^__> R-CH₂OH + R-COOH

Could gylcollic acid, HOCH₂-COOH then react with piperonal in an Aldol side reaction?

Aldol Rxn: R-CHO + HOCH₂-CHO ^__> R-CH=C(OH)-CHO - H₂O  (usual conditions warm concn. alkali).

My questions are:

1) Do people think this oxidation Piperic acid is possible without the alkali? i.e. by filtering the CuO, produced from CuSO₄ + NaOH, and then using that under neutral conditions. Or do you think that the reaction would then not go at all?

I suppose that those side reactions would be even more possible in the original reactions, given in Antoncho's post and the patent, so I suppose things should be OK.

2) Do people think that 8 hours would be too long for piperic acid?, given that piperic acid is much more soluble in aq. alkali than 3,5-dimethoxy-4-hydroxypropenylbenzene.

3) I can't understand the remark about Fehling's Solution made in the patent [Other alkaline copper oxidizing agents such as Fehling's Solution ...  may be used with the same results.]. Finar (vol 1, 5e, p185) says: "Aldehydes reduce Fehling's Solution to red cuprous oxide. The aldehyde is oxidised to the corresponding acid." Any comments on that?

I didn't want to start a new thread but the previous one seems to be closed.

[Sredni_Vashtar]

"3) I can't understand the remark about Fehling's Solution made in the patent [Other alkaline copper oxidizing agents such as Fehling's Solution ...  may be used with the same results.]. Finar (vol 1, 5e, p185) says: "Aldehydes reduce Fehling's Solution to red cuprous oxide. The aldehyde is oxidised to the corresponding acid." Any comments on that?"

Hi Mellow

I don't understand what you mean by point 3). Are you saying that the Cu(II) is strong enough to further oxidise the piperonal to piperonylic acid?

[mellow]

Are you saying that the Cu(II) is strong enough to further oxidise the piperonal to piperonylic acid?

This is what Finar says in his book "Organic Chemistry" (vol 1, 5e, p185):

It is worth noting that Fehling's solution and Tollen's reagent are both weak oxidising agents, and can be used to oxidise readily oxidisable groups, e.g., they will oxidise unsaturated aldehydes to the corresponding unsaturated acids; they are not sufficiently strong oxidising agents to attack the ethylenic bond.

So I've given a bit more thought to it and I see my mistake. Finar says that Fehlings oxidises "unsaturated aldehydes" not "saturated aldehydes". Piperonal is a saturated aldehyde.

But it's interesting. The benzyllic double bond is more readily oxidised than ordinary double bonds so it could be that the reaction is not:

Ar-CH=CH-CH=CH-COOH + 8CuO ^__> Ar-CHO + CHO-CHO + CHO-COOH + 4Cu₂O

but rather only

Ar-CH=CH-CH=CH-COOH + 4CuO ^__> Ar-CHO + CHO-CH=CH-COOH + 2Cu₂O

Further oxidisation of the "unsaturated aldehyde" should then take place:

CHO-CH=CH-COOH + 2CuO ^__> HOOC-CH=CH-COOH + Cu₂O

Making the final sheme:

Ar-CH=CH-CH=CH-COOH + 6CuO ^__> Ar-CHO + HOOC-CH=CH-COOH + 3Cu₂O

The expected by-produce being maleic adic. This is interesting because (1) it will be easier to purify piperonal as there will be no mixture of aldehydes and (2) this will need 6 moles of copper sulphate per mole of piperic acid.

Finar could be wrong but I hope not. I suggest budding experimentors take that into account.

[Sredni_Vashtar]

Fehling's and Benedict's reagents, which are both complexed Cu(II) ions, will oxidise aliphatic but not aromatic aldehydes - this is a standard test. Looks good.

Just a question of whether malic acid produced or not. If the oxidising agent is not strong enough to attack the aldehyde, then an excess would not directly affect yield.