Eugenol Demethylation Proposal

[DRIVEN]

The following demethylation procedure is being considered on a small scale to explore the demethylation of eugenol.  Critisms a plenty are welcome.

Vanillin Demethylation (by Startinout; from Rhodiums site):

"Abstract: Reflux aluminium powder with iodine in benzene or cyclohexane for 90 min to produce AlI3. Add equimolar amounts of vanillin, 0.25 eqv PTC (tetrabutylammonium iodide, chloride may also work), reflux for 20 min, and isolate the protocatechualdehyde (88%)"

Proposed changes:

1) Toluene will be exchanged for benzene and where the procedure indicates reflux, the temperature will be held at the boiling point temperature of cyclohexane.

2) The PTC will be HTAB

Many thanks.

Driven

[Rhodium]

That would be excellent. Make sure that you maintain excellent stirring throughout the reaction as you aren't refluxing.

[eXcidium]

Wich PTC would give the best yelds?
tetrabutylammonium iodide, chloride, or bromide?

And, also, for demethylation, wich is best (again regarding to yelds): pyridine HCL/microwave, or AlI3?

[Rhodium]

Wich PTC would give the best yelds? tetrabutylammonium iodide, chloride, or bromide?

As TBAI will eventually be formed in the reaction, it will make most sense chemically to use that to begin with. But as it is significantly more expensive per mole than TBACl, it might make sense economically to test if the difference in yield is worth  the cost (at least for larger batches).

And, also, for demethylation, wich is best (again regarding to yelds): pyridine HCL/microwave, or AlI3?

I believe that is more a question of scale than absolute yields. You won't be processing mole batches in a microwave even if it somehow seems implicated in

https://www.thevespiary.org/rhodium/Rhodium/chemistry/methylenation.html

- the amounts given are molar ratios rather than being actual molar amounts.

[eXcidium]

You sure TBAI is more expensive that TBACl? SWIM saw the price for both, and for 100g, the TBACl costs $80, and TBAI costs $25

Btw : wound anyone be so kind as to help me with the weights? (instead of using moles)?
e.g. - absolute mesures (ml, grams) for the above mentioned reactions (demethylation) since I got a bit stuck there

[praeseodymium]

Specifying the amount in moles is useful in chemistry becuase the number of each type of molecule in a reaction is generally extremely important. This unit gives an idea of the precision exercised in measuring out- it is a reference to the exact number of molecules of a given substance.
1 mole is a very large, fixed number of molecules of any substance- in fact the exact number, called Avagadro's constant, is 6.0228x10²³.  

To work out how much a mole of something weighs, then, consult your friendly periodic table for the masses of the atoms comprising the molecule in question, and add up all the masses of all the atoms in your molecule... and voila, that number is the Molecular Weight of that molecule, and the mass in grams of 1 mole of molecules of your substance.

The Molecular Weight is usually written on the label of lab reagents, and you may also find the M.W. of your substance in many online chemical databases.

For example, one mole of carbon weighs 12.011 grams, a mole of water weighs 18 g and

tetrabutylammonium iodide,
formula
C₁₆H₃₆IN has
Molecular Weight 12x16+1x36+126.9x1+14x1 = 368.9

Therefore 1 mole, or 6.0228x10²³ molecules of TBAI, weighs 368.9 g.

Correction: Avagadro's constant = 6.0228 x 10²³

[eXcidium]

Thanks, man
that cleared some ideias
Specifically, my question was related to this


Aluminum Triiiodide preparation

1 mol Al turnings (or foil or powder all work well, powder is the quickest closely followed by foil then turnings a distant third) is combind with 1.5mol (note the excess of Al) of Iodine in 1000mL of cyclohexane (exchange solvent as needed but not diethyl ether!). This mix is stirred under reflux until the red colour has turned to a grey (about 1.5hrs with powder, 2-3 with foil, 4+ with turnngs). At this point the heat is turned off and the AlI3 in solvent allowed to cool to room tempeature with stirring.

1 mol Of Al = 26.98g
and 1 mol of I = 126.90  OR 1 mol of I₂ = 253.80

That's what got me confused (should I use I or I₂)

But I guess to make them equimolar It's the I₂ that counts (hence 2AL + 3I₂ = 2AlI₆)

Am I right?

[praeseodymium]

1 mol Of Al = 26.98g    and
1 mol of I = 126.90 OR 1 mol of I2 = 253.80

And your intended product AlI₃ reacts in the ratio of Al:I =1:3 so for every mole of Aluminum atoms there are 3 moles of Iodine atoms.

2Al + 3I₂ -> 2AlI₃

Molarity has been covered before. UTFSE.  

Chemfinder is good but I find it helps to obtain a clear picture of what you're doing if you just work it out. Keep a periodic table handy. I have one on my desk 24x7  

See

Post 319381

(Rhodium: "molar yield!", Newbee Forum)

Post 245494 (missing)

(AMINATOR: "Calculating moles", General Discourse)

Post 85091 (missing)

(smokemouth: "Mole Weights - Scorpion Junior", General Discourse)

Post 57798 (missing)

(Mister_Clean: "Time to learn about moles.", Stimulants)

[eXcidium]

Sorry, my typo, i meant 2Al + 3I2 -> 2AlI3
but thanks for it, and the links

[Rhodium]

This procedure has been discussed before in the following threads (just adding them here for reference):

Post 68536

(startinout: "Success, finally", Novel Discourse)

Post 455261

(nitrous351: "a few more questions", Methods Discourse)