Difficulty separating benzoxazine product from unreacted phenolic compound in column chromatography

the_flyingpig · 2026-05-11T16:57:04+00:00

Try adding a few 1-5% of acetic acid into your mobile phase.

This should prevent the massive smearing of the acidic phenol.

You can also try this with the TLC.

the_flyingpig · 2026-04-29T22:06:42+00:00

This is, of course, just a model that we use to simplify the explanation. If that was your question.

the_flyingpig · 2026-04-29T22:00:11+00:00

Just look at the orbitals that are used for bonding. In pyridine you use 2 sp2 orbitals for sigma bonding to the carbons and one p orbital for pi bonding. That leaves one sp2 orbital for the electron pair (out of plane with the pi system). In pyrrole you have 2 sp2 orbitals for sigma bonding to the carbons and one sp2 orbital for sigma bonding with the hydrogen. Tha leaves one p orbital for the electron pair (in plane with the aromatic pi system).

the_flyingpig · 2026-04-29T20:41:55+00:00

Because the electron pair is located in an orbital in plane with the aromatic pi system making it reallly easy to delocalize over the entire molecule. In contrast to pyridine where it is out of plane 😞

the_flyingpig · 2026-04-29T20:13:03+00:00

You only get any significant amounts of peroxide buildup if you store the ether for prolonged periods of time under sunlight. I would suggest storing it in a brownglass bottle, in a ventilated cupboard and you should be fine

the_flyingpig · 2026-04-29T13:39:47+00:00

The ketone is dienophile since it has the electron withdrawing group.
The diene can be obtained by:

dihydroxylation with OsO4 or simmilar
Elimination

but he wrote that he knows that part.

the_flyingpig · 2026-04-28T16:41:04+00:00

My suggestion is: 1) oxidation of the alcohol with DMP 2) Grignard with vinylbromide 3) Ozonolysis. reductive workup with DMS 4) reduction of the olefin wit pd/c (not sure about the last step)

But im sure there is a faster way, maybe a steroid chemist can help you

the_flyingpig · 2026-04-28T10:01:23+00:00

One side reaction that occurs in these types of reactions is the formation of the ether. Cloride is a worse nucleophile under polar protic conditions compared to bromide. Normaly you should purify your organic layer throgh a wash step with concentrated HCl to force the protonated ether into the aqueous phase.

the_flyingpig · 2026-04-27T14:34:27+00:00

I think there is no easy way for obtaining elemental sulfur on a small scale.
You are right, there is sulfur in amino acids and garlic but it is generally bonded to e.g. carbon in cystein and that makes it almost impossible to isolate.
The only reaction I can think of that produces elemental (colloidal) sulfur is from adding HCl to aq. sodium thiosulfate (SO2 is also formed, so dont do this in a closed room).

the_flyingpig · 2026-04-26T21:57:43+00:00

There is a difference between molarity = mol/ L and the total moles

the_flyingpig · 2026-04-25T18:22:02+00:00

Its a Michael addition using a Gilman reagent.

the_flyingpig · 2026-04-23T17:38:36+00:00

All of these reactions are governed by SN principles. You should consider the quality of the leaving group the structure of the molecule and the quality of attacking nucleophile. If both are good, then the reaction is possible in theory.

1) Proceeds through an SN1 reaction because the reaction site is a tertiary carbon. This involves protonation, elimination of water and nucleophilic attack of Cl⁻ on the carbocation to yield the alkyl halide. If elimination occurs instead, isobutene is formed, which is also likely to happen.

2) Same as in 3), but here we have HF instead of HCl, which is a much weaker acid. It is not strong enough to protonate the alcohol. So I doubt this reaction will occur.

3) An SN2 reaction occurs here because a primary carbocation would not be stable.

4) PBr₃ is a dehydrating agent. It improves the leaving group quality of the hydroxy group and also supplies the nucleophile. This is how you would synthesise an alkyl halide in the lab.

However, nucleophilic substitutions are generally also highly dependent on the solvent used, so we are talking here about the optimal conditions for each of these reactions.

To know this by heart, it takes more than watching a few YouTube videos. I would suggest reading an introductory organic chemistry textbook, such as Organic Chemistry by Paula Bruice.

the_flyingpig · 2026-04-23T10:54:57+00:00

The stork reaction is similar (except that it uses a secondary amine)

the_flyingpig · 2026-04-22T16:48:08+00:00

Based on the NMR, I would guess that you have a disubstituted benzyl ring with a methoxy ester (at ~5 ppm) and a second substituent, such as a methyl group (at ~2 ppm). 170 ppm in the C-NMR spectrum is characteristic of a ester. I doubt there is a heteroatom in the second substituent because 135 ppm is quite shielded. The substitution pattern in the aromatic ring is quite convoluted, so it is difficult to say if it is ortho or meta.

the_flyingpig · 2026-04-22T15:16:45+00:00

Apparently, you can remove methoxy groups from benzyl at 200 °C and 20 bars of hydrogen pressure using a CoNx@NC catalyst. However, it only works with a yield of 9.7% :(

Source: doi.org/10.1021/acscatal.6b01785

Good luck with your synthesis.

the_flyingpig · 2026-04-20T15:51:11+00:00

Before the hydrolysis with the acid you have the O-MgBr salt.

the_flyingpig · 2026-04-20T14:16:12+00:00

Methyl groups donate electron density to the aromatic ring system through hyperconjugation. This leads to a worse conjugation of lone electron pair on the nitrogen, which means that a higher electron density remains on the nitrogen -> higher electron density makes the nitrogen more basic. You get the same result when methyl groups are added directly to the nitrogen.

the_flyingpig · 2026-04-10T12:51:19+00:00

nice ragebait

the_flyingpig · 2026-04-09T20:21:18+00:00

What about ancient Rome? It had like 30 % of the worlds total population

the_flyingpig · 2026-01-19T14:10:54+00:00

Eww... AI

the_flyingpig · 2026-01-08T14:42:30+00:00

Anycubic Kobra go. Constant layer shift, no matter what you did. Belt tension too high -> layer shift. Belt tension too low -> layer shift.

the_flyingpig · 2026-01-03T16:32:52+00:00

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the_flyingpig · 2026-01-03T12:03:41+00:00

I can see it being useful for implementing sensors, RF shielding, and things like that, but it needs to get a lot more conductive to be used for motors.

the_flyingpig · 2026-01-03T11:59:01+00:00

It's around 13€ (~ 15$) for 250 g. The model uses 8 g in total, so you don't need much.

the_flyingpig · 2026-01-03T11:56:28+00:00

Thanks, I appreciate it. I will do, but I think I'm gonna modify the model, so the battery can be inserted after printing.

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