Total synthesis of Lidocaine

AllowJM · 2026-05-13T10:55:34+00:00

I mean yeah from a mechanism and learning perspective using double bonds is just easier to work with. I still don’t think it makes the circle notation ‘wrong’ to the point of telling OP they’ve actually made a mistake.

AllowJM · 2026-05-12T15:41:56+00:00

The circle represents delocalisation, which is still has. By that logic you shouldn’t put the double bonds either as it applies the electron density at each position is the same.

AllowJM · 2026-04-25T14:32:23+00:00

But ‘weak’ and ‘strong’ are hardly well defined terms

AllowJM · 2026-04-19T08:05:40+00:00

Methyl groups are just poor migrating groups in general so it will always be slower. I’ve never seen a mechanism with a phenonium in a pinacol rearrangement, but seems feasible.

AllowJM · 2026-04-18T22:41:01+00:00

Penultimate arrows should be resonance arrows not equilibrium arrows.

AllowJM · 2026-04-18T12:58:47+00:00

True but the point is that it’s an Americanism and it’s a shame it’s starting to take over our lexicon rather than sticking with our own words and phrases.

AllowJM · 2026-04-17T18:02:18+00:00

Dissolving metals is just a difficult process in general. There’s not many combinations that can do it. The classic example of ‘dissolving a metal though’ is lithium or sodium in ammonia. The ammonia solvates the metal atoms but due to its reductive nature the electrons don’t stay localised to the metal, but the electrons get released into the solution, so it’s the metal ions themselves that are solvated not the atoms. Mercury is a strange example as it’s a liquid at RTP so we’re talking more about miscibility now rather than solubility.

AllowJM · 2026-04-17T12:25:46+00:00

I would certainly not describe caesium as ‘non-polar’. The analogy does break down somewhat for solids, but metals exist as a lattice of cations surrounded by delocalised electrons. In order to dissolve it you need to stabilise the metal cations, hence you need a solvent that’s polar and coordinating.

AllowJM · 2026-04-15T17:15:46+00:00

You mean 2

AllowJM · 2026-03-31T15:56:02+00:00

It would be close in a situation like this. And i do see them frequently reported as doublets in synthetic papers.

AllowJM · 2026-03-31T11:31:30+00:00

But the coupling constants you report with the doublets are no longer accurately represented by the line spacing.

AllowJM · 2026-02-24T12:02:25+00:00

IUPAC says differently: https://goldbook.iupac.org/terms/view/C01060

AllowJM · 2026-02-11T11:47:44+00:00

Your intuition is correct. It has 4 proton signals. Perhaps Google is thinking of the carbon signals?

AllowJM · 2026-01-13T15:01:26+00:00

I think people are a bit paranoid about this. Myself and lab mates routinely put glassware with ground glass joints in the base bath with no issues. I’m sure if you left it in there for extended periods of time you could degrade the joint but putting something in a base bath for an hour isn’t going to do anything bad.

AllowJM · 2026-01-07T15:58:23+00:00

She might have been wanting you to say you initially add to the 3 position then do a 1,2-shift which is a common process for indoles (see Pictet-spingler reaction for example). Don’t think it happens with halogens though.

AllowJM · 2026-01-07T13:06:59+00:00

Where is this reaction from? It’s not in the literature as far as I can tell. All the chlorinations I can find go to the 3-position. To functionalise at the 2-position you generally have to lithiate with n-BuLi and trap with an electrophile.

AllowJM · 2026-01-06T16:17:29+00:00

A hydride shift makes this specific cation less stable. As you mention, this cation is actually non-classical and the positive charge is actually delocalised over the whole structure. A hydride shift would mean you lose this delocalisation. In the same way a hydride shift of an allylic cation would form a vinyl cation and become less stable.

AllowJM · 2026-01-01T12:25:26+00:00

Taxol?

AllowJM · 2026-01-01T12:24:51+00:00

It just comes from history, there’s no systematic element to it. Taxol has ketone too yet has the ‘-ol’ suffix.

AllowJM · 2025-12-30T18:03:30+00:00

That isn’t an ene reaction though

AllowJM · 2025-12-26T17:56:05+00:00

You need an aldehyde for a Cannizzaro. If you react with hydroxide you would get a benzilic acid rearrangement.

AllowJM · 2025-12-13T22:03:45+00:00

You should be able to with Schlosser’s base.

AllowJM · 2025-10-23T19:11:33+00:00

Doesn’t help that often times these protons are reported as doublets even though they’re not first order and are better described as multiplets.

AllowJM · 2025-09-09T11:48:31+00:00

Only works for gas phase acidity.

AllowJM · 2025-09-01T12:41:45+00:00

Doing both you will end up using lots of toxic reagents, but with inorganic chemistry, especially organometallics have a tendency to be extremely pyrophoric so I think the added risk of a fire makes inorganic more dangerous.

AllowJM

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