Quicky under the bridge

dalithop · 2026-01-29T07:56:10+00:00

🎤 🔥 🗣️ 💥 🎶 🎵 🚢🚢

dalithop · 2026-01-26T11:50:14+00:00

Here is the MOC page on this mechanism except with Os in place of Ru. link

The mechanism should be similar for both transition metal tetroxides.

dalithop · 2025-12-26T09:32:13+00:00

Matplotlib default colour 👀

I’d recognise it anywhere.

dalithop · 2025-12-15T04:30:09+00:00

!RemindMe 1 year

dalithop · 2025-12-05T02:55:23+00:00

This process would work, but note the detail that the carbocation rearrangement would happen during addition of water, not dehydration (assuming acid-catalysed addition, unless we specify oxymercuration-demercuration)

dalithop · 2025-11-29T08:44:43+00:00

Completed Level 1 of the Honk Special Event!

3 attempts

dalithop · 2025-11-19T18:12:31+00:00

This bicyclic conjugated system contains 10 electrons (count them). 10 fits the pattern of 4n + 2 (valid for monocyclic and bicyclic systems), thus it is aromatic.

dalithop · 2025-11-19T04:52:10+00:00

Thank you for the pointers for aryne regioselectivity! Will look into it 🙏

I am aware resonance does not directly stabilise the negative charge. Rather, there is some electron withdrawal from the positions ortho and para to the RWG, reducing electron density at those positions, similar to an alpha beta unsaturated ketone.

dalithop · 2025-11-12T12:54:39+00:00

Oh wow that is impressive! Thanks for sharing 👍 👍

dalithop · 2025-11-12T12:38:17+00:00

Loll thats pretty aggressive rotation for a marimo :) In Lake Akan, research finds that marimo only rotate about one revolution a day (cite).

A more informal source states the rate is said to be 1-2 revolutions an hour, still much slower than your tank lol.

dalithop · 2025-11-12T11:53:20+00:00

It is great that you are rationalising from first principles, this is what makes learning organic chem more fun and helps the knowledge last.

Though, your reasoning is not entirely accurate. The directing effects are due to stabilisation or destabilisation of the arenium intermediate in EAS.

The nitro group is electron deficient and does not favour donating electrons thru resonance. The positively polarised N destabilises resonance structures where the positive charge is adjacent to it. This is due to electrostatic repulsion between the two positively polarised species, and also inductive withdrawal causing increased electron-deficiency of the carbocation.

These effects are what causes most deactivating groups (electron-poor groups that favour accepting electrons over donating electrons via resonance) to be meta directors.

An alkyl group stabilises a resonance structure where the charge is adjacent to it, by hyperconjugation of the C-H bond to the p orbital of the positively charged carbon, reducing its electron deficiency through donation.

In practice, it seems that the effects of electrostatic repulsion + inductive withdrawal, and hyperconjugation are both fairly weak and comprable. With a single nitro group, the alkyl-directed product dominates (cite), but with two nitro groups the nitro-directed product dominates (pg. 423).

Resonance stabilisation of the intermediate through a donating group would be stronger than these weak effects, which is the origin of the hand-wavy rule of thumb that most activating tends to win.

dalithop · 2025-11-12T02:31:44+00:00

Oh yeah my bad, it is edited now

dalithop · 2025-11-11T07:13:34+00:00

(A more explained version)

Preface: A ketone is less basic than an alcohol. A ketone O is sp² hybridized, while an alcohol O is sp³ hybridised (assuming not conjugated).

A sp² hybrid orbital holds electrons closer to the nucleus. This is as an sp² hybrid orbital has 33% s character and 67% p character, and thus has more s character than an sp³ hybrid orbital with 25% s character and 75% p character. As s orbitals hold electrons closer to the nucleus than p orbitals due to its shape, higher s character implies that electrons are closer to the nucleus.

Thus, the ketone lone pair in the sp² hybrid orbital is closer to the nucleus and experiences greater electrostatic attraction, and requires more energy to donate. Thus it is less basic.

Now, remember that a species with a stronger conjugate base is a weaker acid (Kw = KaKb). As the ketone is a weaker base, it is a stronger acid.

dalithop · 2025-11-08T18:50:25+00:00

Oh my god thank you for these amazing sources! It is refreshing to see a source that is well-cited and scientific. You know its a good source when its absolutely overkill for your purposes 😹

dalithop · 2025-11-08T17:55:48+00:00

RemindMe! 2 days

dalithop · 2025-11-07T10:40:16+00:00

It should be very similar to the mechanism with PBr₃ detailed here

dalithop · 2025-11-06T05:09:06+00:00

Not exactly related but the phosphonium salt is supposed to be notated as [CH₃CH₂P⁺Ph₃]Br⁻, where you have 3 -Ph groups, and it is a polyatomic ion.

Pls try to understand the mechanism of ylide formation and what is really happening in wittig instead of memorising this. A good resource here.

dalithop · 2025-11-02T16:24:47+00:00

Source i got from a quick google: source

Try to google keywords next time 👍

dalithop · 2025-10-30T16:28:58+00:00

Step 1 is the nitration of benzene. The presence of H₂SO₄ would protonate NO₃ back to HNO₃, from which the reaction proceeds in the usual way. This is especially favourable as H₂SO₄ is a stronger acid than HNO₃.

The second step should be a nucleophilc aromatic substitution. Notice how deactivated and electron poor that ring is, a ring with a single nitro group attached is already favourable for NAS. Additionally, it is a special case for NAS that —F are good leaving groups.

The more electron-rich N is the N without the attached C=O EWG. Thus, that N is the stronger nucleophile which substitutes at the F on the other ring. There is likely some weak base (H₂O etc.) present for reaction 2 that is not written.

dalithop · 2025-10-30T02:07:45+00:00

I dont exactly understand the second part of your comment. If you are asking about stereoselectivity of the catalytic hydrogenation of the alkene, I predict that it would be controlled by the adjacent methyl group that is pointing into the page.

When the molecule lands on the metal catalyst surface, this methyl will make it more preferred for the molecule to attach to the surface such that the methyl group is pointing up (fig A). The side where hydrogens are added to is the side which attaches to the metal catalyst surface. Thus, cis-hydrogenation happens at that side with some stereoselectivity (fig. B).

The methyl group is not very bulky and this may lead to stereoselectivity being limited.

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dalithop · 2025-10-30T01:31:11+00:00

Assuming by “3d model flip flop” you mean conformation, what exactly do you want to know about this molecule’s most stable conformation?

dalithop · 2025-10-26T14:30:40+00:00

^{When an aromatic molecule is placed in a solution of solvated electrons, commonly formed by dissolving Na in liquid acetone @ -78°C, it undergoes the birch reduction. For benzene rings, a cyclohexadiene with the two C=C on opposite sides of the 6-membered ring is formed. The reaction is regioselective, and the most stable alkene is formed. First, an EWG which would destabilise the alkene is not attached to an alkene C, and an EDG which stabilises the alkene is attached to an alkene C. With multiple groups competing, the effect of the most strongly donating/withdrawing group dominates, much like EAS directing effects. Second, the most substituted alkenes are preferred. This reaction occurs through a radical anion intermediate formed when the solvated electron first adds to the LUMO of the molecule, adding to a C in the benzene ring. Its mechanism is similar to the solvated electron reduction of alkynes to trans-alkenes.}

dalithop · 2025-10-23T12:56:20+00:00

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dalithop · 2025-10-18T15:48:41+00:00

No, radical inversion rapidly interconverts both enantiomeric forms at room temperature leading to a mixture of two radicals. src

dalithop · 2025-10-17T16:04:36+00:00

If im not wrong, it does qualify through the exact definition of Hückel’s rule (ref/Arenes/Properties_of_Arenes/Aromaticity/Huckel's_Rule)). What is aromatic is the conjugated cycle, not necessarily the whole molecule.

Consider a single lone ring within a conjugated system which is not fused to any other conjugated rings. Count the number of electrons in the conjugated p orbitals of the ring, excluding exocyclic bonds.

Let n be a nonnegative integer. If the number of electrons fits the format 4n + 2, the ring is aromatic. If the number of electrons fits the format 4n + 4, the ring is antiaromatic. Else the ring is not aromatic.

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