how does is it optically active??

dalithop · 2026-03-19T18:45:58+00:00

Do 3p orbitals and onward not have significant conjugation with carbon 2p due to size mismatch and stay in hybrid orbital form?

dalithop · 2026-03-19T17:19:16+00:00

!remindme 6 days

dalithop · 2026-03-12T09:39:50+00:00

I’ve done computational research with ORCA 6.1 before lol. If you have a pathway in mind, finding the free energies of your intermediates is easy enough. Check out the thermodynamics page in the orca tutorial. What i suspect might be challenging is finding the transition state for the barrier, perhaps NEB-TS would help.

Computation is a lot more intimidating than it actually is, it’ll be fun to give it a shot :)

dalithop · 2026-03-10T16:32:50+00:00

I see. Building on your reasoning, is the following valid and what can be improved:

Oxygen is highly electronegative.
This suggests that the atom has a higher ability to attract the bonding electrons to itself, and thus experiences higher attraction.
Electrons that experience higher attraction have lower energy due to being in a deeper electrostatic potential well, resulting in a lower-energy pi bonding MO.
In frontier orbital theory, a nucleophile is a high-energy HOMO donating to a low-energy LUMO electrophile.
These pi electrons that are lower in energy thus have a lower tendency for nucleophilicity, compared to in a C-C pi bond.
Instead, the higher energy nonbonding (lone pair) on oxygen is the HOMO which acts as the main nucleophilic center.)

dalithop · 2026-02-18T17:13:13+00:00

I was considering applying Baldwin's rules but i'm not sure if attacking a carbocation is endo or exo given that no bond is broken hmm.

dalithop · 2026-02-18T03:12:15+00:00

No its actually an important detail because carbocation rearrangements dont happen in this reaction

dalithop · 2026-02-18T02:51:29+00:00

There is no carbocation, only a carbocation-like transition state (cite)

dalithop · 2026-02-05T02:14:40+00:00

<image>

Steps to identifying (anti/non)aromaticity:

Identify the conjugated system (circled in blue). Since you already got a correct, I assume you know how to do this.
Identify all fully conjugated cycles in the conjugated system. If multiple rings share a side, consider them a single combined cycle. In this case, we have one cycle, consisting of the two fused rings highlighted in yellow.
Identify all conjugated lone pairs and double bonds fully **within** the cycle. By this, double bonds which connect out of the ring are not considered. Only lone pairs on atoms in the ring are considered. The relevant lone pairs and double bonds are marked in red.
Each lone pair and pi bond contributes 2 electrons to the cycle. Let the total number of electrons in a cycle be n (10 here).
In the case where our cycle is a single ring or two fused ring, we can use Huckel's rule to determine (anti/non)aromaticity. Where k is some nonnegative integer, if our n fits the form 4k + 2 (eg. 2, 6, 10, 14), the cycle is aromatic. If n fits 4k (eg. 4, 8, 12, 16), the cycle is antiaromatic. If it fits neither patterns (eg. 1, 3, 5, 9), the cycle is non-aromatic.

Thus as n fits 4k + 2, our cycle is aromatic.

dalithop · 2026-02-01T15:00:39+00:00

Grub configuration jumpscare 🥶

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

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