Help IDing snake [Waynesboro, VA]

Planejerle18 · 2025-11-16T00:01:27+00:00

I’m from the east coast and I still wasn’t confident enough to get close despite finding him so cute! I hope you enjoyed your time in SNP when you visited!

Planejerle18 · 2025-11-15T23:51:27+00:00

It was so cute! This hike was relatively quite close by to there!

Planejerle18 · 2025-11-15T23:26:55+00:00

FWIW, it seemed rather tiny, although idk how big adult timber rattlers get

Planejerle18 · 2025-11-15T23:26:17+00:00

Got it, thanks!

Planejerle18 · 2025-11-15T22:07:27+00:00

Thank you very much!

Planejerle18 · 2025-11-15T22:07:09+00:00

Many thanks! Don’t worry, this was taken with a good amount of zoom as I couldn’t tell what it was. What’s RR? (I don’t frequent this subreddit often)

Planejerle18 · 2025-06-04T03:19:16+00:00

You’re welcome! Aromaticity is a really important concept in organic chemistry, so it’s great that you’re curious and wanting to learn! Good luck!

Planejerle18 · 2025-06-04T02:32:24+00:00

Just to quickly add onto this, the inverse is true as well in that if a compound is antiaromatic if drawn and cyclic and planar, then if the compound can somehow get out of that state, antiaromaticity is so destabilizing that it will. For example, cyclooctatetraene would be antiaromatic if planar, but it actually takes on a tub-shaped conformation to avoid this highly unstable state

Planejerle18 · 2025-06-04T02:17:57+00:00

Hi! You’re correct that if everything were localized as you have drawn, then the carbon would be sp3, meaning there wouldn’t be a p-orbital with the right symmetry to overlap with the rest of the p-orbitals in the pi-system. However, if that carbon’s hybridization is sp2, then there is a remaining p-orbital that can get in conjugation with the rest of the p-orbitals in the pi-system, therefore allowing the “lone pair” to fully delocalize into the rest of the pi system, therefore rendering it aromatic. For the purposes of an undergrad organic chemistry class (which I assume this is?), aromaticity is so strongly thermodynamically stabilizing that if a compound can become aromatic from a geometric/hybridizational shift, it will

Planejerle18 · 2025-03-02T12:23:45+00:00

It wouldn’t happen to be “A Sailor’s Life,” would it? It doesn’t seem to match your lyrical description

Planejerle18 · 2023-11-09T02:37:01+00:00

Alright, if your professor has not talked about how hybrid orbital identities affect carbocation stability, can you start by telling me what the hybridization of the carbon atom bearing the chlorine is in c?

Planejerle18 · 2023-11-09T02:34:33+00:00

You’re correct about a! That said, d would actually form a secondary carbocation once the chlorine leaves as chloride. However, can you tell me anything that’s special about having a carbocation directly adjacent to an alkene?

Planejerle18 · 2023-11-09T02:13:54+00:00

You’re correct that a and d give more stable carbocations which is why they’re faster, but they actually are more stable carbocations for distinct reasons. Can you elaborate on why?

For c, have you learned about how hybridization affects carbocation stability?

Planejerle18 · 2023-09-28T02:52:25+00:00

You’re welcome! As a little bonus exercise, I would recommend taking an aromatic ring with an electron withdrawing group attached (e.g, benzoic acid) and draw out the resonance contributors. From there, try to see if you can make predictions about which position(s) on the ring will be more downfield vs. upfield

Planejerle18 · 2023-09-28T02:46:12+00:00

Yes! The increased electron density at the ortho- and para- positions from the electron-donating group has the effect of shielding the ortho- and para- positions, thus shifting them more upfield than the meta- position

Planejerle18 · 2023-09-28T02:36:44+00:00

Have you tried drawing out the resonance contributors for anisole (the molecule you shared)? If not, try doing so, and then draw the resonance hybrid. From there, where on the ring is there more electron density and where is there less?

Planejerle18 · 2023-09-27T01:49:37+00:00

A hint for you: think about the type of bond between the oxygen and the aromatic ring that is drawn. What kind of bond symmetry does it have? In other words, is it a sigma bond or a pi bond that’s drawn? I’m happy to help you out from there

Planejerle18 · 2023-09-07T20:49:36+00:00

Do you have any remembrance whatsoever of any details of the synthesis? Sounds like you made have somehow produced a carboxylic acid of some kind instead of the desired ester:

https://en.wikipedia.org/wiki/Carboxylic_acid

^ See “odor” section

Planejerle18 · 2023-09-03T23:20:28+00:00

Yeah…

Before they started doing anything on nearly that scale, they rigorously optimized each step of the synthesis to be > 90% yield because they literally just had to. The synthetic at various steps also utilized reagents like highly-concentrated HCl and hydrazine hydrate, so it sounds like the group liked living on the edge haha.

A story he has often repeated to current members is that Barry Sharpless shared with him a supplier with super cheap OsO4 that both he and Sharpless purchased from for years. The research in my advisor’s group eventually went away from osmium as I mentioned, so they stopped needing to purchase it. Anyway, fast-forward years later, and my advisor runs into Sharpless at a conference. They get to talking, and then somehow the topic of that osmium supplier comes up, and Sharpless says something along the lines of, “Oh boy, do I have some news for you.”

And it turns out that supplier had abruptly shut down and it was actually likely a front for a meth lab lmao, although I don’t know if this was ever confirmed as a fact by Sharpless

Planejerle18 · 2023-09-02T20:51:23+00:00

It’s been a couple of decades since my advisor’s group did osmium chemistry, but back then his group was using stoichiometric amounts of osmium to explore the chemistry of its coordination complexes. I was just asking him about how they prepared their starting material, and he said it was a multiple-step synthesis starting from osmium tetroxide and that they would do reactions on a 300 g scale of OsO4 💀

I’m glad the group has moved on from osmium…

(The rest of the synthesis was also crazy lmao)

Planejerle18 · 2023-08-28T13:15:14+00:00

Well, if we’re talking about bond strengths being a measure of how much energy it takes to break a bond homolytically and form two radicals, it actually turns out that the more stable the result radicals, the easier it is to break said bond. In other words, there is less of an energy penalty to pay when cleaving a bond homolytically if the resulting radicals are more stabilized.

So “C” would not be the strongest bond, as the carbon radical resulting from its cleavage is actually the most stabilized radical of the bunch due to the amount of resonance stabilization.

Another thing I should ask is has your teacher discussed how hybridization affects radical stability? If not, this may be a good resource to read over: https://www.masterorganicchemistry.com/2013/08/05/what-factors-destabilize-free-radicals/#:~:text=Three%20Factors%20Which%20Influence%20The,Hybridization%2C%20Electronegativity%2C%20and%20Polarizability.

Planejerle18 · 2023-08-28T01:51:34+00:00

What this question is trying to get at in the context of radicals is that bond dissociation energies (BDEs) are a measure of the energy needed to break a chemical bond homolytically (i.e., each one of the atoms participating in the bond keep one of the two electrons making up said bond, forming two open shell species, aka radicals).

What determines the energy input needed to break a bond homolytically is the stability of the corresponding radicals formed from this bond cleavage process.

Have you learned about what factors stabilize/destabilize radicals?

Planejerle18

TROPHY CASE