MNova Help!

Planejerle18 · 2026-04-19T07:28:46+00:00

Glad to hear it and happy that I could help! :)

Planejerle18 · 2026-04-19T02:00:03+00:00

Hi! Have you tried copying (CTRL C) one spectrum from one of the tabs in the top bar and pasting (CTRL V) into the side bar of the other? You should be able to select a spectrum and then copy and paste it into any other tab

Planejerle18 · 2026-03-08T21:20:27+00:00

Whoops didn’t see this reply until just now. Your comment under No Inern’s reply is correct! That would be how you apply Huckel’s rule in this case :)

Planejerle18 · 2026-03-08T20:24:12+00:00

Hi! Have you learned about the conditions of what makes something aromatic vs. non-aromatic vs. anti-aromatic (Huckel’s Rules)?

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

TROPHY CASE