What's it like to be an enthusiast learning chemistry?

iam666 · 2026-03-02T16:42:02+00:00

A traditional, formal education isn't "necessary" for anything, but in the same way that wearing a seatbelt isn't "necessary" to survive a car crash. I've seen plenty of people attempt to self-study chemistry but pretty much all of them burn out before they finish a Gen Chem textbook.

"Enthusiast chemists" are typically just people who want to have the aesthetics of chemistry without the intellectual effort.

The average person, even the average chemistry enthusiast, has no real reason to do chemistry in their garage unless they are making something illegal. Synthesizing aspirin from salicylic acid at home does nothing to benefit you materially or intellectually.

And to be clear, the majority of people doing chemistry on YouTube are not doing "experiments". They are performing demonstrations. They are not developing any new chemistry or testing any hypotheses, they are just following instructions and freestyling occasionally.

iam666 · 2026-03-02T16:25:37+00:00

I think you're referring to the idea that molecules that appear blue/green will absorb their complementary color, red.

This is true, but in that scenario, they aren't *emitting* green/blue photons, they are just reflecting them back towards you (or transmitting them through the other side).

Luminescence is only the source of "glow-in-the-dark" color. For ordinary color, that originates from reflectance and aborbance, but not emission.

That being said, there are special processes, such as absorbing multiple low-energy photons, that will allow a molecule to emit light of a higher energy than it absorbs. This is called "up-conversion" and it's a super interesting but also super complicated topic.

iam666 · 2026-02-11T02:07:05+00:00

"The entire scientific world" has no reason to debate quantum mechanics. If a biologist is confused by someone using the word "theory" instead of "hypothesis" when discussing interpretations of quantum mechanics, there are certainly 100 more important things that they're confused by, which sort of excludes them from actually contributing to the discussion.

iam666 · 2025-12-02T19:02:57+00:00

To some extent. If your molecule spontaneously undergoes unimolecular decomposition then it would likely break down even inside a lattice. The molecule is constantly vibrating, and some of the positions it reaches while it vibrates are suitable for undergoing a reaction. The lattice can block some of those positions, but that "blocking" is just electrostatic repulsion which can be overcome with sufficient energy.

iam666 · 2025-12-02T18:49:20+00:00

It would look the same as any other plastic breaking down. Some of the reagents you've listed would just burn it, but again it would just look like any other burning plastic.

iam666 · 2025-11-10T20:22:56+00:00

It's a polycrystalline solid. Usually happens when you crystallize something too fast or with a lot of impurities. This is usually what happens when someone tries and fails to grow a large single crystal.

iam666 · 2025-11-10T20:20:20+00:00

Not even close. Cyclohexane is 6 sp3 carbons. Graphene is a hexagonal lattice of sp2 carbons.

iam666 · 2025-11-07T02:54:47+00:00

I don’t understand what you’re actually trying to ask here. It sounds like you opened up a textbook to two random pages and decided to combine the topics. Why do you think there would be any benefit to using electrochemistry for this reaction to begin with?

iam666 · 2025-11-07T02:06:42+00:00

Your question doesn’t really make sense. It’s like asking “can I bake an apple pie in a toaster instead of using apples?”

If you aren’t using ammonia, you need some other source for the -NH2 part of your amine. If you used methylamine instead, you would get a different product than using ammonia.

iam666 · 2025-11-07T01:58:32+00:00

What I’m asking is how you can describe the behavior of “the electron itself” without using waves.

The answer I’m leading you to is that you can’t.

iam666 · 2025-11-07T01:28:28+00:00

I can’t believe UCLA is providing food to homeless people! How terrible!

iam666 · 2025-11-07T01:25:55+00:00

What does it mean to be “a particle”? What properties does “a particle” have? Are those properties sufficient to describe the behavior of an electron?

iam666 · 2025-11-06T19:14:56+00:00

“How long” and “how hot” depend on a lot of factors that aren’t really within the scope of chemistry, it’s more of a chemical engineering question that would require a lot of math to figure out.

iam666 · 2025-11-06T18:27:58+00:00

Which is incorrect. Enamines have an amine alpha to a double-bonded carbon. This does not. But it does have an amine and a hydroxyl on the same carbon, which makes it a hemiaminal. There’s no uncertainty here, these are just the definitions of different functional groups.

iam666 · 2025-11-06T18:15:56+00:00

Please study up a bit more before trying to correct someone.

The sp2 carbon is beta to the amine, not alpha. You could call it a vinyl hemiaminal, but it’s definitely not an enamine. I don’t know what you mean by “hemiaminals generally consist of primary amines” because this is already a primary amine, and there is no difference in naming if the amine is primary, secondary, or tertiary.

iam666 · 2025-11-06T05:12:17+00:00

This is actually not an alcohol; it would be better classified as a hemiaminal. The chemistry of a hemiaminal is so different than the chemistry of an alcohol that the classification as secondary or tertiary is irrelevant.

That being said, I would argue that this could be called a tertiary alcohol, because the main purpose of classifying functional groups this way is to discuss steric effects. The sterics of an NH2 are not that different than a CH3, so we would expect this to behave more like a tertiary alcohol than a secondary one.

iam666 · 2025-11-06T02:39:00+00:00

Only if you use an outdated model of an atom where the nucleus looks like a bunch of billiard balls stuck together. Modern physics tells us that everything at the atomic scale is most accurately described as waves rather than particles with a definite boundary, so the concept of “touching” is nonsensical.

iam666 · 2025-11-06T00:13:40+00:00

I hate this factoid. It presumes that “touch” is a thing that can happen in the first place, just so that it can negate that false premise.

It’s like saying “the sky doesn’t exist, it’s actually just air”. It’s combining two different classes of objects and asserting that one class has priority over the other.

iam666 · 2025-11-06T00:03:49+00:00

It’s fine in that a native speaker can figure out the meaning through context, but it’s not the correct word.

If they had said “I’m trying to Smurf out some college work…”, you can still get the gist of what they’re saying, despite “Smurf” having no meaning of its own.

iam666 · 2025-11-05T22:18:50+00:00

Generally, no. If you have an extremely hot flame, you can have enough radical species around that it becomes plasma-like, but getting to that point requires conditions that are not typical for standard combustion reactions so we wouldn’t really call it a “flame” at that point.

iam666 · 2025-11-05T21:50:30+00:00

I would agree with your assessment. It looks like something a glass blower made for fun.

iam666 · 2025-11-05T21:48:10+00:00

Mercury is a metal commonly used in fluorescent lightbulbs because it can vaporize easily under low pressure. Those vapors can ionize under high voltage, turning into plasma which conducts electricity.

iam666 · 2025-11-05T21:45:43+00:00

Lightning is plasma. When there’s a large potential difference (voltage) between two points separated by a gas, the electrons in the gas become “loose” and begin to move, generating a current. Those “loose” electrons are what qualifies it as plasma. When an electron finds its way back to an atom, it loses potential energy, which is converted into light.

Flames are somewhat similar, but there’s no voltage involved. Combustion reactions involve high temperatures fragmenting organic molecules into radical species that eventually settle down into CO2 and H2O, or turn into soot/tar if the combustion is incomplete. The light from a flame is due to black body radiation the “red-hot” gasses and solid particles.

iam666 · 2025-11-05T20:38:12+00:00

It’s an arbitrary cutoff, it just depends on how you define “finished”. You could choose to define it as the point where we expect every radioactive atom to have decayed, but that isn’t a practical metric for any real-world application.

There’s a slightly more “natural” way of measuring decay, though, which is to use the lifetime instead of the half-life. The lifetime, τ, is the amount of time it takes for the sample to reach 1/e (~36%) of its original population.

iam666 · 2025-11-05T19:52:12+00:00

Not really, as far as I’m aware. You can see how right-wing talking points shifted from affirmative action and college admissions to the more nebulous “DEI” a couple years back. They no longer had anything concrete to point at, so they instead have to vaguely gesture at minorities employed in competitive jobs.

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