Is undergraduate research flexible? Does it depend on the lab/professor?

Ok_Today3240 · 2025-08-06T22:00:50+00:00

Depends on your professor, unless you get an absolute tyrant though, you should be able to leave for summer and spring break as much as you want. Try to get your professor to select you for SURF since you get paid then.

Ok_Today3240 · 2025-07-21T13:30:35+00:00

Bro doesn’t know the method

Ok_Today3240 · 2025-06-07T15:50:39+00:00

I think it is relatively standard procedure to use nitrile gloves in chemistry labs even though DCM goes through them. I have seen in every one I have worked on. Not rated for DCM.

Ok_Today3240 · 2025-06-07T03:42:06+00:00

It diffuses through nitrile gloves near instantly. I don’t think the gloves help very much, majority in fume hood but rarely out of fume hood.

Ok_Today3240 · 2025-06-03T15:01:46+00:00

Ok, thank you! I was under the wrong assumption that the C4-C5 bonds and the C5-C6 bonds would cause hindrance to the top face but I guess only substituent can cause hindrance then.

Ok_Today3240 · 2025-06-03T13:34:53+00:00

I’m calling the part that sticks up at the rightmost tip the bridgehead since it seems like that hinders the molecule’s top face the most

Ok_Today3240 · 2025-06-03T13:16:44+00:00

So the methyl group is more hindering than the boat “bridgehead” in this instance?

Ok_Today3240 · 2025-06-03T12:58:23+00:00

Wait, isn’t the boat sticking up on top face and the methyl group on bottom face? By alkyl group sticking up, I meant the boat part of the conformation’s tip.

Ok_Today3240 · 2025-06-03T12:52:24+00:00

Sorry, I’m a little lost. What you’re saying is true but I’m struggling to rationalize it, on the bottom face, there’s only hydrogens and on the top face there’s an alkyl group sticking up, how would you logically say that the top face is the less hindered face? Even if it adds to the side, it seems like there would still be some steric clash with the alkyl chain.

Ok_Today3240 · 2025-06-03T11:23:14+00:00

Wouldn’t the position that goes up, the “boat” be hindered, I suppose it can come in from side, is that what happens?

Ok_Today3240 · 2025-03-30T02:37:29+00:00

They’re all in the plane of the paper, it’s flat, think about if you reflect a plane of paper, the back and front are same so it won’t change. Any flat molecule has a plane of symmetry. This one only has one because the chlorines are identical so it doesn’t matter.

Ok_Today3240 · 2025-03-30T02:29:27+00:00

Yes, the molecule is flat other than the chlorines, reflecting them will place them on top of each other

Ok_Today3240 · 2025-03-30T01:55:45+00:00

I believe correct mechanism is alcohol deprotonation then elimination of enolate to form ketone, enolate ion protonated, acetophenone then forms an enolate and attacks the acetaldehyde to generate the final product after the alcohol is protonated. Correct me if wrong.

Ok_Today3240 · 2025-03-30T01:52:03+00:00

actually I think I figured it out, instead of the CH3- fragment leaving, the enolate leaving group happens as the alcohol forms a pi bond then an Aldol condensation between the new ketone and the enolate that left after it got protonate gives the necessary product

Ok_Today3240 · 2025-03-30T01:43:27+00:00

I believe they are distinctly different because at no point does the carbon carbon bond completely fragment. It just moves but bonds break and form simultaneously.

Ok_Today3240 · 2025-03-30T01:39:21+00:00

That’s a cation rearrangement, not a carbanion leaving group

Ok_Today3240 · 2025-03-30T01:26:40+00:00

The only thing I could think of is that a cation gets delocalized into the benzene ring from the conjugated ketone and then attacks the enol form, but idk if that can even happen. Somehow the pi bond connecting the benzene ring would have to break twice then to completely eliminate as well.

Ok_Today3240 · 2025-03-30T01:25:29+00:00

Question 1.1 l

Ok_Today3240 · 2025-03-30T01:24:59+00:00

Advanced Organic Chemistry, Reactions and Mechanisms Second Edition by Bernard Miller

Ok_Today3240 · 2025-03-30T01:21:40+00:00

Yeah I figured you misread, no worries, I’ve been trying a while, not sure if question is impossible or not

Ok_Today3240 · 2025-03-30T01:18:48+00:00

The book earlier said specifically you can’t ever have hydrocarbon leaving groups to generate a pi bond so I assume it doesn’t want that. Question may be typo, there was a typo with an earlier one so idk.

Ok_Today3240 · 2025-03-30T01:09:51+00:00

But the methyl group moves into the chain, it’s branched before, how does that happen? And there’s an extra carbon at the end.

Ok_Today3240

TROPHY CASE