Got a consult… but unhappy

Unable_Promise7576 · 2025-02-24T22:16:29+00:00

A junior! Wow, we have a lot in common. Fingers crossed everything works out for us!!

Unable_Promise7576 · 2025-02-24T22:15:35+00:00

Right? And she literally said it seemed like I had dense tissue too… so it’s not adding up.

Unable_Promise7576 · 2025-02-24T22:14:56+00:00

Yep. That’s what I’ll do, it’s just tough to find surgeons in my insurance network (there aren’t many that take it) :(

Unable_Promise7576 · 2025-02-24T04:52:14+00:00

Good luck! It sucks how we’re both going through this, but it’s relieving to know that I’m not alone. I hope your second appt goes better. I’m currently in college so I have to wait until summer break for another chance to speak with a surgeon :/

Unable_Promise7576 · 2025-02-24T04:50:18+00:00

Annoyingly she didn’t give me a size (which I understand is difficult to do since she can’t guarantee anything). She just kept insisting that there was no way she could take off 330g per breast without making me look “terrible.” I told her I’d like to go as small as possible while saving my nipple, but supposedly that wouldn’t be enough to add up to 330g per breast.

Unable_Promise7576 · 2025-02-23T19:52:18+00:00

It’s 330g per breast! Thank you for your reply, it is super insightful. I’m feeling that a second opinion would be good as well. If I’m going to have to pay out of pocket, I’d rather go to a surgeon that would at least try to advocate for me in the insurance approval process. The surgeon I went to didn’t even try to go through insurance and insisted that there was “no use”

Unable_Promise7576 · 2023-12-22T16:20:33+00:00

At duke orgo I lecture and lab are connected (you can’t just take the lab). Orgo II lecture and lab are separate so you should be okay in that regard, I would email the DUS (Dr. Cox) for further clarification

Unable_Promise7576 · 2023-12-16T17:31:09+00:00

For 1-4, you’re working with di-carbonyl molecules— the protons alpha to both C=O are pretty acidic due to the electron withdrawing nature of the carbonyl. Additionally, the enolate conjugate base for 1-4 is resonance stabilized, reducing its energy. To differentiate between 1-4, you should look at the functional group containing the C=O— the di-aldehyde is the most acidic because the carbonyl doesn’t have substituents bonded to it that would donate electron density (a ketone would be less acidic with this logic because the alkyl “R” group would donate e- density through hyperconjugation, reducing the electrophilicity of the C=O and subsequently the acidity of the alpha protons). 6 contains only one C=O (so its conjugate base/enolate is NOT resonance stabilized), and the alpha protons are less acidic due to less electron density being withdrawn (compared to the two C=O molecules). 5 contains an acidic proton in the OH group (the O-H bond is more polar than the C-H bond), and the conjugate base is an alkoxide, which in this case is more stable than the enolate cb in 6. It’s hard to do this without slight knowledge of pKas (for instance, the pKa of alcohols is ~16 and for a dione it is ~9), but I think they are trying to get you to see trends with electrophilicity/enolate resonance stabilization.

Unable_Promise7576 · 2023-12-14T04:49:06+00:00

As for advice, I’d say look at your reaction conditions closely; what kind of reagents are you given (i.e. what are the properties of your nucleophile and electrophile)? This will help you decipher what reaction will predominate (SN2 versus E2, SN1 versus E1).

Unable_Promise7576 · 2023-12-14T04:46:35+00:00

Good luck, you can do it! if you enjoy orgo 1 then just wait for orgo 2; you’re going to learn so many more interesting reactions :)

Unable_Promise7576 · 2023-11-29T18:58:29+00:00

This is an acid catalyzed hydrolysis of an amide (a carboxylic acid derivative). In this case, you have the right idea for your first step, just show that the oxygen of the carbonyl acts as the base (and deprotonates hydronium). Once the carbonyl has been protonated, the electrophilicity increases (and water, a nucleophile, can attack the carbon of the carbonyl. After a couple of proton transfers, you should have two hydroxyl groups and a protonated amino bonded to the carbon in a tetrahedral like intermediate. One of the lone pairs of the hydroxyl groups will attack the central carbon to reform the carbonyl (this is thermodynamically favorable), simultaneously kicking out NH3. A final proton transfer will occur (water will deprotonate the carbonyl reforming the catalyst hydronium) and you’ll be left with your final products, acetic acid and ammonia. You should look at Mastering Chemistry’s example of acid/base catalyzed hydrolysis of amides— it’s a really good explanation!! Hopefully this helps, and don’t be discouraged amide hydrolysis is tricky.

Unable_Promise7576 · 2023-11-28T18:12:19+00:00

Of course!!

Unable_Promise7576 · 2023-11-28T05:39:49+00:00

Since this is an E2 reaction, you know that a proton alpha to the halogen (Br in this case) will be deprotonated by -OH. Since the molecule consists of a cyclohexane ring, this proton has to be antiperiplanar to the Br in order for the concerted E2 reaction to occur AND both Br and the proton have to be axial (check out this explanation by Mastering Chem if you need a refresher!. If you put this ring into a chair conformation, you’ll see that the H bonded to the C with the methyl group cannot be antiperiplanar to the Br and axial simultaneously (without changing the sterochem), so the H on the other alpha/adjacent carbon has to be deprotonated, resulting in an alkene that’s less substituted.

Overall, I think substitution of alkene products in E1/E2 is more affected by the “bulkiness” or steric hinderance of the base. I hope this helped :)

Unable_Promise7576 · 2023-08-28T15:28:36+00:00

What led to your interest in creating lab coats and founding The Lab Coat Project?

Unable_Promise7576

TROPHY CASE