Orgo Madness

<p>I have an orgo 2 final in four days. I mastered most of the one-step syntheses covered. I'm attempting to do as many synthesis problems as I can. But I'll only be able to try small number of questions because every single one of them takes up so much time and my time is very limited because I have to prepare for two other finals on the same day. :( </p>

<p>Synthesis problems involving more than 4 steps just seem impossible to do! Problems with less than those number of steps is okay but still hard..:(
I use retrosynthetic analysis but this method doesn't always help.
It's hard to know if my answers are right because textbook answers present only one combination of steps even though there are many possible correct combinations of different one-step synthesis.
Any tips on beating synthesis problems?</p>

<p>Count your carbon skelatons. Have you tracked that?
Know and track your carbon to carbon rxns.</p>

<p>That’s one of the essential things I’m doing, but that doesn’t seem to be very helpful. Thanks for the tip, though.</p>

<p>What textbook are you using?</p>

<p>I always work backwards (retrosynthetic analysis); often times, there are multiple ways to even go about doing this (ie, final product can be made from a number of different compounds, which in turn can be made from a whole mess of other compounds), but one of those paths should lead to the right answer. Hopefully you’ll be able to see when you come up with the best synthesis.</p>

<p>Review all the old reactions, right from the beginning. What I do is make a note card for each new reaction that I come across, along with pertinent information (reagents needed, limitations, etc). At this point, I remember most of them, but if I ever forget one, there’s a stack of note cards I can go to for any reaction.</p>

<p>At this point, there are so many reactions, that I try to remember general concepts rather than specific reactions. For example, I’ll say “hm… I could make this product by combining an ester and an alcohol”, or “I could make this compound from a ketone and an amine… do I know of any reactions that allow this?”, and THEN I’ll try to think of the reaction/reagents necessary to perform it, rather than the other way around (ie, trying to go through every reaction in your head until you stumble on the one needed).</p>

<p>Other than that, you just need to practice, I guess.</p>

<p>this thread makes me glad I never have to think about organic again…</p>

<p>That’s actually a question that I’ve wondered. Am I going to use orgo in medical school, or is it just required for getting in? I can’t quite decide which alternative is worse. I’d like to never think about organic again, but that’d make it all the worse that I had to do it at all.</p>

<p>Well I suppose you’ll be encountering biological (organic) molecules and stuff a lot in med school, but I cannot confirm seeing as how I am not in med school.</p>

<p>Three days are left until the final on Thursday. What are the best things to do if you have three days remaining? The two main things I’m doing right now are doing synthesis problems and constantly reviewing all the reactions covered.</p>

<p>You will see organic molecules in med school, residency and, yes, in practice.</p>

<p>Every year pharmaceutical companies introduce hundreds of “new” and “improved” drugs for clinical use. Clinicians need to able to understand pharmacology and therefore organic chemistry to make informed decisions.</p>

<p>Consider these two drugs:</p>

<p>5-methoxy-2-[(4-methoxy-3,5-dimethyl-pyridin-2-yl)methylsulfinyl]-3H-benzimidazole.
and
(S)-5-methoxy-2-[(4-methoxy-3,5-dimethylpyridin-2-yl) methylsulfinyl]-3H-benzoimidazole.
</p>

<p>One is a racemate and the other is the S enantiomer.</p>

<p>Despite no clear benefit from the S enantiomer preparation, Astra-Zeneca has pushed this drug (esomeprazole) over its older sibling (omeprazole) to achieve annual sales over five billion dollars.</p>

<p>perhaps it’s because of the way my organic prof hated pre-meds, but I consider “organic” chem much more like what the OP has been complaining about - crazy multi-step synthesis problems or detailed C14 NMR interpretations and the like - things that aren’t close to pharmacology as I’ve learned it. Biochem as taught at my undergrad and as a first year medical student corresponds much more closely to the situation that 2 pennies describes. Even then, the important thing - what differentiates medicine from basic science - is the application…you don’t have to be able to draw the molecule from the name to understand the evidence-based medicine results. One could even argue that understanding basic patent law could better explain the difference between esomeparzole and omeprazole (and that’s often the exact method I use when talking to patients about it)…</p>

<p>Yeah, it’s the incredibly long mechanisms and the spectroscopy (omg, the spectroscopy) that I’d love to leave behind for good. I’m okay with most of the theory.</p>

<p>I also have my ochem 2 final coming up, in a week!</p>

<p>I’m just doing hundreds of practice problems. My school (umich) provides a ‘courespack,’ which basically has every question from every test for the past few years.</p>

<p>CCRunner brought up a good point. Definitely buy the course pack if there is one. My chem department doesn’t make them, but the math department does, and they’re definitely a good way to study for exams. Final exams (especially departmental ones) don’t have a whole lot of variation from year to year, so if you’ve got a school-endorsed sample of an old test, it’s definitely worth the money.</p>

<p>Happily, long mechanisms, spectroscopy and NMR are unneeded for clinical practice. Most discussions of drug metabolism are limited to single operations and most structural discussions are limited to recognizing functional groups or differences in lipophilicity.</p>

<p>Unfortunately no such thing exists at my school’s chem dept (though there are tons for math, biochem, etc.) and this is the first time that my instructor is teaching this course so there are no past exams to look at. </p>

<p>Two days are left until the exam. I’m scared to death! :(</p>

<p>Aw, why the hate for ochem? :(</p>

<p>If you need help dont hesitate to PM me, especially with the reactions that are troubling you the most.</p>

<p>piccolojunior, in case you didn’t check my PM:</p>

<p>Why do you use NaBH(OAc)3 instead of NaBH3 in reducing an iminium ion? Doesn’t NaBH3 work as well?
Also, how is it possible to convert an amide into carboxylic acid by using proton and water? Isn’t it that you can only convert a molecule into something that’s less reactive than the precursor? Clearly, a carboxylic acid is more reactive than an amide. It seems strange to me how such a conversion could happen.</p>

<p>Thanks!</p>

<p>Whew, I just had my orgo final. It was hard, but I’m sure I got almost all the questions right except the last two synthesis problems, which were impossible. Anyway, I’m glad this course is over. I hope I’d get an A.</p>

<p>One question in the final asked which of 4-nitrophenol or 2-nitrophenol is more polar. Then the question asked you to write down the structure of the predicted product (tylenol) when you mix a methanoyl chloride with whichever you decided as more polar. I put 2-nitrophenol was more polar, and so my product ended up as 2-hydroxyacetanilide. But when I looked up the structure of tylenol the acetanilide group was actually para to the phenol!</p>

<p>Does anyone know how 4-nitrophenol could be more polar than 2-nitrophenol?</p>

<p>See the discussion “Reactivity and Orientation in Electrophilic Aromatic Substitution” at <a href=“http://people.uis.edu/gtram1/organic/aromatics/reactivityAndOrientation.htm[/url]”>http://people.uis.edu/gtram1/organic/aromatics/reactivityAndOrientation.htm&lt;/a&gt;

</p>

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<p>If the hydroxyl group bears a partial positive charge due to electron donation to the ring and the nitro group bears a partial negative charge due to electron withdrawal from the ring, positioning the two functional groups as far apart as possible (para) would maximize the dipole moment and polarity of the molecule.</p>