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hypervalent_iodine

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Everything posted by hypervalent_iodine

  1. This was my exact point when I had this discussion previously with Genecks. There is a logic behind all the curly arrows, and while there is a degree of 'you just have to know it' when getting through the foundations of o chem (as with any subject), the reactions and mechanistic stuff really shouldn't need to be rote learned for the most part. Indeed, you're going to find yourself in a world of hurt if that's the approach you choose to take with it. Edit: I believe this was the discussion. http://www.scienceforums.net/topic/54900-question-about-structure-of-carbonyl
  2. I don't think you get hypochlorite from the decomposition of ClO3. I believe your first one was correct, but I have not checked that it is balanced properly (you would want to balance via half equations).
  3. You could try dilute bleach solution followed by soap.
  4. Then I'm afraid you learned wrong.
  5. The most important thing is to have your general chemistry basics down pat. Stoicheometry has been mentioned, and this is obviously very important. I would add to this and say that you will also want to know about periodic trends (particularly as this applies to valence electrons, size, and electronegativity), very basics concepts of redox, intermolecular attractive forces, acids and bases (and the difference between strong and weak ones, as well as some of the more common examples of each category), the different types of chemical bonds, trends in bond strength, how to predict the outcomes of your more general reactions (acid plus base, acid plus carbonate, combustion, acid plus metal, etc.), and probably something else I am forgetting. These are all topics covered in high school chemistry to some degree, and in my experience is usually taught prior to o chem. By first year university o chem, almost all of it is assumed knowledge. First year o chem varies pretty wildly depending on where you go, but high school modules will generally begin with alkanes, alkenes, and alkynes, isomerism, their physical property trends, and nomenclature. This segues into more complex functional groups, their physical property trends and naming. Sometimes you will also see electronic configuration, quantum numbers, stereochemistry, hybridisation and molecular geometry thrown in the beginning. After that is usually reactions, electron pushing, and mechanisms. The exact reactions you do will vary, but the standards are things like alkene addition reactions, esterification / ester hydrolysis, oxidation and reduction of alcohols, and substitution reactions. I would not worry too much about leaping into that stuff until you have mastery of the other basics under your belt. Kahn academy is a great resource for this, as has been mentioned.
  6. Because that hydrogen is electronically equivalent to the other hydrogen attached to it and to the CH2 to the left. It won't couple to the CH2 side on the left due to this, but it will couple to the other CH2. This means n = 2, so you get a triplet.
  7. I believe so, it's just more often SN1.
  8. There are plenty of examples of this. Benzylic CH2's are great examples. You will sometimes see these called AB or ABX systems, but I wouldn't get too bogged down in the details of that as it can get very confusing. It's enough to know that diastereotopic hydrogens will quite often have different chemical shifts and therefore couple to one another. Structure three you seemed fine with? Your explanation was correct in your previous post.
  9. Your study group is wrong. Diastereotopic protons will have different chemical shifts due to their different environments. The two hydrogens in the CH2 of structure one are diastereotopic. If they have different chemical shifts, do you think they would couple to each other or not? What would n equal in this case?
  10. Yes, but it's benzylic. They happen to be quite stable due to the aromatic ring.
  11. Okay, there appears to be some miscomprehension going on, or you simply didn't read my post. I didn't mention this previously because I thought it would give you the chance to figure it out for yourself, but it appears I was wrong. I didn't say your answer was incorrect, I said your working out was, even if it gave you the right answer. Moreover, I did in fact explicitly state the answer for structure one in my post. Please reread my post carefully, and look up what diastereotopic hydrogens are.
  12. Given my above post, what do you think would be the correct answer?
  13. blueglass2 has been placed into the mod queue for 1 year, or until he starts making sense.
  14. pKa is related to how favoured the product of the acid dissociation is. In other words, how stable the conjugate base is. Think about how electronegativity might effect that.
  15. No. Could you perhaps make a little more sense? If you have your own questions not on topic, you might like to start your own thread.
  16. I'm glad you figured it out, but your working for structure one is off. The term n represents the total number of neighbouring protons. You do not do the calculation twice. By your logic, that would give n(total) = 0 and therefore only 1 peak, which has to be wrong based on the answers provided and given that structure two is split into 6 peaks. In fact, the coupling that gives the doublet in structure one is due to the other proton attached to the same carbon. These CH2 hydrogens appear to be something known as diastereotopic hydrogens. This occurs when the rotation around one or more of the single bonds is effectively locked in place, providing different electronic environments for the two protons on the one carbon. This is a well known phenomenon in benzylic systems, for example.
  17. Okay, then I would agree with your mechanism except that it would be SN1, not 2.
  18. Your picture hasn't attached. Could you try again?
  19. I think you are misunderstanding the question, or I am misunderstanding your issue. It doesn't matter how many peaks the t butyl gives. The question is asking about the splitting of the circled protons only. Side note: we are not here to do your homework for you, so being rude to those kind enough to offer assistance will do you no favours.
  20. Do you understand the difference between polar bonds and polar molecules (specifically, what causes a molecule with polar bonds to be non polar)?
  21. AbstractDreamer, It is not that simple if you are not just working with strong acids and bases, as is the case here since there is carbonate present. The equivalence point in this case is not pH 7, so you cannot use that math to work out a volume.
  22. I don't have time to look through the bulk of your post, but you might be interested in dimethyldioxirane (DMDO). It's made from acetone, NaHCO3 and Oxone in water. The mechanism isn't too complicated.
  23. It can be a little arbitrary I guess. We consider it from the point of view of the main reaction centre. In carbonyl additions, the carbonyl carbon is an electrophilic and the main reaction centre. It's only possible for a nucleophile to attack this carbon, therefore it is nucleophilic attack. In normal alkene addition, the alkene will act as the nucleophile, but it is the main reaction centre. Thus the thing adding to it is the electrophilic, hence electrophilic addition. While this is the normal case, it can be complicated by the presence of electron withdrawing groups attached to double bonds. Aromatic systems are also different. In summary: it depends on what the main reactant is acting as. If it is the nucleophile, it's electrophilic addition, and if it's acting as an electrophile, it's nucleophilic addition.
  24. Depending on how long ago you're talking, I doubt it was carbon tet. That stuff is next to impossible to get a hold of these days.
  25. I believe it would depend on the species of algae. I used to work with some chlorella species, which I think we measured in the high 600 nm range. You should be able to find literature that specifies a wavelength. However, for the best results using this method you will want to make a standard curve that measures actual cell count against absorbance, which would require a hemocytometer and light microscope.
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