hypervalent_iodine

Specifically what natural means? Could you provide citations?

What exactly are you defining as artificial and natural?

I knew of a guy who once attempted fire breathing while drunk, having not done it before. Instead of spitting out the fuel, he inhaled it and burnt the inside of his lungs quite badly. A good lesson in what not to do when intoxicated.

For inspiration and new ideas it's always been the shower. It's particularly good if I'm up all night writing and am stuck with something. For more general things, I find writing it out to be very clarifying. I think it's a very personal thing, though. By the range of responses here, what works for you may be quite different to what works for another. Getting away and finding a place where you can be introspective seems to be the common theme, however.

! Moderator Note I removed it because it was completely off topic. Start a new thread for your questions.

That's not quite the question being asked. Net and total ionic equations are a step after predicting the reaction itself. You have to determine whether something is a molecule or an ion. It's a little more convoluted for acids and bases. Generally, strong acids are written as the ions because they dissociate fully, whereas weak acids are written as the molecule. H2SO3 is not the same as the sulfite ion, hence it does not cancel.

! Moderator Note This appears to be a homework problem. I have moved the thread. Tink_75, we do not do people's homework for them. Please detail where you are stuck and what your thoughts are so far. Fiveworlds, I removed your post. Since this seems to be homework, you should not be giving out answers.

! Moderator Note And with that, this thread is closed. Overtone, you are not to reintroduce this topic.

Well yes,I figured that much. What type of calculations?

Given that NaCl forms colourless solutions, I'd say it's around 0. Why would you need to know that?

Your carbocation rearrangement appears to be halfway there. If you used HCl as your source of H+, you'd end up with a Cl in the benzylic position. I can think of one fairly straight forward way to get from there to your product.

Well, it's not highly unstable, just less stable than the benzylic position. The plot looks fine otherwise. Personally, I would draw the lines so flat at the product and reactant ends and I'd make the ascent and descent more sloped, but this is a very minor point and one you'd not get marked down for ignoring.

Small note about your first question: your equatorial bonds almost look axial. Either draw them a bit more angled out from the ring or draw in the hydrogens to make it a bit clearer. Your second one is also right. Yes, the t-Bu group locks the ring, and reaction requires backside attack of the brominated carbon from the deprotonated OH. In 1, the OH is axial and can easily attack from above this position. In 2, the OH is, as you have correctly drawn, equatorial. It is repelled by the axial hydrogen from top-down attack and it doesn't have the flexibility or rotational freedom to move up to where it needs to be.

For the third one, the HCl acts to protonate the alcohol to generate a water leaving group. The oxygen on the methanol is the nucleophile. Note that this is not the same as -OMe. You don't generate the alkoxide in this (it's an acidic solution, so you shouldn't have negative charges floating about). For number 4, I'd say it's more to do with steric repulsion.

! Moderator Note Just FTR: the post that iNow responded to in the reply directly before this was removed, as the poster was a spam bot. This was the post:

! Moderator Note Did you have something you wanted to discuss?

That's because you appear to be using mass to determine how much product is formed as opposed to moles. The ratios out the front of the chemical species are mole ratios, not mass ratios. Convert everything to moles work out the limiting reagent first. Edit: I assume you also used mass to determine limiting reagent? This is also incorrect. Remember that when we talk about things reacting, we're talking about individual molecules colliding to form product. Mass =/= molecules (1 gram of Cl2 will not have the same number of molecules as 1 gram of H2O, for instance) and so it makes no sense to use mass to determine something like a limiting reagent in the way that you have, primarily because the ratios in the reaction are (as mentioned) mole ratios.

! Moderator Note This seems to be homework. I have moved the thread to the appropriate forum. Please note that we try not to do other people's homework for them. Please show your working and detail exactly where you need help.

! Moderator Note No. I said you could ask genuine questions, not reintroduce your hypothesis. Thread closed.

! Moderator Note Indeed. Thread moved to the trash.

! Moderator Note No, what you need is to learn some basic physics. Thread closed. If you have specific and real questions, please ask them in a new thread. You are not, however, permitted to reintroduce this particular nonsense.

! Moderator Note You haven't asked the community anything. All you've managed to do is write a few barely comprehensible sentences as a statement of fact, with accompanying links that do absolutely nothing to support whatever it is you're trying to say. Please outline, in detail, what you are proposing and the science behind it. Failure to do this is a breach of the forum rules and this thread will be closed as a result.

Why does it need to be coloured? UV-Vis works outside the visible spectrum as well. Unless there's some other reason for needing the products to be coloured, all you really need is something you can easily make and that absorbs differently to the starting material. A quick search of "Diels Alder reaction rates UV-Vis" in Google provides a few examples you could try, but some of these may be precluded depending on what context you're doing it in. For instance, it is unlikely that you would be able to play with anything especially toxic or potentially explosive if you were doing this for an undergrad course. Edit: alternatively (and this may be easier), you could look for a reaction where the diene absorbs in the UV or visible region, where the product does not (or at least, does not absorb in the same region).

! Moderator Note overtone, My initial reaction to seeing this thread was to close it. You have had numerous chances to have genuine discussion about GMO's in almost as many threads and every time is has come back to the same lines and the same problems. Indeed, you were told not to reintroduce these topics. However, I thought I would give you the benefit of the doubt since you (somewhat surprisingly) posted links to back up your claims. Alas, not even one response in and we are back to what staff have come to expect of you when discussing GMO's. Your attitude here leaves much to be desired. Acting hostile and bizarrely smug, implying that readers and / or their posts are ignorant or stupid (or both) is no way to go about things if having intelligent debate is what you are after. If it isn't, that's too bad, because it's what this forum attempts to foster. I'm giving you one final chance to keep this thread going. Drop the snarky tone and back up your claims. If you can't or won't do that, this gets closed. Do not respond to this note within the thread. Please report the post or PM staff if you take issue with it.

A small point: it's pKa, not Pka. The letters and their capitalisation (or lack thereof) have specific meanings. K is for equilibrium, k is used to represent reaction rates and p means -log10 of whatever it is to the left of. A pKa of 9.5 is not low. Ammonia (NH3) is a weak base and its conjugate acid, ammonium (NH4+), a weak acid. Whether the equilibrium outlined goes towards the right or left depends on the pH of the solution. Understanding why it goes to the left can be explained in a few ways, but it is first necessary to understand what is meant by pKa in reference to pH. At the point where pH = pKa, exactly half of the acid has dissociated into its conjugate base. When the pH is lower than the pKa, it means that there is more H+ on the product side and so the reaction is pushed towards the left (according to Le Chatilier's principle). Similarly, when the pH of the solution is above the pKa, there is less H+ and so the reaction will go to the right. Since the OP referenced soil, we can probably assume the pH is lower than 9.5. Thus, the reaction would be pushed to the left and you have more of the ammonium ion.