How to figure out the exact organic reaction mechanism?How any comp reacts with each other?
605888
There are many different ways of proving a chemical mechanism.  The methods can be different depending on how fast the reaction is, and the conditions of the reaction.\n\nI personally have used deuterium labelling studies.  This is where you prepare your substrate with a deuterium label in place of an hydrogen.  This product undergoes the chemical reaction and you identify in your product which hydrogen in your product has been replaced with a deuterium.  This then tells you how atoms are interacting.  You may need more than one experiment.  You can use 13C, 19F, 31P labelling etc also if necessary.  The trans-elimination of water from an alcohol to give and alkene was further proven using these techniques.\n\nOther methods are to use Mass Spectrometry to obtain the MS data on your intermediates.  By using accurate mass analysis you can determine the molecular formula of your intermediates and the postulate structures, based on experience.  This has been used for example to demonstrate some palladium coupling mechanisms, and proving that the oxidative addition process takes place.\n\nOther methods use NMR (nuclear agnetic resonance) data.  This is where you look at the environment for all H atoms in your molecule.  If your intermediates are stable enough you can see the environments change as the reaction progresses and from experience with NMR interpretation you can make postulates as to what the intermediates and hence mechanisms are.\n\nIn some cases the nature of the conditions you use for the reaction to be successful are enough of an indication as to what mechanism is going on.  \n\nBefore trying to prove a mechanism though you need some information.  You must know what your starting materials are, and you need to know what your product is.  So correct identification of the product is crucial.  NMR/MS/HPLC-MS/X-ray etc are all used to accomplish this, although you wouldn't need to use all the techniques for every molecule.\nYou must also have created a postulate of what the mechanism is that you are going to prove or disprove.  If you don't then you don't know where to look and which technique to use most effectively.\n\nI hope this helps.
940990
http://www.chem.purdue.edu/courses/chm333/enzyme_kinetics.swf\nnow be a good boy, study the link, and then explain it to  me....  :)
904652
CH3CH2C(CH3)3.\nI guess you forgot to type the C before the parenthesis. If you didn't forget, then none of the answers is correct (you would be missing a carbon atom).\n\nChem-man I would have contacted you in private about this but you do not give us the option to e-mail you. I have seen several of your answers prompting askers to verify your answer through echemistryhelp. I checked the website and it is not a free one; you post a question and pay for the answer. I'd say that this is a form of indirect advertisement  which is against the community guidelines. Please refrain from doing that or I will have to report you.
664298
molecular: \nHCl + K2CrO4 + SnCl2 --> CrCl3 + SnCl4 + KCl + H2O\n\nnet ionic: H+ + CrO42- + Sn2+ --> Cr3+ + H2O + Sn4+\n\nreduction half reaction: \n3 e- + 3 H+ + 5 HCl + K2CrO4 --> CrCl3 + 2 KCl + 4 H2O\noxidation half reaction:\n2 HCl + SnCl2 --> SnCl4 + 2 H+ + 2 e-\n\nBalancing them:\n6 e- + 6 H+ + 10 HCl + 2 K2CrO4 --> 2 CrCl3 + 4 KCl + 8 H2O\n6 HCl + 3 SnCl2 --> 3 SnCl4 + 6 H+ + 6 e-\n-----------------------------------------------------------\n6 H+ + 16 HCl + 2 K2CrO4 + 3 SnCl2 --> 2 CrCl3 + 4 KCl + 8 H2O + 3 SnCl4
475338
1) FIND THE REDOX-REACTION\n\nChromate (+VI) is reduced to 2 Cr(III) meaning this that 6 electrons are needed.\n\nThe corresponding oxidation on the alpha-C-atom in Ethanol goes from oxidation-level (-I) to oxidation-level (+III) in acetic acid, meaning it delivers 4 electrons per molecule.\n\nSo we find a stochiometric electron equilibrium if 12 electrons are exchanged:\n\n2 Cr2O7(2-) + 3 CH3CH2OH  =  4 Cr(3+) + 3 CH3COOH\n\n2) THE PROCESS MUST BE ELECTRICALLY NEUTRAL\n\nWe add the K-Ions having the left side of the equation neutral and the right side with 16 loads positive:\n\n2 K2Cr2O7 + 3 CH3CH2OH  = 4 Cr(3+) + 4 K(+) + 3 CH3COOH\n\nIn order to have the left side also with 16 loads positive, we add 16 protons, meaning this, that the reaction needs acid. Since we use sulfuric acid we also add 8 sulfat-ions on both sides and hereby can neutralize the product side:\n\n2 K2Cr2O7 + 8 H2SO4 + 3 CH3CH2OH  =\n                         2 Cr2(SO4)3 + 2 K2SO4 + 3 CH3COOH\n\n3) STOCHIOMETRIC BALANCE OF THE ELEMENTS H and O\n\nTaking into consideration that unused protons finally end up as water, we add on the right side 11 H2O and hereby is also reached the balance of oxigen:\n\n2 K2Cr2O7 + 8 H2SO4 + 3 CH3CH2OH  =\n                  2 Cr2(SO4)3 + 2 K2SO4 + 3 CH3COOH + 11 H2O\n\nSo now good luck for counting the atoms!
958445
http://en.wikipedia.org/wiki/Glycolysis\n\nJust scroll down and look at reaction number 8 for the mechanism.
799202
for those questions, all you have to do is translate them into the equation\ncheck chemtutor.com for hellp, it explains it much clearer than i can
522370
Well, when I was an organic chemist, I mostly mixed and poured.  So, what you do is, take something that's kinda like what you're making, add something that's very reactive (so toxic and/or explosive), mix, and heat.  Wash out the impurities, then either crystallize or distill-crystallization-kinda fun, distilling-monumentally boring(although those explode sometimes too).  Then analyze by NMR, TLC, HPLC,GC and IR.  Some people just do the analysis.\nThat's about it.  It was kinda fun, but it got old, and it was dangerous.  And scale varies depending on where you work.
605005
NaHCO3(s) + CH3COOH(aq) → CH3COONa(aq) + CO2(g) + H2O(l) \nThis is the real chemical reaction.
887756
Basic balancing of chemical reactions. You can always add CO2 O2 and H2O to either side. Start with the Carbon. balance the Oxygen last since it is contained in CO2 and H2O.\n\n1) 2 C6H6 + 15 O2 ==> 12 CO2 + 6 H2O\n2) C12H22011 + 12 O2 ==> 12 CO2 + 11 H20\n3) C3H6O + 4 O2 ==> 3 CO2 + 3 H2O\n4) C4H10O + 6 O2 ==> 4 CO2 + 5 H2O\n5) C4H10O + 6 O2 ==> 4CO2 + 5 H2O
867786
Let's take potassium representing group 1.  All the others will do the same thing.  Here is the reaction:\n\n2K(s) + 2HCl(aq) --> 2KCl(aq) + H2(g)\n\nThe reaction for group 2 will be similar.  Let's take calcium  representing group 2:\n\nCa(s) + 2HCl(aq) → CaCl2(aq) + H2(g)\n\nHope this helps.
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
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The site pointed to above does a good job explaining the mechanism. Just remember that with any unimolecular reaction, there must be a rate determining step.  Since it is always harder to form a carbocation, the ionization step will be your rate determinant.\n\nA few other pointers on carbocation formation:\n\nYou want an ionizing solution.  This basically means a polar protic solvent like water will do the job much better than a polar aprotic one like an ether.  Hydrogen-bonds will help stabilize the carbocation intermediate really well so you're much more likely to produce a C+ in H-bonding solutions.\n\nYou will also form the most substituted product.  So look at all the elimination possibilities and pick the one that's got the fewest H's on your SP2 Carbons.  And anytime a carbocation is involved, you gotta watch out for skeletal rearrangement.
750179
NaCl2 + NaCO3 ==> NaCO3 + NaCl2 NOTE: these two reactants will not react because the reactants and the products are identical\n\nCu + O2 ==> CuO2   This is already balenced\nCuSO3 + Fe ==> FeSO3 + Cu   The charge is not given so procuts could be different\n\n2HCl + 2Zn ==> 2ZnCl + H2\nMnO2 + H2O2 ==> No reaction: similar to NaCl2+NaCo3\n\nGood luck
253931
In the production of esters:\n\nR1-OH + R2COOH -> R2COOR1 + H2O\n\nin this case BuOH + EtCOOH -> EtCOOBu + H2O\n\nThis reaction can be catalysed by acidifying (H+(aq)) the reactants.\n\nAn electron pair on the hydroxyl group [of the alkanol (Butanol)] attacks the delta positive carbon of the carbonyl group [alkanoic acid (Propanoic acid)]. I'm sure you can figure out the rest!
445261
Try it yourself, its not that bad. \nHere is how:\n1) Set up a stoichometry ratio.\n2) Using the mass that they give you, calculate the theoretical yield of each product.\n3) Whichever product has a smaller theoretical yield, analyze the reactants which caused it to be such.\n4) The one that 'limits' the amount of products is the limiting reagent.\n5) Using that value, calcuate the amount of the other reactant that is necessary. Subtract from the orginal value to find the excess reactant.
836198
OK....You need to understand how this is all working, and why.  So, the formula equation for this reaction would be something like:\n\nKOH(aq) + HNO3(aq) --> H2O(l) + KNO3(aq)\n\nSince KOH, HNO3 and KNO3 are all strong electrolytes, they are totally ionized when they're dissolved in water.  So, written in complete ionic form, that equation would be:\n\nK+(aq) + OH-(aq) + H+(aq) + NO3-(aq) --> H2O(l) + K+(aq) + NO3-(aq)\n\nTo get to the net ionic equation, you just cross out the things that are the same on both sides of the arrow.  K+(aq) and NO3-(aq) are on both sides, so the net ionic equation is:\n\nH+(aq) + OH-(aq) --> H2O(l)