Can anyone plz help me i need to find the mechanism for the preparation of a grignard's reagent.?
151615
you can find it in most college organic chemistry textbooks. A couple of my favorites are McMurry or Solomons.\nGrignard reagents are also called organometallic compounds where you take an organic compound(usually an alkyl halide) and react it with a metal, most of the time we use magnesium.\nso for example you could combine bromobenzene reacts with magnesium metal(using an organic solvent like ether) to produce phenylmagnesium bromide.\nfor more look at those textbooks. you'll find lots more.
104161
You could rig a "Dean-Stark Trap" on the side of your condensor. That will eliminate water from your esterification (assuming its similar to the Fisher esterification) and force the production of more water and product, until one of the reagents is used up or an equilibrium is achieved.\n\nAnother method would be TLC, comparing the starting reaction mixture to the final product.\n\nSpectroscopy (NMR, IR) could be used.\n\nFinally, you could just put some of your sample into a GC-MS.
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.
230380
um i don't know if this site has it, but you might want to try chemguide.co.uk :) it's a pretty good site.
949367
First you have to get the molecular weight of those.\n\nits\n180 for C6H12O6\n192 for 6O2\n264 for 6CO2\n108 for H2O\n\na. Since you have the same amount of C6H12O6 and O2, your limiting reagent in this case is the O2.\n\nSo you use it to determine the CO2 produced.\n\nb. if 192 grs of O2 produce 264 grs of CO2\nthen 100 grs will do: (100 * 264)/180 = 137.5 grs of CO2.\n\nc. Now, to calculate the percent yield, in ideal conditions you would have gotten 137.5 grs (which is your 100% theorically) but practically you got 97.1 grs. so\n\n137.5 ___ 100%\n97.1______ ?\n\n(97.1*100)/137.5 = 70.62% is the perecnt yield of CO2.\n\nd. You do the same thing backwards to determine how many excess of the non-limiting reagent (C6H12O6) you have.\n\nto produce 264 grs of CO2 you need 180 grs of C6H12O6\nfor 137.5...\n\n(137.5*180)/264 = 93.75 grs of C6H12O6\n\nso the excess is 100 - 93.75 = 6.25 grs of C6H12O6
644051
See if this article has a diagram\n\nhttp://jchemed.chem.wisc.edu/Journal/Issues/1999/Apr/abs542.html
1187807
a phd student in chem looking for basic chem material on the web.\n\nright.\n\nstart here: http://www.carleton.ca/~rburk/chem1000\n\na textbook might also be useful.
749646
Try doing a Google search under the Hall or the Hall-Heroult process.    That or check out a General or an Inorganic Chemistry book from the library.  That's a good launch point.
115234
It's going to be hard to show a mechanism using this website, so I suggest you look up "esterification" and learn the mechanism.
667881
Try searches at the following.  You may need to hit the library.  YOu might try a general web search.\n\nhttp://chemfinder.cambridgesoft.com/\n\nhttp://www.scirus.com/srsapp/\n\nhttp://www.scienceresearch.com/search/
292071
What class was this in? What grade level? What text were you using?  I may be able to help you if I know the context in which the question was raised. I have a feeling that I am not getting all the information you had. Please tell me and maybe I can help. I am an old chemist and microbiologist.
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!
42193
http://www.itt.edu/staff/boyter/links/index.html\n\nLook under chemistry equipment.  There are some chemicals you will not be able to get.  Posting what you are looking for would help.
84780
!!!!!!!!!!!!!!!! \nI think rather u wanted to ask how can u predict the formula of a chemical reaction!!\nFirstly, we have to determine the moles of reactant taking part.(It can be done easily, think A reacts with B, just mix 1 mole of A with the same amount of B.)\nThen determine what are the different products and their amounts.\nAll u have to do is some calculations..................\nRemember its not as easy as it sounds.
239993
Links at\n\nhttp://www.itt.edu/staff/boyter/links/index.html#CHEMISTRY_CHROMATOGRAPHY
1165218
you appear to be confused, when a grignard reagent reacts with a carbonyl carbon (such as in RCOOH) the R of RMrBr (the most common grig) is bonded to the carbon. however when the electrons move up the oxygen to form RCO- OH, the remaining MrBr+ ion interacts with this O- forming a salt, i believe this reacts further, eliminating the OH. hope this was clear enough.
518616
Check this out....\n\nhttp://www.sciencestuff.com/prod/Chem-Rgnts/C2043