synthesis of fenfluramine by trifluoromethyl-benzene?
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Search http://www.scirus.com
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The trichloride is a little more complicated than borane (BH3).\n\nCheck out the following:
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Methcathinone can be synthesized by oxidation of (pseudo)ephedrine with (di)chromate salts, or via Jones reagent.Here is the synthesis via Jones reagent:\nOxidizing the Alcohol to a Ketone\n1.068 grams (just over 1 gram) CrO3 are added to 0.92ml H2SO4, and distilled water is added to bring the total volume to 4ml. 2 grams of pseudoephedrine are dissolved in 20ml acetone, and cooled in an ice bath. The Jones reagant is slowly added to the pseudoephedrine, dropwise and with stirring. The mixture becomes a reddish brown.Since we used just about the exact number of molecules required, and under very mild conditions, the reaction will proceed slowly, but very smoothly. Just let the ice melt and the stirred solution warm over night.\nExtracting the Ketone\nAbout 12-18 hours later the reaction is complete, the color is now a very dark green. We now have to neutarlize and extract our ketone. Add about 20mL of water, and chill in an ice bath. Dropwise and with stirring add about 1 gram of lye (Sodium Hydroxide - NaOH) in a solution with as little water as possible (highly concentrated). Check the pH regularly, and stop at pH=12. The now basic solution is shaken with 50ml of toluene. Allow to stand and two layers will form. The ketone will move into the toluene layer. Let the mix settle and remove the acetone/toluene/ketone and put it aside. Add another 50mL portion of toluene to the aqueous (water-based) portion, shake, filter (which breaks the emulsion and removes some chromium salts), and remove and put aside the toluene. Ensure that pH is still 12, and repeat again with another 50mL portion of toluene.Combine the three toluene extractions and evaporate them with a hair dryer. The result is the impure ketone of pseudoephedrine, also know as methcathinione or ephedrone.We used very gentle oxidation, neutralization, extraction, and evaporation conditions. Shortcuts can be used (no ice, evaporate with some heat, etc.), but good lab practice is never to stress the substrate needlessly. So we were careful.The raw "dirty" methcathinone weighs about 1.7 grams.It is purified by adding just enough tolene to redissolve it. Some particles will not dissolve, and are left in the drying dish. The toluene is filtered to remove any additional particles, poured back into the separation funnel and shaken with an equal amount of distilled water. This removes any small amount of water solubles. The toluene is put into a clean dish and dried once again. The impurities are more soluble in acetone than the ephedrone, so a very small amout of acetone is added, and sloshed around. Pour this off into the waste (original green mix). What's left is a little over 1 gram (about 60% yield) of relatively pure, slightly off-white methcathinone.\nBetter chemists can expect a much higher yield. This is a general procedure. The example given -Pseudoephedrine or Ephedrine to Methcatinone - is for illustrative purposes only. Oxidizing cold pills is illegal, and makes a lousy drug anyway
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I am not a chemistry student but I make this research for you and maybe it helps you http://www.chem.latech.edu/~deddy/chem104/104Antacid.htm\nhttp://www.chemcool.com/regents/acidsbasessalts/aim3.htm\n\njust copy those links and you should be on your way to solve ur problem.
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CH3CH2CH2CH3 + CL2 -hV--> \nCH3CH2(CL)CHCH3 + HCL ..\n\nwhat are the ways to get the CL at the end of the chain..so that\n\nCH3CH2CH2CH2CL + 2NA + CLCH2CH2CH2CH3\n--->CH3--------CH3\n\nThat's a radical reaction, the chlorination of an alkane with chlorine in UV light. The intermediate is a carbon radical and secondary radicals are more stable than primary. There are also more secondary carbons on this molecule. So probability and reaction kinetics dictates that the primary alkyl halide will be a minor product.\n\nI guess, if this were my synthesis, I'd:\n\n1. Put the glucose in HCl to cleave the anomers it likes to make.\n\n2. Esterify the carboxlic acid with whatever's available (Fisher esterification with MeOH, for example)\n\n3. Reduce the ester with Diisobutyl aluminium hydride (DIBALH)\n\n4. React the aldehyde with 1 equivalent of Ethyl Magnesium Bromide (Grignard)\n\n5. Reduce the hydroxy groups with a Barton-McCombie deoxygenation, although I've never seen it done in a case like this.\n\nThen again:\n\nI'd go to the stockroom and save my research advisor a lot of money. :)
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If your compound has the formula C6H10, then the molecular weight would be 82 grams/mole\n\nYou could have one of 13 different compounds -- \n2,3-dimethyl-1,3-butadiene\n3-3-dimethylbutyne\nCyclohexene\nany one of the five hexadienes\nany one of three hexynes\n4-methyl-1-pentyne\n4-methyl-2-pentyne\n\nWhat is the rest of the question regarding the 5.3*10^-23 odr?
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Hydrobenzamide (1) was made by reacting concentrated ammonia with benzaldehyde, had been discovered by a French scientist in 1836. When it is heated to about 130C it rearranges to amarine (2). Organic analysis at this time was very primitive and organic chemistry was still not very organised. Borodin (chemist)used different structural formulae to the ones we are used to using now. Nevertheless, Borodin correctly concluded from studying the action of ethyl iodide on the two compounds that hydrogen is attached to nitrogen in amarine but not in hydrobenzamide. Many years later he published a paper describing the decomposition of nitrosoamarine to lophine (3).
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bromocyclohexane with FeBr3 as a catalyst. Elimination reaction occurs only as a side reaction. If you want to eliminate hydroxyl group, uses POCl3,
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As you know, benzene is rather unreactive. Its conjugate pi system is entirely impervious to conventional nucleophilic and electrophilic attack. For example, you can react HBr with it forever, and you won't get Bromobenzene.\n\nWhat you have to do is to find a way to "pump up" the electrophile that is being used to attack the ring. When you run a bromination, you do:\n\nPh + Br2 ---> (w/ FeBr3 catalyst) PhBr\n\nAn Iron bromide catalyst works as a Lewis Acid to create Br+ in situ. This Br+ is electrophilic enough to promote the benzene's pi electrons to attack, forming the sigma complex and leading to bromobenzene.\n\nNitration works the same way. NO3 is not a strong enough nucleophile to promote the reaction. The H2SO4 is a stronger acid than NO3 and it protonates it, yielding NO2+ (HSO4 + H20 as spectators). This NO2+, called nitronium, is enough to cause the nitration of the benzene ring.
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The compound is probably made in a lab.  It probably has a starting reagent that looks similar to the product. The image from "the nothing"  shows the structure of the material.  The way that it work it prevents inflammation from occurring in the nasal passages by blocking certain biological processes from happening.   Basically it is an anti-inflammatory drug.
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Both CH3F and CF4 are tetrahedrons, but due to the substitution each has a different symmetry (C4v and Td respectively), as a rule the lower the symmetry the higher the polarity of the molecule. Actually, any Td (fully substituted tetrahedron), Oh (fully substituted octahedron), D4h (fully substituted square planar), D3h (fully substituted trigonal bi-pyramidal) will be non-polar molecules. This is due to the cancellation of the dipole moments.\n\nThus in order of the strongest dipole moment the answer is\n\nCH3F > CF4 (C3v < Td)\nXeO3 > XeO4 (C2v < D4h)\nSO2 > CO2 (C2v < D(inf)H)\nCOCl2 > BF3 (C2v < D3h)\n\nIf you are having a hard time understanding the Schoedinger group types, try a 3D modeling program (eg Spartan Build or ChemDraw 3D) to see the differences in the symmetry of the molecules.\n\nI hope that this helps. Please feel free to contact me if you have any further questions.\n\nKen
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Chemical laboratory supply houses carry it.  Try Alfa Aesar, they have it:\n\nhttp://www.alfa.com
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None.  The sell chemicals with alkyl groups.  In CH3Cl, CH3 is the alkyl.  They don't exist on their own.  Start with\n\nhttp://www.sigmaaldrich.com/Local/SA_Splash.html
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This might help :)\nhttp://www.chemicalforums.com/index.php?board=8;action=display;threadid=3263
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Below is a link to the MSDS (Material Safety Data Sheet) for this compound, which is actually 2,4-Dihydroxy-5-iodopyrimidine, but curiously it does not list the density.\n\nHave you looked in the Merck Index?
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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.
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try:\nhttp://lb.chemie.uni-hamburg.de/static/data/50_zzk46n8n.html