Grignard Synthesis of 6-Bromo,2-methoxy Napthalene Via Magnesium turning?
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The Grignard Synthesis is usually used to convert a brominated ring compound to the corresponding phenol (Hydroxy group).\n\nThe Magnesium turnings provide metallic magnesium which inserts between the Bromine and the ring.  This is then hydrolyzed to the Hydroxy group.\n\nSee the site below for details:
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Diethyl Malonate looks like this:\n\nhttp://en.wikipedia.org/wiki/Diethyl_malonate\n\nWhat's strange is that 3-phenyl propanoic acid is quite simpler than that structure.\n\nMy first move would be to reduce those esters, because it's quite difficult to do a Grignard reaction without getting multiple additions.\n\nSo you react the ester with 2 equivalents of Lithium diisobutyl aluminium hydride (DIBALH), to yield 1,3 propan-di-al (malonaldehyde), a compound with an aldehyde at the end of each side of the propyl group.\n\nThen you react your product with 1 equivalent of phenyl magnesium bromide, with an acidic workup. That will add a phenyl group to one of the aldehydes, making it an alcohol. (the two are equivalent, but you will lose product because the Grignard may reduce both ... chromatography would be needed for this step)\n\nNext, oxidize the remaining aldehyde with Tollen's reagent, to get a carboxylic acid.\n\nThen, to remove the alcohol from the Grignard side of the aldehyde, you use the Barton reduction:(http://en.wikipedia.org/wiki/Barton-McCombie_deoxygenation)\n\nPhenylacetic acid would be hard to prepare in this way, because the carbon-carbon bonds would be difficult to break in the center of the molecule. Any ideas with anyone else?
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Grignard Synthesis of Benzoic Acid\n\n\nPreparation:  McMurry, sections 10.8 and 20.6.\n\nConversion of an organic halide into a Grignard reagent completely transforms the reactive character of the organic compound. Carbon bears partial positive charge when bonded to halogen but partial negative charge when the carbon is bonded to magnesium. The Grignard reagent is a very reactive nucleophile and base. \n\nIn this experiment you will make the Grignard reagent phenyl magnesium bromide; this reagent will then be treated with carbon dioxide to form benzoic acid. The sequence (Equation I) is explained in McMurry Sect 20.6.\n\n\nAs in any work, this experiment may not succeed the first time; be prepared to repeat it from the beginning. On the second try, you should be able to work much more efficiently and with more understanding. \n\n\nProcedure: Obtain a 150 mm piece of magnesium ribbon. Hold one end with forceps and with the other hand remove the residue on the ribbon by scraping it with a spatula, scissors or pliers to obtain a shiny surface. Cut at least 0.12 g of the shiny ribbon into 2-3 mm pieces (do not handle the pieces) and place on a folded piece of paper to obtain the mass on the balance. Transfer the magnesium pieces to the 5 mL conical vial, add the dry spin vane, add one small crystal of iodine and add 300 µL (0.3 mL) of anhydrous ether. Be prepared to add more ether as necessary to maintain the liquid level at a minimum of 300 µL.\n\nAssemble the rest of the apparatus as in Mayo, et al., Fig 3.24. Clamp the apparatus firmly to a stand.\n\nIn the hood, prepare a solution of 510 µL of bromobenzene in 700 µL of anhydrous ether in the 3 mL vial using the delivery pipettes provided. Cap this vial with a polyethylene top. \n\n(*Note:  Be certain the dry ice is ready before you initiate the reaction below.)\n\nDraw half of the bromobenzene solution into the syringe and insert the needle through the polyethylene cap in the Claisen head as in the figure. While stirring the mixture below, add 6-8 drops of the bromobenzene solution. Evidence of reaction is the formation of tiny bubbles. When the reaction has started, add the remainder of the bromobenzene solution slowly over 3-5 min. The temperature should be maintained at about 40 oC to allow gentle refluxing (Mayo, et al., p. 29) of the ether. To the 3 mL vial add an extra 300 µL of ether, then transfer this to the reaction via the syringe. Adjust the total volume to 3 mL with the addition of more anhydrous ether. Reflux the mixture for 10 minutes, then allow the gray-brown solution of the Grignard reagent to cool to room temperature. What do you observe?\n\nThe Dry Ice (at least 1.0 g) is added in small chunks to the Grignard solution. Note the reaction as this is added. To convert the C6H5C(=O)-OMgBr, to benzoic acid (last step, Equation I), use a pipet to VERY SLOWLY add 1 mL of 6 N HCl and again observe the result. Test the aqueous layer with pH paper; it should be acidic. Remove the ether by gently heating the vial for 5 minutes on the steam bath. Cap the vial, mark it and set it in your drawer until the next laboratory period. \n\nPreparation for the 2nd laboratory period:  The benzoic acid has been synthesized, but it must be purified. Water soluble magnesium salts are removed by extracting away the water layer; but, a number of organic side products remain with the benzoic acid.  These are side products of the Grignard reaction; small amounts of benzene are formed (Equation II), unreacted bromobenzene remains, and biphenyl, H5C6-C6H5, is formed from the reaction (Equation III). \n\nThe benzoic acid can be isolated from this mixture by taking advantage of its property as an acid. Only it can dissolve in aqueous NaOH (Equation IV) while the side products remain in the ether. After the ether-soluble side products have been removed, strong acid is again added to reverse this reaction in order to restore the benzoic acid.\n\nRecrystallization of the crude product in
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http://en.wikipedia.org/wiki/Reaction_mechanism\nhttp://en.wikipedia.org/wiki/Bromomethane
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Oleic acid ((Z)-9-octadecenoic acid) + Br2 ---> 9,10-dibromooctadecanoic acid \n\nOr: \n\nC18H34O2 + Br2 ----> C18H34O2Br2
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This might not be the simplest way possible, but here goes:\n\n(Starting with propanone)\n1. H3O+/I2 (to form 1-iodopropanone)\n2. (Ph)3P (triphenylphosphine forms the ylide)\n3. butanone* (Wittig reaction, now you should have hepta-3-ene-2,6-one)\n4. H2/Pd/C (hydrogenate the C=C; hopefully this'll leave the C=O's alone)\n5. H3O+ (intramolecular aldol condensation to close the ring, with dehydration to form the C=C double bond)\n\n*can be made from propylaldehyde and CH3MgBr\n\nNote: No guarantees this'll work as desired. :)
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ChemDraw from CambridgeSoft Corporation may be help you. Look at www.cambridgesoft.com
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Mg + 2 CH3COOH = Mg(CH3COO)2 + H2
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The easiest way to approach problems like these, my friend, is to look at your alkyl halide first and then look at your nucleophile or base.\n\nT-butyl bromide is a tertiary halide.  Tertiary halides undergo E1 and SN1 reactions because of their ability to form a stable carbocation intermediate.  If you recall, tertiary carbocations are more stable than primary and secondary.\n\nYour -OCH3 is a strong base known as methoxide, the conjugate base of methanol.  Strong bases favor elimination reactions and thus the elimination product is your answer.
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It might help if you spelled "naphthalene" correctly. \nhttp://www.madsci.org/posts/archives/may98/894550073.Gb.r.html \nThis website has the structural pictures you can use to do your Lewis dot structure.  \nNaphthalene is an aromatic compound.  It's two aromatic rings sharing a side.
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any chemical trading co. in Egypt has this chemical available. i mean if you know one company, they can provide it to you, or at least will tell you who sell it. I know one company in Alex. but i don't know addresses for sure.\nanyway, go to egyptian yellow pages and search by trade; chemicals, it will give you tons of companies that you can call and ask
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Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
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para-nitroacetophenone, in other words.\n\nMy first inclination is a Friedel-Craft acylation followed by a nitration with nitric acid in sulfuric acid (the nitronium ion). But the addition of the acetyl group would direct the nitro group to the meta position.\n\nMaybe if you nitrate the ring first and reduce the nitro group to an amino group. Then you could Friedel-craft to get the ketone at the para position (with some ortho isomerism). You'd have to find a way to oxidize the amino group to a nitro group ... I don't know how.\n\nAnother method would involve oxidizing toluene with KMn04, to get benzoic acid. You could then esterify the benzoic acid. Using DIBALH (Diisobutyllithium aluminium hydride), you could get an aldehyde (1-phenyl-methanal). You could use a Grignard reagent (methyl magnesium bromide) to get the alcohol with a methyl group added. You could then oxidize the alcohol with PCC to get a ketone. Finally, you could nitrate the ring with nitric acid and sulfuric acid.\n\nI'm not sure if either synthesis is perfect; My second one is long and rather expensive, in terms of reagents. Also, PCC is toxic (chromium) and many of the reducing agents are water sensitive.
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2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
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(NH4)2CO3+MgSO4->(NH4)2SO4+MgCO3
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This is a Double Replacement Rxn:\n\nMgCl2 (aq) + Li2CO3 (aq) --> MgCO3(s) + LiCl(aq)\n\nSince the reaction of two aq reactants yield solid MgCO3, it can also be considered Metathesis reaction.  Remember, Metathesis reaction is a double replacement reaction that forms Solid, Liquid or Gas.
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1\na) H2C = CH - CH2 - CH2 - CH = CH2\n ( 1,6-hexadiene )\n\nb) H2FC - CHF - CFH2\n ( 1,2,3-Trifluoropropane )\n\nc) H3C - CH2 - CH2 - CH2OH\n  ( 1-Butanol )\n\nd) H3C - CH2 - CH2 - CHO\n  ( Butanal )\n\n2.\na) Usually FeBr3 is used in addition to Br2 to brominate benzene, but for this:\n\nProduct: Bromobenzene\n\nb) \nProduct: Ethyl butanoate\n\nc) Dunno this one.\n\nd) \nProduct: 1,2-Dichloropropane