what is the role of each, FeSO4, nitrobenzene & boric acid in the last step of Skraup quinoline synthesis?
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The Iron (II) sulfate seems to be an Oxygen carrier.  It tends to prolong the reaction time.\n\nThe nitrobenzene is used as an oxidizing reagent.\n\nThe Boric acid is a reaction moderator.  It tends to keep the reaction slow and under control.
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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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Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
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There are many you could choose.  These may not be the most efficient, but I'm just going with initial thoughts here.\n\n1.  Aniline -> benzamide\nDiazotise the aniline and trap out with CO2 to form benzoic acid, then make an amide.  Probably easiest way is to make the acid chloride with (COCl)2 and then treat with NH3.\n\n2. Bromobenzene -> 1-phenylethanol\nForm phenylmagnesium bromide by treating bromobenzene with magnesium in diethyl ether, and react with ethanal.\n\n3. Butanal -> 1-chloropropane\nMany ways to do this, none of them nice.\n\nReduce butanal to butanol, de-hydrate, perform a reductive oxonolysis to give 1-propanol and then do a chloro-dehydroxylation with SOCl2 or something like that.\n\nAlterantively you could oxidise and do a Barton-Crich de-carboxylation and trap out the radical with NCS, although this method may not be too successful.\n\nHope these quick ideas help.
854318
One needs toluene and concentrated nitric and sulfuric acids, plus "fuming" sulfuric acid (H2SO4 + SO3). As the reaction proceeds, the first nitro groups usually go on either in the position next to the -CH3 (2-, ortho) or diametrically opposite (4-, para). All three mostly end up in the 2-, 4-, and 6-positions. The operator nitrates in three stages, with the temperature raised gradually during the three stages from 35degC to 110degC. However, there is also formation of small amounts of the 2,3,4- and 2,4,5-trinitro compounds. These unsymmetrical isomers make the product dangerously shock-sensitive to handle. The crude product is therefore stirred with 15% aqueous sodium sulfite solution at 100degC. This treatment converts the two impurities to sulfonic acids, which are water soluble and poured away from the oily product in the water.
729557
_NaOH(aq) + _Fe(ClO4)3(aq) => _Fe(OH)3(s) + _NaClO4(aq)\n\n3  NaOH(aq) + 1  Fe(ClO4)3(aq) => 1  Fe(OH)3(s) + 3  NaClO4(aq)\nThe coefficients of one could be excluded but I put one which is also equivalent.\n\n_((NH4)2)SO4(aq) + _Pb(NO3)2(aq) => _PbSO4(s) + _NH4NO3(aq)\n\n1  ((NH4)2)SO4(aq) + 1  Pb(NO3)2(aq) => 1  PbSO4(s) + 2  NH4NO3(aq)\n\nThese are only the balanced equations.\n\nThe following is the ionic equations:\n(Na)+ (aq) + (OH)- (aq) + (Fe)3+ (aq) + (ClO4)- (aq) => Fe(OH)3(s) + (Na)+ (aq) + (ClO4)- (aq)\nand\n(NH4)+ (aq) + (SO4)2- (aq) + (Pb)2+ (aq) + (NO3)- (aq) => PbSO4(s) + (NH4)+ (aq) + (NO3)- (aq)\n\nThe ionic equations are already balanced.  It is simply the breaking down of the equations into ions.  The solid remains intact while the compounds in aqueous solution become ions.\n\nThe + and - signs are the charge of the ions.  I can't find any other way to write them so I separated the compound with parentheses.  Sorry if it's confusing.\n\nThe net ionic equations are "simplified" ionic equations.  Basically, the things that are repeated on both sides get crossed off.  What is remaining is simply the net ionic equation.\n\n(OH)- (aq) + (Fe)3+ (aq) => Fe(OH)3(s)\nand\n(SO4)2- (aq) + (Pb)2+ (aq) => PbSO4(s)
15429
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).
844624
FeCl3   +    3NaOH  ----->    Fe(OH)3   +   3NaCl\n\nthats it
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FeCl3 + NaCl -> FeCl2 + NaCl3\n\n\nFeCl3 + NaCl -> FeCl2 + NaCl3\n\nthese reactions are not possible \nas this reaction can take place if iron(fe) was more reactive than sodium or fe was less electropositive than na\nhence \nthese reactions are not possible
561728
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
744530
It is an electrophilic aromatic substitution reaction.  Because the -COO- acts as an electron withdrawing group to one phenyl group and a moderate electron donating group to the other pheny group the nitration will occur on the pheny group where the COO- is acting as an electron donating group.  The major product of the nitration will have the -NO2 in the para position because the ortho position is partially blocked by the large COPh group.\n\nI hope this helps!
845828
Here you go...\n\nhttp://home.planet.nl/~skok/techniques/laboratory/pka_pkb.html
370859
Presuming that bromine is getting reduced to Br(-) and bisulfite is getting oxidized to sulfate (not including the spectator sodium ion):\n\nReduction half-reaction\nBr2 + 2e ---->  2Br(-)\n\nOxidation half reaction\nHSO3(-) + H2O ----->  SO4(2-) + 2e + 3H(+)\n\nOverall Reaction:\nBr2 + HSO3(-) + H2O ---->  SO4(2-) + 2Br(-) + 3H(+)
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Chemisorption of propionic acid vapors by an iron impregnated MCM-41 porous silica leads to the formation of iron-propionate.\n\nSource is from American Chemical Society Journal.
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Cu2+ (aq)  +  Zn (s) -->  Cu (s) + Zn2+ (aq)
904363
(NH4)2CO3+MgSO4->(NH4)2SO4+MgCO3
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Sodium carbonate\n\nBarium difluoride\n\nCopper(II) nitrate\n\nNickel (II) sulfate