hydrolysis of p-nitroacetanilide?
116320
I am not 100% sure but it seems reasonable that by hydrolysis, you will cleave the bond from the carbonyl to the nitrogen.  Yielding, Acetic acid (CH3CO2H) + para-Phenylenediamine  ( H2N-Ph-NH2).
1087032
try Vogel's "Qualitative Inorganic Analysis", I'm pretty sure it has all the answers 4 u.\n\ndont care editions, but for 7th ed, ISBN is 0582218667\n\n****\nokey then, from Vogels\n\n1) when heated, lead nitrate decomposes to white solid lead oxide (PbO), brown gas (NO2) and oxygen gas. Brown gas decolorizes when contact with water. \n\n2) ammonia odur. ammonium compounds are all volatile towards heat and decomposes to ammonia and an acid but quickly recombine if the acid, in this case, hydrochloric acid, is a strong acid.\n\n3) anhydrate copper sulphate is white, hydrate copper sulphate is blue. I have no idea which one you actually heated. If it's anhydrate, no change; if hydrate, blue ---> white.\n\n4) blue precipitate forms initally, but resolve upon excess reagent, clear blue soln results due to the formation of copper complex [blue precipitate: Cu(OH)2*CuSO4, complex: [Cu(NH3)4]^2+]\n\n5) brown precipitates froms [Cu(OH)2], no change with excess reagents.\n\n6) metal-gray stipe (magnesium stipe) gains a redish brown color due to formation of copper metal. Possible decolorization of the soln (depends on how much magnesium u actually put in)\n\n7) volatile reaction, solid metal solves,ignitable gas is formed. (if phenolphthalein is added to water initally, soln becomes pink)
684025
Are you thinking of tRinitrotoluene -- TNT?\n\nThe trick is getting the nitro groups in the right place on the toluene base -- not an easy thing, becuase the methyl group deactivates the ortho and para positions.
585025
I don't know that you can, different solvents react differently to pressure, so I don't think it's a one eqaution fits all kind of process.  That's why they all have different graphs for phase vs temperature/pressure.  I'm sorry.\nSounds like you have a pretty high boiling point though, the things I've taken off that had to be real high temp were water, toluene, and pyridine.  Could it be one of them?  They smell pretty different, so you'd know which.  \nThat's an interesting question though, I've never stripped a solvent that I didn't know what it was.
491215
-Fluorinate toluene with F2 (highly controlled) or FCl or F5Cl, or DAST or one of these other nice fluorinators - gives the trifluoride on the methyl carbon\n-Nitrate it with nitric acid/sulfuric acid - CF3 is withdrawing so nitro goes on the 3-position\n-Reduce the nitro with Sn/HCl or H2/Pd\n-Nitrate again - the amino protonates and directs to the 5-position\n-Titrate the water to basic pH and extract with solvent like ether.  Concentrate and purify - probably crystallizes nicely.
852453
The previous answer is partially correct.\n\nWater reacts with the excess acetic anhydride which is converted to acetic acid.\n\nSince the product is only slightly soluble in water, this also makes it possible to collect the product as a solid.
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
316857
Part 2: Synthesis of Benzil \n\nStarting with the a-hydroxyketone benzoin (prepared in Part 1), you will prepare an oxidation product, benzil, which is an a-diketone. \n\n \n\nThis oxidation can easily be done with a variety of mild oxidizing agents, including Fehling's solution (an alkaline cupric tartrate complex) or copper(II) sulfate in pyridine.  In addition, benzoin could be oxidized by sodium dichromate, but the yield of benzil is lower because some of the benzoin is converted back into benzaldehyde following cleavage of the bond between the two oxidized carbon atoms, which is activated by the phenyl rings, producing benzoic acid as the final product.  In this experiment, due to ease of use and consistent results, we will use nitric acid as the oxidizing agent. \n\nCaution: Concentrated nitric acid is highly corrosive and causes severe burns if spilled onto your skin. Nitrogen dioxide (NO2) fumes are highly toxic and can damage the lungs due to inflammation. Do not breathe NO2 fumes, and perform this part of the experiment in the hood. \n\nProcedure \n\nHeat a mixture of 4.0 g of benzoin (prepared in Part 1; save enough for analysis, but proportion reagents as needed) into a round-bottom flask with 14.0 mL of concentrated nitric acid.  Add a stirring bar and attach a condenser to the top of the flask. Set up a reflux in the hood to vent NO2 produced during the reaction. With stirring, heat the reaction mixture. Begin timing the reaction when NO2 (red-brownish colored gas) are visible above the reaction mixture and gas bubbles are present on the stir bar. Reflux for at least 30 minutes, or until no more NO2 gas is apparent. Do not stop the reaction until the reaction is complete. Stop the reaction by removing the heating mantle, and letting the reaction mixture cool for about 5 minutes.  Add about 75 mL of cold water to the reaction mixture, cool to room temperature, and swirl for a minute or two to coagulate the precipitated product: collect and wash the yellow solid using vacuum filtration.  Continue drawing air through the crystals on the funnel by suction for about 5 minutes to assist in drying the crystals..  The crude product (dry weight 3.7-3.9 grams) need not be dried but can be crystallized at once from ethanol.  Dissolve the product in 10 mL of hot ethanol, add water dropwise to the cloud point (mixture goes from clear to cloudy, or solids being formed), and set aside to crystallize.  Record the yield, crystalline form, color, and melting point of the purified, and recrystallized, benzil.   \n\nTest for the presense of unoxidized benzoin: Dissolve about 0.5 mg of crude or purified benzil in 0.5 mL of 95% ethanol or methanol and add one drop of 10% NaOH.  If benzoin is present the solution soon acquires a purplish color owing to a complex of benzil with a product of autoxidation of benzoin.  If no color develops in 2-3 min, and indication that the sample is free from benzoin, add a small amount of benzoin, observe the color that develops, and not that if the test tube is stoppered and shaken vigorously the color momentarily disappears; when the soluton is then let stand, the color reappears. \n\nCleaning up: The aqueous filtrate (containing HNO3) should be neutralized with sodium carbonate, diluted with water, and flushed down the drain.  Ethanol used in crystallization should be placed in the organic solvents container.
515166
Hmmm, since you're only getting one product, and I assume you have an excess of NaCN, I would suspect that you have a double displacement reaction, and you're getting pentanedinitrile, NC-(CH2)3-CN
329297
decomposition reaction\n\n2H2O2 -------- O2   +  2H20
757438
CaC2 + 2 H2O ---> Ca(OH)2 + HCCH
948313
http://www.chemguide.co.uk/physical/catalysis/hydrolyse.html   \n                                                                                                    this will help ya
821017
(CH3)2 N2H2+2N2O4-->3N2+4H2O+2CO2
695634
d) use a nitrate ester of glycerin. The body will hydrolyze the nitrates leaving glycerin, the active chemical.
963233
Lucas Reagent replaces the OH group by a Cl group.\nTherefore CH3-C5H8-OH + HCl -> CH3-C5H8-Cl + H2O
1062523
2NaI + Hg(NO3)2 --> HgI2 + NaNO3\nwe have;\n52 ml and a 0.78M Hg(NO3)\ntherefore we can find # of Moles\nn = CV\nn = 0.78 * 0.052                       Concentration should be in Litres\nn = 0.04056 mol\n\naccording to the equation;\n2 moles of NaI react with 1 mole of Hg(NO3)2\ntherefore 2(0.04056) moles of NaI needed in the reaction\ni.e. 0.08112\nwe know V i.e. Volume equals the ratio of #of moles and Concentration \ntherefore ; \nV= n/C\nV= 0.08112 / 1.60\nV= 0.0507 L\ni.e. 50.7 ml of NaI required in the reaction !!
832115
N2H4 + H2O -> N2H5+ + OH-\n\nKb = c[N2H5+] * c[OH-] / c[N2H4]\n\nAssuming that c[N2H5+] = c[OH-] and that c[N2H4] >> c[OH-] (so that we can use the starting concentration for c[N2H4]), we get approximately\n\nc[OH-] = sqrt( Kb * c[N2H4])\n\nor logarithmized\n\npOH = 1/2 ( pKb + 10^-(c[N2H4])\n\npH= 14 - pOH\n\n\ncalculator time.