synthesis of hydrazone?
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Which hydrazone you want will determine how you synthesize it.
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C3H7NO2 \n\nwww.chemfinder.com is a useful resource.  \n\nI hope this helps :)
1144601
The products may vary depending on the physical conditions(temperature and pressure) as well as the amount of reactants, catalyst etc. So the final product can be either ammonia or hydrazine (or no reaction at all just like in the atmosphere).
665494
It can be made by dehydration of acrylamide\nCH2=CH-C(O)NH2 --> CH2=CH-CN + H2O\nP2O5 is a good reagent for that.
464400
An aldehyde is either a functional group consisting of a terminal carbonyl group or a compound containing a terminal carbonyl group.\n\nCommon reactions\naldehydes react with alcohols and an acid or base to a hemiacetal, the hemiacetal and alcohol and more acid catalyst further react to an acetal and water \nSimple hemiacetals are usually unstable, although cyclic ones such as glucose do exist. Acetals are stable, but these revert to the aldehyde in the presence of aqueous acid. \nTreating aldehydes with oxidizing agents such as potassium permanganate, nitric acid, chromium(VI) oxide or acidified potassium dichromate), will yield a carboxylic acid. \nTreating aldehydes with Tollens' reagent (which is prepared by adding a drop of sodium hydroxide solution into silver nitrate solution to give a precipitate of silver(I) oxide, and then adding just enough dilute ammonia solution to redissolve the precipitate in aqueous ammonia to produce [Ag(NH3)2]+ complex) will convert aldehydes to carboxylic acids without attacking carbon-carbon double-bonds. This reaction is also known as the silver mirror test. \nAldehydes can react with water (under acidic or basic conditions) to form hydrates, R-C(H)(OH)(OH), although these are only stable when strong electron withdrawing groups are present, as in chloral hydrate. The mechanism is identical to hemiacetal formation. \nAldehydes can react with HCN to form cyanohydrins, R-C(H)(OH)(CN). \nTreating an aldehyde with a Grignard reagent can yield an alcohol with a substituted group from the Grignard reagent. \nTreating aldehydes with NH3 derivatives (like NH2OH, H2NNH2, 2,4-dinitrophenylhydrazine) will result in the formation of an oxime or hydrazone. This is classed as an addition-elimination reaction or addition-condensation reaction. If the simple hydrazone (RCH=NHNH2) of an aldehyde is heated with a base such as KOH, the terminal carbon is fully reduced via the Wolff-Kishner reaction to a methyl group. The Wolff-Kishner reaction may be performed as a one-pot reaction, giving the overall conversion RCH=O → RCH3. \nReaction of aldehydes with reducing agents such as magnesium gives diols in a Pinacol coupling reaction \nThe Wittig reaction takes aldehydes to alkenes and the Corey-Fuchs reaction takes aldehydes to alkynes both with a triphenylphosphine reagent. The Corey-Chaykovsky reagent is a sulfonium ylide which convert aldehydes to epoxides. \n\nExamples of Aldehydes:\n\nMethanal (Formaldehyde) \nEthanal (Acetaldehyde) \nPropanal \nButanal \nPentanal \nGlucose \nBenzaldehyde
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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.
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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
821017
(CH3)2 N2H2+2N2O4-->3N2+4H2O+2CO2
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2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
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So, usually the best way to do these is to just fiddle with them until you get them right, but for the sake of it, let's go with a systematic approach using algebra.  First we assign a variable in front of each of the terms on the left and right like:\n\nx Pb  +  y H2O  +  z O2  -->  a Pb(OH)2\n\nSo now we'll write down equations to make sure that we have the same numbers of both sides of each *element*.  I'll do this by writing down the variable in front of each compound in your equation (so x, y z and a) and then multiplying those by the number of times an element exists in that compound.  (So when I'm writing the equation for hydrogen, the H20 term will be 2*y, because y is the coefficient in front of H20 and H occurs twice in that compound).\n\nRemember, the number of any element on the reactants (left) side equals the number of that element on the products (right) side.\n\nFor Pb: x = a, because Pb shows up only once on each side with a subscript of 1, those two compounds will have to be present in the same molar amounts.\n\nFor Hydrogen: 2y = 2a\n\nFor Oxygen: y + 2*z = 2 * a\n\nSo now we have three equations:\n\nx = y\ny = a\ny + 2z = 2a\n\nBut four unknowns, which is fine because we'll just solve for them in terms of one of the variables and then simply choose the last variable to make all the numbers nice.  So solving the equations in terms of a:\n\nx = a\ny = a\nz = (2a - y)/2 = (2a - a)/2 = a/2\n\nSo now we have x, y and z in terms of a free parameter, a.  We'll just choose a = 2, for example, to make x, y and z all turn out as integers.  So now we've chosen:\n\nx = 2\ny = 2\nz = 1\na = 2\n\nSo the balanced equation is:\n\n2 Pb + 2 H2O + O2 --> 2 Pb(OH)2\n\nMake sense?
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The problem with trying to get this to happen is that you need the 4-hydroxy to enolise.  This would allow both your imine synthesis and S-alkylation.  Without this neither is going to happen.\n\nPyrones have some alkenic and some aromatic characteristics na dyou need the alkenic characterisitics to dominate.  Unfortunately the C3-C4 double bond is shorter than standard C=C double bonds and is closer to the length of a C-C bond fond in benzene, suggesting it is more aromatic in nature and therefore less likely to enolise.\n\nMy opinion is that this cyclisation won't be very effective.  As I say this is only my opinion and performing the actual experiment may prove me wrong.
215604
C6H12O6(s) ---H2SO4--> 6C(s) + 6H2O(in H2SO4 solu.) + Heat
814611
CdSo4+NH4OH ---->Cd(OH)2 + (NH4)2SO4
462663
C4H9OH + 6O2 --> 4CO2 + 5H2O
810244
If the reaction is correct it should be :\n  \n              2ZnS + 3O2  -------->  2ZnO + 2SO2
65886
Synthesis - a compound forms from simpler reactants
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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.