How do i obtain benzalacetone from Benzaldehyde and acetone?
913417
In a mixture of acetone and freshly distilled benzaldehyde, add aqueous NaOH from dropping funnel - while the solution is stirred, and cooled by a water bath (such that temp of 25–31°C is maintained). Then, stir the mixture further for two and one-quarter hours at room temperature. Then, add HCl until the mixture is acid to litmus. The two layers which form are separated in a separatory funnel. The lower aqueous layer is extracted with benzene. The benzene is then removed by distilling from a steam bath, and the residue is distilled under reduced pressure. The material which distils at 148–160° /25 mm., 133–143° /16 mm., or 120–130° /7 mm. weighs 430–470 g.; it solidifies to a crystalline mass on standing, and is pure enough for all practical purposes. This is your benzalacetone.
294819
acetol?\n\nany of various compounds characterized by the grouping C(OR)2 and obtained especially by heating aldehydes or ketones with alcohols
574036
The first step, find the moles of both.\nYou should get 25 mmol benzoic acid, and 15.6 mmol NaOH.\n\nThen you must find after the reactions, how many moles remain.  \nYou should get 9.4 mmol benzoic acid, 0 moles NaOH, and 15.4 mmol benzoate.\n\nYou have some benzoic acid and some benzoate ion, a conjugate acid/base pair.  This is a buffer solution, so use the Henderson-Hesselbach equation.  You should get pH = 4.4.
1150323
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
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
403780
GHB got a bum rap.  it's just more nonsense hysteria to justify the bloated budgets of the police state.\n\nThat aside, you don't convert beta to gamma.  suggesting an elimination failed to consider that the double bond formed will be more stable in conjugation with the carbonyl group; you'll get crotonic acid.  Subsequent hydrolysis would give you a mixture of alpha and beta.\n\nyou'd have much better luck trying to figure out a way to do it starting with GABA.\n\nOf course, if you have to ask, you lack the skilz.
366491
different ways to do it.  steam distillation, expression, solvent extraction and efleurage. the first is the easiest and what you want to know if you are looking for a how to. Use hot steam, like a steamer, you'll end up with the oil and water.  then separate the oil.  \n\nthe other methods are more complicated, but you do end up with more oil.  but at what price?  second method the is to use pressure or centrifugation. At home, not likely you have anything to make this work.   The third method you use a solvent, but then the oil and solvent are mixed.  You then evaporate the solvent (you can get at a whole foods store to do this) when the solvent is evaporated you are left with a heavy, waxy substance called concrete. Then you have to separate the liquid from the was.  it's great because it is most concentrated from of aroma, but very time consuming and complicated.  That last, efleurage is a really long process using animal fats and alcohol. Although essential oils main usage is in cosmetics and perfumery, many of them do have proved therapeutic properties.
48368
2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
806753
Are there any tests you can run? Usually if you mix a substance with something else, the product has different chemical and physical properties than the initial substance.\nFor example, let's say that you would mix isoamyl acetate with "X" the product will have certain carachteristics gained only after it was mixed with "X".... I don't know how much this will help...\n\nLook it up on the internet, I am sure you'll find something.\nGood luck.
719339
You can synthesize an 'amide' (has a carbonyl-nitrogen linkage) from a carboxylic acid in one step by reacting it with an amine.  In order to convert a carboxylic acid into an amine you would first have to convert the acid into another derivative (i.e. aan alcohol, etc.) before converting to the amine.  Regardless of how you do it, you will most likely have to employ multiple steps.\n\nIf anyone knows of a way to do this in one step, I would really like to know as well.
555664
The formula could be (C6H12OH)n, in which case it could be synthesised from the corresponding aldehyde. Expensiveness depends on how much and how pure of a product you want.
781306
Jade,\n\nConsider using Acetone if you can get your hands on some.  Ethanol (alcohol) should be able to dissolve some fat but it is miscible with water so you need to watch out for some water-soluble stuff eluding out.\n\nThinner might not be a good idea as it might be reactive to fat (though I am not certain).\n\nVinegar is not an organic solvent (Vinegar is polar and water soluble).\n\nGasoline is indeed a mixture so you do not want that.  Would you be able to get Benzene or Toulene?
929369
I sure hope you have more information than you're giving us.  Typically profs will narrow the choices to one example of each major class of organic compound.  If you have the whole universe of canidates to choose from, you will need a lot more than these simple test.\n\nTLC:  In one of your textbooks or handouts, you should have a short list of organic candidates vs their Rfs.  Assuming you are using silica gel or alumina plates, ethyl acetate is about twice a strong a chromatographic mobile phase as dichloromethane (DCM) (see linked pdf, p. 15).  The fact that your unknown moved at all in ethyl acetate but hardly at all with DCM lets you know that your compound is fairly polar.  This is not an alkane or a fatty acid/ester.\n\nSolubility:  Not soluble in water?  It's not a saccharide, short-chain acid, alcohol or phenol.  Not soluble in base means it's definitely not a carboxylic acid or a phenol.  Soluble in acetone means it's not a fat or long-chain fatty acid.\n\nSo, what are we left with?  Ethers, ketones, esters and aldehydes.  I suggest you look in the chromatographic tables you have for one of these compounds that has the same Rf values and melting point you've quoted us.\n\nGood luck.
532583
benzioc acid in its neutral form is non-polar. So you cannot separate it from ether (also non-polar). you want to make benzoic acid polar by deprotonating it so you can separate it from ether.\n\nsodium bicarbonate is a weak base, so it would deprotonate the acidic hydrogen on benzoic acid and makes it polar. However, since sodium bicarbonate is a weak base, it does not deprotonate the benzoic acid very efficiently. To do that, you can you sodium hydroxide, which is a much stronger base and can deprotonoate benzoic acid almost completely. so sodium hydroxide is a better choice.
817601
try the local chemist....
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
1141478
I don't know about chloroform but acetone works well