synthesis for Benzamide starting from Aniline-BROMOBENZENE TO GIVE 1PHENYLETHANOL-BUTANAL TO GIVE 1CHLOROPROPA
342187
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.
131270
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.
233816
None.  The sell chemicals with alkyl groups.  In CH3Cl, CH3 is the alkyl.  They don't exist on their own.  Start with\n\nhttp://www.sigmaaldrich.com/Local/SA_Splash.html
298764
that can be done either just by the density diffrences (one will float on top of the other and can be skimmed off)\n\nOR\n\nyou can disolve one of them and make the liquid layers seperate that way. I think the benzophenone will disolve in an organic layer and the naphthol will disolve in water.
156987
OH- is not substituted in haloalkanes using water it is done using NaOH and KOH.\nRX + NaOH--->ROH +NaX\nHalo arenes can be converted to phenols using water if they contain many electron withdrawing groups like NO2+.(eg)picric acid.\n2,4,6 Trinitro chloro benzene + H2O--->picric acid \n                                     (2,4,6 trinitro phenol
151307
N-(4-hydroxyphenyl)acetamide\n\nC8H9NO2
34972
A Synthesis of Safrole and o-Safrole\nW.H. Perkin & V.M. Trikojus\nJ. Chem. Soc. 1663 (1927)\nHTML by Rhodium\nNo direct synthesis of safrole has been recorded, the most recent attempt being that of Baker and Robinson (JCS, 1925, 127, 1424), who distilled the product of the action of silver oxide and water upon gamma-piperonylpropyltrimethylammonium iodide, but obtained isosafrole, the double bond moving into the position of greater stability. Kawai (Sci. Papers Inst. Phys. Chem. Res., 1925, 3, 263) has shown that the monoallyl ether of pyrocatechol undergoes the Claisen rearrangement to give an oil which he considered to be a mixture of 3- and 4-allyl-1,2-dihydroxybenzene.\n\nThis matter has now been reinvestigated and, by repeated fractionation of the product of the rearrangement, both of the above isomerides have been isolated in a pure state as colourless, crystalline solids melting at 28°C and 48°C respectively. As further confirmation of the course of the Claisen migration, it has been found that the properties of the higher-melting isomeride (II) are in accord with those assigned to the "4-allylbrenzcatechin" which Schimmel and Co. (Centr., 1907, II, 1741) isolated in small quantity from betel-leaf oil from Java. The second isomeride, 1,2-dihydroxy-3-allylbenzene (I), has not previously been described.\n\nIn order to convert these substances into safrole (III) and o-safrole (IV), respectively, methylenation was carried out, with moderately good yields, by gently refluxing them with methylene iodide and anhydrous potassium carbonate in dry acetone solution. By employing such mild conditions, any change into the isomeric isosafroles during the course of the reaction was inhibited. The identity of the synthetic product with naturally occurring safrole has been established by preparing the pentabromo-derivative (mp 169°C) and the acetamido-derivative (mp 162-163°C) and comparing these with the same derivatives from natural safrole. The properties of the hitherto unknown isomeride o-safrole (IV) (which resembles safrole in odour and in many other ways) have been recorded. The process of methylenation described above takes place very smoothly without formation of tarry products and may prove to be valuable in other cases of this kind.
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
198113
This might work:\n\n1. pyrrolidine (to form the enamine)\n2. Chloromethanol*\n3. water\n\n*Don't know what reactants you have, but you could start with ethene:\n\n1. ozone\n2. peroxide in HCOOH\n3. NaBH4\n4. water
172774
your formula refers to several different structures which of course would give different NMR spectra. It could be 2-bromo-X-methylpyridine where X is the 3,4,5 or 6 position or Bromoaniline, with the various positional isomers. If you don't know the name go to the sigma website and find the structure/name (it may even have the NMR).
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.
523447
Most industrial acrylonitrile is produced through the Sohio process, the catalytic ammoxidation of propylene:\n\nCH3-CH=CH2 + NH3 + 1.5 O2 → CH2=CH-C≡N + 3 H2O
571583
Chloramine (NH2Cl) with p-toluenesulfonamide
613727
try:\nhttp://lb.chemie.uni-hamburg.de/static/data/50_zzk46n8n.html
695914
Chemical laboratory supply houses carry it.  Try Alfa Aesar, they have it:\n\nhttp://www.alfa.com
903140
Although the economics would never qualify for an industrial process, there are a number of ways that this could be accomplished.   \n\nSince alkanes themselves are not very reactive, the first step would be to halogenate the methane.  This could be accomplished by reacting the methane with elemental bromine (Br2) with light to initate free radical halogenation.\n\nThere are a number of ways that the methyl bromide produced could be coupled.  \n\nReaction with elemental sodium, the "Wurtz reaction", would be the simplest.\n\nAnother method would be to prepare (CH3)2CuLi by first reacting the methyl bromide with Li to convert it into methyl lithium (CH3Li).   The methyl lithium could be converted into  \n(CH3)2CuLi by reaction with CuI.\n\n(CH3)2CuLi will react with methyl bromide to produce ethane.
900980
A 1,2-diol would form.\n\nEthene would form ethan-1,2-diol, (CH2OH)2.