hydrolysing halogenoalkanes?
959110
You pass the Halogenoalkane on hot paladium and add H2 during the process, the amount of acid produced indicates the rate of the reaction.
104161
You could rig a "Dean-Stark Trap" on the side of your condensor. That will eliminate water from your esterification (assuming its similar to the Fisher esterification) and force the production of more water and product, until one of the reagents is used up or an equilibrium is achieved.\n\nAnother method would be TLC, comparing the starting reaction mixture to the final product.\n\nSpectroscopy (NMR, IR) could be used.\n\nFinally, you could just put some of your sample into a GC-MS.
771931
Starting material:\nCH3CH2CHBrCH2Br ?\n\nProduct wanted:\nCH3CH2CH2CH=CHCH3 ?  Cis or Trans?\n\nIf the starting material is 1,2 Dibromobutane, then treating it with base (eg. NaNH2) will give:\n\n CH3CH2CHBr–CH2Br   +   base   ——>  CH3CH2CH=CHBr   +   base   ——>  CH3CH2C≡CH\n\nIf this is done with excess NaNH2 in liquid ammonia or dry diethyl ether, then the sodium salt will be produced:\n\n(CH3CH2C≡C)- Na+\n\n(CH3CH2C≡C)- Na+ + BrCH2CH3  ——>  CH3CH2C≡CCH2CH3 \n\nCH3CH2C≡CCH2CH3 when heated gives the more stable compound CH3CH2CH2C≡CCH3  (this is 2-hexyne).  The path will be different from here depending on if you want a cis or trans product.  A controlled catalytic addition (Lindlar catalyst) of H2 will give the Cis form.  Na in liquid ammonia will give the Trans.\n\nThere might be an easier way, but this path is a classic.
166454
I can supply you said chemicals.\n(C2H5NH)(CH3)C6H3OH \nPlease contact me by email: halsliu@gmail.com
101358
I think the only pratical way ot do this will be to distill the kerosene.\n\nBut be aware that you are probably breaking the law by doing so. The colouration is added to make it comply with safety laws.
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.
575847
Any amine hydrochloride, ammonium chloride, or even hydrochloric acid can be used.  You may need to actually call a company to get more specific information, if they will even tell you.  If you look at MSDSs, you can gleam some information.  \n\nhttp://www.efdsolder.com/lit_msds.htm\n\nYou will see that nonionic organics can be used.  Because they would be bought in bulk, there are no commercial names other than the chemical name.
196696
One possibility is fractional distillation.  This procedure takes the boiling points of each component of a mixture into account.  The mixture is connected to the proper apparatus and placed over heat.  That component with the lower boiling point will evaporate off first and can be collected in a recovery flask.  That material with a higher boiling point will evaporate latest.
985077
chlorine, Cl2, bromine, Br2, or iodine, I2 or fluorine
381243
Are you actually working on silazanes? They are hot stuff. Wanted to do them too but had problems with shipping of some reagents. Anyway, I was tryig to access this article but I was unsuccessful. Maybe you ca try it yourself: "Rates of methanolysis of several organosilazanes" by William L. Budde, J. Phys. Chem.; 1969; 73(11); 3718-3721. Hope this helps. Or click here: http://pubs.acs.org/cgi-bin/abstract.cgi/jpchax/1969/73/i11/f-pdf/f_j100845a028.pdf\n\nIf you were not able to access this, then contact me so i can send you the pdf file.
506807
I'm guessing this is a question for class. In practice, working with acetylene gas as a synthetic reagent isn't a good idea. It's a flammable, explosive gas and most synthetic labs really don't have the necessary safety equipment to handle it. It's much safer to buy the butyne you need and work from there.\n\nHowever, on paper you would first make the 2-butyne. You would react 1 eq of acetylene with 2 eq of n-BuLi followed by quenching with 2 eq of MeI and standard workup and purification. If you want the 1-butyne, you can isomerize 2-butyne to 1-butyne with an amide base like NaNH2 or KHMDS.  The reason you can't just alkylate with iodoethane is that you are going to end up with a mix of acetylene, 1-butyne and 3-hexyne since you can't selective monodeprotonate every acetylene molecule. \n\nHope this helps.
584861
LiAlH4 is a bit harsh to reduce an azide.  Pd/C- H2 is just as effective and the reaction work-up is a lot easier.  Azides are very easily reduced.\n\nIf chlorination is too tricky because you are dealing with a gas, then you can do the same reaction using bromine / red phosphorous to give bromoacetic acid.  All other subsequent steps will remain the same.\n\nAlternative methods would be to take the chloroacetic acid, and treat with potassium phthalimide, followed by hydrazine to remove the phthalimide protecting group.  This is the Gabriel synthesis of amino acids.\n\nAnother way would be to nitrate with amyl nitrate, and then reduce the nitro to the amine.  Hydrogenation should be sufficient to do this.  Metal reducing agents for this functional group interconversion are best used with nitrobenzenes, in my experience.\n\nYou could aminate your chloroacetic acid with diphenylmethylamine, and then treat with TFA to remove the benzhydryl protecting group.  I think it is stable to HCl so you do need to used TFA. (At least from memory I think this is the right way round).\n\n\nThese are a few more ideas but I am sure there are many more.
903981
It depends on the alkyl halide.\n\nA primary alkyl halide would undergo an SN2 reaction with sodium hydroxide because the only time an elimination product is given is with a bulky base like t-butoxide.\n\nA secondary alkyl halide would undergo E1/E2.  If E1, a carbocation intermediate would occur.\n\nA tertiary alkyl halide would undergo E1 with a carbocation intermediate.  No rearrangement would occur because it is already stable.\n\nMy guess is this:  You have a secondary halide that is undergoing E1, which is possible and would result in a rearrangement.\n\nDid you learn about E1?  E1 is basically like SN1, where the rate determining step is the formation of a carbocation intermediate via the leaving of the halide.\n\nI hope this helps.
1053808
Gee, I know you're a student but you are dull and lazy.\n\nHaven't you heard of google??  \n\nStep 1 - I went to www.google.com\nStep 2 - I used the following search  "electroclorination +equation"\n\nhttp://www.google.com/search?hl=en&lr=&q=electrochlorination+%2Bequation\n\nStep 3 - I followed the first link (listed below) on the google.com search results.\n\nAnd you've waited how many days for this answer?  It took me less than one minute to find.
750457
Dehydrogenation is the removal of hydrogen.  Two processes by which this can be done is by:\n\n1.  Free Radical Halogenation\nCH3-CH3  +  X2  ---->CH3-CH2-X  +  H-X\n(wher X =  F, Cl, Br, or I)\n\n2.  Catalytic Dehydrogenation\nCH3-CH3 ---->CH2=CH2 + H2\n(a catalyst such as platinum needs to be added when performing this reaction).
140226
ethanol and oil\nwater and chloroform
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.