What would be the result of treating 1-methylcyclopentanol with a Lucas Reagent?
963233
Lucas Reagent replaces the OH group by a Cl group.\nTherefore CH3-C5H8-OH + HCl -> CH3-C5H8-Cl + H2O
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.
717741
I'll talk through it, since I can't draw structures.  Presumably, this is an acid catalyzed process.  The first step is protonation of the oxygen on acetone.  This creates a more electrophilic carbonyl.  One of the hydroxyls of the starting material will attack the carbonyl carbon, generating a hemiacetal.  Protonation of the hydroxyl of this hemiacetal, followed by elimination of water with assistance from the lone pairs on the neighboring oxygen will generate an oxonium ion suited for another nucleophilic attack by the other hydroxyl of the hydrobenzoin.  A final proton transfer will generate the acetal.
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.
182435
Well, there are a lot.\n\nSubstiution:\n\nR-OH + SOCl2 ---> R-Cl\n\nElimination\n        \nR-R'-OH ---->  R=R' + H20 (with an acid catalyst and heat)\n\nReduction:\n\nR-OH ---> R-CH3 (Barton Reduction)\n\nOxidation:\n\nR-OH ---> R-COOH\nR-OH ---> R=O\n\nThen alcohols can act as nucleophiles, so there is a another set of reactions. But I've listened enough, imho.
385291
You'll get CH3--CHBr--CH3. It's a markovnikov reaction with rearrangement.
93201
To go from 1-butanol to 1-pentanol, the easiest path to go is,\n1. Treat 1-butanol with PBr3, to yield 1-Bromobutane.\n2. Use a grignard reagent to produce CH3CH2CH2CH2MgBr.\n3. React this compound with formaldehyde, HCOH.\n4. Protonate to form 1-pentanol.\n\nCH3CH2CH2CH2-OH + PBr3 --> CH3CH2CH2CH2Br + grignard (Mg) ->\nCH3CH2CH2CH2MgBr + HCOH --> / H+ CH3CH2CH2CH2CH2-OH\n\n-----------------------------------------\nPropanal cannot undergo cannizzaro reaction because it has acidic alpha hydrogens to the carbonyl group of the aldehyde, instead it will undergo aldol condensation.\n\nhttp://en.wikipedia.org/wiki/Cannizzaro_reaction
570617
KMnO4 is an oxidizing agent.\n\nit will turn ethanol into ethanal (or acetylaldehyde).  if there is a great excess of potassium permanganate, it will turn ethanal into ethanoic acid (or acetic acid).\n\nbasically, both steps involve decreasing the number of bonds from carbon to hydrogen and increasing the number of bonds from carbon to oxygen.
864893
If I remember correctly it is there to wash away any un-reacted reactants and also "neutralize" them. So if your proportions are not exact it will react away either of the reactants left since both would also be soluble in the ethyl acetate.
445261
Try it yourself, its not that bad. \nHere is how:\n1) Set up a stoichometry ratio.\n2) Using the mass that they give you, calculate the theoretical yield of each product.\n3) Whichever product has a smaller theoretical yield, analyze the reactants which caused it to be such.\n4) The one that 'limits' the amount of products is the limiting reagent.\n5) Using that value, calcuate the amount of the other reactant that is necessary. Subtract from the orginal value to find the excess reactant.
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
791347
1. Methyl iodide: would involve a primary carbocation, which do not exist. Methyl iodide only reacts by an SN2 mechanism.\n2. 2-iodopropane: secondary carbocation\n3. 2-chloro-2-methylpropane--i.e. t-butyl chloride: tertiary carbocation\n4. 2-iodo-2-methylpropane--i.e. t-butyl iodide: tertiary carbocation. Iodide is a better leaving group than chloride, so this substrate is better for SN1.
635418
CH3CH2CH2CH3 + CL2 -hV--> \nCH3CH2(CL)CHCH3 + HCL ..\n\nwhat are the ways to get the CL at the end of the chain..so that\n\nCH3CH2CH2CH2CL + 2NA + CLCH2CH2CH2CH3\n--->CH3--------CH3\n\nThat's a radical reaction, the chlorination of an alkane with chlorine in UV light. The intermediate is a carbon radical and secondary radicals are more stable than primary. There are also more secondary carbons on this molecule. So probability and reaction kinetics dictates that the primary alkyl halide will be a minor product.\n\nI guess, if this were my synthesis, I'd:\n\n1. Put the glucose in HCl to cleave the anomers it likes to make.\n\n2. Esterify the carboxlic acid with whatever's available (Fisher esterification with MeOH, for example)\n\n3. Reduce the ester with Diisobutyl aluminium hydride (DIBALH)\n\n4. React the aldehyde with 1 equivalent of Ethyl Magnesium Bromide (Grignard)\n\n5. Reduce the hydroxy groups with a Barton-McCombie deoxygenation, although I've never seen it done in a case like this.\n\nThen again:\n\nI'd go to the stockroom and save my research advisor a lot of money. :)
615612
1.  Oxidise toluene to benzoic acid using strong oxidising agents.  KMnO4 (aq) is strong enough to do this.\n\n2.  Convert you benzoic acid to benzoyl chloride, using (COCl)2 - oxalyl chloride/DMF(cat.) or SOCl2 or Ghosezs reagent or PPh3/CCl4 or PPh3/NCS etc - the former two being by far the most common methods.\n\n3.  Nitrate your malonic ester in the 2 position using isoamyl nitrate in the presence of a base, ie NaH or NaOMe.\n\n4.   Reduce your nitro to an amine (H2-Pd/C)\n\n5.  Acylate your amine with your prepared benazoyl chloride (DIPEA as a base, DCM as a solvent for example)\n\n6.  Saponify both your esters with 1M NaOH.\n\n7.  Acidify your sodium carboxylates to the free carboxylic acids and heat for a couple of hours to promote de-carboxylation.  This is a standard procedure so will work well\n\n8.  Isolate your prepared hippuric acid!\n\nThere will be many other methods - especially to prepare your methyl 2-aminomalonate.  You don;t need to form the acid chloride also.  You could just perform an amidation between your amine and benzoic acid in the presence of a coupling agent such as DIC/HOBt, CDI, HATU, PyBrOP etc etc - there are too many to mention.\n\nYou could also brominate your methyl malonate with Br2, to give methyl 2-bromomalonate.  Treat your benzoyl chloride with ammonia gas to give benzamide.  Then treating your benzamide with NaH to de-protonate your amide, and then treat with your methyl 2-bromomalonate will generate the same product as described in step 5 above.  Following the rest will give you your product.\n\nThere are more methods but hopefully this has given you some useful information.
610332
carwash's ideas are pretty good, particularly TLC and spectroscopy.\n\nI don't think I'd do a titration against a base, because esters will also react with bases (hydrolysis). For a synthesis, this doesn't matter. But for an analytical method, it invalidates the technique, unless you can account for it with a standard or something. You might end up getting a result higher than what it should be because of it. (It depends on the chemical environment of the ester)
762929
CH3CONH2 + Na+-OH ---->  CH3COO -Na+ +  :NH3\n\nThis is a amide hydrolysis reaction, similar to the hydrolysis of an ester. Hydroxide attacks carbonyl to give tetrahedral intermediate. Explulsion of an amide ion gives a carboxylic acid, which is deprotonated quickly to give the carboxylate ion and ammonia.
958454
To visualize (C6H5)2CHBr, you have to identify the "side groups".  Obvious Br is one side group and so is one H.  Then you have two phenyl side groups so that leaves you that carbon to build on.  The molecule one carbon bound to two phenyl groups, one hydrogen and one bromine, each with a single bond.\n\nOK, let's build on that.  You are now going to react it with ethanol and you want to know what mechanism to expect.  Since Teach didn't show you the products, you're going to have to think about this one.  First, analyze what type of molecules you have.  The first molecule we spoke of, what type of carbon is the main carbon (primary, secondary, tertiary)?  What type of reaction (Sn1 or E1) does that type of carbon favor?  How about the ethanol?\n\nActually, the alpha carbon in ethanol, being a primary carbon, would favor Sn2 but that's not an option.  OH is an awful leaving group and alkanes don't like to carry a charge.  So we can rule the carbon out.\n\nThe problem must revolve around diphenylmethylbromide.  Having two phenyls would stabilize a positive charge fairly well and that is the first step for both E1 and Sn1 mechanisms.  So, which one is it?  The big giveaway is that En reactions are called elimination for a reason.  They lose a pair of substituents to form a double or triple bond, always a higher order bond.  Can there be any higher order bonds formed in this molecule?  How about Sn1, then?  That would result in (C6H5)2CH(OC2H5) + HBr.  Is this more likely than elimination?  Would this be endothermic or exothermic?\n\nThat's as much as I'm going to do.  The rest is for you, as you only really needed to know the structure of the first molecule.  But I hope this helps you understand what you are doing.\n\nGood luck.