mechanism of Synthesis of a tertiary alcohol via a Grignard Reagent?
369946
The reaction is generally thought to proceed through a nucleophilic addition mechanism.Grignard reagents such as methylmagnesium bromide are sources of a nucleophile that can attack the + end of the C=O double bond in aldehydes and ketones.If we treat the product of this reaction with water, we get an tertiary alcohol.Sterically hindered substrates may react according to an SET (single electron transfer) mechanism.
251170
Diethyl Malonate looks like this:\n\nhttp://en.wikipedia.org/wiki/Diethyl_malonate\n\nWhat's strange is that 3-phenyl propanoic acid is quite simpler than that structure.\n\nMy first move would be to reduce those esters, because it's quite difficult to do a Grignard reaction without getting multiple additions.\n\nSo you react the ester with 2 equivalents of Lithium diisobutyl aluminium hydride (DIBALH), to yield 1,3 propan-di-al (malonaldehyde), a compound with an aldehyde at the end of each side of the propyl group.\n\nThen you react your product with 1 equivalent of phenyl magnesium bromide, with an acidic workup. That will add a phenyl group to one of the aldehydes, making it an alcohol. (the two are equivalent, but you will lose product because the Grignard may reduce both ... chromatography would be needed for this step)\n\nNext, oxidize the remaining aldehyde with Tollen's reagent, to get a carboxylic acid.\n\nThen, to remove the alcohol from the Grignard side of the aldehyde, you use the Barton reduction:(http://en.wikipedia.org/wiki/Barton-McCombie_deoxygenation)\n\nPhenylacetic acid would be hard to prepare in this way, because the carbon-carbon bonds would be difficult to break in the center of the molecule. Any ideas with anyone else?
900980
A 1,2-diol would form.\n\nEthene would form ethan-1,2-diol, (CH2OH)2.
932033
Use a retro synethic method.\n\nTarget Molecule <-----  1,2-dibromo- 1,3-dibenzylpropane   (M2) (use 3eq. of NaNH3 in ammonia)\nM2 <---- 1,3-diphenyl-1-propene (M3) (use Br2 in CCl4)\nM3 <---- 1,3-diphenyl-2-propanol (M4) (use KOH/EtOH)\nM4 <---- 2-phenylethanal (M5) + benzyl bromide (M6) (Grignard rxn)\nM5 <---- 2-phenylethanol (M7) + Cr6+ (use chromic acid)\nM7 <----  Styrene (starting material #1) + i)BF3 ii) H2O2, HO-, H2O (anti-Markovnikov hydroboration)\nM6 <---- Benzyl alcohol (starting material #2) + Mg(s) in Et2O (make your Grignard to be condensed in a step later on)\n\nNow, you do the connections. Note my nomenclature is rusty and I'm not 100% sure about the phenyl nomenclature as Benzen side chains to a higher priority parent chain.
799054
Consider a primary alcohol that has a structure like C--OH\n\nthen consider a tertiary alcohol that has a structure similar to \n          CH3\n           |\nH3C--C--OH\n           |\n          CH3\n\nIn a primary alcohol such as the one shown above, the distribution of charge is relatively polar due to the presence of only one other carbon atom with one very electronegative oxygen atom. In a tertiary alcohol, there are four carbon atoms versus only one electronegative oxygen atom. Therefore, in the tertiary carbon, the polarity is greatly less because the polarity is balanced by the presence of so many carbon atoms around the molecule.
663295
Chemistry relating to Carbon derivatives.
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.
2865
In synthetic organic chemistry, water free ethanol can  be produced by adding elementary sodium wire under a nitrogen atmosphere. The sodium will react with water forming NaOH and with the ethanol forming sodium ethanolate, which in turn would react with any water that should make its way into the liquid.\n\nTo remove the salts, you can distill in a water free environment.\n\nSodium wire is highly inflammable. You won't be able to get it outside of laboratories, for a good reason.
788034
Once the hydroxyl group leaves the alcohol, the structure would lose its pyramidal shape and flatten into a two-dimensional shape.  The halide ion would add from either side, having a 50% chance of attaching from either side.  Once halogenated, the new molecule would again be optically active but you would have an equal abundance of both R and L enantiomers.  This follows one of the rules of stereochemistry in that optically inactive reactants (in this case HX) always yield optically inactive products.
57499
the best way to do it is by oxidation of alcohols.
518616
Check this out....\n\nhttp://www.sciencestuff.com/prod/Chem-Rgnts/C2043
798882
no it doesnt.........its IUPAC NAME is 2,3-dihydroxypropanal which suggets that it has a linear structure with  only three carbon atoms with an alcoholic group at the 2 n 3 carbon atom and with a carbonyl group (aldehyde gropu in this case) as the terminal carbon atom\nA compound will have a cyclic structure only if its a cycloalkane , cycloalkene or an aromatic compound.........since glyceraldehyde is none of the above..........its evident that its got a linear structure.
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.
6121
Hey cable_ray,\n     One of the most easiest way to get the recipie you require or any other synthesis can be found on the following website. It is the organic synthesis website, maintained by cambridgesoft.com. It is a free service they provide. You can use it. Just put in the name of the ether you want to produce eg. dimethyl ether, diethyl ether, etc. I am an MSc by research student. So you can also mail me. I will surely reply. my mail add is jchemie@gmail.com
137706
H3O+(the acid) attacks the carbonyl oxygen, attaching a proton to it. \nthe pi electrons in the carbonyl bond move over to the oxygen atom.\n then a nucleophilic attack on the carbon in the carbonyl bond(it now carries a positive charge and is attracted to the nonbonding electrons on the nucleophile). \nas the nucleophile(usually an alcohol) forms a new bond with the carbon atom a water molecule is kicked off.\nthis is a classic SN2 reaction mechanism.
412197
It is a nucleophilic substitution reaction, where the nucleophile -OH from KOH replaces -Cl from 1-chlorobutane; this forms KCl and 1-butanol as products.\n\nKOH + CH3-CH2-CH2-CH2Cl ---> KCl + CH3-CH2-CH2-CH2OH
1005052
Typically you can do most regular organic chemistry reactions with radiolabelled substrates. So for instance\n\nacetaldehyde + NaT4 -> CH3CTHOH\n\nCH3CTHOH + Na(129)I -> CH3CTH(129)I would work