grignard reagent problem
Started by TimeforDAT
I believe the answer is right. Grignard reagents are very basic and will react readily with alcohol groups as is present in the starting reagent. Therefore, CH3- will go to methane and the subsequent acid workup will only re-protonate the alcohol, forming the starting material. Therefore, this synthesis scheme is not a very well-thought-out one.
the grignarg regent will attack the carbony carbon to give c-o with a negative charge.
but, since the c-o double-bond is very stable it will be regenaratad, and will give off a primary carbocation that will combine to MgBr to become stable, and will react with H+ to give CH4.
try to do it on a piece of paper, and draw out the entire molecule and assaign charges. remember, c-o double bond is very stable.
but, since the c-o double-bond is very stable it will be regenaratad, and will give off a primary carbocation that will combine to MgBr to become stable, and will react with H+ to give CH4.
try to do it on a piece of paper, and draw out the entire molecule and assaign charges. remember, c-o double bond is very stable.
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Actually the C=O group is not that stable--that's why there is so much chemistry surrounding the carbonyls in the aldehydes, ketones, and acids. Here, in this problem, the faux pas was to use the Grignard reagent with an alcohol, which are not compatible. This is why standard Grignard reactions are carried out in aprotic solvents such as diethylether and THF. Here, a secondary alkoxide anion is generated, and the subsequent acid workup restores the alcohol to its original form.
you mean primary alkoxide anion is generated right?
so..
when are grignard reagents going to work?
in aprotic solvents, aldehydes, ketones, and esthers?
so if there is an (OH) somewhere on those molecules, will it work in an Aprotic solvent?
Your original alcohol (perhaps aldol would be a better term) is secondary, so when the CH3- pulls off the proton it will be a secondary alcohol. As far as I know all Grignard reactions are used in aprotic solvents, diethylether, THF, etc.. I don't believe that even in aprotic solvents an alcohol is a suitable reactant, because the acidic (pKa ~17) is still there.
J
Joey413
If only one eq is used it will be for the proton since it carries the highest partial positive charge and the CH3MgBR carries 1-(one full negative charge really unstable). Electrons of the carbon will most likely go to to this proton and stip it off, and the H2O and H+ will just reprotanate it regenerating the same thing. Why can't it attack the carbonyl carbon? It can because it does carry a positive charge on the carbon, but it will not attack the carbonyl carbon over the most partial positive hydrogen. If 2 eq's were to be used, then yeah, it would be possible to attack the carbonyl carbon only after depronating the H of the OH first. I hope this helps.
so in this case, the product would be CH3COOH (the starting material because the R- would strip the H+ off the alcohol, and then it would be protanated by the H+/H2O?
can you double check for the answer, because in the second step the H+ seems like to combine to the o- that resulted by attacking c=o to give alocohol functional group, and will form a diol.
can you check it please.
the answer i originally posted is right
Yes, I believe that is true. That is why Grignard Reagent are not to be used with solvents containing acidic protons, such as carboxylic acid or even water.
