I need 1 M Sodium Methoxyde solution.
Can I simply add NaOH crystal ( 40 grams) into CH3OH (pa) reagent (1 L)?
To prepare Sodium Methoxyde Solution (1 M) , take 4.6 ml in dry methanol , add 2.3 gram of freshly cut sodium metal till it is slowly dissolved. Use this turbid solution as 1M sodium methoxide for the reaction. If you want more sodium methoxide use Methanol as well as Na metal weight accordingly in 1:1 molar proportion.
Please give us (the answerers) the following infos prior you start a dangerous preparation.
Do you need sodium methoxyde for WHAT ??
- For saponification of ester or another reaction (e.g. Michael-addition, C-H activation ...?
FOR SAPONIFICATION read this link:
http://link.springer.com/article/10.1007/BF02531175
Abstract
Evidence is offered to support the contention that methanolysis precedes the saponification of esters in methanolic solutions of sodium hydroxide. This results from the hydroxide-alkoxide equilibrium which greatly favors methoxide formation even in the presence of rather considerable amounts of water. Saponification-reesterification methods of methyl ester formation are shown to be actually extensions of methanolysis procedures. A simplified method is proposed for the preparation of fatty acid methyl esters.
Scientific Journal Series 7706, Agricultural Experiment Station, University of Minnesota, St. Paul, Minnesota 55101.
For the sodium methoxide synthesis according to :
Organikum (German / pages 704 and 715) and Houben-Weyl Bd. VI/2 pages 1-70 (1963)
you need for the required sodium quantity (freshly cut) the tenfold quantity of methanol pro analysis .
You need a three-necked flask with magnetic stirring, thermometer and condenser and dripping funnel. Give the required quantity of Na in the flask and adjust the drop vlocity for a gentle boiling of MeOH.
please read :
https://en.wikipedia.org/wiki/Sodium_methoxide
JRG
PLEASE NEVER GIVE SODIUM ON METHANOL (STRONG EXOTHERMIC REACTION / DANGEROUS)
I agree with my colleagues, I prepared also my sodium methoxide with sodium metal. This reaction is very exothermic so get your reflux condenser ready as well as a cooling bath just in case. I would advice you maybe to do several small batches, and not on 1 liter scale for safety reason.
2 Na + 2 CH3OH → 2 CH3ONa + H2
Sodium methoxide is prepared by carefully treating methanol with sodium:
Take 4.6 ml in dry methanol , add 2.3 gram of freshly cut sodium metal till it is slowly dissolved. Use this turbid solution
The reaction is so exothermic that ignition is possible. The resulting solution, which is colorless, is often used as a source of sodium methoxide, but the pure material can be isolated by evaporation followed by heating to remove residual methanol. The solid hydrolyzes in water to give methanol and sodium hydroxide; commercial-grade product can be contaminated with the hydroxide. The solid and especially solutions absorb carbon dioxide from the air, thus diminishing the effectiveness of the base.
In the solid form, sodium methoxide is polymeric, with a sheet-like array of Na+ centers, each bonded to four oxygen centers.
The structure, and hence its basicity, of sodium methoxide in solution depends on the solvent. It is a significantly stronger base in DMSO where it is more fully ionized and free of hydrogen bonding.
Follow Professor Deshmukh's advice. (Mixing NaOH and MeOH just gives you an high pH aqueous MeOH solution.). This is very elementay chemistry!!
Thank you all, for the answers of my questions also the link attached.
Especially @JRG.
I want to analyze (quantitavely) the content of lauric acid in biscuit product fortified with virgin coconut oil and microlgae using Gas Chromatgraphy. The procedure I read from some references, that we can use NaOMe 1 M to make an ester of the FA. I wondered whether I could simply mix NaOH and MeOH :-) .
The recipes given here are for 0.1M, correct? Or does it go without saying this is made in small 100mL batches?
TY
Madhukar Baburao Deshmukh Your answer does not take into consideration that almost all of your methanol will be consumed in the reaction, and you'll be left with something that is so concentrated or even dry from the reaction heat itself. Assuming that you don't lose any methanol to evaporation, you will be left with .55 mL.
@JY..cool your reaction vessel as the reaction is highly exothermic..you will succeed..
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