Why must grignard reactions be anhydrous?

1 Answer
anor277
Jun 3, 2016

Because the natural enemy of the Grignard reagent, of any organometallic reagent, is the water molecule.

Explanation:

#RH_2CX + Mg rarr RH_2C^(delta-){""^(delta+)MgX}#

The ipso carbon in a Grignard reagent has carbanionic character; it reacts as would #RH_2C^-#. With water, its reaction wold be rapid and irreversible:

#RH_2C^(delta-){""^(delta+)MgX} + H_2O rarr RCH_3 +MgX(OH)#

Sometimes you can exploit this reactivity. Suppose, for another experiment, you wanted to have the deuterium (i.e. #""^2H=D#) or tritium labelled (i.e. #""^3H#) alkane.

#RH_2C^(delta-){""^(delta+)MgX} + D_2O rarr RCH_2D +MgX(OD);#
#D=""^2H#

Making the Grignard and quenching it with isotopically labelled water would be a convenient method of making the labelled alkane. Deuterium and tritium oxides (heavy waters) are fairly cheap; certainly it would be a lot cheaper than buying the deuterated alkanes.

Note also that when you make a Grignard reagent you go to some unlittle effort in drying your ethers and drying your halocarbon. Typically, an ether would be distilled from sodium or potassium or sodium/potassium alloy.

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