# What would happen if water reacted with Grignard reagent?

Dec 21, 2015

A fizz and a bang and a pop, and the ruination of your hard work.

#### Explanation:

Organometallic reactions in general (and lithium reagents reactions in particular) usually involve some effort to make the reaction water-free. Solvents (ether and THF) are dried over alkali metal, and freshly distilled; glassware is sometimes dried under vacuum with a soft bunsen flame (a heady mix where you have ethereal solvents!); the alkyl halide reactant is passed thru alumina to dry it and remove HCl.

Why bother with all these precautions? Well, because water is the natural enemy of the Grignard reagent, and of organometallic reagents in general . Not only will water react irreversibly with your Grignard, the molecular mass of water is so low that a little goes a long way, and ineptly dried solvent or glassware will seriously reduce your yield. Consider (finally)! the reaction of a Grignard with water:

$R - M g X + {H}_{2} O \rightarrow R - H + M g X \left(O H\right)$.

This is very facile, and will occur even with atmospheric moisture.

Note that sometimes we can actually exploit this reactivity. Suppose (for some other experiment) you wanted to label an alkane with ""^2H, or ""^3H.

$R - M g X + {D}_{2} O \rightarrow R - D + M g X \left(O D\right)$
D = ""^2H

The simplest way would be to take the corresponding alkyl halide, make the Grignard, and quench the Grignard (carefully!) with heavy water or tritiated water (as shown above). This would be much cheaper than buying the ${d}_{1}$-alkane. It would also get your lazy graduate students off the computer, and into the laboratory doing stuff. Deuterium oxide is certainly very cheap, much less sponds than perdeuterated organic solvents.