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I want to react my ketone with methylmagnesium iodide (the Grignard reagent). I then need to use a dilute acid solution to protonate the $\ce{MgI-O-R'}$ product to produce my alcohol: $\ce{H-O-R'}$

Now I was trying to figure out what acid to use. The first thing that came to mind was sulfuric acid. What would the products be?

In general the reaction should look like: $$\ce{R'=O + XMgI + H-A -> X-R'-OH + MgI+ + A-}$$

Now if I use sulfuric acid the conjugate base reacted with the Grignard products is something like $\ce{MgIHSO4}$. Is this correct?

This molecule doesn't look very stable so my thinking is that the $\ce{Mg}$ donates an electron to $\ce{I}$ in $\ce{MgI+}$ to form elemental iodine and $\ce{Mg^2+}$, then this reacts with $\ce{HSO4}$ to produce $\ce{Mg(HSO4)2}$ and $\ce{MgSO4}$.

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  • $\begingroup$ Your product is that X-R'-OH. What's the point in worrying about byproducts? Also, that Mg-something is an ionic salt and hence is not made of molecules. Also, you are not going to crystallize it anyway, so there is not even a definitive formula. It's just such-and-such ions in the solution. $\endgroup$ – Ivan Neretin Sep 2 '15 at 20:50
  • $\begingroup$ I would like to recycle as much of my byproduct as possible since the iodine is expensive. I agree about the Mg salt solution... Although I am not sure about the elemental iodine presence, I guess I'll just see when I do the experiment. $\endgroup$ – jakerz Sep 2 '15 at 20:54
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Your reaction scheme doesn't make much sense. $\ce {R'=O}$ is not a ketone. Is $\ce X$ supposed to be $\ce Me$? This is confusing since $\ce X$ is most commonly used as the symbol of a halogen. I would rewrite as: $$\ce {R'RC=O + MeMgI +H_2O-> R'RC(Me)OH + Mg(OH)I}$$ $\ce {R'(R)C=O}$ is implied.

Make sure you use very dilute aqueous sulphuric acid since your product is a tertiary alcohol. I would rather use saturated aq. $\ce {NH_4Cl}$. Even water will do but I suspect that the work-up might not be as smooth (a less water soluble magnesium salt will be formed). There will be no iodine formed and in any case I don't find the idea of recycling iodine cost effective. It will cost you more to isolate iodine that can be pure enough to work in the notoriously capricious Grignard reagent formation. In fact I would suggest you to buy a commercially available solution of $\ce {MeMgX \ (X=Cl, Br, I)}$ since preparing a Grignard can be very challenging and you will also have to handle toxic and volatile $\ce {MeI}$, have access to dry ether/THF etc. I assume you follow a detailed experimental procedure from the literature.

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  • $\begingroup$ I like the idea of using a weak acid so I don't need to worry about how dilute my acid is. Doesn't the Mg(OH)I ionize in the solution? And in order to form Mg(OH)I, doesn't the NH4+ -> H+ + NH3 where then the NH3 +H2O -> OH- +NH4? But then that process would self eliminate? $\endgroup$ – jakerz Sep 3 '15 at 0:28

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