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What are the products of the Grignard reaction with methyl-magnesium iodide?
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Luis Abreu
What are the products of the Grignard reaction with methyl-magnesium iodide?
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.
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.
I like the idea of using a weak acid so I dont need to worry about how dilute my acid is. Doesnt the Mg(OH)I ionize in the solution? And in order to form Mg(OH)I, doesnt the NH4+ -> H+ + NH3 where then the NH3 +H2O -> OH- +NH4? But then that process would self eliminate?More
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.
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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