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Mechanism of MOMylation of alcohol with P2O5 and dimethoxymethane
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Mike Miller
Mechanism of MOMylation of alcohol with P2O5 and dimethoxymethane
I don't think you will find a nice and tidy mechanism for this reaction, though I don't think one cannot come up with a reasonable mechanism. The protection step should be a simple acid catalyzed addition of the (+)CH2OCH3 cation to the alcohol being protected. So you need an acid to generate it from dimethoxymethane. If you used pTsOH, this would work, but it will also give a competing mole of methanol. I anticipate that any methanol being generated can be removed by reacting it with P2O5, the anhydride of phosphoric acid. Its reaction will give some phosphoric acid (or its equivalent) to acid as the mineral acid reacting with the demethoxymethane.
This reaction is an alternative to using chloromethylmethyl ether, a highly carcinogenic reagent.
I don't think you will find a nice and tidy mechanism for this reaction, though I don't think one cannot come up with a reasonable mechanism. The protection step should be a simple acid catalyzed addition of the (+)CH2OCH3 cation to the alcohol being protected. So you need an acid to generate it from dimethoxymethane. If you used pTsOH, this would work, but it will also give a competing mole of methanol. I anticipate that any methanol being generated can be removed by reacting it with P2O5, the anhydride of phosphoric acid. Its reaction will give some phosphoric acid (or its equivalent) to acid as the mineral acid reacting with the demethoxymethane.
This reaction is an alternative to using chloromethylmethyl ether, a highly carcinogenic reagent.
I doubt they add water. That would negate the purpose of using P2O5. What would happen if only three molecules of water were present? (I only suggest this so you can rationalize the reaction using a mineral acid.) If water can react with P2O5, can methanol? If you are protonating to get dimethoxymethane, can't the methanol also react with the cation to reform dimethoxymethane? How might a reaction of methanol with P2O5 affect the reaction? Think le Chatliere.
I doubt they add water. That would negate the purpose of using P2O5. What would happen if only three molecules of water were present? (I only suggest this so you can rationalize the reaction using a mineral acid.) If water can react with P2O5, can methanol? If you are protonating to get dimethoxymethane, can't the methanol also react with the cation to reform dimethoxymethane? How might a reaction of methanol with P2O5 affect the reaction? Think le Chatliere.
Thanks for the posts. To recapitulate (correct me if I'm wrong): P2O5 + 3H2O --> 2H3PO4. This is the acid that will 'donate' a proton to one of the methoxy groups of dimethoxymethane. Elimination of MeOH leads to the formation of the (+)CH2OCH3 cation, which is then trapped by the alcohol. Water is still around and can trap the cation too with formation of HOCH2OCH3. The molecule of methanol (generated from dimethoxymethane) is trapped by P2O5.
It's a bit weird to state that on the one hand you need water to form the acid that initiates the reaction and on the other hand you don't want water to be there because it competes with the alcohol for (+)CH2OCH3. No?
Thanks for the posts. To recapitulate (correct me if I'm wrong): P2O5 + 3H2O --> 2H3PO4. This is the acid that will 'donate' a proton to one of the methoxy groups of dimethoxymethane. Elimination of MeOH leads to the formation of the (+)CH2OCH3 cation, which is then trapped by the alcohol. Water is still around and can trap the cation too with formation of HOCH2OCH3. The molecule of methanol (generated from dimethoxymethane) is trapped by P2O5.
It's a bit weird to state that on the one hand you need water to form the acid that initiates the reaction and on the other hand you don't want water to be there because it competes with the alcohol for (+)CH2OCH3. No?
This reaction is an alternative to using chloromethylmethyl ether, a highly carcinogenic reagent.
This reaction is an alternative to using chloromethylmethyl ether, a highly carcinogenic reagent.
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To recapitulate (correct me if I'm wrong):
P2O5 + 3H2O --> 2H3PO4. This is the acid that will 'donate' a proton to one of the methoxy groups of dimethoxymethane. Elimination of MeOH leads to the formation of the (+)CH2OCH3 cation, which is then trapped by the alcohol. Water is still around and can trap the cation too with formation of HOCH2OCH3.
The molecule of methanol (generated from dimethoxymethane) is trapped by P2O5.
It's a bit weird to state that on the one hand you need water to form the acid that initiates the reaction and on the other hand you don't want water to be there because it competes with the alcohol for (+)CH2OCH3. No?
Thanking you for the help you can provide
To recapitulate (correct me if I'm wrong):
P2O5 + 3H2O --> 2H3PO4. This is the acid that will 'donate' a proton to one of the methoxy groups of dimethoxymethane. Elimination of MeOH leads to the formation of the (+)CH2OCH3 cation, which is then trapped by the alcohol. Water is still around and can trap the cation too with formation of HOCH2OCH3.
The molecule of methanol (generated from dimethoxymethane) is trapped by P2O5.
It's a bit weird to state that on the one hand you need water to form the acid that initiates the reaction and on the other hand you don't want water to be there because it competes with the alcohol for (+)CH2OCH3. No?
Thanking you for the help you can provide
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Think about acetal formation and reversibility.
Think about acetal formation and reversibility.
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