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What are the products of the reaction of 1-methoxycyclohexene with dilute hydrochloric acid?
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Aidan Doherty
What are the products of the reaction of 1-methoxycyclohexene with dilute hydrochloric acid?
Vinyl ethers are "masked" carbonyls, therefore in acidic aqueous environment the following pathway takes place: alkene protonation assisted by the methoxy group followed by nucleophilic attack of water with hemiacetal formation. Subsequent intramolecular proton transfer to the methoxy group makes it a good leaving group and cyclohexanone (the product) is formed while methanol is released.
Vinyl ethers are "masked" carbonyls, therefore in acidic aqueous environment the following pathway takes place: alkene protonation assisted by the methoxy group followed by nucleophilic attack of water with hemiacetal formation. Subsequent intramolecular proton transfer to the methoxy group makes it a good leaving group and cyclohexanone (the product) is formed while methanol is released.
Methoxy cyclohexane could undergo protonation on the methoxy group but the reaction conditions (aqueous HCl) are not harsh enough to promote cleavage of such ethereal bonds. However, use of HBr or HI in organic solvents (since bromide and iodide are much more nucleophilic than chloride anions), one could achieve ether cleavage thus leading to the formation of cyclohexanol from methoxycyclohexane. The mechanism consists of initial protonation of methoxy group and nucleophilic attack of bromide on the methyl group with subsequent release of methyl bromide and cyclohexanol.More
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You're close. You are right in thinking that the first step involves protonation of the double bond. But which end of the double bond will you protonate? If you protonate the carbon bearing the methoxy group you wind up with a secondary carbocation. However if you protonate the other end of the double bond you wind up with a carbocation that has the methoxy group attached to it. Draw a resonance structure and convince yourself that the methoxy group can stabilize this carbocation.
So now we have a carbocation bearing a methoxy group. The next step is 1) attack by water and ejection of a proton from the water to yield a neutral hemiacetal. Under acidic conditions the hemiacetal will hydrolyze to produce cyclohexanone and methanol.
Methoxycyclohexane won't undergoe this reaction since there is no double bond to protonate.
You're close. You are right in thinking that the first step involves protonation of the double bond. But which end of the double bond will you protonate? If you protonate the carbon bearing the methoxy group you wind up with a secondary carbocation. However if you protonate the other end of the double bond you wind up with a carbocation that has the methoxy group attached to it. Draw a resonance structure and convince yourself that the methoxy group can stabilize this carbocation.
So now we have a carbocation bearing a methoxy group. The next step is 1) attack by water and ejection of a proton from the water to yield a neutral hemiacetal. Under acidic conditions the hemiacetal will hydrolyze to produce cyclohexanone and methanol.
Methoxycyclohexane won't undergoe this reaction since there is no double bond to protonate.
The reaction is being run in dilute acid. SN1 (protonation of the methoxy group followed by elimination of methanol to generate a secondary cyclohexyl carbocation) is unlikely under these weak acid conditions because the cyclohexyl carbocation is not particularly stable (e.g. it is a high energy intermediate). SN2 is also unlikely since water is a weak nucleophile and methoxide is not a great leaving group.More
Vinyl ethers are "masked" carbonyls, therefore in acidic aqueous environment the following pathway takes place: alkene protonation assisted by the methoxy group followed by nucleophilic attack of water with hemiacetal formation. Subsequent intramolecular proton transfer to the methoxy group makes it a good leaving group and cyclohexanone (the product) is formed while methanol is released.
Vinyl ethers are "masked" carbonyls, therefore in acidic aqueous environment the following pathway takes place: alkene protonation assisted by the methoxy group followed by nucleophilic attack of water with hemiacetal formation. Subsequent intramolecular proton transfer to the methoxy group makes it a good leaving group and cyclohexanone (the product) is formed while methanol is released.
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You're close. You are right in thinking that the first step involves protonation of the double bond. But which end of the double bond will you protonate? If you protonate the carbon bearing the methoxy group you wind up with a secondary carbocation. However if you protonate the other end of the double bond you wind up with a carbocation that has the methoxy group attached to it. Draw a resonance structure and convince yourself that the methoxy group can stabilize this carbocation.
So now we have a carbocation bearing a methoxy group. The next step is 1) attack by water and ejection of a proton from the water to yield a neutral hemiacetal. Under acidic conditions the hemiacetal will hydrolyze to produce cyclohexanone and methanol.
Methoxycyclohexane won't undergoe this reaction since there is no double bond to protonate.
You're close. You are right in thinking that the first step involves protonation of the double bond. But which end of the double bond will you protonate? If you protonate the carbon bearing the methoxy group you wind up with a secondary carbocation. However if you protonate the other end of the double bond you wind up with a carbocation that has the methoxy group attached to it. Draw a resonance structure and convince yourself that the methoxy group can stabilize this carbocation.
So now we have a carbocation bearing a methoxy group. The next step is 1) attack by water and ejection of a proton from the water to yield a neutral hemiacetal. Under acidic conditions the hemiacetal will hydrolyze to produce cyclohexanone and methanol.
Methoxycyclohexane won't undergoe this reaction since there is no double bond to protonate.
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