10
$\begingroup$

When 1-methoxycyclohexene is treated with dilute aqueous hydrochloric acid, methanol is produced along with another organic product. Identify this product. Explain why methoxycyclohexane is inert under these reaction conditions.

Methoxycyclohexene and methoxycyclohexane

I don't understand why it isn't just a simple addition to the double bond. I get that there's conjugation but is this what makes things different? I don't get why methanol is the leaving group and therefore I cannot predict the other product. Also, it says that methoxycyclohexane (the same molecule without the double bond) is inert under these conditions. Why does it not undergo substitution? The oxygen could surely be protonated and then methanol would be a good leaving group which would give rise to substitution (probably SN1). Why is this not the case?

$\endgroup$
7
$\begingroup$

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.

enter image description here

$\endgroup$
  • $\begingroup$ Why wont methoxycyclohexane undergo substitution? $\endgroup$ – RobChem Jan 28 '15 at 22:53
  • 1
    $\begingroup$ 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. $\endgroup$ – Elena Cosimi Jan 29 '15 at 15:21
4
$\begingroup$

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.

I'll add a figure in a minute.

$\endgroup$
  • $\begingroup$ Why wont methoxycyclohexane undergo substitution? $\endgroup$ – RobChem Jan 28 '15 at 22:38
  • $\begingroup$ 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. $\endgroup$ – ron Jan 28 '15 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.