Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about reactivity of substituted furan in H3O

  1. Jun 28, 2005 #1
    My textbook doesn't discuss this and the ACS exam study guide asks about it.

    I don't know how to name this, but basically it is almost a tetramethoxyfuran, except that one of the methoxy groups is actually dimethyl ether (methylmethoxy?) So, i'm going to take a stab at this and call is 1-methylmethoxy-2,3,4-methoxy furan. I hope this is correct or at least nearly correct.

    Anyhow, in acidic solution (H3O) what happens? The book says that only one of the methoxy groups reacts to give a hydroxy group there, the other two and the methylmethoxy remain unreacted. I thought that any ether (ROR) needed more extreme conditions to react. Obviously I am mistaken. Can someone explain?

  2. jcsd
  3. Jun 28, 2005 #2


    User Avatar
    Science Advisor

    Are you familiar with acetals? How do acetals react in the presence of acid? What makes them different from a regular ether?
  4. Jun 28, 2005 #3
    I thought acetal when I first saw it but then it looked weird so I wasn't 100% on it and then thought of reactions of ethers. I definitely needed some acetal review, and i've spent the last 20 minutes going through the reactions that would create this compound, and that would hydrate it.

    Oh, and on the questions of why acetals react differently, is it because of the electron withdrawing affect of two oxygen atoms instead of one makes the carbon much more reactive toward attack by a nucleophile (H2O)?

    Thanks for the help!
    Last edited: Jun 28, 2005
  5. Jun 28, 2005 #4


    User Avatar
    Science Advisor

    I think of the difference between acetals and ethers as in both cases you have a decent leaving group (alkoxide) but with an acetal you have the lone pairs from the other oxygen to assist in kicking out the leaving group (making an oxocarbenium ion).

    Does that help any?
  6. Jun 28, 2005 #5
    I am seeing that this reaction can go one of three ways. It can go acetal. It can go hemi-acetal through cleavage of the ring oxygen, or it can go hemi-acetal by cleavage of the -OCH3 group in question. Protonation of the -OCH3 group makes it a good leaving group, If it were SN1 the methanol would leave a carbocation adjacent to the ring oxygen. I am not really sure how this works because i'd think that the ringed oxygen pulled electron density away from the adjacent carbon making it impossible to give a carbocation. Maybe this is then SN2, and the carbon adjacent to the ring oxygen is then a good candidate for SN2 nuclephilic attack (by H2O) due to it being very positive. That is all I could come up with to explain this reaction.

    What is oxocarbenium? Is that an anion where H+ or +OR group leaves electrons behind or something? My book has no reference to this can you explain what it is and what mechanism you are proposing here? I am confused.

    Thanks again!
    Last edited: Jun 28, 2005
  7. Jun 28, 2005 #6


    User Avatar
    Science Advisor

    Nucleophilic attack on oxygen is almost always a bad thing. There are very few reagents which permit this.

    An oxocarbenium ion is a resonance structure of for a carbocation adjacent to an oxygen atom. In this resonance structure, a lone pair from the oxygen has formed a double bond with the carbon so that the oxygen bears a formal positive charge. This is actually a reasonably good resonance structure. Like carbonyl groups, nucleophiles will attack oxocarbenium ions at carbon to ultimately quench the positive charge.

    They best way to think of why oxocarbenium ions relative to regular carbocations is that you can draw another reasonable resonance structure and therefore oxocarbenium ions are actually lower in energy than unstabilized carbocations!

    Does this help your understanding?
  8. Jun 28, 2005 #7
    Yes, I can see what you are saying. Let me play around with it and see if I can explain it to myself through drawings. Thanks so much!

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook