I came across this question enter image description here

I thought that the answer should be option (a)but the answer is given option (d)

I thought that other products might be a result of the following mechanismenter image description here

But this cannot be the case as a positive charge still remains on the carbon atom. Can you please help me figure out where I am going wrong?
Or is this a possible mechanismenter image description here But I am still unclear about how the double bond will reappear between 3 position and the carbon that carried Chlorine

  • $\begingroup$ FWIW, in my browser both links say "The image cannot be displayed because it contains errors". That might just be me and my somewhat odd browser, but it's good if you can insert images into your question rather than links. Anyway, you got an answer, so I guess someone can see them ;) $\endgroup$
    – airhuff
    May 2, 2017 at 19:17
  • $\begingroup$ en.wikipedia.org/wiki/Arene_substitution_pattern $\endgroup$
    – Mithoron
    May 2, 2017 at 20:32
  • $\begingroup$ @airhuff I am rather new to this site and so I don't know how to add an image in my question. I admit adding links of images is a cumbersome process. $\endgroup$ May 3, 2017 at 5:32

1 Answer 1

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All the reactions eliminate chloride (a) by the obvious direct attack, the other products b and c arise from methoxide attack on the 3 and 5 positions of the furan and chloride loss via an exomethylene intermediate. The system is somewhat similar to a polyallyl choride where attack can be direct or at the 3 or 5 position.

  • $\begingroup$ I'm sorry I don't know what an exomethylene intermediate is. $\endgroup$ May 2, 2017 at 11:44
  • $\begingroup$ Double band from carbon at 2 position (next to oxygen) to carbon that carried chlorine $\endgroup$
    – Waylander
    May 2, 2017 at 12:11
  • 1
    $\begingroup$ I have no access to a drawing package. Chlorine leaves as Cl- just as in the direct attack. Consider product c to be result of Michael attack at the 3 position and move the bonds around to kick out Cl- $\endgroup$
    – Waylander
    May 2, 2017 at 12:58
  • 1
    $\begingroup$ The mechanism you added is fundamentally correct. As TAR86 says methoxide will remove the proton leading to re-aromatisation. The same mechanism applies to attack at the 5 position. $\endgroup$
    – Waylander
    May 3, 2017 at 9:21
  • 1
    $\begingroup$ The exo carbon will pick up a proton from the MeOH. You could visualise it as a 1,3 proton shift $\endgroup$
    – Waylander
    May 3, 2017 at 12:01

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