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I recently came across the following in my OChem class:

Ethylene dichloride on reaction with NaI in acetone reacts via E2 mechanism to give ethene.

I have three questions:

  1. Since it is an E2 mechanism, a strong base should be present, but the iodide ion is a weak conjugate base of its corresponding acid. So why is the mechanism E2?

  2. Bases are thought to be as proton acceptors. According to the above reaction, however it seems that the iodide ion attacks the chlorine on one of the carbons. Why is that so?

  3. Chloride ion is less stable than iodide anion as more charge dispersion takes place in the latter. Then why is the chlorine a leaving group here, i.e. why does the reaction even occur?

I have just started OChem, so please correct me if my question has mistakes.

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    $\begingroup$ NaCl is insoluble in acetone, this is what drives Finkelstein $\endgroup$
    – Waylander
    Dec 5 '20 at 14:15
  • $\begingroup$ Ohh right.. completely overlooked that part about Le Chatlier's ...that does answer the third part. What about the E2 part, however how does the reaction proceed via E2 when iodide ion is not a strong base? $\endgroup$
    – sfsg
    Dec 5 '20 at 16:00
  • $\begingroup$ @Waylander can you please share some insights $\endgroup$
    – sfsg
    Dec 8 '20 at 16:19
  • $\begingroup$ I would if I could find a reference to this reaction. 1,2-Diodoethane is a stable compound see chemistry.stackexchange.com/questions/123561/… $\endgroup$
    – Waylander
    Dec 8 '20 at 18:23

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