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Topic: Reaction of a chloroether with an iodide ion in a SN2 reaction  (Read 3797 times)

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Offline edwinksl

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Let's say we take the rate of the SN2 reaction of the n-BuCl with KI in acetone at 50°C to be 1.

My notes say the relative rate of the reaction between MeOCH2Cl (called methoxymethyl chloride or MOM chloride for short) and KI is then 920.

Quite clearly, the transition state (TS) is being stabilized. The actual full quote from my notes is: "Conjugation with oxygen lone pair accelerates reaction." Presumably, the conjugation between the oxygen lone pair with the filled p orbital on the pentavalent carbon atom in the TS stabilizes the TS.

I am interested in how exactly the conjugation works. More specifically, the filled p orbital at the carbon atom in the TS is overlapping with which other orbital on the oxygen atom?

A case in point: If you react an α-bromo carbonyl compound (MeCOCH2Br for example) with KI, the SN2 reaction is accelerated because the TS is stabilized by the overlap of the π* MO (empty) of the C=O bond with the filled p orbital of the pentavalent carbon in the TS. What would be the case for a reaction between MeOCH2Cl and KI? My guess is that the σ* orbital of the C-O bond (this C belongs to the -Me group!) in MeOCH2Cl overlaps with the p orbital of the pentavalent carbon.

Thanks.

Offline kiwi

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Re: Reaction of a chloroether with an iodide ion in a SN2 reaction
« Reply #1 on: May 30, 2007, 05:28:41 PM »
just an idea, but i would think the activation wouldn't be due to stabilisation of the transition state, but due to bond weakening by the oxygen lone pair. this could potentially go all the way and form an oxocarbenium ion, providing a new, easier pathway:

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