The question is to find the product in the following reaction-

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In the answer given, they have replaced Cl with I and inverted the configuration at that carbon(Finkelstein reaction).

But $\text{I}^-$ being a good base, wouldn't it first take away O's acidic H and then $\text{O}^-$ will show Neighboring Group Participation attacking the carbon with chlorine and finally resulting in retention of configuration?

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    $\begingroup$ I- is not a good enough base to deprotonate OH $\endgroup$ – Waylander Jul 2 '20 at 8:39
  • $\begingroup$ How do you say that? $\endgroup$ – Robin Singh Jul 2 '20 at 8:42
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    $\begingroup$ I- is really not a good base. The pKa of the OH proton is around 16. Iodide in EtOH does not give EtO- and HI $\endgroup$ – Waylander Jul 2 '20 at 8:59
  • $\begingroup$ Ok, now it makes sense $\endgroup$ – Robin Singh Jul 2 '20 at 9:13
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    $\begingroup$ Iodide is the very opposite of a good base. HI is the strongest hydrohalogenic acid. $\endgroup$ – Jan Jul 3 '20 at 6:18

You are thinking too far ahead for a simple reaction. First of all, $\ce{I-}$ is not a very strong base to abstract a protone from alcohol. $\ce{I-}$ is the conjugate base of very strong acid, $\ce{HI}$. Yet, $\ce{I-}$ is a very good nucleophile. Therefore, the condition given ($\ce{NaI}$ in anhydrous acetone; $\ce{NaI}$ dissolves in anhydrous acetone and therefore exists as $\ce{I-}$ in solution) is suitable only for nucleophilic substitution reaction and also favored $\mathrm{S_N2}$ mechanism. Since the byproduct form ($\ce{NaCl}$) is not soluble in acetone, the reaction favors only the forward reaction:

$$\ce{R-Cl + Na+ + I- -> R-I + NaCl_{(s)}}$$

Since it is a $\mathrm{S_N2}$ reaction, inversion is the result ($(1R,2S)$ to $(1R,2R)$ with closed to $100\% e.e.$).

However, note that there is a neighboring group participation (NGP) possible in 1,2-halohydrins in certain reactions (Ref.1). As OP suggested, if NGP is contributed here (without strong base I doubt it though), the product from NGP is going to be the retention product with $(1R,2S)$ configuration. Thus, total $\%e.e.$ of final product is $\lt 100\% e.e.$.


  1. T. A. Geissman, Richard I. Akawie, "Rearrangements of Halomagnesium Derivatives of Halohydrins," J. Am. Chem. Soc. 1951, 73(5), 1993–1998 (https://doi.org/10.1021/ja01149a029).
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    $\begingroup$ While OP's mistaking iodide for a strong base is mind-boggling, he has a point - why NGP wouldn't change the picture here? $\endgroup$ – Mithoron Jul 2 '20 at 17:11
  • $\begingroup$ @Mithoron: I agree with you. There is a possibility but it is very faint. But worth mentioning it. :-) Accordingly I change my answer $\endgroup$ – Mathew Mahindaratne Jul 2 '20 at 19:47
  • $\begingroup$ @Mithoron What do you want it to do? Form an epoxide? $\endgroup$ – Zhe Jul 3 '20 at 2:31
  • $\begingroup$ @Zhe -OH can well enough participate in SN1, but it's effect is much more transient then that, more like first step on the road to epoxide and then step backwards. $\endgroup$ – Mithoron Jul 3 '20 at 14:52

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