# NGP in Finkelstein reaction

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

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?

• I- is not a good enough base to deprotonate OH – Waylander Jul 2 '20 at 8:39
• How do you say that? – Robin Singh Jul 2 '20 at 8:42
• 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 – Waylander Jul 2 '20 at 8:59
• Ok, now it makes sense – Robin Singh Jul 2 '20 at 9:13
• Iodide is the very opposite of a good base. HI is the strongest hydrohalogenic acid. – 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 $$\lt 100\% e.e.$$.

References:

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).
• While OP's mistaking iodide for a strong base is mind-boggling, he has a point - why NGP wouldn't change the picture here? – Mithoron Jul 2 '20 at 17:11
• @Mithoron: I agree with you. There is a possibility but it is very faint. But worth mentioning it. :-) Accordingly I change my answer – Mathew Mahindaratne Jul 2 '20 at 19:47
• @Mithoron What do you want it to do? Form an epoxide? – Zhe Jul 3 '20 at 2:31
• @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. – Mithoron Jul 3 '20 at 14:52