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The mechanism of the reaction between 2-chloroethanol ($\ce{CH2ClCH2OH}$) and hydroxyl ions in aqueous solution forming ethylene oxide ($\ce{CH2CH2)O}$ probably contemplates two steps:

  1. $\ce{CH2ClCH2OH + OH- <=> CH2ClCH2O- + H2O}$

  2. $\ce{CH2ClCH2O- <=> (CH2CH2)O + H2O + Cl-}$

Is the mechanism valid for the following experimental reaction rate?

$\frac{-d[\ce{CH2ClCH2OH}]}{dt} = K_{exp}[\ce{CH2ClCH2OH}][\ce{OH-}]$

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According to Ballinger and Long (1959), the reaction mechanism is as such:

$\ce { (1) CH2ClCH2OH (aq) + OH^- (aq) -> CH2ClCH2O^- (aq) + H2O (aq)}$ (equilibrium)

$\ce { (2) CH2ClCH2O^- (aq) -> (CH2CH2)O (aq) + Cl^- (aq)}$ (slow)

This is not surprising since acid-base reactions, such as these proton transfers, take place very quickly. Although the second step may appear to be intramolecular, it may be slower since we are forming a strained epoxide ring and the $\ce {C-O^-}$ and the $\ce {C-Cl}$ bonds also need to be in a trans orientation for most effective orbital interaction of the relevant orbitals. Based on the above mechanism, your rate law seems to be correct, based on the assumption that the first step is much slower than the subsequent step and has come to establish an equilibrium.

Reference

Ballinger, P.; Long, F. A. The Reaction of Chlorohydrins and Hydroxide Ion in the Solvents H2O and D2O. J. Am. Chem. Soc., 1959, 81 (10), 2347–2352.

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Yes it is. Because alcohol is a weak acid and it is reasonable to consider first step is irreversible because of the second step almost occuring simultaneously. And also because that the second step is much faster than the first.

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