When $\ce{OH-}$ interacts with a halogenoalkane, for instance $\ce{CH3CH2Cl}$, it acts as a nucleophile and undergoes $\mathrm{S_N2}$ mechanism, which results in the following reaction:

$$\ce{CH3CH2Cl + OH- -> CH3CH2OH + Cl-}$$

Since $\ce{Cl-}$ has a free pair of electrons, can it act as a nucleophile that will reverse the previous reaction?

$$\ce{CH3CH2OH + Cl- -> CH3CH2Cl + OH-}$$

Since an $\ce{OH}$ group is more electronegative than $\ce{Cl}$ group, the carbon atom to which it attaches would have a greater partial positive charge ($\delta+$), does this mean it's more prone to nucleophilic substitution?

  • $\begingroup$ Can such nucleophilic substitution occur on halogenoalkanes with fluorine groups? My textbook says it occurs very slowly, does this mean it occurs? $\endgroup$
    – Cheng
    Apr 20, 2020 at 5:57
  • $\begingroup$ A nucleophilic substitution will occur only when the nucleophile approaches the substrate and also attack at the carbon to displace the leaving group. More partial positive charge means that the chloride might solvate the substrate slightly better as compared to OH- in the alkyl halide case, but due to higher electronegativity and charge density of the former,hydroxide will be the species which "actually attacks" better $\endgroup$ Apr 20, 2020 at 5:58
  • $\begingroup$ Fluorine is a very bad leaving group,so you can't really use simple SN2 on fluoroalkanes for getting an appreciable yield $\endgroup$ Apr 20, 2020 at 6:00

1 Answer 1


In general, yes, $\mathrm{S_N2}$ reactions are reversible. But not in this particular case.

Rates of $\mathrm{S_N2}$ reactions depend on several factors: the nucleophile, the leaving group, the alkyl group undergoing substitution, and so on. In the context of your question, the leaving group ability is quite possibly the most important. Iodide, $\ce{I-}$, is an excellent leaving group. Fluorine, $\ce{F-}$, is an extremely poor leaving group. Again, there are several reasons which dictate leaving group ability, but at the simplest level, it has to do with the stability of the negative charge on the leaving group. This broadly correlates with basicity: iodide is hardly basic at all (its conjugate acid, $\ce{HI}$, is a very strong acid) and is quite happy to exist as $\ce{I-}$. Conversely, fluoride is reasonably basic (its conjugate acid, $\ce{HF}$, is a weak acid) and is not all that happy existing as $\ce{F-}$.

Using this logic you will find that hydroxide ion $\ce{OH-}$ is a very poor leaving group. Consequently, it's almost impossible to get ethanol (or other alcohols) to directly react in an $\ce{S_N2}$ fashion. So in the particular case you have raised, no, the reaction is effectively irreversible.*

The way around the leaving group problem is to convert the hydroxyl group to something that leaves better. Because alcohols are common, lots of people have devoted effort into making sophisticated ways of doing this. But by far the simplest way is to simply protonate it:

$$\ce{ROH + H+ <=> ROH2+}$$

The leaving group now is not hydroxide $\ce{OH-}$; it's now water $\ce{H2O}$, which is a much, much better leaving group. Consequently, if you treat alcohols with $\ce{HCl}$, you have a much better chance of getting an alkyl chloride, as opposed to if you were to just add chloride ion (in the form of e.g. $\ce{NaCl}$). The $\ce{H+}$ protonates the alcohol, and the $\ce{Cl-}$ does the actual substitution.

For an example of a reversible $\mathrm{S_N2}$ reaction, the best example is one where the nucleophile is the same as the leaving group. For example, consider a chiral alkyl iodide $\ce{R1R_2R_3CI}$. Let's say we have an enantiomerically pure sample of this alkyl iodide. Since $\mathrm{S_N2}$ reactions occur with inversion, if we treat it with sodium iodide, can we effect an $\mathrm{S_N2}$ reaction which will convert it entirely to the opposite enantiomer ?

The answer is no, and that's precisely because the reverse reaction will also occur. Both the forward and reverse reactions have the same nucleophile and the same leaving group, so in this case there aren't any considerations which would make one less effective than the other.

* "effectively" because technically speaking every reaction is reversible.

  • $\begingroup$ Hoe is every reaction reversible? Are reversible reactions ideal by definition and thus not possible to realise in real life? Like so for reversible processes $\endgroup$
    – Micelle
    Apr 20, 2020 at 15:23
  • 5
    $\begingroup$ no, the usage of reversible here is not the same as the idea of a reversible process. Reversible reactions simply refer to those which can happen in either the forward or backward direction, which in principle, is true of every chemical reaction (en.wikipedia.org/wiki/Microscopic_reversibility). Sometimes the rate of one far outstrips the other and the position of equilibrium lies very far to the left or to the right. Colloquially we call those "irreversible". Technically speaking they are still reversible in the original sense, though. $\endgroup$ Apr 20, 2020 at 16:07

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