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We know that haloalkanes are able to undergo $\mathrm{S_N1}$ or $\mathrm{S_N2}$ in alkaline hydrolysis, and the hydroxide ion is the nucleophile responsible for attack. But when we talk about hydrolysis, isn't it a reaction with water? Where does the hydroxide ion that attacks the haloalkane really come from; the water, or the alkaline catalyst?

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Consider the reaction of a haloalkane with aqueous sodium hydroxide to produce the alcohol plus sodium chloride. The net reaction consumes one hydroxide ion for each alcohol molecule produced, so it is not really correct to refer to it as a catalyst.

(1) If this reaction occurs via the $\mathrm{S_N2}$ mechanism, a one-step displacement mechanism, you are correct to say that hydroxide (being a much better nucleophile than neutral water) is the nucleophile attacking carbon.

(2) If the reaction occurs via the $\mathrm{S_N1}$ mechanism (note that which mechanism is preferred depends on the structure of the haloalkane, and in some cases a haloalkane can react via both mechanisms), you have a two-step process in which the haloalkane initially dissociates (a slow step) to form a carbocation plus halide ion. The carbocation then reacts very rapidly with whatever nucleophile is handy to form the alcohol. Water is likely more abundant than hydroxide, and is a strong enough nucleophile to react instantly with a carbocation, so most of the product will be formed that way, with release of $\ce{H+}$, rather than by combination of the carbocation directly with hydroxide. The $\mathrm{S_N1}$ reaction of a haloalkane in this manner with excess water or other polar solvent is often referred to as a "solvolysis" reaction.

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  • $\begingroup$ thanks for your information :) So from your explanation, can I conclude that 1. in SN2, it is the hydroxide ion(which is not from water but from alkaline) that acts as a nucleophile and in SN1, it is the water itself. If the statement for SN1 is justified, does this mean that the hydroxide ion is actually not required in the hydrolysis process? $\endgroup$ – txn Sep 17 '15 at 14:12
  • $\begingroup$ Both conclusions are correct. However, the reaction might proceed faster in the presence of ions than in pure water. Also, please note that hydroxide ion in water exchanges protons with water at a diffusion-controlled rate, so the OH groups that end up in the product could still have originated from the water rather than from the NaOH that you started with. $\endgroup$ – iad22agp Sep 17 '15 at 14:47

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