for my etherification of an alcoxide with alkylbromide I used TBAI as a catalyst. The catalyst transfers the alkylbromide into an alkyliodide which is more reactive.

I was wondering, what is the driving force for the Iodide to bond to the carbonchain instead of the bromide? Is there a mechanism to explain that?

  • $\begingroup$ Tell us more about the conditions you used $\endgroup$
    – Waylander
    Jan 18, 2019 at 19:40
  • $\begingroup$ Classic conditions as usally described in literature. NaH in THF at room teperature, then add the alcohol (cooling because is quite exothermic) and TBAI, then add the alkylbromide. Stir overnight at RT. $\endgroup$
    – Inselino
    Jan 18, 2019 at 20:07
  • $\begingroup$ iodide replacing bromide is usually driven by solubilities, the iodide is the more soluble is organic media $\endgroup$
    – Waylander
    Jan 18, 2019 at 20:51

2 Answers 2


I was wondering, what is the driving force for the Iodide to bond to the carbonchain instead of the bromide?

It's not so much a driving force as much as it is a dynamic equilibrium of sN2/sN1 reactions. Halides can exchange each other in equilibrium if one is not removed thus processes that remove the unwanted product such as the Finkelstein reaction are use to achieve complete substitution (think Le'Chatlier's Principle).

$$\ce{R-Br +I- <=> R-I + Br-}\qquad\quad\ \ \ \text{(Equillibrium)}$$ $$\ce{R-Br +I- +Na+ <=> R-I + NaBr (s)}\qquad \text{(Finkelstein Reaction)}$$

Is there a mechanism to explain that?

Given the information you provided, Your reaction is proceeding as follows:

$$\ce{R-Br + I- -> R-I + Br-}\tag{1a}$$

$$\ce{R-I + R'-O- -> R-O-R' +I-}\tag{1b}$$ Where as the reaction: $$\ce{R-Br + R'-O- -> R-O-R' +Br-}\tag{2}$$

cannot proceed or does not proceed at a fast enough rate and thus the catalyst can improve the reaction rate by first substituting the less reactive halide. The Wikipedia article on Williamson ether synthesis agrees with this:

However, if an unreactive alkylating agent is used (e.g. an alkyl chloride) then the rate of reaction can be greatly improved by the addition of a catalytic quantity of a soluble iodide salt (which undergoes halide exchange with the chloride to yield a much more reactive iodide, a variant of the Finkelstein reaction).


You don't form the iodide from all of the bromide, and then the iodide reacts.

The iodide is formed from the bromide and then consumed to form the bromide via the reverse reaction. The bromide is more stable, so there's more of it and probably very little of the iodide sitting around, especially if you use the TBAI in catalytic quantities.

This reaction likely is under conditions that could reasonably described as Curtin-Hammett. The iodide and bromide are in equilibrium and the more reactive species is the one that reacts, regardless of their relative concentrations.

This provides a faster overall pathway to the product by performing substitution on the iodide, but that doesn't mean that the iodide is more preferred overall.


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