What are the products when tert-butyl ethyl ether is cleaved with concentrated $\ce{HI}$?

My answer is iodoethane and tert-butanol, but the correct answer is ethanol and 2-iodo-2-methylpropane.

I know that the ethyl side goes by an SN2 mechanism because it is a primary R group, and the t-butyl group would be SN1 because it is tertiary. However, why would the SN2 reaction not happen first?


3 Answers 3


One "side" of a molecule doesn't react by one mechanism and the other "side" by another mechanism. Generally speaking, a molecule reacts by a single mechanism. Now, the mechanism can change depending upon the other reactants that are present as well as the reaction conditions. In the case of t-butyl ethyl ether, if the reaction were carried out with $\ce{KI}$ under non-acidic conditions an $\mathrm{S_N2}$ reaction would occur and your answer (iodoethane and tert-butanol) would be correct. On the other hand, if the reaction were carried out under acidic conditions then an $\mathrm{S_N1}$ reaction would occur. As you pointed out a tertiary carbocation is more stable than a primary carbocation so the expected products in this case would be ethanol and 2-iodo-2-methylpropane. Apparently, the reaction conditions in your reaction must have been such that the reaction occurred by the $\mathrm{S_N1}$ mechanism.

  • $\begingroup$ Please see page 21, second slide in this pdf. One side is cleaved via SN2, and the other is cleaved via SN1. Furthermore, no conditions were stated in the question beyond having KI. I don't understand why I should presume the mechanism is SN1 and SN1 only. $\endgroup$
    – halcyon
    Jan 25, 2014 at 0:39
  • $\begingroup$ I looked at p.21. While it may be literally true that"One side is cleaved via SN2, and the other is cleaved via SN1", it is not true in a chemical reaction sense. T-Butyl methyl ether is only cleaved via an SN1 process on p.21. A resultant product, methanol, can then go on to react further via an SN2 reaction in a second, independent reaction. If it happens like this in the lab, then you get a mixture of products from the separate SN1 and SN2 reactions. Usually reagent concentrations and reaction conditions are chosen to favor one pathway over the other. But the key is that the original $\endgroup$
    – ron
    Jan 25, 2014 at 1:01
  • $\begingroup$ ether reacts only by the SN1 mechanism $\endgroup$
    – ron
    Jan 25, 2014 at 1:02
  • $\begingroup$ Why does the acidic condition favour the SN1 mechanism? $\endgroup$ Dec 30, 2017 at 11:00
  • $\begingroup$ @TanYongBoon It is not so much that acidic conditions "favor" the SN1 pathway as that protons are necessary for the pathway to occur. Without a proton source there is no easy way to protonate the ether oxygen and then generate SN1 carbocation intermediate. $\endgroup$
    – ron
    Dec 30, 2017 at 15:52

I would like to elaborate more on Ron's answer to make it easier to understand. In particular, I would like to elaborate on the significance of acidity on the reaction mechanism.

Ron mentions that:

Acidic conditions favour the $\mathrm{S_N}1$ mechanism while non-acidic conditions favour the $\mathrm{S_N}2$ mechanism.

Acidic conditions

Initially, I had a lot of problems understanding this too. Upon discussion with my friend Ian, I was able to understand this fully. I was puzzled as to why there was a preference for one mechanism over the other in the different conditions. Let me explain why this preference exists in detail.

When using the reagent $\ce{HI}$, the conditions are acidic, meaning that there is an abundance of proton-donating species in the solution. Under such conditions, the oxygen atom in the ether is protonated. This gives the oxygen atom a formal positive charge, causing it to be more electron-withdrawing, resulting in larger partial positive charges on the two carbon atoms bonded to it. This also provides impetus for the breaking of the $\ce{C-O}$ bond to break. Now, we have two options, the $\ce{C-O}$ bond to the t-butyl group or that to the ethyl group. Since this bond breaks to form a carbocation, it necessarily means that a more stable carbocation would be preferred. Thus, this $\ce{C-O}$ bond would break to form the t-butyl carbocation. Eventually, the products ethanol and 2-iodo-2-methylpropane are formed.

This reaction follows the $\mathrm{S_N}1$ mechanism due to the protonation event which allowed the $\ce{C-O}$ bond to break more easily.

Non-acidic conditions

When using the reagent $\ce{KI}$ in non-acidic condiitions, there is no abundance of proton-donating species in solution. Under such conditions, the oxygen atom in the ether is not protonated significantly. Thus, the impetus to break the $\ce{C-O}$ bond to form the carbocation is not present. The mechanism of the nucleophilic substitution reaction would thus be a concerted one (i.e. $\mathrm{S_N}2$) as the $\ce{C-O}$ bond cannot break on its own but only breaks as a nucleophile forms a bond to the carbon. It thus follows that the nucleophile attacks from the opposite side. Now, note that the t-butyl group is rather bulky and thus, it would sterically hinder the approach of the nucleophile. However, the ethyl group is less sterically-hindering. Thus, the nucleophile would attack the carbon of the ethyl group, forming iodoethane and tert-butanol.

Without the protonation, the reaction can only proceed in a concerted manner.

  • $\begingroup$ I hope I got what you meant right? @ron $\endgroup$ Dec 31, 2017 at 3:46
  • $\begingroup$ Sn1 and sn2 are extreme cases. In reality there will always be a mixture of these mechanisms. And if you have competing mechanisms, they will still both happen, one to a lesser degree, because of energetics of the potential energy surface. $\endgroup$ Dec 31, 2017 at 7:20
  • 1
    $\begingroup$ @Martin-マーチン Sorry I did not intend to mean that they react only via a particular pathway. Notice that I used the phrasing "favour the [mechanism]". $\endgroup$ Dec 31, 2017 at 8:30

This answer is based on what I have learnt in class $12$ of Indian curriculum (which has not been updated for over several decades, so might be slightly erroneous).

There are a few (non-exhaustive) "rules" for cleavage of ethers by $\ce{HI}$, in general reaction conditions -

  1. If both sides have primary or secondary alkyl carbons (dimethyl ether, ethyl isopropyl ether), favoured pathway is $\mathrm{S_N2}$.
  2. If any one side has a tertiary alkyl group (di tert-butyl ether, ethyl tert-butyl ether), favoured pathway is $\mathrm{S_N1}$ in aqueous medium. This is due to high stability of carbocation + solvation by water molecules.
  3. If any one side has a benzyl group, favoured pathway is $\mathrm{S_N1}$ in aqueous medium. (Same reason as in 2)
  4. With anhydrous $\ce{HI}$, favoured pathway is $\mathrm{S_N2}$ (I think it is the only pathway) regardless of substrate, as it is much less polar than $\ce{H2O}$.

Source - NCERT for class 12, part 2 and Organic Chemistry for JEE Advanced, part 2 by Dr. KS Verma (Cengage publications)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.