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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?

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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.

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  • $\begingroup$ 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 '14 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 '14 at 1:01
  • $\begingroup$ ether reacts only by the SN1 mechanism $\endgroup$ – ron Jan 25 '14 at 1:02
  • $\begingroup$ Why does the acidic condition favour the SN1 mechanism? $\endgroup$ – Tan Yong Boon Dec 30 '17 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 '17 at 15:52
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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.

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  • $\begingroup$ I hope I got what you meant right? @ron $\endgroup$ – Tan Yong Boon Dec 31 '17 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. $\endgroup$ – Martin - マーチン Dec 31 '17 at 7:20
  • $\begingroup$ @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]". $\endgroup$ – Tan Yong Boon Dec 31 '17 at 8:30

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