When methyl tert-butyl ether (MTBE) is treated with one equivalent of hydrogen iodide, the ether C–O bond is cleaved.

I read that if the oxygen in MTBE is isotopically labelled, it is possible to work out the mechanism, but I am not able to understand the function of $\ce{^{18}O}$.

$$\ce{CH3-^{18}O-C(CH3)3 + HI ->[S_N1/S_N2] products}$$

How does isotopic labelling help determining the mechanism?


2 Answers 2


If your products are simply an alcohol plus an alkyl iodide, then you really don't need the isotopic labelling. If you have an $\mathrm{S_N1}$ mechanism operating, then iodide ion attacks the tert-butyl cation, and you get the products

$$\ce{Me-O-^tBu + HI ->[S_N1] MeOH + ^tBuI}$$

If you have an $\mathrm{S_N2}$ mechanism, then iodide ion attacks the methyl group, displacing t-butanol:

$$\ce{Me-O-^tBu + HI ->[S_N2] MeI + ^tBuOH}$$

In both cases it should be trivial to take a proton NMR of the crude reaction mixture and determine which set of products are present.

The isotope label is more pertinent if you are forming two alcohols as the products. In that case, if the $\ce{^18O}$ remains attached to the methyl group (for example), it means that the $\ce{^tBu-O}$ bond was cleaved: this indicates an $\mathrm{S_N1}$ mechanism. In other words:

$$\ce{Me-^{18}O-^tBu ->[S_N1] Me^{18}OH + HO-^tBu}$$


$$\ce{Me-^{18}O-^tBu ->[S_N2] MeOH + H^{18}O-^tBu}$$

However, you would have to be careful when running this experiment: if you leave it too long, especially under acidic conditions, the labelled oxygen will simply be exchanged with unlabelled oxygen from water by a degenerate nucleophilic substitution reaction, destroying any precious evidence that you had obtained:

$$\ce{R^{18}OH + H3^{16}O+ <=>> R^{16}OH + H3^{18}O+}; \quad \ce{R} = \ce{Me}\text{ or }\ce{^tBu}$$

where the equilibrium position lies to the right because of the larger concentration of $\ce{H3^{16}O+}$. Equilibrium tells us nothing about kinetics, though, so as stated earlier, you will have to stop the reaction (or monitor it using e.g. mass spectrometry, since NMR won't tell you the difference between oxygen isotopes) before this happens.


The function of the oxygen isotope is to determine what mechanism the reaction undergo. The mechanism of ether cleavage is SN1 or SN2 depending on the nature of the carbons bonded to oxygen. If both carbons are primary, cleavage involves a SN2 reaction, the halide ion being the nucleophile. With secondary or tertiary carbons, the $\ce{C-O}$ bond is cleaved by an SN1 mechanism.

In your case you have primary and tertiary carbons bonded to oxygen, but the tertiary carbons are particularly susceptible to cleavage by acid. So your reaction with one equivalent of $\ce{HI}$ will undergo cleavage with SN1 mechanism, forming $\ce{(CH3)3CI}$ and $\ce{H3COH}$.

The SN1 or SN2 mechanism starts with the formation of the oxonium cation: start of mechanism

And then, the characteristics of the carbon atoms bonded to the oxygen select between the SN1 or SN2. If the mechanism is SN2 the nature of the nucleophile also is important, otherwise not. The solvation energy, the strength of the bond with carbon, electronegativity, polarizability and sterics involving the nucleophile are considering factors.

Taking the case if the reaction undergo SN2; the transition state would have a partial $\ce{H3C-I}$ bond and a partial $\ce{H3C-O}$ bond. The free energy of this transition state is higher in energy because the $\ce{C-I}$ bond is weak ($\pu{213 kJ/mol}$), compared to $\ce{C-O}$ ($\pu{336 kJ/mol}$). In general the $\ce{C-I}$ bond is the weaker of the 14-17 groups general elements in organic chemistry (the source is a table from the organic chemistry class that I had). So, the activation energy of the SN2 mechanism is high.

However, if the reaction is SN1, the relative nucleophilicity has no effect on the rate of the reaction. Moreover, the substitution proceeds by rate-determining heterolytic dissociation of the reactant to a carbocation and the leaving group. So, the stability of the carbocation, the nature of the leaving group and the solvent's ionizing power is important. The electronic effect that stabilizes is the electron release, the accepting ability of the leaving group and the capacity of the solvent to stabilize the charge separation in the transition state. The mechanism would be:

end of mechanism

where, like I said in the comment, the $\ce{-O(H+)CH3}$ part of the oxonium cation is a good leaving group. For the other hand, the carbocation intermediate is tertiary, i.e. the most stable carbocation. This stability is due to hyperconjugation.

  • 4
    $\begingroup$ That does not at all mention why the oxygen needs to be labelled. $\endgroup$ Sep 12, 2018 at 16:53

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