How can isotope labelling be used to determine the mechanism of ether cleavage with hydrogen iodide?
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

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

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

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

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:

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

Likewise

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

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:

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

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.