Consider the cleavage of tert-butyl methyl ether using concentrated hydroiodic acid (). Which of the — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

When one of the alkyl groups is tertiary (like tert-butyl), the cleavage shifts from $S_N2$ to $S_N1$ because the intermediate tertiary carbocation is highly stable. Following protonation (Step 1), the $O-C(tert)$ bond breaks (rate-determining step) to form a $3^\circ$ carbocation and methanol. The $I^-$ then attacks the carbocation, yielding tert-butyl iodide and methanol. Therefore, statement D is incorrect.