See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: HI is a strong acid and a nucleophilic reagent that cleaves C-O bonds, particularly ethers. The reaction of HI with ethers proceeds via protonation of the oxygen followed by nucleophilic attack by iodide, cleaving the C-O bond. Step 1: Identify the ether linkages in the molecule. The structure shown is a bicyclic system where a benzene ring is fused via two oxygen bridges (forming a 1,4-benzodioxane or similar scaffold) to a second ring, which itself bears two O-Et (ethoxy) groups. Step 2: Classify each ether. The two O-Et groups are alkyl aryl or alkyl-alkyl ethers attached to the saturated ring carbons. These are simple dialkyl (or alkyl) ethers that can be cleaved by HI. Step 3: The cyclic ether oxygens that form the dioxane bridge between the two rings are part of the ring system. In the context of this molecule, the ring oxygens connecting the benzene to the saturated ring are aryl ethers (ArO-C bonds). HI can cleave aryl ethers, but under normal conditions the selectivity matters. However, the question focuses on how many moles of HI are consumed total. Step 4: The two ethoxy (O-Et) groups are the reactive sites under mild HI conditions. Each O-Et group requires one mole of HI for cleavage (ROEt + HI → ROH + EtI). There are exactly 2 such ethoxy groups on the molecule. Step 5: The cyclic dioxane-type oxygens forming the ring bridges are less reactive alkyl aryl ethers or are part of an acetal-like system. In many such problems from M.S. Chauhan, acetal-type C-O bonds or simple alkyl ether C-O bonds are counted. Here the two O-Et groups are the ones cleaved, each consuming 1 mole of HI, giving x = 2. Step 6: The ring ether oxygens (the bridging oxygens of the dioxane ring) are not cleaved under these conditions or are not counted in this context, consistent with the answer of 2. Therefore, the correct answer is 2.