Problem 1
New catalysts for stereoregular polymerisation prompted the study of the system A–B, where A = Te(cryst.) and B = (an analogue of , which cannot be isolated). Interaction of A and B gives three compounds I, II, III for initial mixtures containing 77.8, 87.5 and 91.7 mol % of A respectively. For II and III no side products form; formation of I releases 1 mol of volatile per 2 mol of I. I and II are pinkish-purple, dissociate into three ions in molten , and have molar masses and g mol⁻¹ (cryoscopy in ). Each shows a single IR band at 133 cm⁻¹ assigned to a Te–Te bond, and a single NMR signal of tetrahedral Al, at different shifts for I and II. (1.1) Determine the minimal Te : Al : Cl ratio for I, II, III. (1.2) Write the molecular formulae of I and II. (1.3) Give the cation and anion formulae in I and II. (1.4) Describe the stereochemistry of the ions, assuming the cations are inorganic aromatic systems. (1.5) Which compound is thermally more stable, given is volatile? (1.6) Write the reaction converting one compound into the other on heating.
Step 4 of 4: Thermal conversion
Analysis
Heating I drives off volatile , yielding II; therefore II — lacking removable — is the thermally more stable compound.