See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: The solubility of a salt (zwitterion pair) in methanol depends on its stereochemical relationship — specifically whether the cation and anion components are enantiomers or diastereomers of each other, and whether the overall salt is a racemic mixture, a meso compound, or a pair of diastereomeric salts. Two salts have identical solubilities in methanol if they are enantiomers of each other (mirror images), since enantiomers have identical physical properties in achiral solvents like methanol. Step 1: Identify the structure of each salt as a combination of a cationic ammonium component and an anionic carboxylate component, both derived from the amino acid Ph-CH(NH2)-CH3 (i.e., alpha-methylbenzylamine / alanine analog: 2-amino-1-phenylpropane or similar). Step 2: Analyze Salt I. The cation has configuration: central C with Ph, H, CH3, NH3+ — assign R or S. The anion has central C with Ph, CH3, CO2-, H — assign configuration. Salt I represents one diastereomeric combination (e.g., R-cation paired with S-anion, or a specific RS combination). Step 3: Analyze Salt III. The cation has Ph (up), NH3+ (left), CH3 (right), H (down) and the anion has Ph (up), H (left), CH3 (right), CO2- (down). Comparing Salt I and Salt III: the wedge-dash arrangements of both the cation and anion in Salt III are the mirror images of those in Salt I. This means Salt III is the enantiomer of Salt I. Step 4: Since Salt I and Salt III are enantiomers of each other (every stereocenter in the cation and anion is inverted), they have identical physical properties in an achiral solvent such as methanol, including identical solubilities. Step 5: Check other pairs. Salt II has a different connectivity arrangement (NH3+ on top, Ph on right for cation; H on top, Ph on bottom for anion), making it a diastereomer relative to Salt I, not an enantiomer — different solubility. Salt IV similarly is not the enantiomer of Salt I or II in the same way. Options (a), (c), and (d) involve pairs that are diastereomers and thus have different solubilities in methanol. Therefore, the correct answer is B.