See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

CIP priority rules assign priority based on atomic number of the first atom of each group, then proceed to the next atoms if there is a tie. Step 1: Identify the first atom of each group. - Group I: -SCH3 → first atom is S (atomic number 16) - Group II: -NO2 → first atom is N (atomic number 7) - Group III: -C≡CH → first atom is C (atomic number 6) - Group IV: -CH2C6H5 → first atom is C (atomic number 6) Step 2: Rank by first atom atomic number. - S (16) > N (7) > C (6) = C (6) - So Group I (S) has highest priority, Group II (N) has second highest, and Groups III and IV (both C) must be compared further. Step 3: Break the tie between Group III (-C≡CH) and Group IV (-CH2C6H5). For CIP rules, multiple bonds are treated as phantom atoms (duplicate atoms). - Group III: -C≡CH → the triple bond carbon is treated as C bonded to (C, C, C) phantom atoms [i.e., (C, C, C) at first carbon] - Group IV: -CH2C6H5 → the CH2 carbon is bonded to (H, H, C) [two H's and one C of the phenyl ring] At the first carbon of Group III: attached atoms are C, C, C (from triple bond expansion: the actual bonded C plus two phantom C's) At the first carbon of Group IV: attached atoms are C, H, H Comparing the substituents at the first carbon in decreasing order: - Group III: (C, C, C) - Group IV: (C, H, H) At the first comparison: C vs C → tie; second: C vs H → C wins. Therefore Group III > Group IV in priority. Step 4: Compile the priority order from highest to lowest: I (S,16) > II (N,7) > III (C≡CH) > IV (CH2C6H5) Step 5: Increasing order (lowest to highest priority): IV < III < II < I This corresponds to option (b): IV, III, II, I. Why other options fail: - (a) I, III, II, IV: Places I lowest and IV highest, which is incorrect since S > N > C. - (c) II, IV, I, III: Places S (I) in the middle, which is wrong since S has the highest atomic number. - (d) III, IV, II, I: Places IV before II, but N (atomic number 7) > C (atomic number 6), so II should be higher than both III and IV. Therefore, the correct answer is B.