See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

To classify substituents attached to a benzene ring by their electronic effects: **Concepts:** - **+M (positive mesomeric/resonance effect):** The substituent donates electron density INTO the ring via resonance. This requires a lone pair on the atom directly attached to the ring that can overlap with the pi system. Examples: -NR2, -OR, -NH2, -OH. - **-M (negative mesomeric effect):** The substituent withdraws electron density FROM the ring via resonance. This requires a pi bond (C=O, N=O, etc.) conjugated with the ring. Examples: -CHO, -COCH3, -COOH, -CONH2, -NO2. - **-I (negative inductive effect):** Electronegative atoms (N, O, halogens) attached to or near the ring withdraw electrons through sigma bonds. Nearly all heteroatom-containing groups show -I. - **+I (positive inductive effect):** Electron-donating groups through sigma bonds, typically alkyl groups. None of these substituents are pure alkyl groups directly on the ring. **Row-by-row analysis:** **Row 1: N-phenylpiperidine** — The piperidine nitrogen has a lone pair and is directly bonded to the benzene ring with no C=O adjacent to N. The lone pair donates into the ring (+M). Nitrogen is more electronegative than carbon, so -I also applies. No +I, no -M. → +M: true, -M: false, -I: true, +I: false **Row 2: 1-phenyl-2-piperidinone** — The nitrogen is attached to the ring AND has a C=O in the piperidine ring adjacent to it. However, the nitrogen lone pair is still available to donate into the phenyl ring (+M). The nitrogen is electronegative (-I). Even though the amide C=O pulls on N, the N still acts as +M toward phenyl. The C=O of the lactam is not conjugated directly to the phenyl ring in the same way. The group as a whole still shows +M to phenyl via N lone pair, and -I via electronegativity of N. → +M: true, -M: false, -I: true, +I: false **Row 3 & 4: -C(=O)-NH-CH3 (N-methylbenzamide group on benzene)** — The carbonyl (C=O) is conjugated with the benzene ring. This withdraws electrons from the ring by resonance (-M). The carbonyl carbon and oxygen are electronegative (-I). No lone pair directly on the carbon attached to ring to give +M. → +M: false, -M: true, -I: true, +I: false **Row 5: -C(=O)-CH3 (acetyl/acyl group)** — C=O conjugated with ring withdraws electrons (-M). Electronegative oxygen (-I). No +M. → +M: false, -M: true, -I: true, +I: false **Row 6: -NO2** — The nitro group has N=O bonds conjugated with the ring, strongly withdrawing electrons by resonance (-M). Nitrogen and oxygens are electronegative (-I). No +M. → +M: false, -M: true, -I: true, +I: false **Row 7: -OCH3 (methoxy)** — Oxygen has lone pairs that donate into the ring (+M). Oxygen is electronegative (-I). No -M (no pi bond withdrawing from ring), no +I. → +M: true, -M: false, -I: true, +I: false **Row 8: -CH=O (aldehyde/formyl)** — The C=O is conjugated with the ring, withdrawing electrons by resonance (-M). Electronegative oxygen (-I). No +M. → +M: false, -M: true, -I: true, +I: false **Row 9 (answer key row 8): -C(=O)-OH (carboxyl/benzoic acid)** — The C=O conjugated with ring gives -M. Electronegative O atoms give -I. No +M. → +M: false, -M: true, -I: true, +I: false **Row 10 (answer key row 9): -NH-CH3 (methylamino)** — Nitrogen lone pair donates into ring (+M). Nitrogen is electronegative (-I). No -M, no +I. → +M: true, -M: false, -I: true, +I: false Therefore, the correct answer is {"1": {"+M": true, "-M": false, "-I": true, "+I": false}, "2": {"+M": true, "-M": false, "-I": true, "+I": false}, "3": {"+M": false, "-M": true, "-I": true, "+I": false}, "4": {"+M": false, "-M": true, "-I": true, "+I": false}, "5": {"+M": false, "-M": true, "-I": true, "+I": false}, "6": {"+M": true, "-M": false, "-I": true, "+I": false}, "7": {"+M": false, "-M": true, "-I": true, "+I": false}, "8": {"+M": false, "-M": true, "-I": true, "+I": false}, "9": {"+M": true, "-M": false, "-I": true, "+I": false}}.