See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

To match each substituent with its electronic effects, we analyze inductive (I) and mesomeric/resonance (m) effects: (a) -NO2 (nitro group): - Inductive effect: -NO2 contains electronegative nitrogen and oxygen atoms; it withdraws electrons inductively → -I effect (s) - Mesomeric effect: -NO2 is an electron-withdrawing group by resonance (the nitrogen has a positive formal charge and withdraws π electrons from the ring) → -m effect (p) - Therefore: (a) → (p, s) (b) -O^- (oxide/oxyanion): - Inductive effect: The negatively charged oxygen pushes electrons toward the attached carbon → +I effect (r) - Mesomeric effect: The lone pairs on O^- can donate into a conjugated system → +m effect (q) - Therefore: (b) → (q, r) (c) -O-CH3 (methoxy group): - Inductive effect: Oxygen is electronegative, so it withdraws electrons inductively from the ring → -I effect (s) - Mesomeric effect: Oxygen has lone pairs that can donate into the π system via resonance → +m effect (q) - Therefore: (c) → (q, s) (d) -C≡N (cyano group): - Inductive effect: The electronegative nitrogen at the end of the triple bond withdraws electrons inductively → -I effect (s) - Mesomeric effect: The C≡N group can accept π electrons through the triple bond (electron-withdrawing by resonance) → -m effect (p) - Therefore: (d) → (p, s) Why other options fail: - -NO2 cannot have +I or +m because it is entirely electron-withdrawing. - -O^- cannot have -I or -m because the negative charge makes it a strong electron donor both inductively and by resonance. - -O-CH3 has -I (not +I) because neutral oxygen is still electronegative, but +m due to lone pair donation. - -C≡N cannot have +I or +m because it withdraws electrons through both mechanisms. Therefore, the correct answer is {"a": ["p", "s"], "b": ["q", "r"], "c": ["q", "s"], "d": ["p", "s"]}.