See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The acidity of phenols depends on the stability of the phenoxide ion (conjugate base) formed after loss of the proton. Electron-withdrawing groups (EWGs) stabilize the phenoxide ion by delocalizing the negative charge, thereby increasing acidity. Electron-donating groups (EDGs) destabilize the phenoxide ion and decrease acidity. Step 1: Identify substituents and their electronic effects. - (a) Phenol: no substituent, baseline acidity (pKa ≈ 10.0) - (b) 4-methylphenol (p-cresol): CH3 is an EDG (hyperconjugation/induction), donates electron density into the ring, destabilizes the phenoxide ion → less acidic than phenol - (c) 3-nitrophenol: NO2 is a strong EWG; at meta position it stabilizes the phenoxide primarily through inductive/field effect (resonance cannot directly delocalize into the O- from meta position) → more acidic than phenol (pKa ≈ 8.4) - (d) 4-nitrophenol: NO2 is a strong EWG; at para position it stabilizes the phenoxide through both inductive effect AND direct resonance delocalization of the negative charge onto the oxygen of the nitro group (through the ring) → maximum stabilization of phenoxide ion → most acidic (pKa ≈ 7.1) Step 2: Rank acidity. Order of acidity: (b) p-cresol < (a) phenol < (c) 3-nitrophenol < (d) 4-nitrophenol Step 3: Why para > meta for nitrophenol? In 4-nitrophenol, the negative charge on the phenoxide oxygen can be delocalized via resonance all the way onto the nitro group's oxygen atoms (para resonance pathway), providing extra stabilization not available in the meta isomer. This makes 4-nitrophenol more acidic than 3-nitrophenol. Step 4: Why other options fail. - (a) Phenol: no EWG to stabilize phenoxide, baseline acidity. - (b) p-Cresol: EDG (CH3) decreases acidity below phenol. - (c) 3-Nitrophenol: EWG stabilizes via induction only, less effective than para resonance. Therefore, the correct answer is D.