See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The acidic strength of substituted phenols depends on the ability of substituents to stabilize or destabilize the phenoxide anion formed after deprotonation. Electron-withdrawing groups (EWG) stabilize the phenoxide ion and increase acidity, while electron-donating groups (EDG) destabilize the phenoxide ion and decrease acidity. Position (para vs meta vs ortho) also matters. Step 1: Identify the substituents and their electronic effects. - Compound I: 4-methylphenol. CH3 is an electron-donating group (EDG) by hyperconjugation and induction. At para position, it donates electrons into the ring and onto oxygen, destabilizing phenoxide. Decreases acidity relative to phenol. - Compound II: 4-cyanophenol. CN is a strong electron-withdrawing group (EWG) by resonance and induction. At para position, it strongly stabilizes the phenoxide ion. Greatly increases acidity. - Compound III: 4-methoxyphenol. OCH3 is an EDG by resonance (stronger donor than CH3 by resonance). At para position, OCH3 donates electrons strongly into the ring via resonance, destabilizing phenoxide more than CH3 does. Decreases acidity more than compound I. - Compound IV: phenol (plain, no substituent shown as meta-OH benzene but effectively treated as phenol itself). No substituent, baseline acidity. Step 2: Rank the compounds. - III (4-methoxyphenol) is least acidic because OCH3 at para donates electrons most strongly by resonance, making phenoxide least stable. - I (4-methylphenol) is next least acidic because CH3 donates electrons but less strongly than OCH3 by resonance. - IV (phenol, unsubstituted) is more acidic than both I and III since there is no EDG to destabilize phenoxide. - II (4-cyanophenol) is most acidic because CN strongly withdraws electrons, stabilizing phenoxide. Step 3: Therefore increasing order of acidity: III < I < IV < II. Step 4: Check options: - Option (a): III < I < IV < II — matches our ranking. - Option (b): II < I < IV < III — incorrect, places strongest acid as weakest. - Option (c): I < III < IV < II — incorrect order of I and III. - Option (d): I < III < II < IV — multiple errors. Therefore, the correct answer is A.