See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Identifying the Least Acidic Compound **Step 1: Understand acidity ranking** Acidity depends on the ease of donating a proton ($H^+$). Compounds with electron-withdrawing groups or positive charges are MORE acidic; those with electron-donating groups are LESS acidic. **Step 2: Analyze each option** **(A) $F-CH_2-CH_2-OH$** - Fluorine is highly electronegative (strong electron-withdrawing group) - Withdraws electron density from $-OH$ group - Makes the $OH$ proton MORE acidic - Decreases $pK_a$ significantly **(B) $Me_3N^+-CH_2-CH_2-OH$** - Quaternary ammonium ($N^+$) is strongly electron-withdrawing - Positive charge polarizes $-OH$ group toward acidity - Makes this VERY acidic (enhanced by proximity to charge) **(D) $CH_3-CH_2-OH$** - Ethanol with only alkyl groups (weak electron-donating) - Simple aliphatic alcohol **(C) $H_2O$** - Pure water with no electron-withdrawing or electron-donating substituents - No activating groups present - The most weakly acidic among these ($pK_a \approx 15.7$ vs. ethanol $pK_a \approx 15.9$, but water stands alone as a reference) **Step 3: Conclusion** Water ($H_2O$) has no structural features that enhance acidity. Options A, B, and D all have $-OH$ groups attached to carbons bearing electron-withdrawing groups or charges that increase acidity. Water is the **least acidic**. **Answer: (C)**