See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: The pKa of an alcohol measures its acidity. A lower pKa means a stronger acid, which means the O-H bond donates a proton more readily. Electron-withdrawing groups (EWGs) stabilize the conjugate base (alkoxide ion) by dispersing negative charge through inductive effects, thereby increasing acidity and lowering pKa. Step 1 - Identify substituents affecting acidity: (a) Ethanol (CH3CH2OH): No EWG; baseline alcohol acidity (pKa ≈ 15.9) (b) 1-Propanol (CH3CH2CH2OH): No EWG; slightly more electron-donating alkyl chain than ethanol; similar or slightly higher pKa (~16.1) (c) 2,2,2-Trifluoroethanol (CF3CH2OH): Three fluorine atoms on the carbon adjacent to CH2OH; fluorine is the most electronegative element, and three fluorines exert a very strong inductive electron-withdrawing effect, strongly stabilizing the alkoxide conjugate base (pKa ≈ 12.4) (d) 2-Chloroethanol (ClCH2CH2OH): One chlorine atom two carbons away from OH; chlorine is EWG but less electronegative than fluorine, and only one halogen is present (pKa ≈ 14.3) Step 2 - Rank acidity (lowest pKa = most acidic): 2,2,2-Trifluoroethanol >> 2-chloroethanol > Ethanol > 1-propanol Step 3 - Why other options fail: - Ethanol and 1-propanol have only electron-donating alkyl groups, making them weaker acids (higher pKa). - 2-Chloroethanol has only one EWG (Cl) and it is one carbon further from OH compared to CF3CH2OH, so stabilization of the alkoxide is much less. - 2,2,2-Trifluoroethanol has three highly electronegative fluorines directly on the carbon bearing the OH group, providing the maximum inductive stabilization of the conjugate base among all options. Therefore, the correct answer is C.