See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: This reaction is an esterification (Fischer-type) between an alcohol R-OH and p-nitrobenzoic acid under acid catalysis (H+). The mechanism proceeds via nucleophilic acyl substitution (AAC2 mechanism), where the alcohol oxygen attacks the carbonyl carbon of the acid. Step 1 - Identify the rate-determining factor: In acid-catalyzed esterification, the alcohol acts as the nucleophile attacking the protonated carbonyl carbon. The rate depends on the nucleophilicity and steric accessibility of the -OH group in R-OH. Step 2 - Steric effects: As the R group becomes more bulky (more substituted), steric hindrance around the oxygen increases, making it harder for the alcohol to approach and attack the electrophilic carbonyl carbon of the acid. This slows the reaction. Step 3 - Electronic effects: More alkyl groups on the carbon bearing -OH increase electron density on oxygen via induction, slightly increasing nucleophilicity. However, for esterification, steric hindrance dominates over electronic effects. Step 4 - Compare the options: - (a) CH3-OH (methanol): primary, no steric hindrance, smallest R group -> fastest rate - (b) CH3CH2-OH (ethanol): primary but slightly more steric bulk than methanol -> slightly slower - (c) (CH3)2CH-OH (isopropanol): secondary alcohol, more steric hindrance -> slower - (d) (CH3)3C-OH (tert-butanol): tertiary alcohol, maximum steric hindrance -> slowest rate Step 5 - Conclusion: Methanol (R = CH3-) has the least steric hindrance around the nucleophilic oxygen, allowing the fastest attack on the carbonyl carbon, giving the highest rate of esterification. Why other options fail: Options (b), (c), and (d) have progressively larger R groups that increase steric hindrance, reducing the rate of nucleophilic attack and therefore the rate of esterification. Therefore, the correct answer is A.