An ester forms from a carboxylic acid and an alcohol. $$\mathrm{RCO}_{2}
\mathrm{H}+\mathrm{HOR}^{\prime} \longrightarrow \mathrm{RCO}_{2}
\mathrm{R}^{\prime}+\mathrm{H}_{2} \mathrm{O}.$$ This reaction is
superficially similar to the reaction of an acid with a base such as sodium
hydroxide. The mechanism of the reaction can be followed by using the tracer
\(^{18} \mathrm{O}\). This isotope is not radioactive, but other physical
measurements can be used to detect its presence. When the esterifcation
reaction is carried out with the alcohol containing oxygen-18 atoms, no
oxygen-18 beyond its naturally occurring abundance is found in the water
produced. How does this result affect the perception that this reaction is
like an acid-base reaction?