A) \[C{{H}_{3}}OH\]
B) \[{{C}_{2}}{{H}_{5}}OH\]
C) \[{{(C{{H}_{3}})}_{2}}CHOH\]
D) \[{{(C{{H}_{3}})}_{3}}COH\]
Correct Answer: A
Solution :
As the hydrocarbon part (R) of the alcohol increases, the tetrahedral intermediate (I) formed during esterification becomes more and more crowded and hence the rate of esterification decreases accordingly. Thus, the smallest alcohol, i.e.,\[C{{H}_{3}}OH\]reacts at the fastest rate \[C{{H}_{3}}-\overset{O}{\mathop{\overset{\mathbf{||}}{\mathop{C}}\,}}\,-OH\xrightarrow{{{H}^{+}}}C{{H}_{3}}-\overset{{}^{+}OH}{\mathop{\overset{\mathbf{||}}{\mathop{C}}\,}}\,-OH\xrightarrow{ROH}\]\[\underset{\text{tetrahedral}\,\text{intermediate}\,\text{(i)}}{\mathop{\left[ {{H}_{3}}C\,\,\,\underset{OR}{\overset{OH}{\mathop{\overset{\mathbf{|}}{\mathop{\overset{C}{\mathop{|}}\,}}\,}}}\,\,\,\,\,OH \right]}}\,\xrightarrow{-{{H}_{2}}O}\underset{\text{ester}}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{\mathbf{||}}{\mathop{C}}\,}}\,-OR}}\,\]
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