A) \[MeCOCl\]
B) \[MeCHO\]
C) \[MeCOOMe\]
D) \[MeCOOCOMe\]
Correct Answer: A
Solution :
[a] More the magnitude of positive charge on the carbonyl carbon, higher will be its reactivity toward nucleophilic attack. \[C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-Cl\,\,\,C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-H\text{ }\,\,C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-OCOC{{H}_{3}}\,\,\,\,C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-OC{{H}_{3}}\] Chlorine being more electron-withdrawing than O develops more positive charge on carbonyl carbon. The order of reactivity of the various carboxyl derivatives toward nucleophile is \[\underset{Most\text{ }reactive}{\mathop{RCOCl}}\,>RCHO>{{\left( RCO \right)}_{2}}O>\underset{Least\text{ }reactive}{\mathop{RCOOR}}\,\]
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