A) \[FC{{H}_{2}}COOH>ClC{{H}_{2}}COOH>BrC{{H}_{2}}COOH\]
B) \[ClC{{H}_{2}}COOH>BrC{{H}_{2}}COOH>FC{{H}_{2}}COOH\]
C) \[BrC{{H}_{2}}COOH>ClC{{H}_{2}}COOH>FC{{H}_{2}}COOH\]
D) \[ClC{{H}_{2}}C{{O}_{2}}H>FC{{H}_{2}}COOH>BrC{{H}_{2}}COOH\]
Correct Answer: A
Solution :
[a] \[F-C{{H}_{2}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-OH>\underset{\begin{smallmatrix} (Electron-withdrawing\text{ }character \\ decreasing\text{ }from\text{ }L\text{ }to\text{ }R) \end{smallmatrix}}{\mathop{Cl-C{{H}_{2}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-OH>}}\,\] \[Br-C{{H}_{2}}-\overset{O}{\mathop{\overset{|}{\mathop{C}}\,}}\,-OH\]
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