A) \[N{{O}_{2}}C{{H}_{2}}COOH\text{ }>\text{ }FC{{H}_{2}}COOH\text{ }>CNC{{H}_{2}}COOH\text{ }>\text{ }ClC{{H}_{2}}COOH\]
B) \[CNC{{H}_{2}}COOH>{{O}_{2}}NC{{H}_{2}}COOH\text{ }>FC{{H}_{2}}COOH\text{ }>\text{ }ClC{{H}_{2}}COOH\]
C) \[FC{{H}_{2}}COOH>ClC{{H}_{2}}COOH>N{{O}_{2}}C{{H}_{2}}COOH\text{ }>\text{ }ClC{{H}_{2}}COOH\]
D) \[N{{O}_{2}}C{{H}_{2}}COOH>NCC{{H}_{2}}COOH>FC{{H}_{2}}COOH>ClC{{H}_{2}}COOH\]
Correct Answer: D
Solution :
Acidic strength order \[\propto \,\,EWG\,\,-\,\,{{M}_{1}}\,\,-\,\,{{H}_{1}}-\,I\] In given example only -I of substituent is applicable - I order - \[N{{O}_{2}}\,>\,-\,C\,\,\equiv \,N\,>\,-F\,>\,-\,Cl\] Acidic strength order = \[N{{O}_{2}}-C{{H}_{2}}-COOH\,>\] \[NC-C{{H}_{2}}-COOH>F-CH2-COOH>\] \[Cl\,-\,C{{H}_{2}}\,-\,COOH\]
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