A) \[C{{H}_{3}}CHFCOOH\]
B) \[FC{{H}_{2}}C{{H}_{2}}COOH\]
C) \[BrC{{H}_{2}}C{{H}_{2}}COOH\]
D) \[C{{H}_{3}}CBrCOOH\]
Correct Answer: C
Solution :
Lower the acidic character, smaller is the value of dissociation constant. Acidity\[\propto \] dissociation constant. \[-I\](inductive effect) of \[F\] is more than\[Br\], so \[C{{H}_{3}}CHFCOOH\] is a stronger acid as compared to\[C{{H}_{3}}CHBrCOOH\]. As the distance between electron withdrawing group and \[-COOH\] group increases, acidity decreases. Thus, the order of acidity and thus, of dissociation constant is \[\underset{(smallest\,\,{{K}_{a}})}{\mathop{BrC{{H}_{2}}C{{H}_{2}}COOH}}\,>FC{{H}_{2}}C{{H}_{2}}COOH\] \[<C{{H}_{3}}CHBrCOOH<C{{H}_{3}}CHFCOOH\]
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