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COMEDK · Chemistry · 28. Carboxylic Acid Derivatives

Which of the following is the correct sequence of relative acidic strength?

  1. A \(\mathrm{FCH}_{2} \mathrm{COOH}>\mathrm{ClCH}_{2} \mathrm{COOH}>\mathrm{BrCH}_{2} \mathrm{COOH}\)
  2. B \(\mathrm{ClCH}_{2} \mathrm{COOH}>\mathrm{BrCH}_{2} \mathrm{COOH}>\mathrm{FCH}_{2} \mathrm{COOH}\)
  3. C \(\mathrm{BrCH}_{2} \mathrm{COOH}>\mathrm{ClCH}_{2} \mathrm{COOH}>\mathrm{FCH}_{2} \mathrm{COOH}\)
  4. D \(\mathrm{ClCH}_{2} \mathrm{COOH}>\mathrm{FCH}_{2} \mathrm{COOH}>\mathrm{BrCH}_{2} \mathrm{COOH}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\mathrm{FCH}_{2} \mathrm{COOH}>\mathrm{ClCH}_{2} \mathrm{COOH}>\mathrm{BrCH}_{2} \mathrm{COOH}\)

Step-by-step Solution

Detailed explanation

Acid strength increases as the electronegativity of halogen increases because the anion formed after \(\mathrm{H}^{+}\) ion release, is stabilised by \(-I\)-effect. The order of \(-\mathrm{I}\)-effect is \(\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}\). \(\therefore\) Order of…