1. Principle involved: Inductive effect

For carboxylic acids, acidity depends on the stability of the conjugate base, $\mathrm{RCOO^-}$.

• Electron-withdrawing groups ($-I$ effect) increase acidity by stabilizing the carboxylate ion.
• Electron-donating groups decrease acidity.

So, among substituted acetic acids, stronger $-I$ effect on the $\alpha$-carbon gives stronger acid.

2. Given acids

• $A = \mathrm{CH_3COOH}$ (acetic acid)
• $B = \mathrm{CF_3COOH}$ (trifluoroacetic acid)
• $C = \mathrm{ClCH_2COOH}$ (chloroacetic acid)
• $D = \mathrm{FCH_2COOH}$ (fluoroacetic acid)
• $E = \mathrm{BrCH_2COOH}$ (bromoacetic acid)

3. Compare electron-withdrawing power

For halogens, the $-I$ effect follows:

$\mathrm{F > Cl > Br}$

Also, three fluorine atoms withdraw much more strongly than one fluorine atom. Therefore:

$\mathrm{CF_3COOH} > \mathrm{FCH_2COOH}$

And among mono-haloacetic acids:

$\mathrm{FCH_2COOH} > \mathrm{ClCH_2COOH} > \mathrm{BrCH_2COOH}$

Finally, unsubstituted acetic acid is the weakest because $\mathrm{CH_3}$ has a $+I$ effect:

$\mathrm{BrCH_2COOH} > \mathrm{CH_3COOH}$

4. Overall descending order of acidity

Combining all comparisons:

$\mathrm{CF_3COOH > FCH_2COOH > ClCH_2COOH > BrCH_2COOH > CH_3COOH}$

In terms of labels:

$B > D > C > E > A$

5. Match with options

Option C gives:

$B > D > C > E > A$

So, the correct option is C.

6. Comparison with stored answer

Stored correct answer = C.

My derived answer = C.

Hence, they agree.