The correct order of acidic strength of the major products formed in the given reactions, is:

Choose the correct answer from the options given below :

For every carboxylic acid, its acidic strength depends mainly on the nature of the groups attached to the -COOH moiety.
(i) Electron-withdrawing (-I/-M) groups stabilise the conjugate base $$\left(RCOO^{-}\right)$$ and therefore increase acidity.
(ii) Electron-releasing (+I/+M) groups do the opposite and decrease acidity.

The four reactions given in the paper convert different aldehydes/ketones into four major carboxylic-acid products. From the reagents supplied (Cl2/OH-, HVZ conditions, Tollens oxidation, etc.) the structures of those products are:

Case A: $$CH_{3}COOH \xrightarrow[\text{red P}]{Cl_{2}} CH_{2}ClCOOH$$ (chloro-acetic acid)

Case B: $$CH_{3}CH_{2}CHO \xrightarrow[\text{Tollens}]{[Ag(NH_{3})_{2}]^{+}} CH_{3}CH_{2}COOH$$ (propionic acid)

Case C: $$CH_{3}COCH_{3} \xrightarrow[OH^{-}]{3Cl_{2}} CHCl_{3}+ CCl_{3}COO^{-} \xrightarrow{H^{+}} CCl_{3}COOH$$ (trichloro-acetic acid)

Case D: $$CH_{3}CHO \xrightarrow[\text{Tollens}]{[Ag(NH_{3})_{2}]^{+}} CH_{3}COOH$$ (acetic acid)

Now compare their electron-withdrawing or electron-donating effects:

• C Trichloroacetic acid, $$CCl_{3}COOH$$: three highly electronegative Cl atoms exert a very strong -I effect, greatly stabilising $$CCl_{3}COO^{-}$$. Hence it is the strongest acid of the set.
• A Chloroacetic acid, $$CH_{2}ClCOOH$$: contains one Cl atom. Its -I effect is strong, but weaker than that of three Cl atoms, so it is less acidic than C but more acidic than those lacking Cl altogether.
• D Acetic acid, $$CH_{3}COOH$$: the adjacent methyl group has a weak +I effect, giving an acidity close to that of a simple aliphatic acid.
• B Propionic acid, $$CH_{3}CH_{2}COOH$$: the ethyl group exerts a slightly stronger +I effect than a methyl group, so this conjugate base is the least stabilised and B is the weakest acid.

Therefore the acidic strength decreases in the order

$$CCl_{3}COOH \; (C) \; \gt \; CH_{2}ClCOOH \; (A) \; \gt \; CH_{3}COOH \; (D) \; \gt \; CH_{3}CH_{2}COOH \; (B)$$

The correct option is
Option D which is: C > A > D > B