Among the following haloacetic acids, which one is the strongest acid: ClCH2COOH, BrCH2COOH, FCH2COOH or ICH2COOH?

Introduction / Context:
This organic chemistry question examines your understanding of how substituents influence the acidity of carboxylic acids. Haloacetic acids are acetic acid derivatives where a hydrogen on the alpha carbon is replaced by a halogen. The electron withdrawing power of the halogen through the inductive effect stabilises the conjugate base and changes acid strength. Recognising which substituent makes the strongest acid is a classic exam problem.

Given Data / Assumptions:
- Four haloacetic acids are given: fluoroacetic, chloroacetic, bromoacetic and iodoacetic acids.
- All have the general formula XCH2COOH where X is a halogen (F, Cl, Br, I).
- We assume that solvent and other conditions are comparable, so we focus on inductive effects of the halogens.

Concept / Approach:
The acidity of a carboxylic acid depends on the stability of its conjugate base (carboxylate anion). Electron withdrawing substituents near the carboxyl group increase acidity by pulling electron density away, stabilising the negative charge. Halogens exert a strong -I (negative inductive) effect, and the strength of this effect decreases down the group: F > Cl > Br > I. Therefore, the acid with fluorine attached, FCH2COOH, should have the strongest -I effect and the most stabilised conjugate base, making it the strongest acid among the options.

Step-by-Step Solution:
Step 1: Identify the substituent in each acid: F in fluoroacetic acid, Cl in chloroacetic acid, Br in bromoacetic acid and I in iodoacetic acid. Step 2: Recall the electron withdrawing ability of halogens via the inductive effect: fluorine, being the most electronegative, has the strongest -I effect, followed by chlorine, bromine and iodine. Step 3: Recognise that a stronger -I effect stabilises the carboxylate anion more effectively, increasing the acid strength. Step 4: Arrange the acids in decreasing order of acidity based on inductive effect: FCH2COOH > ClCH2COOH > BrCH2COOH > ICH2COOH. Step 5: Conclude that FCH2COOH (fluoroacetic acid) is the strongest acid among the given options.

Verification / Alternative check:
You can confirm this reasoning by remembering standard pKa trends for haloacetic acids. Fluoroacetic acid has the lowest pKa value (strongest acid), chloroacetic acid is slightly weaker, bromoacetic weaker still and iodoacetic the weakest in this series. This sequence matches the electronegativity trend of the halogens. The closer and more electronegative the substituent is to the carboxyl group, the stronger its inductive effect and the greater the acid strength.

Why Other Options Are Wrong:
ClCH2COOH: Chloroacetic acid is quite strong but weaker than fluoroacetic acid because chlorine is less electronegative than fluorine.
BrCH2COOH: Bromoacetic acid has a weaker -I effect than both fluoro and chloro derivatives, thus it is less acidic.
ICH2COOH: Iodoacetic acid has the weakest electron withdrawing effect among the four and is therefore the weakest acid of the series.

Common Pitfalls:
A common mistake is to think that heavier halogens always make stronger acids because they are larger or more polarisable. While polarisability plays a role in some systems, in this simple alpha substituted carboxylic acid series, the dominant factor is electronegativity and inductive effect. Another pitfall is to forget that the substituent is on the alpha carbon, which makes the inductive effect particularly strong compared with more distant positions.

Final Answer:
The strongest acid among the given haloacetic acids is FCH2COOH (fluoroacetic acid).