Problem 80 - A Complete Resource Book in Chemistry for JEE Main

Question

Here the strongest acid is (a) \(\mathrm{CH}_{3}-\mathrm{COOH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{COOH}\) (c) \(\mathrm{Cl}-\mathrm{CH}_{2}-\mathrm{COOH}\) (d) \(\mathrm{Br}-\mathrm{CH}_{2}-\mathrm{COOH}\)

Step-by-Step Solution

Answer

The strongest acid is (c) Cl-CH₂-COOH due to the strong electron-withdrawing effect of Cl.

Step 1: Identify the Functional Group

All the compounds listed are carboxylic acids due to the presence of the -COOH group. In an acidic reaction, the ability to donate a proton (H^+) determines the strength of the acid.

Step 2: Analyze Electron-Withdrawing Groups

Electron-withdrawing groups attached to the carbon chain increase acid strength by stabilizing the conjugate base. Among the options, (c) Cl and (d) Br are halogens known for their electron-withdrawing ability through inductive effects.

Step 3: Compare Inductive Effects

The inductive effect decreases with distance and strength. Chlorine (Cl) has a stronger -I effect (electron-withdrawing) than bromine (Br) due to its higher electronegativity. This makes (c) Cl-CH_2-COOH more acidic than (d) Br-CH_2-COOH.

Step 4: Exclude Compounds Without Electron-Withdrawing Groups

Compounds (a) CH_3-COOH and (b) CH_3CH_2-COOH do not have strong electron-withdrawing groups, making them weaker acids compared to (c) and (d).

Key Concepts

Carboxylic Acids

Carboxylic acids are a group of organic compounds known for their acidic properties, primarily due to the presence of the carboxyl group (\(-\text{COOH}\)). This functional group is capable of donating a proton (\(\text{H}^+\)), making it a key player in acid-base chemistry. The ability of a carboxylic acid to donate its proton is what determines its strength as an acid.

When surveying carboxylic acids, it's essential to consider the nature of the substituents attached to the carbon atom adjacent to the carboxyl group. These substituents can significantly influence acid strength by either stabilizing or destabilizing the conjugate base formed after the acid donates the proton.

Conjugate Base Stabilization

A conjugate base is formed when a carboxylic acid donates a proton. Its stability is crucial in determining the overall strength of the acid. In simple terms, the more stable the conjugate base, the stronger the acid. But what contributes to this stability?

Several factors can enhance the stabilization of a conjugate base. Delocalization of electrons, where negative charge is spread over several atoms, is one of the most effective ways to stabilize a conjugate base. Resonance, which allows for multiple structures, also plays a significant role in this stabilization. When the conjugate base is well-stabilized, it implies the original acid was strong, as it can more readily donate its proton.

Inductive Effect

The inductive effect can be thought of as the transmission of charge through a chain of atoms in a molecule, resulting from the electronegativity differences between atoms. Specifically, it's about the electron-withdrawing or electron-donating nature of substituents attached to the main chain.

In the context of acid strength, groups that withdraw electrons primarily exhibit a \(-I\) effect, making the carboxylic acid stronger. This effect diminishes rapidly with distance from the affected group, which means substituents need to be close to the carboxyl group to significantly influence the acid's behavior. In our specific problem, both chlorine (\(\text{Cl}\)) and bromine (\(\text{Br}\)) exert such effects, but chlorine is more effective due to its higher electronegativity.

Electron-Withdrawing Groups

Electron-withdrawing groups are critical in affecting the strength of carboxylic acids. These groups pull electron density away from the rest of the molecule, enhancing the acid's ability to donate a proton.

Halogens, such as chlorine and bromine, are common electron-withdrawing groups that impact acidity through the inductive effect. Chlorine, being more electronegative, exerts a stronger \(-I\) effect compared to bromine, thus making \(\text{Cl}-\text{CH}_2-\text{COOH}\) a stronger acid than its bromine counterpart. Without electron-withdrawing groups, the acid can't stabilize the conjugate base as effectively, leading to a weaker acid.