Question

that is, select 1 in the second column for the most acidic molecule, 2 for the next most acidic molecule, and so on.

Explanation:

To determine the acidity of these carboxylic acid molecules, we use the inductive effect (electron-withdrawing groups like $\ce{Cl}$ stabilize the conjugate base by dispersing negative charge). The key factors are:

1. Proximity of $\boldsymbol{\ce{Cl}}$ to the carboxyl group (closer $\ce{Cl}$ atoms have a stronger inductive effect).
2. Number of $\boldsymbol{\ce{Cl}}$ atoms (more $\ce{Cl}$ atoms increase electron withdrawal).

Step 1: Analyze Each Molecule

• Top molecule (first row): Two $\ce{Cl}$ atoms on the carbon adjacent to the $\ce{CH2}$ group (one carbon away from the carboxyl group, $\ce{-COOH}$).
• Second molecule (second row): Two $\ce{Cl}$ atoms, but one $\ce{Cl}$ is two carbons away from $\ce{-COOH}$, and the other is three carbons away.
• Third molecule (third row): Two $\ce{Cl}$ atoms on the carbon adjacent to the $\ce{CH2}$ group (one carbon away from $\ce{-COOH}$).
• Bottom molecule (fourth row): Two $\ce{Cl}$ atoms on the carbon directly attached to the carboxyl group (the $\alpha$-carbon, adjacent to $\ce{-COOH}$).

Step 2: Rank by Acidity

Acidity increases with:

• Closer $\ce{Cl}$ to $\ce{-COOH}$ (stronger inductive effect).
• More $\ce{Cl}$ atoms (greater electron withdrawal).

The bottom molecule has $\ce{Cl}$ atoms on the $\alpha$-carbon (closest to $\ce{-COOH}$), so it is the most acidic (rank 1).

The third and top molecules have $\ce{Cl}$ atoms one carbon away from $\ce{-COOH}$, but the third molecule’s $\ce{Cl}$ atoms are on a more substituted carbon (but inductive effect depends on distance, not substitution). However, the bottom molecule is still closest.

The second molecule has $\ce{Cl}$ atoms farther away, so it is the least acidic (rank 4).

Final Ranking (Most to Least Acidic)

1. Bottom molecule (fourth row: $\ce{Cl}$ on $\alpha$-carbon).
2. Third molecule (third row: $\ce{Cl}$ one carbon from $\ce{-COOH}$).
3. Top molecule (first row: $\ce{Cl}$ one carbon from $\ce{-COOH}$).
4. Second molecule (second row: $\ce{Cl}$ farther from $\ce{-COOH}$).

Answer:

(Example Selection)

• Bottom molecule: Select $\boldsymbol{1}$ (most acidic).
• Third molecule: Select $\boldsymbol{2}$.
• Top molecule: Select $\boldsymbol{3}$.
• Second molecule: Select $\boldsymbol{4}$.

(Adjust based on exact molecular structure, but the key is proximity of $\ce{Cl}$ to $\ce{-COOH}$.)