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Question #45048

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following statements regarding the acidity, tautomerism, and structural features of organic compounds at $298\text{ K}$ :

\begin{itemize} \item[\textbf{P.}] The $pK_{a1}$ of 2-hydroxybenzoic acid (salicylic acid) is lower than that of 4-hydroxybenzoic acid, but its $pK_{a2}$ is significantly higher due to the stabilization of the monoanion via intramolecular hydrogen bonding. \item[\textbf{Q.}] 1,3-Cyclopentanedione ( $pK_a \approx 4.5$ ) is significantly more acidic than 2,4-pentanedione ( $pK_a \approx 9.0$ ) because the conjugate base of the cyclic diketone is stabilized by strong intramolecular hydrogen bonding. \item[\textbf{R.}] The $\alpha$ -hydrogens of 1-fluoropropan-2-one are less acidic than those of 1-chloropropan-2-one because the strong $+M$ effect of fluorine destabilizes the enolate intermediate, overriding its superior $-I$ effect. \item[\textbf{S.}] In aqueous solution, the enol content of 2,4-pentanedione is higher than that in non-polar solvents (like hexane) because water molecules facilitate intermolecular hydrogen bonding with the enol $-\text{OH}$ group. \end{itemize}

Which of the given statements are CORRECT?

Options

A.
P and R only
B.
P, Q, and R only
C.
Q and S only
D.
P, R, and S only

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Inductive Effect (I-Effect)|Alpha-Hydrogen Acidity

Why This Question Is Tricky

Common Mistakes

This question tests the boundaries of inductive effects, resonance, and hydrogen bonding thermodynamics.

Trap 1 (Statement Q): Students often blindly associate high acidity in $\beta$ -dicarbonyls with intramolecular H-bonding, failing to recognize that cyclic 1,3-diketones cannot form them due to ring strain and transoid geometry.

Trap 2 (Statement R): Students typically rank acidity strictly by the $-I$ effect ( $F > Cl > Br$ ).

The hidden variable is the $+M$ effect in the enolate intermediate, where Fluorine's $2p$ lone pairs strongly repel the enolate's $\pi$ -electrons.

Trap 3 (Statements P & S): Misapplication of H-bonding thermodynamics.

Stabilizing a monoanion increases $K_{a1}$ but decreases $K_{a2}$ , and polar protic solvents destroy intramolecular H-bonds, favoring the diketo tautomer over the enol.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
Concept	Inductive Effect (I-Effect)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free