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Question #44516

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The Question

A student attempts to synthesize the Schiff base of 2,2,4,4-tetramethyl-3-pentanone using an aqueous solution of methanamine at 298 K. To maximize the yield, the total analytical concentration of methanamine is increased to 5.0 M. However, the yield of the imine remains negligible at both pH = 2.0 and pH = 4.5. Which of the following statements correctly identifies the primary chemical principles responsible for this failure?

Options

A.
At pH = 2.0, the nucleophile is deactivated via complete protonation ( $\ce{CH3NH3+}$ ). At pH = 4.5, severe B-strain in the $sp^3$ -hybridized carbinolamine intermediate renders the equilibrium thermodynamically unfavorable ( $\Delta G^\circ > 0$ ), overriding any Le Chatelier concentration effects.
B.
At pH = 2.0, the carbonyl oxygen is insufficiently protonated to undergo nucleophilic attack. At pH = 4.5, the high molarity of methanamine decreases the activity of water, shifting the equilibrium towards the reactants due to steric hindrance.
C.
The reaction is thermodynamically favorable at both pH levels, but the high concentration of methanamine increases the solvent viscosity, kinetically freezing the formation of the tetrahedral intermediate.
D.
At pH = 4.5, methanamine acts as a base rather than a nucleophile, enolizing the ketone. At pH = 2.0, B-strain prevents the elimination of water from the carbinolamine intermediate.

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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

Concentration Units Dependency|Schiff Base pH Dependence|Steric Hindrance (B-Strain)

Why This Question Is Tricky

Common Mistakes

This question tests the limits of Le Chatelier's principle and the pH-dependence of nucleophilic addition.

Students often fall into the trap of assuming that increasing the total analytical concentration (Molarity) of a reactant will universally increase the product yield.

The 'Hidden Variable' here is twofold: First, the active concentration of the nucleophile is strictly pH-dependent, rendering high molarity useless at low pH (pH = 2.0).

Second, the 'Chemical Exception' of B-strain (Back-strain) in highly hindered ketones like 2,2,4,4-tetramethyl-3-pentanone creates a massive thermodynamic barrier ( $\Delta G^\circ > 0$ ) for the $sp^3$ tetrahedral intermediate.

This demonstrates that kinetic optimization (pH 4.5) and concentration adjustments cannot overcome a fundamentally unfavorable thermodynamic equilibrium dictated by steric hindrance.

Option D also contains a structural trap, as 2,2,4,4-tetramethyl-3-pentanone lacks alpha-hydrogens and cannot enolize.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Concentration Units Dependency
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free