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Question #45076

Difficulty: 8/10 | 🟢 Free

The Question

Consider the reaction of ethyl 6-oxoheptanoate with $1.0$ equivalent of sodium ethoxide ( $\ce{NaOEt}$ ) in anhydrous ethanol at $298\text{ K}$ , followed by careful acidification with dilute $\ce{HCl}$ . Identify the major organic product formed.

Options

A.
2-Acetylcyclopentan-1-one
B.
Ethyl 2-methylcyclopent-1-ene-1-carboxylate
C.
Cycloheptane-1,3-dione
D.
Ethyl 2-methylcyclopent-2-ene-1-carboxylate

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

IUPAC Nomenclature Priority|Alpha-Hydrogen Acidity|Aldehydes, Ketones, and Carboxylic Acids

Why This Question Is Tricky

Common Mistakes

The question tests the 'Hidden Variable' of competing $\alpha$ -hydrogen acidities in multifunctional compounds.

Students often fall into the trap of assuming the ester enolate will attack the more electrophilic ketone carbonyl (Aldol pathway) to form ethyl 2-methylcyclopent-1-ene-1-carboxylate.

However, they overlook that the 'oxo' substituent's $\alpha$ -protons are $10^4$ times more acidic.

The reaction is thermodynamically driven by the formation of a highly stable $\beta$ -diketone enolate (Dieckmann pathway), overriding the inherent electrophilicity difference between the ketone and ester.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
Concept	IUPAC Nomenclature Priority
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free