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Question #45064
The Question
The electromeric effect () involves the complete transfer of -electrons to one of the bonded atoms under the influence of an attacking reagent, such as during the nucleophilic addition of to a carbonyl group. However, this transient polarization does not imply the structural rearrangement required for keto-enol tautomerism.
Consider the following 8 compounds: I. Methanal II. Benzaldehyde III. 2,2,2-Trichloroethanal IV. 3,3-Dimethylbicyclo[2.2.1]heptan-2-one V. 2,2-Dimethylpropanal VI. 2,2-Diphenylethanal VII. Acetophenone VIII. Cyclohexa-2,4-dien-1-one
How many of these compounds will exhibit the electromeric effect upon the approach of a nucleophile at , but will strictly NOT exhibit keto-enol tautomerism in an aqueous medium?
Options
- A.
4
- B.
5
- C.
6
- D.
7
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
Electromeric Effect|Tautomerism Conditions|Acidity and Basicity (pKa/pKb)
Why This Question Is Tricky
Common Mistakes
This problem traps students by conflating the transient electronic polarization of the C=O bond (Electromeric Effect) with the permanent structural rearrangement of Tautomerism.
The key hidden variables are the structural prerequisites for tautomerism: recognizing the absence of alpha-hydrogens in compounds like chloral and pivalaldehyde, and crucially, applying Bredt's Rule to realize that the bridgehead alpha-hydrogen in 3,3-dimethylbicyclo[2.2.1]heptan-2-one cannot participate in enolization despite its presence.
It also tests the recognition of aromatization as a powerful driving force for tautomerism in cyclohexa-2,4-dien-1-one.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2 |
| Concept | Electromeric Effect |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |