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Question #43956
The Question
An organic compound () is identified as 2,2-dimethyl-3-oxo-3-phenylpropanal.
Reaction I: is treated with an excess of in anhydrous ether, followed by mild acidic workup, to yield the major product .
Reaction II: is heated with concentrated aqueous , followed by acidic workup, to yield a major organic product containing a benzene ring, along with a one-carbon byproduct.
Identify and .
Options
- A.
: 2,2-dimethyl-1-phenylpropane-1,3-diol; : 2-methyl-1-phenylpropan-1-one
- B.
: 2,2-dimethyl-1-phenylpropane-1,3-diol; : 3-hydroxy-2,2-dimethyl-3-phenylpropanoic acid
- C.
: 2,2-dimethyl-1-phenylpropane-1,3-diol; : 2,2-dimethyl-3-oxo-3-phenylpropanoic acid
- D.
: 3-hydroxy-2,2-dimethyl-3-phenylpropanal; : 3-hydroxy-2,2-dimethyl-3-phenylpropanoic acid
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
Reduction Selectivity (Hydrides)|Alpha-Hydrogen Acidity|Cannizzaro Mechanism
Why This Question Is Tricky
Common Mistakes
The question tests the kinetic competition between hydride transfer (Cannizzaro reaction) and enolate expulsion (retro-Claisen cleavage).
Students often blindly apply the heuristic 'no alpha-hydrogens + strong base = Cannizzaro', leading them to incorrectly predict an intramolecular Cannizzaro product (Option B).
The hidden variable is the leaving group ability: an enolate is a significantly better leaving group than a hydride ion, making C-C bond cleavage (deformylation) the dominant pathway for alpha,alpha-disubstituted beta-keto aldehydes.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Aldehydes, Ketones, and Carboxylic Acids - Class 12 |
| Concept | Reduction Selectivity (Hydrides) |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |