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Question #43959
The Question
Consider the compound \textbf{X}, 1-(4-hydroxyphenyl)pent-4-en-1-one. \textbf{X} is subjected to two different reduction conditions to yield major products \textbf{P} and \textbf{Q}:
Reaction 1: \textbf{X} \textbf{P}
Reaction 2: \textbf{X} \textbf{Q}
Identify the major products \textbf{P} and \textbf{Q}.
Options
- A.
\textbf{P}: (rac)-8-methyl-5,6,7,8-tetrahydronaphthalen-2-ol ; \textbf{Q}: 4-(pent-4-en-1-yl)phenol
- B.
\textbf{P}: 4-(pent-4-en-1-yl)phenol ; \textbf{Q}: (rac)-8-methyl-5,6,7,8-tetrahydronaphthalen-2-ol
- C.
\textbf{P}: 4-pentylphenol ; \textbf{Q}: 4-(pent-4-en-1-yl)phenol
- D.
\textbf{P}: 4-(4-hydroxypentyl)phenol ; \textbf{Q}: 4-(pent-4-en-1-yl)phenol
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)|Clemmensen vs. Wolff-Kishner|EAS in Phenols
Why This Question Is Tricky
Common Mistakes
This problem tests the 'Hidden Variable' of reagent pH in standard reductions.
Students often memorize Clemmensen and Wolff-Kishner as interchangeable carbonyl-to-alkane reductions.
The trap is failing to recognize that the harsh acidic conditions of the Clemmensen reduction will trigger an intramolecular electrophilic aromatic substitution (Friedel-Crafts alkylation) when an alkene and an activated aromatic ring (phenol) are present.
Wolff-Kishner is the required pathway to avoid this exception and cleanly reduce the ketone while preserving the alkene.
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 |