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Question #44416
The Question
Consider the compound \textbf{X}, 5-hydroxyhexan-2-one. It is subjected to three different reducing conditions to yield major organic products \textbf{P}, \textbf{Q}, and \textbf{R}:
Reaction 1: \textbf{X} \textbf{P}
Reaction 2: \textbf{X} \textbf{Q}
Reaction 3: \textbf{X} \textbf{R}
Identify the major products \textbf{P}, \textbf{Q}, and \textbf{R}.
Options
- A.
\textbf{P}: 2,5-dimethyltetrahydrofuran; \textbf{Q}: hexan-2-ol; \textbf{R}: hexane-2,5-diol
- B.
\textbf{P}: 2-chlorohexane; \textbf{Q}: hexan-2-ol; \textbf{R}: hexane-2,5-diol
- C.
\textbf{P}: hexan-2-ol; \textbf{Q}: hex-2-ene; \textbf{R}: hexane-2,5-diol
- D.
\textbf{P}: 2,5-dimethyltetrahydrofuran; \textbf{Q}: 2,5-dimethyltetrahydrofuran; \textbf{R}: hexane-2,5-diol
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
Oxidation State Traps|Reduction Selectivity (Sn/HCl vs LiAlH4)|Clemmensen vs. Wolff-Kishner
Why This Question Is Tricky
Common Mistakes
This question tests the chemoselectivity and condition-dependent side reactions of standard carbonyl reductions.
Students often memorize Clemmensen () and Wolff-Kishner () as simple 'ketone to alkane' transformations, ignoring the solvent/pH conditions.
The 'Hidden Mechanism' is the acid-catalyzed intramolecular cyclization of the -hydroxy ketone to a hemiacetal during the Clemmensen reduction.
Because the hemiacetal forms rapidly, the reduction acts on the oxocarbenium ion, leading to a cyclic ether rather than the expected open-chain alkane or alkyl chloride.
Conversely, the basic conditions of Wolff-Kishner preserve the secondary alcohol, demonstrating orthogonal chemoselectivity.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Alcohols, Phenols, and Ethers - Class 12 |
| Concept | Oxidation State Traps |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |