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Question #44401
The Question
An optically pure alcohol, -3-methylhexan-3-ol (Compound P), is treated with sodium metal to form an intermediate Q. Q is subsequently reacted with excess methyl iodide to yield an ether, R. Compound R is then treated with one equivalent of anhydrous HI at 298 K to produce an alkyl halide S and an alcohol T. Which of the following correctly identifies the major products S and T, and the stereochemical nature of S?
Options
- A.
- B.
- C.
- D.
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
Williamson Ether Synthesis Constraints|Ether Cleavage Patterns
Why This Question Is Tricky
Common Mistakes
The question tests the student's understanding of the mechanistic switch in ether cleavage and stereochemical consequences of SN1 reactions.
A common trap is assuming that ether cleavage by HI always proceeds via an SN2 mechanism at the less hindered alkyl group (which would yield methyl iodide and the starting chiral alcohol, reversing the synthesis).
However, when a tertiary alkyl group is present, the mechanism shifts to SN1 due to the stability of the tertiary carbocation.
This results in the cleavage of the tertiary C-O bond, yielding methanol and a planar carbocation intermediate, which subsequently forms a racemic tertiary iodide.
The student must recognize this 'exception' to the less-hindered rule and correctly predict the loss of stereochemical information.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Alcohols, Phenols, and Ethers - Class 12 |
| Concept | Williamson Ether Synthesis Constraints |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |