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Question #44413
The Question
Let represent the kinetic classification of alcohol in the Lucas test (conc. \ce{HCl} + anhydrous \ce{ZnCl2}) at 298 K: if turbidity appears immediately. if turbidity appears after 5-10 minutes. if no turbidity appears at 298 K.
Let represent the number of -hydrogen atoms in the major alkyl chloride product formed when alcohol is heated with Lucas reagent for a prolonged period.
Consider the following three alcohols: \textbf{P}: 3-Methylbutan-2-ol \textbf{Q}: 2,2-Dimethylpropan-1-ol \textbf{R}: 1-Methylcyclopentan-1-ol
Calculate the value of the expression: .
Options
- A.
47
- B.
31
- C.
23
- D.
15
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
Lucas Test Logic|SN1 Carbocation Rearrangement|Steric Hindrance in SN2
Why This Question Is Tricky
Common Mistakes
Tests the 'Chemical Exception' where students conflate thermodynamic product stability with kinetic reaction rates.
A common trap is assuming that because neopentyl alcohol (Q) and 3-methylbutan-2-ol (P) rearrange to stable tertiary carbocations, they will give immediate turbidity in the Lucas test.
However, the rate-determining step for P is the formation of a secondary carbocation (5-10 mins), and for Q, it is a high-energy concerted rearrangement (no reaction at 298 K).
The problem forces students to independently evaluate the kinetic RDS and the thermodynamic rearranged products.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Alcohols, Phenols, and Ethers - Class 12 |
| Concept | Lucas Test Logic |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |