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Question #44413

Difficulty: 8/10 | 🟢 Free

The Question

Let tit_i represent the kinetic classification of alcohol ii in the Lucas test (conc. \ce{HCl} + anhydrous \ce{ZnCl2}) at 298 K: ti=1t_i = 1 if turbidity appears immediately. ti=2t_i = 2 if turbidity appears after 5-10 minutes. ti=3t_i = 3 if no turbidity appears at 298 K.

Let hih_i represent the number of β\beta-hydrogen atoms in the major alkyl chloride product formed when alcohol ii 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: Σ(ti×hi)=(tP×hP)+(tQ×hQ)+(tR×hR)\Sigma (t_i \times h_i) = (t_P \times h_P) + (t_Q \times h_Q) + (t_R \times h_R).

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

SubjectChemistry
ChapterAlcohols, Phenols, and Ethers - Class 12
ConceptLucas Test Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree