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Question #44375

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequence starting from 2-methylbut-2-ene (Compound A\textbf{A}):

A→2. \ceH2O2,NaOH1. \ceBH3⋅THFB→\ceTsCl,PyridineC\textbf{A} \xrightarrow[\text{2. } \ce{H2O2, NaOH}]{\text{1. } \ce{BH3 \cdot THF}} \textbf{B} \xrightarrow{\ce{TsCl, Pyridine}} \textbf{C}

Compound C\textbf{C} is subjected to elimination under two different conditions: Condition I: \cet−BuOK/t−BuOH,Δ→D (Major)\ce{t-BuOK / t-BuOH}, \Delta \rightarrow \textbf{D} \text{ (Major)} Condition II: \ceEtONa/EtOH,Δ→E (Major)\ce{EtONa / EtOH}, \Delta \rightarrow \textbf{E} \text{ (Major)}

Identify the major products D\textbf{D} and E\textbf{E}.

Options

  • A.

    D\textbf{D} = 3-methylbut-1-ene; E\textbf{E} = 2-methylbut-2-ene

  • B.

    D\textbf{D} = 2-methylbut-1-ene; E\textbf{E} = 2-methylbut-2-ene

  • C.

    D\textbf{D} = 2-methylbut-2-ene; E\textbf{E} = 3-methylbut-1-ene

  • D.

    D\textbf{D} = 2-methylbut-2-ene; E\textbf{E} = 2-methylbut-2-ene

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

Hydroboration-Oxidation Regiochemistry|E2 Regioselectivity (Saytzeff vs Hoffmann)

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to track regiochemistry through a multi-step sequence, specifically linking anti-Markovnikov installation to subsequent elimination preferences.

The first trap is the HBO step: students might incorrectly predict Markovnikov addition or carbocation rearrangement, leading to a tertiary alcohol (which would yield 2-methylbut-1-ene with a bulky base, as seen in Option B).

The second trap is the E2 elimination of the secondary tosylate.

Students must recognize that the choice of base dictates the regioselectivity (Saytzeff vs.

Hoffmann).

A bulky base (t-BuOK) suffers from steric clash during internal proton abstraction, favoring the less substituted kinetic product (Hoffmann).

A small base (EtONa) favors the more stable, substituted alkene (Saytzeff).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterAlcohols, Phenols, and Ethers - Class 12
ConceptHydroboration-Oxidation Regiochemistry
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree