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Question #44625

Difficulty: 8/10 | 🟢 Free

The Question

Consider the enantiomerically pure alkyl halide, (S)-2-bromo-3-methylbutane (\textbf{A}). It undergoes three distinct parallel reaction sequences:

Reaction I: \textbf{A} is treated with \ceNaCN\ce{NaCN} in anhydrous DMSO to yield major product \textbf{D}.

Reaction II: \textbf{A} is treated with potassium tert-butoxide (\cet−BuOK\ce{t-BuOK}) in tert-butanol at 70∘C70^\circ\text{C} to yield major product \textbf{B}. \textbf{B} is then subjected to hydroboration-oxidation (\ceB2H6/THF\ce{B2H6/THF} followed by \ceH2O2/NaOH\ce{H2O2/NaOH}) to yield \textbf{E}.

Reaction III: \textbf{A} is treated with sodium ethoxide (\ceNaOEt\ce{NaOEt}) in ethanol at 70∘C70^\circ\text{C} to yield major product \textbf{C}. \textbf{C} is then subjected to oxymercuration-demercuration (\ceHg(OAc)2/H2O\ce{Hg(OAc)2/H2O} followed by \ceNaBH4\ce{NaBH4}) to yield \textbf{F}.

Identify the correct IUPAC names and stereochemistry for \textbf{D}, \textbf{E}, and \textbf{F}.

Options

  • A.

    \textbf{D} = (R)-2,3-dimethylbutanenitrile; \textbf{E} = 3-methyl-1-butanol; \textbf{F} = 2-methyl-2-butanol

  • B.

    \textbf{D} = (S)-2,3-dimethylbutanenitrile; \textbf{E} = 3-methyl-2-butanol; \textbf{F} = 3-methyl-2-butanol

  • C.

    \textbf{D} = (R)-2,3-dimethylbutanenitrile; \textbf{E} = 2-methyl-2-butanol; \textbf{F} = 3-methyl-1-butanol

  • D.

    \textbf{D} = (S)-2,3-dimethylbutanenitrile; \textbf{E} = 3-methyl-1-butanol; \textbf{F} = 2-methyl-2-butanol

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

E2 Regioselectivity (Saytzeff vs Hoffmann)|Steric Hindrance in SN2|E1/E2 Solvent and Base Effects

Why This Question Is Tricky

Common Mistakes

This problem tests the critical exception to Zaitsev's rule dictated by base bulkiness, alongside stereochemical tracking in SN2S_N2 reactions.

Students often blindly apply Zaitsev's rule to all E2 eliminations.

The trap is failing to recognize that a bulky base like \cet−BuOK\ce{t-BuOK} shifts the regioselectivity to the kinetically favored Hofmann product due to steric hindrance in the transition state.

Furthermore, it tests the ability to correctly assign R/S configurations post-inversion, and apply the correct regiochemistry (Markovnikov vs. anti-Markovnikov) for subsequent hydration reactions on the divergent alkene intermediates.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterHaloalkanes and Haloarenes - Class 12
ConceptE2 Regioselectivity (Saytzeff vs Hoffmann)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree