Consider the following reaction and identify the major product (P): CH3-CH(CH3)-CH(OH)-CH3 ->[HBr] P (3-methylbutan-2-ol)

NEET 2023 Phase 1 · ChemistryPrevious Year Question

Consider the following reaction and identify the major product (P): $CH X_{3} - CH (CH X_{3}) - CH (OH) - CH X_{3} HBr P$ (3-methylbutan-2-ol)

Answer: (A) (A) $CH X_{3} - C (Br) (CH X_{3}) - CH X_{2} - CH X_{3}$ (2-bromo-2-methylbutane). \textbf{Answer:} (A) $CH X_{3} - C (Br) (CH X_{3}) - CH X_{2} - CH X_{3}$ (2-bromo-2-methylbutane) \textbf{Solution:} 3-Methylbutan-2-ol reacts with HBr by an $SN X_{1}$ pathway.

A.(A) $CH X_{3} - C (Br) (CH X_{3}) - CH X_{2} - CH X_{3}$ (2-bromo-2-methylbutane)✓
B.(B) $CH X_{3} - CH = CH - CH X_{3}$ (2-methylbut-2-ene)
C.(C) $CH X_{3} - CH (CH X_{3}) - CH (Br) - CH X_{3}$ (2-bromo-3-methylbutane)
D.(D) $(CH X_{3}) X_{3} C - CH X_{2} Br$ (1-bromo-2,2-dimethylpropane)

Correct Answer

(A) (A) $CH X_{3} - C (Br) (CH X_{3}) - CH X_{2} - CH X_{3}$ (2-bromo-2-methylbutane)

Solution & Explanation

\textbf{Answer:} (A) $CH X_{3} - C (Br) (CH X_{3}) - CH X_{2} - CH X_{3}$ (2-bromo-2-methylbutane) \textbf{Solution:} 3-Methylbutan-2-ol reacts with HBr by an $SN X_{1}$ pathway. Protonation of $- OH$ and loss of water gives a secondary carbocation at C-2. A 1,2-hydride shift from the adjacent C-3 then converts it into the more stable tertiary carbocation (positive charge now on C-3, which bears two methyls). Capture of this $3^{\circ}$ carbocation by $Br X^{-}$ gives the tertiary bromide, 2-bromo-2-methylbutane $CH X_{3} - C (Br) (CH X_{3}) - CH X_{2} CH X_{3}$ . Direct substitution without rearrangement would give 2-bromo-3-methylbutane (C), but the rearranged tertiary product (A) is the major product.

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Consider the following reaction and identify the major product (P): CHX3​−CH(CHX3​)−CH(OH)−CHX3​HBr​P (3-methylbutan-2-ol)

Solution & Explanation

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Consider the following reaction and identify the major product (P): $CH X_{3} - CH (CH X_{3}) - CH (OH) - CH X_{3} HBr P$ (3-methylbutan-2-ol)