Consider the Reaction
CH₃(CH₂)₄CH₂CH(Br)CH₃ + NaOH $\rarr$ CH₃(CH₂)₄CH₂CH(OH)CH₃
(Optically Pure Enantiomer)
the Rate Law for This Reaction

question 5

Multiple Choice

Consider the reaction
CH₃(CH₂) ₄CH₂CH(Br) CH₃ + NaOH $\rarr$ CH₃(CH₂) ₄CH₂CH(OH) CH₃
(optically pure enantiomer)
The rate law for this reaction over a wide range of [OH^-] is R = k₁[RX] + k₂[RX][OH^-] where RX is 2-bromooctane and k₂/k₁ = 20.
At very low hydroxide ion concentration,

• A) the product of the reaction does not rotate the plane of polarized light.
• B) the product formed is CH₃(CH₂)₄CH=CHCH₃ and not the alcohol.
• C) the reaction proceeds by an S_N2 mechanism.
• D) the hydroxide ion attacks the face of the carbocation in a concerted mechanism on the face opposite the leaving group.
• E) the product is optically active and rotates the plane of polarized light in the opposite direction with respect to 2-bromooctane.

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