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

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.

Correct Answer:

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Sign up to view answer