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Choose the best explanation for the proposed curved arrow mechanism for an electrophilic addition reaction of butadiene with HBr.

Choose the best explanation for the proposed curved arrow mechanism for an electrophilic addition reaction of butadiene with HBr.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Choose the best explanation for the proposed curved arrow mechanism for an electrophilic addition reaction of butadiene with HBr.

:Br: Choose the best explanation for the proposed curved arrow mechanism for an electrophilic addition reaction of butadiene with HBr. H Br: A) The terminal carbon of the butadiene moves toward the HBr to form a bond with the proton. At the same time, the bromine atom moves away. The result of this atomic movement is a carbocation and a bromide anion. B) The electronegative bromine draws the electrons of the H-Br bond toward it, resulting in a fully ionized proton and bromide ion. The proton is donated to the butadiene, resulting in a cation on the secondary position. C) HBr donates a proton to the conjugated pi bond of butadiene, leading to formation of a delocalized carbocation. Donation of the proton leaves behind a bromide ion. D) Electrons are attracted from the pi bond to the acidic hydrogen of the electrophile, leading to formation of a carbocation. The electrons of the H-Br bond become a lone pair on the bromide ion.
Transcribed Image Text::Br: Choose the best explanation for the proposed curved arrow mechanism for an electrophilic addition reaction of butadiene with HBr. H Br: A) The terminal carbon of the butadiene moves toward the HBr to form a bond with the proton. At the same time, the bromine atom moves away. The result of this atomic movement is a carbocation and a bromide anion. B) The electronegative bromine draws the electrons of the H-Br bond toward it, resulting in a fully ionized proton and bromide ion. The proton is donated to the butadiene, resulting in a cation on the secondary position. C) HBr donates a proton to the conjugated pi bond of butadiene, leading to formation of a delocalized carbocation. Donation of the proton leaves behind a bromide ion. D) Electrons are attracted from the pi bond to the acidic hydrogen of the electrophile, leading to formation of a carbocation. The electrons of the H-Br bond become a lone pair on the bromide ion.
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