Br b) d) Br Br Br

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Select the least reactive alkyl bromide in an SN1 reaction?

The image displays chemical structures labeled as a), b), c), and d). Each structure is a variation of a cyclohexane ring with different substituents:

a) A cyclohexane ring with a methyl group (CH₃) attached to one carbon and a bromopropyl group (Br-CH₂-CH₂-) attached to another carbon.

b) A cyclohexane ring with a bromomethyl group (Br-CH₂-) at one position and an ethyl group (CH₂-CH₃) at another.

c) A benzene ring (indicated by alternating double bonds) with a bromomethyl group (Br-CH₂-) attached.

d) A cyclohexane ring with a methyl group (CH₃) attached and an isopropyl bromide group (CH(Br)-CH₃) connected to another carbon.

Each compound represents different structural variations and substitutions on a cyclohexane ring, illustrating concepts related to organic chemistry and functional group substitution.
Transcribed Image Text:The image displays chemical structures labeled as a), b), c), and d). Each structure is a variation of a cyclohexane ring with different substituents: a) A cyclohexane ring with a methyl group (CH₃) attached to one carbon and a bromopropyl group (Br-CH₂-CH₂-) attached to another carbon. b) A cyclohexane ring with a bromomethyl group (Br-CH₂-) at one position and an ethyl group (CH₂-CH₃) at another. c) A benzene ring (indicated by alternating double bonds) with a bromomethyl group (Br-CH₂-) attached. d) A cyclohexane ring with a methyl group (CH₃) attached and an isopropyl bromide group (CH(Br)-CH₃) connected to another carbon. Each compound represents different structural variations and substitutions on a cyclohexane ring, illustrating concepts related to organic chemistry and functional group substitution.
Expert Solution
Step 1

The unimolecular nucleophilic substitution reaction (SN1) proceeds in two steps. In the first step, the leaving group removed from the substrate, and the carbocation intermediate is formed. This step is a slow and rate-determining step. The rate of this step depends on the stability of the carbocation intermediate. The higher the stability of the carbonation intermediates the higher will be the rate of reaction.

In the second step, the nucleophile adds to the carbocation intermediate to form the final product.

Step 2

The increasing order of the stability of the carbonation for the even alkyl halide is as follows:

Chemistry homework question answer, step 2, image 1

The primary carbocation is the lowest stable and the secondary allylic carbonation is the highest stable intermediate carbocation.

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