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8 3 3 Identify the product when the alkyl bromide is treated with sodium methoxide in DMSO. OF D B A E Identify the product when the alkylbromide is treated with sodium-butoxide in DMSO. Oc D E F

8 3 3 Identify the product when the alkyl bromide is treated with sodium methoxide in DMSO. OF D B A E Identify the product when the alkylbromide is treated with sodium-butoxide in DMSO. Oc D E F

Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter21: Organic And Biological Molecules
Section: Chapter Questions
Problem 7RQ
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I keep seeing B, E for the answers or EC and E on chegg which doesn't make sense? It can only be one. Someone please help with a non confusing answer 

### Understanding Alkyl Halides and Their Reaction Products #### Structure Analysis Consider the structure of the alkyl halide shown: ![Alkyl Halide](image) The given alkyl halide has the following structure: - A cyclopentane ring. - A central carbon bonded to two different groups: -CH3 (methyl), -CH3 (methyl), and -Br (bromine). #### Possible Reaction Products The image displays compounds labeled A through F, which represent potential substitution or elimination products for the given alkyl halide. ##### Compounds A to F 1. **Compound A**: - Structure: A cyclopentane ring with a branching chain [1-Bromo-2-methylcyclopentane]. - Substitution product due to the substitution of the bromine atom while retaining the cyclopentyl and methyl groups. 2. **Compound B**: - Structure: A cyclopentane ring with double bond [Cyclopentene with an ethylmethyl substituent]. - Likely an elimination product, as shown by the presence of a double bond in the ring. 3. **Compound C**: - Structure: Another cyclopentane with a different chain attached [1-Bromo-1-ethylcyclopentane]. - Represents an alternate substitution product. 4. **Compound D**: - Structure: A cyclopentane with a substitution group [2-ethyl-1-cyclopentanol]. - This is another substitution product moving the methyl group position. 5. **Compound E**: - Structure: Like compound B, but the structure positions the group at a different carbon indicating an isomer [2-bromo-1-cyclopentanol]. - This could be another substitution or elimination product. 6. **Compound F**: - Structure: A cyclopentane ring with varying position substituent [3-ethyl-1-bromocyclopentane]. - Another substitution product showing different position attachment. #### Explanation The image showcases the structure of an alkyl halide along with various potential products formed via substitution or elimination reactions. These reactions involve the loss or replacement of the bromine atom and can either retain the single bond structure of the starting material (substitution) or convert it to involve double bonds through elimination. Each compound A through F represents a unique product structure resulting from these
Transcribed Image Text:### Understanding Alkyl Halides and Their Reaction Products #### Structure Analysis Consider the structure of the alkyl halide shown: ![Alkyl Halide](image) The given alkyl halide has the following structure: - A cyclopentane ring. - A central carbon bonded to two different groups: -CH3 (methyl), -CH3 (methyl), and -Br (bromine). #### Possible Reaction Products The image displays compounds labeled A through F, which represent potential substitution or elimination products for the given alkyl halide. ##### Compounds A to F 1. **Compound A**: - Structure: A cyclopentane ring with a branching chain [1-Bromo-2-methylcyclopentane]. - Substitution product due to the substitution of the bromine atom while retaining the cyclopentyl and methyl groups. 2. **Compound B**: - Structure: A cyclopentane ring with double bond [Cyclopentene with an ethylmethyl substituent]. - Likely an elimination product, as shown by the presence of a double bond in the ring. 3. **Compound C**: - Structure: Another cyclopentane with a different chain attached [1-Bromo-1-ethylcyclopentane]. - Represents an alternate substitution product. 4. **Compound D**: - Structure: A cyclopentane with a substitution group [2-ethyl-1-cyclopentanol]. - This is another substitution product moving the methyl group position. 5. **Compound E**: - Structure: Like compound B, but the structure positions the group at a different carbon indicating an isomer [2-bromo-1-cyclopentanol]. - This could be another substitution or elimination product. 6. **Compound F**: - Structure: A cyclopentane ring with varying position substituent [3-ethyl-1-bromocyclopentane]. - Another substitution product showing different position attachment. #### Explanation The image showcases the structure of an alkyl halide along with various potential products formed via substitution or elimination reactions. These reactions involve the loss or replacement of the bromine atom and can either retain the single bond structure of the starting material (substitution) or convert it to involve double bonds through elimination. Each compound A through F represents a unique product structure resulting from these
**Organic Chemistry Educational Content** **Question 1:** Identify the product when the alkyl bromide is treated with sodium methoxide in DMSO. Options: A) B) C) D) E) **Answer:** B (highlighted) --- **Question 2:** Identify the product when the alkyl bromide is treated with sodium t-butoxide in DMSO. Options: A) B) C) D) E) F) **Answer:** E (highlighted) **Explanation:** The image presents two reaction scenarios where an alkyl bromide undergoes specific treatments with different bases in DMSO, a polar aprotic solvent. 1. **First Reaction:** - **Reagent**: Sodium methoxide (NaOCH3) - This strong base in a polar aprotic solvent like DMSO typically favors an \(S_N2\) reaction mechanism, resulting in a specific product (B) that is the direct result of nucleophilic substitution. 2. **Second Reaction:** - **Reagent**: Sodium t-butoxide (NaO-t-Bu) - This bulky strong base in a polar aprotic solvent like DMSO typically favors an \(E2\) elimination reaction. This leads to the formation of the more substituted alkene (product E), due to the steric hindrance that the bulky base provides. **Structural Diagrams:** - **Compound D**: Cyclopentane ring with a bromine (Br) and a methyl (CH3) group. - **Compound E**: Unsaturated cyclopentane with a double bond (alkene) and a methyl (CH3) group. - **Compound F**: Cyclopentane ring with two substituents, probably indicating a different isomer or substitution pattern than compounds D and E. Understanding the differences between these products helps in predicting the outcomes based on the reaction conditions used.
Transcribed Image Text:**Organic Chemistry Educational Content** **Question 1:** Identify the product when the alkyl bromide is treated with sodium methoxide in DMSO. Options: A) B) C) D) E) **Answer:** B (highlighted) --- **Question 2:** Identify the product when the alkyl bromide is treated with sodium t-butoxide in DMSO. Options: A) B) C) D) E) F) **Answer:** E (highlighted) **Explanation:** The image presents two reaction scenarios where an alkyl bromide undergoes specific treatments with different bases in DMSO, a polar aprotic solvent. 1. **First Reaction:** - **Reagent**: Sodium methoxide (NaOCH3) - This strong base in a polar aprotic solvent like DMSO typically favors an \(S_N2\) reaction mechanism, resulting in a specific product (B) that is the direct result of nucleophilic substitution. 2. **Second Reaction:** - **Reagent**: Sodium t-butoxide (NaO-t-Bu) - This bulky strong base in a polar aprotic solvent like DMSO typically favors an \(E2\) elimination reaction. This leads to the formation of the more substituted alkene (product E), due to the steric hindrance that the bulky base provides. **Structural Diagrams:** - **Compound D**: Cyclopentane ring with a bromine (Br) and a methyl (CH3) group. - **Compound E**: Unsaturated cyclopentane with a double bond (alkene) and a methyl (CH3) group. - **Compound F**: Cyclopentane ring with two substituents, probably indicating a different isomer or substitution pattern than compounds D and E. Understanding the differences between these products helps in predicting the outcomes based on the reaction conditions used.
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