Identify the correct structure of the alkene which gives the two electophilic addition products with HCl/ether.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Identify the correct structure of the alkene which gives the two electophilic addition products with HCl/ether.

The image is presenting a chemical reaction question. The reaction involves an unknown starting alkene, represented by a question mark, that reacts with hydrochloric acid (HCl) in the presence of ether to produce two chlorinated products.

**Chemical Equation:**

- **Reactant:** ? (Unknown alkene)
- **Reagents:** HCl and ether
- **Products:** Two molecules of 3-chloro-3-methylpentane.

**Potential Reactants (Options A, B, C, D):**

1. **Option A:** 2,3-dimethyl-1-butene
2. **Option B:** 3,3-dimethyl-1-butene
3. **Option C:** 3-methyl-1-pentene
4. **Option D:** 2-methyl-1-pentene

The goal is to determine which starting alkene (A, B, C, or D) will lead to the formation of the specified products upon reaction with HCl in an ether solvent.

This reaction likely involves a Markovnikov addition, where the HCl adds across the double bond to form the most substituted (stabilized) carbocation intermediate, resulting in a mixture of 3-chloro-3-methylpentane products.
Transcribed Image Text:The image is presenting a chemical reaction question. The reaction involves an unknown starting alkene, represented by a question mark, that reacts with hydrochloric acid (HCl) in the presence of ether to produce two chlorinated products. **Chemical Equation:** - **Reactant:** ? (Unknown alkene) - **Reagents:** HCl and ether - **Products:** Two molecules of 3-chloro-3-methylpentane. **Potential Reactants (Options A, B, C, D):** 1. **Option A:** 2,3-dimethyl-1-butene 2. **Option B:** 3,3-dimethyl-1-butene 3. **Option C:** 3-methyl-1-pentene 4. **Option D:** 2-methyl-1-pentene The goal is to determine which starting alkene (A, B, C, or D) will lead to the formation of the specified products upon reaction with HCl in an ether solvent. This reaction likely involves a Markovnikov addition, where the HCl adds across the double bond to form the most substituted (stabilized) carbocation intermediate, resulting in a mixture of 3-chloro-3-methylpentane products.
Expert Solution
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First product is a minor product in which chlorine will add on 1°degree carbon

second product is the major product because it is a 3 degree carbon and chlorine will add on that carbon

 

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