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What would be the major product of the dehydrohalogenation of 2-chloropentane with KOH in DMSO? O 2-pentene 1-pentene O 2-methyl-1-butene 2-methyl-2-butene

What would be the major product of the dehydrohalogenation of 2-chloropentane with KOH in DMSO? O 2-pentene 1-pentene O 2-methyl-1-butene 2-methyl-2-butene

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Chemistry
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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
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**Question:** What would be the major product of the dehydrohalogenation of 2-chloropentane with KOH in DMSO? **Options:** - 2-pentene - 1-pentene - 2-methyl-1-butene - 2-methyl-2-butene **Explanation:** This question is focused on the outcome of a dehydrohalogenation reaction. Dehydrohalogenation typically involves the removal of a hydrogen halide from an alkyl halide, resulting in the formation of an alkene. In this scenario, 2-chloropentane is treated with potassium hydroxide (KOH) in dimethyl sulfoxide (DMSO), a solvent that favors the elimination (E2) mechanism. In an E2 reaction, the base (KOH) abstracts a proton from the β-carbon relative to the halide (chlorine, in this case), resulting in the formation of a double bond (alkene) and the expulsion of the halide as the leaving group. The position of the double bond will determine which pentene product is formed. In this case, the possible alkenes are: 1. 2-pentene, where the double bond is between the second and third carbon atoms. 2. 1-pentene, where the double bond is between the first and second carbon atoms. 3. 2-methyl-1-butene and 2-methyl-2-butene, which involve branching at the first or second carbon of butene, respectively, but these are not typically expected from the given starting material, 2-chloropentane. Considering the mechanism and regioselectivity (Zaitsev's rule, which suggests the more substituted alkene is favored), 2-pentene is expected to be the major product of this dehydrohalogenation reaction.
Transcribed Image Text:**Question:** What would be the major product of the dehydrohalogenation of 2-chloropentane with KOH in DMSO? **Options:** - 2-pentene - 1-pentene - 2-methyl-1-butene - 2-methyl-2-butene **Explanation:** This question is focused on the outcome of a dehydrohalogenation reaction. Dehydrohalogenation typically involves the removal of a hydrogen halide from an alkyl halide, resulting in the formation of an alkene. In this scenario, 2-chloropentane is treated with potassium hydroxide (KOH) in dimethyl sulfoxide (DMSO), a solvent that favors the elimination (E2) mechanism. In an E2 reaction, the base (KOH) abstracts a proton from the β-carbon relative to the halide (chlorine, in this case), resulting in the formation of a double bond (alkene) and the expulsion of the halide as the leaving group. The position of the double bond will determine which pentene product is formed. In this case, the possible alkenes are: 1. 2-pentene, where the double bond is between the second and third carbon atoms. 2. 1-pentene, where the double bond is between the first and second carbon atoms. 3. 2-methyl-1-butene and 2-methyl-2-butene, which involve branching at the first or second carbon of butene, respectively, but these are not typically expected from the given starting material, 2-chloropentane. Considering the mechanism and regioselectivity (Zaitsev's rule, which suggests the more substituted alkene is favored), 2-pentene is expected to be the major product of this dehydrohalogenation reaction.
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