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3. Show the reaction and products formed when methyl magnesium iodide (CH3-Mg-I) is reacted with butanal (butyraldehyde), then the alcoholate formed is treated with aqueous acid. What alcohol is formed (draw its structure.)

3. Show the reaction and products formed when methyl magnesium iodide (CH3-Mg-I) is reacted with butanal (butyraldehyde), then the alcoholate formed is treated with aqueous acid. What alcohol is formed (draw its structure.)

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
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**3. Reaction and Product Formation with Methyl Magnesium Iodide and Butanal** **Objective:** To demonstrate the reaction and identify the products formed when methyl magnesium iodide (CH₃-Mg-I) reacts with butanal (butyraldehyde). The subsequent step involves treating the alcoholate formed with aqueous acid to yield an alcohol. **Reaction Steps:** 1. **Initial Reaction:** - **Reagents:** Methyl magnesium iodide (CH₃-Mg-I) and butanal - **Process:** In the presence of butanal, the methyl group from the Grignard reagent (CH₃-Mg-I) attacks the carbonyl carbon of butanal. 2. **Intermediate Formation:** - **Intermediate:** An alkoxide ion is formed as an intermediate. - **Structure Explanation:** The alkoxide ion structure results from the attachment of the methyl group to the carbonyl carbon, converting the carbonyl group into an alkoxide. 3. **Final Step:** - **Treatment:** The alkoxide is treated with aqueous acid. - **Product:** The final product is an alcohol, specifically 2-pentanol. - **Alcohol Structure:** The structure can be represented as CH₃-CH(OH)-CH₂-CH₂-CH₃, with the hydroxyl group attached to the second carbon. This reaction is a classic example of a Grignard reagent addition to a carbonyl compound, illustrating the formation of new carbon-carbon bonds and the synthesis of alcohols.
Transcribed Image Text:**3. Reaction and Product Formation with Methyl Magnesium Iodide and Butanal** **Objective:** To demonstrate the reaction and identify the products formed when methyl magnesium iodide (CH₃-Mg-I) reacts with butanal (butyraldehyde). The subsequent step involves treating the alcoholate formed with aqueous acid to yield an alcohol. **Reaction Steps:** 1. **Initial Reaction:** - **Reagents:** Methyl magnesium iodide (CH₃-Mg-I) and butanal - **Process:** In the presence of butanal, the methyl group from the Grignard reagent (CH₃-Mg-I) attacks the carbonyl carbon of butanal. 2. **Intermediate Formation:** - **Intermediate:** An alkoxide ion is formed as an intermediate. - **Structure Explanation:** The alkoxide ion structure results from the attachment of the methyl group to the carbonyl carbon, converting the carbonyl group into an alkoxide. 3. **Final Step:** - **Treatment:** The alkoxide is treated with aqueous acid. - **Product:** The final product is an alcohol, specifically 2-pentanol. - **Alcohol Structure:** The structure can be represented as CH₃-CH(OH)-CH₂-CH₂-CH₃, with the hydroxyl group attached to the second carbon. This reaction is a classic example of a Grignard reagent addition to a carbonyl compound, illustrating the formation of new carbon-carbon bonds and the synthesis of alcohols.
Expert Solution
Step 1: Introduce the reaction

When methyl magnesium iodide is treated with butanal there occurs nucleophilic addition reaction of aldehyde.Thealdehyde.Then we get an alcoholate ,after the acidification we get 2-pentanol as the Major product.

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