Questions: Determine the products formed in each reaction. If there is no reaction, label the reaction as "no reaction". xrightarrow[[2] CH3CH2 ][[1] LDA, 78°C]

Determine the products formed in each reaction. If there is no reaction, label the reaction as "no reaction". xrightarrow[[2] CH3CH2 ][[1] LDA, 78°C]
Transcript text: Determine the products formed in each reaction. If there is no reaction, label the reaction as "no reaction". $\xrightarrow[{\text { [2] } \mathrm{CH}_{3} \mathrm{CH}_{2} \text { ] }}]{\text { [1] LDA, } 78^{\circ} \mathrm{C}}$
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Solution

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Solution Steps

Step 1: Analyze the first reaction

The first reaction involves a ketone with no alpha-hydrogens. LDA is a strong base, but without acidic alpha-hydrogens, it cannot deprotonate the ketone. Thus, there is no reaction.

Step 2: Analyze the second reaction

The second reaction has a ketone with alpha-hydrogens. LDA will deprotonate at the less substituted side, forming the kinetic enolate. Then, the enolate will react with the primary alkyl halide (ethyl iodide) in an SN2 reaction, alkylating the ketone at the less substituted alpha carbon.

Step 3: Analyze the third reaction

The third reaction shows a ketone with LDA again. Like the second reaction, LDA will deprotonate the less hindered alpha carbon.

Final Answer

First reaction: No reaction

Second reaction: \(\boxed{\chemfig{C6CCCC(=O)CC}}\)

Third reaction: \(\boxed{\chemfig{C6CCCC(=O)CC}}\)

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