НО. НО. (above product not formed) The Grignard reagent will deprotonate the carboxylic acid because the carboxylic acid is a stronger base than the conjugate acid of the Grignard reagent. The Grignard reagent will add to the carboxylic acid instead of the ketone because the carboxylic acid is more electrophilic. The Grignard reagent will add to both the ketone and carbonyl, and the carboxylic acid will be reduced to an alcohol with two ethyl groups attached. The wrong product is written. A methyl group should be added to the ketone carbonyl instead of an ethyl group.

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The reaction written below will not create the product that has been written. What reaction happened instead and why?
НО.
O
MgBr
HO
OH
(above product not
formed)
The Grignard reagent will deprotonate the carboxylic acid because the carboxylic acid is a stronger base than the conjugate acid of the
Grignard reagent.
The Grignard reagent will add to the carboxylic acid instead of the ketone because the carboxylic acid is more electrophilic.
The Grignard reagent will add to both the ketone and carbonyl, and the carboxylic acid will be reduced to an alcohol with two ethyl
groups attached.
The wrong product is written. A methyl group should be added to the ketone carbonyl instead of an ethyl group.
Transcribed Image Text:The reaction written below will not create the product that has been written. What reaction happened instead and why? НО. O MgBr HO OH (above product not formed) The Grignard reagent will deprotonate the carboxylic acid because the carboxylic acid is a stronger base than the conjugate acid of the Grignard reagent. The Grignard reagent will add to the carboxylic acid instead of the ketone because the carboxylic acid is more electrophilic. The Grignard reagent will add to both the ketone and carbonyl, and the carboxylic acid will be reduced to an alcohol with two ethyl groups attached. The wrong product is written. A methyl group should be added to the ketone carbonyl instead of an ethyl group.
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