Consider the reaction as described below, then identify what change(s), if any, would be observed between the IR spectrum of the reactant and product.Cyclopentanone was treated with lithium aluminum hydride followed by H3O+. A signal at 1745 cm–1 was observed in the IR spectrum of the starting ketone. In the IR spectrum of the product this signal would Choose your answer here .
Question
Consider the reaction as described below, then identify what change(s), if any, would be observed between the IR spectrum of the reactant and product.Cyclopentanone was treated with lithium aluminum hydride followed by H3O+. A signal at 1745 cm–1 was observed in the IR spectrum of the starting ketone. In the IR spectrum of the product this signal would Choose your answer here .
Solution
The reaction described involves the reduction of a ketone (cyclopentanone) to an alcohol using lithium aluminum hydride (LiAlH4) and then acid workup (H3O+).
In the IR spectrum, the signal at 1745 cm-1 corresponds to the carbonyl (C=O) stretch in the ketone.
When the ketone is reduced to an alcohol, the C=O bond is replaced by a C-O bond. The C-O bond in alcohols shows up in the IR spectrum at a different wavenumber, typically around 1000-1300 cm-1.
Therefore, in the IR spectrum of the product, the signal at 1745 cm-1 would disappear, indicating the loss of the carbonyl group. Instead, a new signal would appear in the range of 1000-1300 cm-1, indicating the presence of the C-O bond in the alcohol.
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