Q18P - Organic Chemistry (Mcmurry)

Question

Assume that you need to prepare 5-Cholesten-3-one from Cholesterol. How could you use IR spectroscopy to tell whether the reaction was successful? What differences would you look for in the IR spectra of starting material and product?

Step-by-Step Solution

Answer

The Cholesterol shows an –OH stretch at around 3300-3600 ${\text{cm}}^{\text{-1}}$ , and the C=O stretch is shown by 5-Cholestene-3-one at 1715${\text{cm}}^{\text{-1}}$ . 

Step 1: IR spectra of alcohols

The IR spectra of alcohols show the –OH stretch at 3300-3600${\mathbf{cm}}^{-1}$, and the exact position of the –OH stretch depends upon the hydrogen bonding to which extent it is present in a molecule. A sharp absorption near 3600 ${\mathbf{cm}}^{-1}$  is for unassociated alcohols.

Step 2: The use of IR spectroscopy to tell whether the reaction is successful

The IR spectra of Cholesterol and 5-Cholestene-3-one each show a unique absorption pattern which makes them easy to differentiate. The Cholesterol shows an –OH stretch at around 3300-3600 ${\text{cm}}^{\text{-1}}$, and the C=O stretch is shown by 5-Cholestene-3-one at 1715${\text{cm}}^{\text{-1}}$ .

When Cholesterol oxidizes to 5-Cholestene-3-one, the –OH band disappears and is replaced by the C=O band. When complete oxidation occurs, no–OH absorption is visible in the spectra.

Oxidation of Cholesterol to 5-Cholestene-3-one