Question

PROBLEM 18-5

Oxidation of cholesterol converts the alcohol to a ketone. Under acidic or basic oxidationconditions, the C=C double bond migrates to the more stable, conjugated position. BeforeIR and NMR spectroscopy, chemists watched the UV spectrum of the reaction mixture tofollow the oxidation. Describe how the UV spectrum of the conjugated product, cholest-4-en-3-one, differs from that of cholesterol.

Short answer

The molecule cholest-4-en-3-one would have longer wavelength or greater value of lambda max as compared to that of cholesterol which is the differentiating property obtained from UV Spectrum.

Step-by-step solution

Detailed Answer:Step 1: Theory of UV spectrum

UV spectrum is a plot of Absorbance versus wavelength. Absorbance is defined as the amount of light absorbed by a particular solution. UV spectrum is used to mainly calculate the value of lambda max. Lambda max refers to the wavelength at which maximum absorbance takes place.

Detailed explanation of the answer 

The molecule cholest-4-en-3-one has a conjugated α,β-unsaturated carbonyl compound where carbonyl group and the double bond are in conjugation. In case of cholesterol, there is no conjugation of the double bond.The energy gap between Highest Occupied Molecular Orbital (HOMO) and Lowest Unoccupied Molecular Orbital (LUMO) decreases with increase in conjugation. Hence, the molecule cholest-4-en-3-one has lower energy gap as compared to cholesterol due to presence of conjugation. Energy gap is inversely proportional to the wavelength. Hence, molecule cholest-4-en-3-one has greater wavelength than cholesterol.