Question

Draw the structural formula for the product formed by treatment of cholesterol with \(\mathrm{H}_{2} / \mathrm{Pd}\); then do the same for \(\mathrm{Br}_{2}\).

Short answer

Question: Draw the structural formulas of the products formed when cholesterol undergoes (1) hydrogenation with \(\mathrm{H}_{2} / \mathrm{Pd}\), and (2) halogenation with \(\mathrm{Br}_{2}\). Answer: 1) The product of the hydrogenation of cholesterol with \(\mathrm{H}_{2} / \mathrm{Pd}\) is a saturated steroid compound, with the double bond between carbons 5 and 6 replaced by a single bond, while retaining the hydroxyl group at carbon 3 and the hydrocarbon side chain. 2) The product of the halogenation of cholesterol with \(\mathrm{Br}_{2}\) is a vicinal dibromide compound. The double bond between carbons 5 and 6 is replaced by two carbon-bromine single bonds, while still having the hydroxyl group at carbon 3 and the hydrocarbon side chain.

Step-by-step solution

Examine the structure of cholesterol

Cholesterol is a complex organic compound containing four fused hydrocarbon rings (known as the steroid nucleus) and a hydrocarbon side chain. The molecule also possesses a hydroxyl group (OH) attached to carbon 3. To carry out the reactions, it's essential to identify the double bonds and hydroxyl group in the molecule.

Hydrogenation reaction with \(\mathrm{H}_{2} / \mathrm{Pd}\)

The \(\mathrm{H}_{2} / \mathrm{Pd}\) reagent is a common hydrogenation catalyst, which is used to reduce double bonds in organic compounds. In cholesterol, there is only one double bond between carbons 5 and 6. During the hydrogenation reaction, the \(\mathrm{H}_{2} / \mathrm{Pd}\) reagent adds two hydrogen atoms to the carbons involved in the double bond, converting it into a single bond. The product of this reaction is a saturated steroid compound.

Draw the structural formula of the hydrogenation product

After adding hydrogen to the double bond between carbons 5 and 6, the structural formula of the product can be drawn. The product has no double bond between carbons 5 and 6, but retains its hydroxyl group at carbon 3 and the hydrocarbon side chain.

Halogenation reaction with \(\mathrm{Br}_{2}\)

Halogenation is the process of adding halogens, in this case, bromine (\(\mathrm{Br}_{2}\)), to a molecule. In the case of cholesterol, the double bond between carbons 5 and 6 will be the primary site of bromination. The \(\mathrm{Br}_{2}\) will add across the double bond, specifically, each bromine atom will attach to carbons 5 and 6, breaking the double bond and forming two new carbon-bromine single bonds.

Draw the structural formula of the halogenation product

After adding the bromine atoms to the double bond, the structural formula of the product can be drawn, which will show the bromine atoms attached to carbons 5 and 6. The product will still have the hydroxyl group at carbon 3 and the hydrocarbon side chain. The double bond between carbons 5 and 6 will be replaced by two carbon-bromine single bonds, which results in a vicinal dibromide compound. Now that we have drawn the structural formula for each product, we have completed the exercise.