Question

When 4-hydroxybutanoic acid is treated with an acid catalyst, it forms a lactone (a cyclic ester). Draw the structural formula of this lactone and propose a mechanism for its formation.

Short answer

Answer: The structural formula of the lactone formed from 4-hydroxybutanoic acid is: O ║ C-O-CH2-CH2-CH2 The mechanism for its formation involves four steps: protonation of the carbonyl oxygen, intramolecular nucleophilic attack by the hydroxyl group, loss of a water molecule, and deprotonation.

Step-by-step solution

Determine the structure of 4-hydroxybutanoic acid

First, let's determine the structure of 4-hydroxybutanoic acid. The name suggests it is a four-carbon chain with a carboxylic acid functional group at one end (butanoic acid) and a hydroxyl group (OH) at the fourth carbon. The structure can be represented as follows: HO-CH2-CH2-CH2-COOH.

Identify the acid-catalyzed esterification reaction to form a lactone

Next, we will focus on the reaction between the carboxylic acid and alcohol groups in the presence of an acid catalyst to form the ester. In this case, the hydroxyl group (-OH) on the 4th carbon reacts with the carbonyl carbon (C=O) of the carboxylic group to form an ester linkage. The reaction will be cyclic, meaning that the ring is formed by the connection of the hydroxyl and carboxylic acid functional groups.

Draw the structural formula of the lactone

Now that we know the basic reaction that forms the lactone, let's draw the structural formula of the lactone. This cyclic ester is formed when the oxygen (O) from the hydroxyl group attacks the carbonyl carbon (C=O) of the carboxylic acid group, closing the ring and forming an ester linkage. The ester linkage is represented as -C(=O)-O-. The 4-hydroxybutanoic acid forms a 5-membered ring called γ-butyrolactone, as there are three carbon atoms between the ester oxygen and the carbonyl carbon. The structure should look like this: O ║ C-O-CH2-CH2-CH2

Propose a mechanism for the lactone formation

Having drawn the lactone, we can now propose a mechanism for its formation: 1. Protonation of the carbonyl oxygen: The acid catalyst donates a proton (H+) to the carbonyl oxygen (C=O) of the carboxylic acid group. This makes the carbonyl carbon more electrophilic, allowing a nucleophilic attack by the hydroxyl group. 2. Intramolecular nucleophilic attack: The now-nucleophilic oxygen from the hydroxyl group (HO-) attacks the electrophilic carbonyl carbon. This forms a cyclic intermediate with a free -OH group attached to the carbonyl carbon. 3. Loss of water molecule: The -OH group attached to the carbonyl carbon picks up a proton (H+) donated by the surrounding acidic environment, forming a water molecule (H2O). The water molecule leaves, and the carbonyl carbon forms a double bond with the oxygen atom to recreate the pi bond in the transition stage. 4. Deprotonation: The final step is the deprotonation of the carbonyl oxygen by the surrounding acidic environment to form the lactone. By following these steps, a student should be able to understand the structure of 4-hydroxybutanoic acid, draw the structure of the resulting lactone, and propose a mechanism for its formation when treated with an acid catalyst.