Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 38

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

Expert verified
Question: Draw the structural formula of the lactone formed when 4-hydroxybutanoic acid is treated with an acid catalyst, and propose a mechanism for this reaction. Answer: The lactone formed from 4-hydroxybutanoic acid has a 5-membered ring structure with an ester linkage between the first and fourth carbons. The mechanism involves acid catalyst activation, intramolecular nucleophilic attack, and proton transfer resulting in ring closure and the formation of the lactone.

Step by step solution

01

Draw the structure of 4-hydroxybutanoic acid

First, we need to draw the structure of 4-hydroxybutanoic acid. The 4-hydroxybutanoic acid molecule consists of a 4-carbon chain, with a hydroxyl group (-OH) attached to the first carbon and a carboxylic acid group (-COOH) attached to the fourth carbon.
02

Identify the reactive sites

Now that we have the structure of 4-hydroxybutanoic acid, we need to identify the reactive sites in the molecule that will be involved in the lactone formation. The key reactive sites are the hydroxyl group (-OH) on the first carbon, the carbonyl carbon (C=O) of the carboxylic acid group, and the acidic hydrogen (H) on the carboxylic acid group.
03

Acid catalyst activation

The next step is to consider the role of the acid catalyst in the formation of the lactone. The acid catalyst can protonate the carbonyl oxygen atom of the carboxylic acid group, which makes the carbonyl carbon more electrophilic and more susceptible to nucleophilic attack.
04

Intramolecular nucleophilic attack

Once the carbonyl carbon is activated by the acid catalyst, the hydroxyl oxygen atom of the hydroxyl group can attack the carbonyl carbon in an intramolecular nucleophilic attack. This forms a tetrahedral intermediate with a positive charge on the oxygen atom that attacked the carbonyl carbon.
05

Proton transfer and ring closure

In the next step, a proton is transferred from the positively charged oxygen atom to the negatively charged oxygen atom of the original carboxylic acid group. This proton transfer results in the formation of a 5-membered ring, which is the lactone.
06

Draw the structural formula of the lactone

Finally, we can draw the structural formula of the resulting lactone. This lactone has a 5-membered ring structure with an ester linkage between the first and fourth carbons, and all other carbons maintaining their original valencies. By following these steps, we have drawn the structural formula of the lactone formed from 4-hydroxybutanoic acid and proposed a mechanism for its formation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Low-molecular-weight dicarboxylic acids normally exhibit two different \(\mathrm{p} K_{\mathrm{a}}\) values. Ionization of the first carboxyl group is easier than the second. This effect diminishes with molecular size, and, for adipic acid and longer chain dicarboxylic acids, the two acid ionization constants differ by about one \(\mathrm{p} K\) unit. $$ \begin{array}{llcc} \hline \begin{array}{l} \text { Dicarboxylic } \end{array} & \text { Structural } & & \\ \hline \text { Oxalic } & \text { Formula } & \mathrm{p} K_{\mathrm{a} 1} & \mathrm{p} K_{\mathrm{a} 2} \\ \text { Malonic } & \mathrm{HOOCCOOH} & 1.23 & 4.19 \\ \text { Succinic } & \mathrm{HOOCCH}{ }_{2} \mathrm{COOH} & 2.83 & 5.69 \\ \text { Glutaric } & \mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{COOH} & 4.16 & 5.61 \\ \text { Adipic } & \mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{COOH} & 4.31 & 5.41 \\ \hline \end{array} $$ Why do the two \(\mathrm{p} K_{\mathrm{a}}\) values differ more for the shorter chain dicarboxylic acids than for the longer chain dicarboxylic acids?

Show how to prepare pentanoic acid from each compound. (a) 1-Pentanol (b) Pentanal (c) 1-Pentene (d) 1-Butanol (e) 1-Bromopropane (f) 1-Hexene

We have studied Fischer esterification, in which a carboxylic acid is reacted with an alcohol in the presence of an acid catalyst to form an ester. Suppose that you start instead with a dicarboxylic acid such as terephthalic acid and a diol such as ethylene glycol. Show how Fischer esterification in this case can lead to a macromolecule with a molecular weight several thousands of times that of the starting materials. O=C(O)c1ccc(C(=O)O)cc1 1,4-Benzenedicarboxylic acid \(\quad 1,2\)-Ethanediol (Terephthalic acid) (Ethylene glycol) As we shall see in Section \(29.5 \mathrm{~B}\), the material produced in this reaction is a highmolecular-weight polymer, which can be fabricated into Mylar films, and into the textile fiber known as Dacron polyester.

Using your roadmaps as a guide, show how to convert propane into propyl propanoate. You must use propane as the source of all carbon atoms in the target molecule. Show all reagents needed and all molecules synthesized along the way. CCCOC(=O)CC Propane Propyl propanoate

On a cyclohexane ring, an axial carboxyl group has a conformational energy of \(5.9 \mathrm{~kJ}\) (1.4 kcal)/mol relative to an equatorial carboxyl group. Consider the equilibrium for the alternative chair conformations of trans-1,4-cyclohexanedicarboxylic acid. Draw the less stable chair conformation on the left of the equilibrium arrows and the more stable chair on the right. Calculate \(\Delta G^{0}\) for the equilibrium as written, and calculate the ratio of the more stable chair to the less stable chair at \(25^{\circ} \mathrm{C}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free