Acid Derivatives and Esters

Question 1

Describe the structure of an acyl chloride.

Answer 1

This is an organic molecule. It consists of an alkane chain with a chlorine atom bound to a terminal carbonyl carbon.

To name an acyl chloride, remove the “-e” from the end of the name of the parent alkane and replace it with “-oyl chloride.” Since an acyl chloride is a terminal group, it is unnecessary to write “1-propanoyl chloride.”

Question 2

What is the name of this molecule?

Answer 2

butanoyl chloride, an acyl chloride.

To name an acyl chloride, remove the “-e” from the end of the name of the parent alkane and replace it with “-oyl chloride.”

Question 3

Describe the structure of an anhydride.

Answer 3

This is an organic molecule containing two carbonyl groups connected by an oxygen atom.

Symmetric anhydrides, in which identical groups are bonded to the two carbonyls, are named by removing the “-e” from the parent alkane and replacing it with “-oic anhydride.” Asymmetrical anhydrides are named by alphabetically arranging the two parent alkanes, replacing the “-e” with “-oic” in both, and ending with the word “anhydride.”

Question 4

What is the name of this molecule?

Answer 4

ethanoic propanoic anhydride

Asymmetrical anhydrides are named by alphabetically arranging the names of the two parent alkanes, replacing the “-e” with “-oic” in both, and ending with the word “anhydride.”

Question 5

Describe the structure of an amide.

Answer 5

This is an organic molecule. It consists of an alkane chain with a carbonyl carbon bound to a nitrogen atom.

Amides are named by removing the “-e” from the parent alkane and replacing it with “-amide.” Additional carbon chains bound to the nitrogen should be listed in front of the alkanamide name without spaces.

Question 6

What is the name of this molecule?

Answer 6

methylethanamide

To name an amide, remove the “-e” from the parent alkane and replace it with “-amide.” Additional carbon chains bound to the nitrogen should be listed in front of the alkanamide name without spaces, as “methyl” is here.

Question 7

What is an ester?

Answer 7

It is an organic molecule. It consists of an alkane chain with a carbonyl carbon bound to an -OR group.

Esters are named by removing the “-e” from the parent alkane chain and replacing it with “-oate,” then adding the name of the R group from the -OR moiety as a prefix.

Question 8

What is the name of this molecule?

Answer 8

propyl butanoate

To name an ester, first name the R group that is bound directly to the oxygen atom. After a space, add the name of the parent alkane, replacing the “-e” with “-oate.”

Question 9

What type of intermolecular forces do all carboxylic acid derivatives exhibit?

Answer 9

All acid derivatives have dipole-dipole interactions due to the polar carbonyl bond.

Additionally, derivatives can be involved in hydrogen bonding, but to varying degrees. For example, carboxylic acids themselves can both donate and accept hydrogen bonds, while esters are only weak H-bond acceptors.

Question 10

How does the boiling point of an acid derivative compare to that of an alkane?

Assume the two molecules have carbon chains of similar length.

Answer 10

They have higher boiling points than alkanes.

Due to its carbonyl group, an acid derivative is able to exert dipole-dipole interactions. These intermolecular forces are stronger than those present on alkanes, which are relatively nonpolar. Stronger intermolecular forces lead to higher boiling points.

Question 11

What acid derivative is generally used to synthesize all of the other acid derivatives?

Answer 11

Acyl chlorides, the most reactive of the derivatives, are often used in synthesis reactions. They can be converted into anhydrides, esters, and amides.

The chlorine atom on an acyl chloride is a good leaving group, and thus can be easily replaced via nucleophilic attack. Reacting with a carboxylic acid creates an anhydride, reacting with an alcohol creates an ester, and reacting with an amine creates an amide.

Question 12

What type of reaction most often occurs at the carbonyl carbon of an acid derivative?

Answer 12

They undergo nucleophilic substitution reactions at their carbonyl carbons.

The carbonyl carbon is fairly electropositive. For this reason, a nucleophile can easily attack and replace the characteristic group bonded to that carbon. For example, a nucleophilic attack on an acyl chloride will replace the chlorine atom with the nucleophile.

Question 13

What product is created in a Hofmann rearrangement?

Answer 13

A primary amine is created from the rearrangement and decarboxylation of a primary amide.

This reaction ocurrs in basic conditions and proceeds through an isocyanate (O=C=N-R) intermediate. Overall, the R group attached to the carbonyl bonds to the amide nitrogen, and the carbonyl group is lost as carbon dioxide gas.

Question 14

What product will this reaction form?

Answer 14

This will produce 1-aminopropane. Carbon dioxide gas will also form as a side product.

This reaction is a Hofmann rearrangement. The R group bonds to the nitrogen atom, and the carbonyl group is lost as carbon dioxide gas.

Question 15

What is the purpose of a transesterification?

Answer 15

The -OR group of an ester is replaced by a different -OR’ group. In other words, one ester is converted into another.

This reactants involved are the original ester and an alcohol in acidic conditions.

Question 16

Briefly describe the steps involved in a transesterification reaction.

Answer 16

1. An alcohol (containing the new -OR’ group) is added to the original ester (containing the old -OR group) in acidic conditions.
2. The alcohol oxygen acts as a nucleophile, attacking the carbonyl carbon and replacing the ester -OR with the alcohol -OR’.
3. The original -OR group is protonated and leaves as an alcohol.

Question 17

What alternative name is given to the hydrolysis of specific esters under basic conditions?

Answer 17

saponification

A triglyceride contains a glycerol backbone bound to three fatty acids through ester linkages.

Question 18

Briefly describe the steps involved in a saponification reaction.

Answer 18

1. A triglyceride and a base (generally NaOH) are combined.
2. One equivalent of OH^- acts as a nucleophile, attacking the carbonyl carbon of one of the triglyceride’s ester linkages.
3. The bond between this ester and the glycerol backbone breaks, allowing part of the molecule to leave as a fatty acid salt.
4. The same process occurs with the remaining two ester linkages, requiring two additional equivalents of OH^-. The final products are glycerol and three fatty acid salts, which can later be acidified to carboxylic acids.

Question 19

What product(s) will be formed in this reaction?

Answer 19

• glycerol
• three fatty acid chains

This reaction is a saponification, also known as a hydrolysis of fats or triglycerides. Three hydroxide ions from the basic solution act as nucleophiles, attacking the carbonyl carbons and allowing the backbone to leave as an alcohol.

Question 20

In the hydrolysis of an amide, what is the leaving group?

Answer 20

The leaving group is an amine. Overall, products of such a reaction include the amine and a carboxylic acid.

In amide hydrolysis, a hydroxide ion acts as a nucleophile and attacks the carbonyl carbon. This allows -OH to bind and replace the amine.

Question 21

What products form when this molecule reacts with water in the presence of acid?

Answer 21

• propanoic acid
• aminomethane (methylamine)

This is an amide hydrolysis. First, the carbonyl oxygen is protonated. Next, water acts as a nucleophile, attacking the carbonyl carbon of the amide. The primary amine is the leaving group.

Question 22

Order the acid derivatives (anhydrides, amides, esters, and acyl chlorides) from most reactive to least reactive.

Assume that “most reactive” means “most likely to react in a nucleophilic substitution.”

Answer 22

acyl chlorides > anhydrides > esters > amides

Acid derivatives with better leaving groups will be more reactive in a nucleophilic substitution. Halogens (like chlorine) are the best leaving groups, while deprotonated amines (like NH₂^-) are the worst.

Question 23

Which would more readily convert to a carboxylic acid: an amide or an ester?

Answer 23

ester

The ester’s OR^- group is a better (more stable) leaving group than the amide’s amino group. This allows the ester to react more readily.

Question 24

How do large substituents affect the reactivity of a carbonyl compound in a nucleophilic attack?

Answer 24

It decrease the overall reactivity.

Large substituents near the carbonyl carbon cause steric hindrance, blocking the nucleophile’s access to the electrophilic carbon.

Question 25

How would the *tert*-butyl groups on this amide **affect** a nucleophilic attack at the carbonyl carbon?

Answer 25

These would **not** affect the reaction. ## Footnote Since the *tert*-butyl groups are far from the carbonyl carbon, they will not cause steric hindrance that would hinder a nucleophilic attack.

Question 26

What main **characteristic** defines a good leaving group?

Answer 26

Able to **stabilize negative charge**. ## Footnote For example, the chloride anion is electronegative and fairly capable of holding a negative charge. Thus, acid chlorides react well in nucleophilic substitution reactions. In contrast, a deprotonated amine does not stabilize negative charge well, making it very reactive and a poor leaving group.

Question 27

Which **stabilizes** negative charge more effectively: a carboxylate anion (COO^-) or a alkoxide anion (RO^-)?

Answer 27

**Carboxylate anion (COO-)** stabilizes negative charge more effectively. Since it displays resonance between its two electronegative oxygen atoms, it is able to "spread out" the charge. ## Footnote This makes the carboxylate anion a better leaving group, which is why anhydrides are more reactive than esters in nucleophilic substitution reactions.

Question 28

How does ring strain **affect** the reactivity of cyclic acid derivatives in ring-opening reactions?

Answer 28

It **increases** the reactivity of a cyclic acid derivative. ## Footnote A ring-opening reaction (hydrolysis, for example) will be very energetically favorable as it eliminates this strain. Such reactions will generally be very exothermic.

Question 29

In a hydrolysis reaction, which is **more reactive**: a straight-chain amide or a cyclic amide? ## Footnote ß-lactam, below, is an example of a cyclic amide.

Answer 29

**Cyclic amide** (such as ß-lactam) is much more reactive due to its significant ring strain that is relieved upon hydrolysis. ## Footnote An amide's carbon-nitrogen bond has partial double bond character. This increases ring strain in cyclic amides because it limits the rings from bending.

Question 30

What **functional groups** are present on a keto acid?

Answer 30

* ketone * carboxylic acid ## Footnote Keto acids with the ketone group on the alpha carbon are called alpha keto acids, also known as 2-oxoacids. Beta keto acids, also known as 3-oxoacids, have the ketone group on the beta carbon.

Question 31

What is the **name** of this molecule?

Answer 31

ß-ketobutanoic acid | 3-oxobutanoic acid ## Footnote To name a keto acid, first identify the location of the ketone group relative to the carboxylic acid carbon (alpha, beta, gamma, etc.). Place this identifier in front of the carboxylic acid name. Keto acids may also be named as oxoacids, with a number indicating the number of carbons from the carboxyl carbon to the ketone carbon. The name of the carboxylic acid follows that number and the prefix "oxo-."

Question 32

Which **type** of keto acids readily undergo decarboxylation when heated?

Answer 32

ß-keto acids ## Footnote The carboxylic acid group will be lost as carbon dioxide, and the resulting reaction intermediate is stabilized in the enol form.

Question 33

What **product(s)** will this reaction form?

Answer 33

* 2-butanone * carbon dioxide gas ## Footnote The ß-keto acid undergoes a decarboxylation, losing its carboxylic carbon in the form of CO_2.

Question 34

Which **proton(s)** on a ß-keto ester is/are the most acidic?

Answer 34

The **alpha protons** of a ß-keto ester are the most acidic. This is also true of other carbonyl-containing molecules. ## Footnote When the alpha proton is deprotonated, the resulting molecule is stabilized by resonance. On a ß-keto ester, the electronic stabilization from the two carbonyl groups makes the alpha proton(s) even more acidic.

Question 35

What is **keto-enol tautomerism**?

Answer 35

It is the **interconversion of a molecule** between a keto form (containing a carbonyl group) and an enol form (containing an alcohol). ## Footnote This type of tautomerism includes the movement of an alpha hydrogen and the bonding electrons of a pi bond. Note that this is distinct from resonance, in which the atoms do not move.

Question 36

In which **tautomeric form** is the ester shown below?

Answer 36

The ester is in the **keto** form. The molecule shown here is the **enol** form of the same ester.

Question 37

How is keto-enol chemistry **involved** in the decarboxylation of a ß-keto acid?

Answer 37

After decarboxylation of a ß-keto acid, the enol form stabilizes the reaction intermediate. The molecule then tautomerizes to form the keto final product.