Chapter 15 Organic Acids and Bases and Some of Their Derivatives

The organic compounds that we consider in this chapter are organic acids and bases. We will also consider two derivatives of carboxylic acids: esters and amides. An ester is derived from a carboxylic acid and an alcohol. Fats and oils are esters, as are many important fragrances and flavors. (For more information about fats and oils, see Chapter 17 "Lipids", Section 17.2 "Fats and Oils".) An amide is derived from a carboxylic acid and either ammonia or an amine. Proteins, often called “the stuff of life,” are polyamides. (For more information about proteins, see Chapter 18 "Amino Acids, Proteins, and Enzymes", Section 18.4 "Proteins".)

15.1 Functional Groups of the Carboxylic Acids and Their Derivatives¶

We introduced the carbonyl group (C=O)---the functional group of aldehydes and ketones---in Chapter 14 "Organic Compounds of Oxygen". The carbonyl group is also found in carboxylic acids, esters, and amides. However, in these compounds, the carbonyl group is only part of the functional group.

A carboxylic acid^[1] is an organic compound that has a carboxyl group^[2]. The carboxyl group is a functional group that contains a carbon--oxygen double bond and an OH group also attached to the same carbon atom, but it has characteristic properties of its own. As with aldehydes and ketones, carboxylic acid formulas can be written to show the carbon-to-oxygen double bond explicitly, or the carboxyl group can be written in condensed form on one line. In general, carboxylic acids are represented by the formula RCOOH, where R is a hydrocarbon group.

Esters are represented by the formula RCOOR’, where R and R’ are hydrocarbon groups. The ester^[3], which is an organic compound derived from a carboxylic acid and an alcohol in which the OH of the acid is replaced by an OR group, looks somewhat like an ether and also somewhat like a carboxylic acid. Even so, compounds in this group react neither like carboxylic acids nor like ethers; they make up a distinctive family. Unlike ethers, esters have a carbonyl group. Unlike carboxylic acids, esters have no acidic hydrogen atom; they have a hydrocarbon group in its place.

An amine^[4] is a compound derived from ammonia (NH₃); it has one, two, or all three of the hydrogen atoms of NH₃ replaced by an alkyl (or an aryl) group. Like NH₃, amines are weak bases. The functional group of an amine is a nitrogen atom with a lone pair of electrons and with one, two, or three alkyl or aryl groups attached.

The amide^[5] functional group has a carbonyl group joined to a nitrogen atom from ammonia or an amine. The properties of the amide functional group differ from those of the simple carbonyl group, NH₃, and amines.

Esters and amides are considered to be derivatives of carboxylic acids because the OH in the carboxyl group is replaced with another group. These functional groups are listed in Table 15.1 "Organic Acids, Bases, and Acid Derivatives", along with an example (identified by common and International Union of Pure and Applied Chemistry [IUPAC] names) for each type of compound.

Table 15.1 Organic Acids, Bases, and Acid Derivatives

Family	Functional Group	Example	Common Name	IUPAC Name
carboxylic acid			acetic acid	ethanoic acid
amine		CH3NH2	methylamine	methanamine (aminomethane)
amide			acetamide	ethanamide
ester			methyl acetate	methyl ethanoate

Concept Review Exercises¶

1. How is the carboxyl group related to the carbonyl group and the OH group?

2. How is the amide group related to the carboxyl group and amines?

Answers¶

1. The carboxyl group has a carbonyl group joined to an OH group.

2. The amide group has a carboxyl group joined to an amino group.

Exercises¶

1. Draw the functional group in each class of compounds.

   1. aldehydes

   2. esters

   3. carboxylic acids

2. How are the functional groups in Exercise 1 alike and different?

3. Draw the functional group in each class of compounds.

   1. amides

   2. ketones

   3. ethers

4. How are the functional groups in Exercise 2 alike and different?

Answers¶

15.2 Carboxylic Acids: Structures and Names¶

Carboxylic acids occur widely in nature, often combined with alcohols or other functional groups, as in fats, oils, and waxes. They are components of many foods, medicines, and household products (Figure 15.1 "Carboxylic Acids in the Home"). Not surprisingly, many of them are best known by common names based on Latin and Greek words that describe their source. What you learn in this chapter about the chemistry of carboxylic acids will help you understand biochemistry (Chapter 16 "Carbohydrates" through Chapter 20 "Energy Metabolism").

The simplest carboxylic acid, formic acid (HCOOH), was first obtained by the distillation of ants (Latin formica, meaning “ant”). The bites of some ants inject formic acid, and the stings of wasps and bees contain formic acid (as well as other poisonous materials).

The next higher homolog is acetic acid, which is made by fermenting cider and honey in the presence of oxygen. This fermentation produces vinegar, a solution containing 4%--10% acetic acid, plus a number of other compounds that add to its flavor. Acetic acid is probably the most familiar weak acid used in educational and industrial chemistry laboratories.

The third homolog, propionic acid (CH₃CH₂COOH), is seldom encountered in everyday life. The fourth homolog, butyric acid (CH₃CH₂CH₂COOH), is one of the most foul-smelling substances imaginable. It is found in rancid butter and is one of the ingredients of body odor. By recognizing extremely small amounts of this and other chemicals, bloodhounds are able to track fugitives. Models of the first four carboxylic acids are shown in Figure 15.2 "Ball-and-Stick Models of Carboxylic Acids".

The acid with the carboxyl group attached directly to a benzene ring is called benzoic acid (C₆H₅COOH).

The common names of carboxylic acids use Greek letters (α, β, γ, δ, and so forth), not numbers, to designate the position of substituent groups in acids. These letters refer to the position of the carbon atom in relation to the carboxyl carbon atom.

In the nomenclature system of the International Union of Pure and Applied Chemistry (IUPAC), the parent hydrocarbon is the one that corresponds to the longest continuous chain (LCC) containing the carboxyl group. The -e ending of the parent alkane is replaced by the suffix -oic and the word acid. For example, the carboxylic acid derived from pentane is pentanoic acid (CH₃CH₂CH₂CH₂COOH). As with aldehydes, the carboxyl carbon atom is counted first; numbers are used to indicate any substituted carbon atoms in the parent chain.

Example 1¶

Give the common and IUPAC names for each compound.

1. ClCH₂CH₂CH₂COOH

2. 

Solution

1. The LCC contains four carbon atoms; the compound is therefore named as a substituted butyric (or butanoic) acid.

   

   The chlorine atom is attached to the γ-carbon in the common system or C4 in the IUPAC system. The compound is γ-chlorobutyric acid or 2-bromobutanoic acid.

2. The LCC contains four carbon atoms; the compound is therefore named as a substituted butyric (or butanoic) acid.

The bromine (Br) atom is at the α-carbon in the common system or C2 in the IUPAC system. The compound is α-bromobutyric acid or 4-chlorobutanoic acid.

Skill-Building Exercise¶

Give the IUPAC name for each compound.

1. ClCH₂CH₂CH₂CH₂COOH

2. (CH₃)₂CHCH₂CHBrCOOH

Example 2¶

Write the condensed structural formula for β-chloropropionic acid.

Solution

Propionic acid has three carbon atoms: C--C--COOH. Attach a chlorine (Cl) atom to the parent chain at the beta carbon atom, the second one from the carboxyl group: Cl--C--C--COOH. Then add enough hydrogen atoms to give each carbon atom four bonds: ClCH₂CH₂COOH.

Skill-Building Exercise¶

1. Write the condensed structural formula for 4-bromo-5-methylhexanoic acid.

Concept Review Exercises¶

1. What is the IUPAC name for the straight-chain carboxylic acid with six carbon atoms?

2. The straight-chain aldehyde with five carbon atoms has the common name valeraldehyde. What is the common name of the corresponding straight-chain carboxylic acid?

Answers¶

1. hexanoic acid

2. valeric acid

Exercises¶

1. Draw the structure for each compound.

   1. heptanoic acid

   2. 3-methylbutanoic acid

   3. 2,3-dibromobenzoic acid

   4. β-hydroxybutyric acid

2. Draw the structure for each compound.

   1. o-nitrobenzoic acid

   2. p-chlorobenzoic acid

   3. 3-chloropentanoic acid

   4. α-chloropropionic acid

3. Name each compound with either the IUPAC name, the common name, or both.

   1. (CH₃)₂CHCH₂COOH

   2. (CH₃)₃CCH(CH₃)CH₂COOH

   3. CH₂OHCH₂CH₂COOH

4. Name each compound with its IUPAC name.

   1. CH₃(CH₂)₈COOH

   2. (CH₃)₂CHCCl₂CH₂CH₂COOH

   3. CH₃CHOHCH(CH₂CH₃)CHICOOH

Answers¶

1. 1. CH₃CH₂CH₂CH₂CH₂CH₂COOH

   2. 

   3. 

   4. 

3. 1. 3-methylbutanoic acid; β-methylbutyric acid

   2. 3,4,4-trimethylpentanoic acid

   3. 4-hydroxybutanoic acid; γ- hydroxybutyric acid

15.3 The Formation of Carboxylic Acids¶

As we noted in Chapter 14 "Organic Compounds of Oxygen", the oxidation of aldehydes or primary alcohols forms carboxylic acids:

In the presence of an oxidizing agent, ethanol is oxidized to acetaldehyde, which is then oxidized to acetic acid.

This process also occurs in the liver, where enzymes catalyze the oxidation of ethanol to acetic acid.

Acetic acid can be further oxidized to carbon dioxide and water.

Concept Review Exercises¶

1. Caproic acid (hexanoic acid) can be prepared in an oxidation reaction from

   1. what alcohol?

   2. what aldehyde?

2. Give the structures of the aldehyde and the carboxylic acid formed by the oxidation of isobutyl alcohol [(CH₃)₂CHCH₂OH].

Answers¶

1. 1. CH₃CH₂CH₂CH₂CH₂CH₂OH

   2. CH₃CH₂CH₂CH₂CH₂CHO

2. 

Exercises¶

1. Caprylic acid (octanoic acid) can be prepared in an oxidation reaction from

   1. what alcohol?

   2. what aldehyde?

2. Give the structures of the aldehyde and the carboxylic acid formed by the oxidation of 1,4-butanediol (HOCH₂CH₂CH₂CH₂OH).

Answer¶

1. 1. CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂OH

   2. CH₃CH₂CH₂CH₂CH₂CH₂CH₂CHO

15.4 Physical Properties of Carboxylic Acids¶

Many carboxylic acids are colorless liquids with disagreeable odors. The carboxylic acids with 5 to 10 carbon atoms all have “goaty” odors (explaining the odor of Limburger cheese). These acids are also produced by the action of skin bacteria on human sebum (skin oils), which accounts for the odor of poorly ventilated locker rooms. The acids with more than 10 carbon atoms are waxlike solids, and their odor diminishes with increasing molar mass and resultant decreasing volatility.

Carboxylic acids exhibit strong hydrogen bonding between molecules. They therefore have high boiling points compared to other substances of comparable molar mass.

The carboxyl group readily engages in hydrogen bonding with water molecules (Figure 15.3 "Hydrogen Bonding between an Acetic Acid Molecule and Water Molecules"). The acids with one to four carbon atoms are completely miscible with water. Solubility decreases as the carbon chain length increases because dipole forces become less important and dispersion forces become more predominant. Hexanoic acid [CH₃(CH₂)₄COOH] is barely soluble in water (about 1.0 g/100 g of water). Palmitic acid [CH₃(CH₂)₁₄COOH], with its large nonpolar hydrocarbon component, is essentially insoluble in water. The carboxylic acids generally are soluble in such organic solvents as ethanol, toluene, and diethyl ether.

Table 15.2 "Physical Constants of Carboxylic Acids" lists some physical properties for selected carboxylic acids. The first six are homologs. Notice that the boiling points increase with increasing molar mass, but the melting points show no regular pattern.

Table 15.2 Physical Constants of Carboxylic Acids

Concept Review Exercises¶

1. Which compound has the higher boiling point---butanoic acid (molar mass 88) or 2-pentanone (molar mass 86)? Explain.

2. Would you expect butyric acid (butanoic acid) to be more or less soluble than 1-butanol in water? Explain.

Answers¶

1. butyric acid because of hydrogen bonding (There is no intermolecular hydrogen bonding in 2-pentanone.)

2. more soluble because there is more extensive hydrogen bonding

Exercises¶

1. Which compound has the higher boiling point---CH₃CH₂CH₂OCH₂CH₃ or CH₃CH₂CH₂COOH? Explain.

2. Which compound has the higher boiling point---CH₃CH₂CH₂CH₂CH₂OH or CH₃CH₂CH₂COOH? Explain.

3. Which compound is more soluble in water---CH₃COOH or CH₃CH₂CH₂CH₃? Explain.

4. Which compound is more soluble in water---CH₃CH₂COOH or CH₃CH₂CH₂CH₂CH₂COOH? Explain.

Answers¶

1. CH₃CH₂CH₂COOH because of hydrogen bonding (There is no intermolecular hydrogen bonding with CH₃CH₂CH₂OCH₂CH₃.)

3. CH₃COOH because it engages in hydrogen bonding with water (There is no intermolecular hydrogen bonding with CH₃CH₂CH₂CH₃.)

15.5 Chemical Properties of Carboxylic Acids: Ionization and Neutralization¶

Water-soluble carboxylic acids ionize slightly in water to form moderately acidic solutions.

Their aqueous solutions exhibit the typical properties of acids, such as changing litmus from blue to red.

Whether soluble in water or not, carboxylic acids react with aqueous solutions of sodium hydroxide (NaOH), sodium carbonate (Na₂CO₃), and sodium bicarbonate (NaHCO₃) to form salts:

RCOOH + NaOH(aq) → RCOO⁻Na⁺(aq) + H₂O 2RCOOH + Na₂CO₃(aq) → 2RCOO⁻Na⁺(aq) + H₂O + CO₂(g) RCOOH + NaHCO₃(aq) → RCOO⁻Na⁺(aq) + H₂O + CO₂(g)

In these reactions, the carboxylic acids act like inorganic acids: they neutralize basic compounds. With solutions of carbonate (CO₃) and bicarbonate (HCO₃) ions, they also form carbon dioxide gas.

Carboxylic acid salts are named in the same manner as inorganic salts: the name of the cation is followed by the name of the organic anion. The name of the anion is obtained by dropping the -ic ending of the acid name and replacing it with the suffix -ate. This rule applies whether we are using common names or International Union of Pure and Applied Chemistry (IUPAC) names:

Example 3¶

Write an equation for each reaction.

1. the ionization of propionic acid in water (H₂O)

2. the neutralization of propionic acid with aqueous sodium hydroxide (NaOH)

Solution

Propionic acid has three carbon atoms, so its formula is CH₂CH₂COOH.

1. Propionic acid ionizes in water to form a propionate ion and a hydronium (H₃O⁺) ion.

   CH₃CH₂COOH(aq) + H₂O(ℓ) → CH₃CH₂COO⁻(aq) + H₃O⁺(aq)

2. Propionic acid reacts with NaOH(aq) to form sodium propionate and water.

   CH₃CH₂COOH(aq) + NaOH(aq) → CH₃CH₂COO⁻Na⁺(aq) + H₂O(ℓ)

Skill-Building Exercise¶

Write an equation for each reaction.

1. the ionization of formic acid in water

2. the ionization of p-chlorobenzoic acid in water

Example 4¶

Write an equation for the reaction of decanoic acid with each compound.

1. aqueous sodium hydroxide (NaOH)

2. aqueous sodium bicarbonate (NaHCO₃)

Solution

1. Decanoic acid has 10 carbon atoms. It reacts with NaOH to form a salt and water (H₂O).

   CH₃(CH₂)₈COOH + NaOH(aq) → CH₃(CH₂)₈COO⁻Na⁺(aq) + H₂O(ℓ)

2. With NaHCO₃, the products are a salt, H₂O, and carbon dioxide (CO₂).

   CH₃(CH₂)₈COOH + NaHCO₃(aq) → CH₃(CH₂)₈COO⁻Na⁺(aq) + H₂O(ℓ) + CO₂(g)

Skill-Building Exercise¶

Write an equation for the reaction of benzoic acid with each compound.

1. aqueous sodium hydroxide (NaOH)

2. aqueous sodium bicarbonate (NaHCO₃)

Concept Review Exercises¶

1. How does the neutralization of a carboxylic acid differ from that of an inorganic acid? How are they similar?

2. What products are formed when a carboxylic acid is neutralized with a strong base? What additional product is formed when a carboxylic acid is neutralized with a carbonate or a bicarbonate?

Answers¶

1. Insoluble carboxylic acids often form soluble carboxylate salts. Both form a salt and water.

2. a carboxylate salt and water; carbon dioxide

Exercises¶

1. Write the equation for the ionization of CH₃CH₂CH₂COOH in water.

2. Write the equation for the neutralization of CH₃CH₂CH₂COOH with sodium hydroxide [NaOH(aq)].

3. Write the equation for the reaction of CH₃COOH with sodium carbonate [Na₂CO₃(aq)].

4. Write the equation for the reaction of CH₃CH₂COOH with sodium bicarbonate [NaHCO₃(aq)].

5. Write the equation for the ionization of propionic acid in water.

6. Write the equation for the ionization of γ-chloropentanoic acid in water.

7. Write an equation for the reaction of butyric acid with each compound.

   1. aqueous NaOH

   2. aqueous NaHCO₃

8. Write the condensed structural formula for each compound.

   1. potassium acetate

   2. calcium propanoate

9. Name each compound.

   1. CH₃CH₂CH₂COO⁻Li⁺

   2. CH₃CH₂CH₂COO⁻NH₄⁺

Answers¶

1. CH₃CH₂CH₂COOH(aq) + H₂O(ℓ) → CH₃CH₂CH₂COO⁻(aq) + H₃O⁺(aq)

3. 2CH₃COOH + Na₂CO₃(aq) → 2CH₃COO⁻Na⁺(aq) + H₂O(ℓ) + CO₂(g)

5. CH₃CH₂COOH(aq) + H₂O(ℓ) → CH₃CH₂COO⁻(aq) + H₃O⁺(aq)

7. 1. CH₃CH₂CH₂COOH(aq) + NaOH(aq) → CH₃CH₂CH₂COO⁻Na⁺(aq) + H₂O(ℓ)

   2. CH₃(CH₂)₂COOH + NaHCO₃(aq) → CH₃(CH₂)COO⁻Na⁺(aq) + H₂O(ℓ) + CO₂(g)

9. 1. lithium butyrate (lithium butanoate)

   2. ammonium butanoate or ammonium butyrate

15.6 Esters: Structures and Names¶

Esters have the general formula RCOOR′, where R may be a hydrogen atom, an alkyl group, or an aryl group, and R′ may be an alkyl group or an aryl group but not a hydrogen atom. (If it were a hydrogen atom, the compound would be a carboxylic acid.) Figure 15.4 "The Structure of Esters" shows models for two common esters.

Esters occur widely in nature. Unlike carboxylic acids, esters generally have pleasant odors and are often responsible for the characteristic fragrances of fruits and flowers. Once a flower or fruit has been chemically analyzed, flavor chemists can attempt to duplicate the natural odor or taste. Both natural and synthetic esters are used in perfumes and as flavoring agents.

Names of Esters¶

Although esters are covalent compounds and salts are ionic, esters are named in a manner similar to that used for naming salts. The group name of the alkyl or aryl portion is given first and is followed by the name of the acid portion. In both common and International Union of Pure and Applied Chemistry (IUPAC) nomenclature, the -ic ending of the parent acid is replaced by the suffix -ate (Table 15.3 "Nomenclature of Esters").

Table 15.3 Nomenclature of Esters

Condensed Structural Formula	Common Name	IUPAC Name
HCOOCH3	methyl formate	methyl methanoate
CH3COOCH3	methyl acetate	methyl ethanoate
CH3COOCH2CH3	ethyl acetate	ethyl ethanoate
CH3CH2COOCH2CH3	ethyl propionate	ethyl propanoate
CH3CH2CH2COOCH(CH3)2	isopropyl butyrate	isopropyl butanoate
	ethyl benzoate	ethyl benzoate

Example 5¶

Give the common and IUPAC names for each compound.

Solution

1. The alkyl group attached directly to the oxygen atom is a butyl group (in green).

   

   The part of the molecule derived from the carboxylic acid (in red) has three carbon atoms. It is called propionate (common) or propanoate (IUPAC). The ester is therefore butyl propionate or butyl propanoate.

2. An alkyl group (in green) is attached directly to the oxygen atom by its middle carbon atom; it is an isopropyl group. The part derived from the acid (that is, the benzene ring and the carbonyl group, in red) is benzoate. The ester is therefore isopropyl benzoate (both the common name and the IUPAC name).

   

Skill-Building Exercise¶

Give the common and IUPAC names for each compound.

Example 6¶

Draw the structure for ethyl pentanoate.

Solution

Start with the portion from the acid. Draw the pentanoate (five carbon atoms) group first, keeping in mind that the last carbon atom is a part of the carboxyl group.

Then attach the ethyl group to the bond that ordinarily holds the hydrogen atom in the carboxyl group.

Skill-Building Exercise¶

1. Draw the structure for phenyl pentanoate.

Concept Review Exercises¶

1. From what carboxylic acid and what alcohol can isopropyl hexanoate be made?

2. From what carboxylic acid and what alcohol can cyclobutyl butyrate be made?

Answers¶

1. hexanoic acid and isopropyl alcohol

2. butyric acid and cyclobutyl alcohol

Exercises¶

1. Draw the structure for each compound.

   1. methyl acetate

   2. ethyl pentanoate

   3. phenyl acetate

   4. isopropyl propionate

2. Draw the structure for each compound.

   1. ethyl hexanoate

   2. ethyl benzoate

   3. phenyl benzoate

   4. ethyl 3-methylhexanoate

3. Name each compound with both the common name and the IUPAC name.

   1. 

   2. 

4. Name each compound with both the common name and the IUPAC name.

   1. 

   2. 

Answers¶

1. 1. 

   2. 

   3. 

   4. 

3. 1. methyl formate; methyl methanoate

   2. ethyl propionate; ethyl propanoate

15.7 Physical Properties of Esters¶

Ester molecules are polar but have no hydrogen atom attached directly to an oxygen atom. They are therefore incapable of engaging in intermolecular hydrogen bonding with one another and thus have considerably lower boiling points than their isomeric carboxylic acid counterparts. Because ester molecules can engage in hydrogen bonding with water molecules, however, esters of low molar mass are somewhat soluble in water. Borderline solubility occurs in those molecules that have three to five carbon atoms. Table 15.4 "Physical Properties of Some Esters" lists the physical properties of some common esters.

Table 15.4 Physical Properties of Some Esters

Concept Review Exercises¶

1. Which compound has the higher boiling point---CH₃CH₂CH₂CH₂OH or CH₃COOCH₃? Explain.

2. Which compound is more soluble in water---methyl butyrate or butyric acid? Explain.

Answers¶

1. CH₃CH₂CH₂CH₂OH because there is intermolecular hydrogen bonding (There is no intermolecular hydrogen bonding in CH₃COOCH₃.)

2. butyric acid because of hydrogen bonding with water

Exercises¶

1. Which compound has the higher boiling point---CH₃CH₂CH₂COOH or CH₃CH₂CH₂COOCH₃? Explain.

2. Which compound is more soluble in water---methyl acetate or octyl acetate? Explain.

Answer¶

1. CH₃CH₂CH₂COOH because there is intermolecular hydrogen bonding (There is no intermolecular hydrogen bonding in CH₃CH₂COOCH₃.)

15.8 Preparation of Esters¶

Some esters can be prepared by esterification^[6], a reaction in which a carboxylic acid and an alcohol, heated in the presence of a mineral acid catalyst, form an ester and water:

The reaction is reversible. As a specific example of an esterification reaction, butyl acetate can be made from acetic acid and 1-butanol.

Concept Review Exercises¶

1. From what carboxylic acid and what alcohol can the ester isopropyl nonanoate be made?

2. From what carboxylic acid and what alcohol can the ester cyclobutyl butyrate be made?

Answers¶

1. nonanoic acid and isopropyl alcohol

2. butyric acid and cyclobutyl alcohol

Exercises¶

1. Write the equation for the reaction of acetic acid with each compound.

   1. ethanol

   2. 1-butanol in the presence of a mineral acid catalyst

2. Write the equation for the reaction of benzoic acid with each compound.

   1. methanol

   2. 1-propanol in the presence of a mineral acid catalyst

Answer¶

15.9 Hydrolysis of Esters¶

Esters are neutral compounds, unlike the acids from which they are formed. In typical reactions, the alkoxy (OR′) group of an ester is replaced by another group. One such reaction is hydrolysis^[7], literally “splitting with water.” The hydrolysis of esters is catalyzed by either an acid or a base.

Acidic hydrolysis is simply the reverse of esterification. The ester is heated with a large excess of water containing a strong-acid catalyst. Like esterification, the reaction is reversible and does not go to completion.

As a specific example, butyl acetate and water react to form acetic acid and 1-butanol. The reaction is reversible and does not go to completion.

Example 7¶

Write an equation for the acidic hydrolysis of ethyl butyrate (CH₃CH₂CH₂COOCH₂CH₃) and name the products.

Solution

Remember that in acidic hydrolysis, water (HOH) splits the ester bond. The H of HOH joins to the oxygen atom in the OR part of the original ester, and the OH of HOH joins to the carbonyl carbon atom:

The products are butyric acid (butanoic acid) and ethanol.

Skill-Building Exercise¶

1. Write an equation for the acidic hydrolysis of methyl butanoate and name the products.

When a base (such as sodium hydroxide [NaOH] or potassium hydroxide [KOH]) is used to hydrolyze an ester, the products are a carboxylate salt and an alcohol. Because soaps are prepared by the alkaline hydrolysis of fats and oils, alkaline hydrolysis of esters is called saponification^[8] (Latin sapon, meaning “soap,” and facere, meaning “to make”). In a saponification reaction, the base is a reactant, not simply a catalyst. The reaction goes to completion:

As a specific example, ethyl acetate and NaOH react to form sodium acetate and ethanol:

Example 8¶

Write an equation for the hydrolysis of methyl benzoate in a potassium hydroxide solution.

Solution

In basic hydrolysis, the molecule of the base splits the ester linkage. The acid portion of the ester ends up as the salt of the acid (in this case, the potassium salt). The alcohol portion of the ester ends up as the free alcohol.

Skill-Building Exercise¶

1. Write the equation for the hydrolysis of ethyl propanoate in a sodium hydroxide solution.

Concept Review Exercises¶

1. How do acidic hydrolysis and basic hydrolysis of an ester differ in terms of

   1. products obtained?

   2. the extent of reaction?

2. What is saponification?

Answers¶

1. 1. acidic hydrolysis: carboxylic acid + alcohol; basic hydrolysis: carboxylate salt + alcohol

   2. basic hydrolysis: completion; acidic hydrolysis: incomplete reaction

2. the basic hydrolysis of an ester

Exercises¶

1. Write an equation for the acid-catalyzed hydrolysis of ethyl acetate.

2. Write an equation for the base-catalyzed hydrolysis of ethyl acetate.

3. Complete each equation.

   1. 

   2. 

4. Complete each equation.

   1. ${\text{(CH}_{\text{3}}\text{)}}_{\text{2}}\text{CHCOOCH}_{\text{2}}\text{CH}_{\text{3}}\text{~+~H}_{\text{2}}\text{O~}\overset{\text{H}^{\text{+}}}{\rightleftarrows}$

   2. CH₃COOCH(CH₃)₂ + KOH(aq) →

Answers¶

1. $\text{CH}_{\text{3}}\text{COOCH}_{\text{2}}\text{CH}_{\text{3}}\text{~+~H}_{\text{2}}\text{O~}\overset{\text{H}^{\text{+}}}{\rightarrow}\text{~CH}_{\text{3}}\text{COOH~+~CH}_{\text{3}}\text{CH}_{\text{2}}\text{OH}$

3. 1. CH₃COONa(aq) + CH₃CH₂CH₂OH

   2. CH₃CH₂CH₂COOH + CH₃CH₂OH

15.10 Esters of Phosphoric Acid¶

Just as carboxylic acids do, inorganic acids such as nitric acid (HNO₃), sulfuric acid (H₂SO₄), and phosphoric acid (H₃PO₄) also form esters. The esters of phosphoric acid are especially important in biochemistry. A phosphoric acid molecule can form a monoalkyl, a dialkyl, or a trialkyl ester by reaction with one, two, or three molecules of an alcohol.

Esters of pyrophosphoric acid and triphosphoric acid are also important in biochemistry.

Esters of these acids are present in every plant and animal cell. They are biochemical intermediates in the transformation of food into usable energy. The bonds between phosphate units in adenosine triphosphate (ATP) are called phosphoanhydride bonds. These are high-energy bonds that store energy from the metabolism of foods. Hydrolysis of ATP releases energy as it is needed for biochemical processes (for instance, for muscle contraction). Phosphate esters are also important structural constituents of phospholipids and nucleic acids. (For more information about phospholipids and nucleic acids, see Chapter 17 "Lipids", Section 17.3 "Membranes and Membrane Lipids", and Chapter 19 "Nucleic Acids", respectively.)

Concept Review Exercise¶

1. What compounds combine to form phosphate esters?

Answer¶

1. phosphoric acids and alcohols

Exercises¶

1. Draw the structure for each compound.

   1. diethyl hydrogen phosphate

   2. methyl dihydrogen phosphate

   3. 1-glycerol phosphate

2. Name each compound.

   1. 

   2. 

   3. 

Answer¶

15.11 Amines: Structures and Names¶

Amines are classified according to the number of carbon atoms bonded directly to the nitrogen atom. A primary (1°) amine^[9] has one alkyl (or aryl) group on the nitrogen atom, a secondary (2°) amine^[10] has two, and a tertiary (3°) amine^[11] has three (Figure 15.5 "The Structure of Amines Compared to Water, an Alcohol, and an Ether").

The common names for simple aliphatic amines consist of an alphabetic list of alkyl groups attached to the nitrogen atom, followed by the suffix -amine. (Systematic names are often used by some chemists.) The amino group^[12] (NH₂) is named as a substituent in more complicated amines, such as those that incorporate other functional groups or in which the alkyl groups cannot be simply named.

Example 9¶

Name and classify each compound.

1. CH₃CH₂CH₂NH₂

2. 

3. CH₃CH₂NHCH₂CH₃

4. CH₃CH₂CH₂NHCH₃

Solution

1. There is only one alkyl group attached to the nitrogen atom, so the amine is primary. A group of three carbon atoms (a propyl group) is attached to the NH₂ group through an end carbon atom, so the name is propylamine.

2. There are two methyl groups and one ethyl group on the nitrogen atom. The compound is ethyldimethylamine, a tertiary amine.

3. There are two ethyl groups attached to the nitrogen atom; the amine is secondary, so the compound is diethylamine.

4. The nitrogen atom has a methyl group and a propyl group, so the compound is methylpropylamine, a secondary amine.

Skill-Building Exercise¶

Name and classify each compound.

1. 

2. 

3. CH₃CH₂CH₂CH₂NH₂

4. CH₃CH₂CH₂NHCH₂CH₂ CH₃

Example 10¶

Draw the structure for each compound and classify.

1. isopropyldimethylamine

2. dipropylamine

Solution

1. The name indicates that there are an isopropyl group (in red) and two methyl groups (in green) attached to the nitrogen atom; the amine is tertiary.

   

2. The name indicates that there are two propyl groups attached to the nitrogen atom; the amine is secondary. (The third bond on the nitrogen atom goes to a hydrogen atom.)

   CH₃CH₂CH₂NHCH₂CH₂CH₃

Skill-Building Exercise¶

Draw the structure for each compound and classify.

1. ethylisopropylamine

2. diethylpropylamine

The primary amine in which the nitrogen atom is attached directly to a benzene ring has a special name---aniline. Aryl amines are named as derivatives of aniline.

Example 11¶

Name this compound.

Solution

The benzene ring with an amino (NH₂) group is aniline. The compound is named as a derivative of aniline: 3-bromoaniline or m-bromoaniline.

Skill-Building Exercise¶

1. Name this compound.

   

Example 12¶

Draw the structure for p-ethylaniline and classify.

Solution

The compound is a derivative of aniline. It is a primary amine having an ethyl group located para to the amino (NH₂) group.

Skill-Building Exercise¶

1. Draw the structure for p-isopropylaniline and classify.

Example 13¶

Draw the structure for 2-amino-3-methylpentane.

Solution

Always start with the parent compound: draw the pentane chain. Then attach a methyl group at the third carbon atom and an amino group at the second carbon atom.

Skill-Building Exercise¶

1. Draw the structure for 2-amino-3-ethyl-1-chloroheptane.

Ammonium (NH₄⁺) ions, in which one or more hydrogen atoms are replaced with alkyl groups, are named in a manner analogous to that used for simple amines. The alkyl groups are named as substituents, and the parent species is regarded as the NH₄⁺ ion. For example, CH₃NH₃⁺ is the methylammonium ion. The ion formed from aniline (C₆H₅NH₃⁺) is called the anilinium ion.

Example 14¶

Name each ion.

1. CH₃NH₃⁺

2. (CH₃)₂NH₂⁺

3. (CH₃)₃NH⁺

4. (CH₃)₄N⁺

Solution

The ions have one, two, three, and four methyl (CH₃) groups attached to a nitrogen atom. Their names are as follows:

1. methylammonium ion

2. dimethylammonium ion

3. trimethylammonium ion

4. tetramethylammonium ion

Skill-Building Exercise¶

Name each ion.

1. CH₃CH₂NH₃⁺

2. (CH₃CH₂)₃NH⁺

3. (CH₃CH₂CH₂)₂NH₂⁺

4. (CH₃CH₂CH₂CH₂)₄N⁺

Concept Review Exercises¶

1. To what inorganic compound are the amines related?

2. How are amines classified?

Answers¶

1. ammonia

2. by the number of hydrocarbon groups on the nitrogen atom: primary amine, one group; secondary amine, two groups; tertiary amine, three groups

Exercises¶

1. Draw the structure for each compound and classify the amine as primary, secondary, or tertiary.

   1. dimethylamine

   2. diethylmethylamine

   3. 2-aminoethanol

2. Draw the structure for each compound and classify the amine as primary, secondary, or tertiary.

   1. 3-aminopentane

   2. 1,6-diaminohexane

   3. ethylphenylamine

3. Draw the structure for each compound.

   1. aniline

   2. m-bromoaniline

4. Draw the structure for each compound.

   1. 2-chloroaniline

   2. 3,5-dichloroaniline

5. Name each compound.

   1. CH₃CH₂CH₂NH₂

   2. 

   3. 

6. Name each compound.

   1. (CH₃CH₂)₃N

   2. (CH₃CH₂)₂NCH₃

7. Draw the structure for each compound.

   1. dimethylammonium chloride

   2. anilinium chloride

8. Draw the structure for each compound.

   1. ethylmethylammonium chloride

   2. anilinium nitrate

9. Name each compound.

   1. [CH₃CH₂NH₂CH₂CH₃]⁺Br⁻

   2. [(CH₃CH₂)₃NH]⁺I⁻

10. Name each compound.

    1. [(CH₃)₃NH]⁺NO₃⁻

    2. [(CH₃CH₂)₂NH₂]⁺Cl⁻

Answers¶

1. 1. CH₃NHCH₃; secondary

   2. tertiary

      

   3. HOCH₂CH₂NH₂; primary

3. 1. 

   2. 

5. 1. propylamine

   2. isopropylmethylamine

   3. 2-aminopentane

7. 1. [(CH₃)₂NH₂⁺]Cl^--

   2. 

9. 1. diethylammonium bromide

   2. triethylammonium iodide

15.12 Physical Properties of Amines¶

Primary and secondary amines have hydrogen atoms bonded to a nitrogen atom and are therefore capable of hydrogen bonding (part (a) of Figure 15.6 "Hydrogen Bonding"), although not as strongly as alcohol molecules (which have hydrogen atoms bonded to an oxygen atom, which is more electronegative than nitrogen). These amines boil at higher temperatures than alkanes but at lower temperatures than alcohols of comparable molar mass. For example, compare the boiling point of methylamine (CH₃NH₂; −6°C) with those of ethane (CH₃CH₃; −89°C) and methanol (CH₃OH; 65°C). Tertiary amines have no hydrogen atom bonded to the nitrogen atom and so cannot participate in intermolecular hydrogen bonding. They have boiling points comparable to those of ethers (Table 15.5 "Physical Properties of Some Amines and Comparable Oxygen-Containing Compounds").

Table 15.5 Physical Properties of Some Amines and Comparable Oxygen-Containing Compounds

All three classes of amines can engage in hydrogen bonding with water (part (b) of Figure 15.6 "Hydrogen Bonding"). Amines of low molar mass are quite soluble in water; the borderline of solubility in water is at five or six carbon atoms.

Concept Review Exercises¶

1. Which compound has the higher boiling point, CH₃CH₂CH₂CH₂CH₂NH₂ or CH₃CH₂CH₂CH₂CH₂CH₃? Explain.

2. Which compound is more soluble in water, CH₃CH₂CH₂CH₂CH₃ or CH₃CH₂NHCH₂CH₃? Explain.

Answers¶

1. CH₃CH₂CH₂CH₂CH₂NH₂ because the nitrogen-to-hydrogen (N--H) bonds can engage in hydrogen bonding; CH₃CH₂CH₂CH₂CH₂CH₃ cannot engage in hydrogen bonding

2. CH₃CH₂NHCH₂CH₃ because amines can engage in hydrogen bonding with water; alkanes cannot engage in hydrogen bonding

Exercises¶

1. Which compound of each pair has the higher boiling point? Explain.

   1. butylamine or pentane

   2. CH₃NH₂ or CH₃CH₂CH₂CH₂CH₂NH₂

2. Which compound of each pair has the higher boiling point? Explain.

   1. butylamine or butyl alcohol

   2. trimethylamine or propylamine

3. Which compound is more soluble in water---CH₃CH₂CH₃ or CH₃CH₂NH₂? Explain.

4. Which compound is more soluble in water---CH₃CH₂CH₂NH₂ or CH₃CH₂CH₂CH₂CH₂CH₂NH₂? Explain.

Answers¶

1. 1. butylamine because the N--H bonds can engage in hydrogen bonding; pentane cannot engage in hydrogen bonding

   2. CH₃CH₂CH₂CH₂CH₂NH₂ because it has a greater molar mass than CH₃NH₂

3. CH₃CH₂NH₂ because amines can engage in hydrogen bonding with water; alkanes cannot engage in hydrogen bonding

15.13 Amines as Bases¶

Recall that ammonia (NH₃) acts as a base because the nitrogen atom has a lone pair of electrons that can accept a proton. Amines also have a lone electron pair on their nitrogen atoms and can accept a proton from water to form substituted ammonium (NH₄⁺) ions and hydroxide (OH⁻) ions:

As a specific example, methylamine reacts with water to form the methylammonium ion and the OH⁻ ion.

Nearly all amines, including those that are not very soluble in water, will react with strong acids to form salts soluble in water.

Amine salts are named like other salts: the name of the cation is followed by the name of the anion.

Example 15¶

What are the formulas of the acid and base that react to form [CH₃NH₂CH₂CH₃]⁺CH₃COO⁻?

Solution

The cation has two groups---methyl and ethyl---attached to the nitrogen atom. It comes from ethylmethylamine (CH₃NHCH₂CH₃). The anion is the acetate ion. It comes from acetic acid (CH₃COOH).

Skill-Building Exercise¶

1. What are the formulas of the acid and base that react to form (CH₃CH₂CH₂)₃NH⁺I⁻?

Heterocyclic Amines¶

Looking back at the various cyclic hydrocarbons introduced in Chapter 12 "Organic Chemistry: Alkanes and Halogenated Hydrocarbons" and Chapter 13 "Unsaturated and Aromatic Hydrocarbons", we see that all the atoms in the rings of these compounds are carbon atoms. In other cyclic compounds, called heterocyclic compounds^[13] (Greek heteros, meaning “other”), nitrogen, oxygen, sulfur, or some other atom is incorporated in the ring. Many heterocyclic compounds are important in medicine and biochemistry. Some compose part of the structure of the nucleic acids, which in turn compose the genetic material of cells and direct protein synthesis. (For more information about nucleic acids, see Chapter 19 "Nucleic Acids".)

Many heterocyclic amines occur naturally in plants. Like other amines, these compounds are basic. Such a compound is an alkaloid^[14], a name that means “like alkalis.” Many alkaloids are physiologically active, including the familiar drugs caffeine, nicotine, and cocaine.

Concept Review Exercises¶

1. Explain the basicity of amines.

2. Contrast the physical properties of amines with those of alcohols and alkanes.

3. What is a heterocyclic compound?

Answers¶

1. Amines have a lone pair of electrons on the nitrogen atom and can thus act as proton acceptors (bases).

2. The solubilities of amines are similar to those of alcohols; the boiling points of primary and secondary amines are similar to those of alcohols; the boiling points of tertiary amines, which cannot engage in hydrogen bonding because they do not have a hydrogen atom on the nitrogen atom, are comparable to those of alkanes.

3. Heterocyclic compounds are ring compounds with atoms other than carbon atoms in the ring.

Exercises¶

1. What salt is formed in each reaction? Write its condensed structural formula.

   1. CH₃NH₂(aq) + HBr(aq) →

   2. CH₃NHCH₃(aq) + HNO₃(aq) →

2. What salt is formed in each reaction? Draw its structure.

   1. 

   2. 

Answer¶

1. 1. CH₃NH₃⁺Br⁻(aq)

   2. [CH₃NH₂CH₃]⁺NO₃⁻(aq)

15.14 Amides: Structures and Names¶

The amide functional group has a nitrogen atom attached to a carbonyl carbon atom. If the two remaining bonds on the nitrogen atom are attached to hydrogen atoms, the compound is a simple amide. If one or both of the two remaining bonds on the atom are attached to alkyl or aryl groups, the compound is a substituted amide.

Simple amides are named as derivatives of carboxylic acids. The -ic ending of the common name or the -oic ending of the International Union of Pure and Applied Chemistry (IUPAC) name of the carboxylic acid is replaced with the suffix -amide.

Example 16¶

Name each compound with the common name, the IUPAC name, or both.

Solution

1. This amide has two carbon atoms and is thus derived from acetic acid. The OH of acetic acid is replaced by an NH₂ group. The -ic from acetic (or -oic from ethanoic) is dropped, and -amide is added to give acetamide (or ethanamide in the IUPAC system).

2. This amide is derived from benzoic acid. The -oic is dropped, and -amide is added to give benzamide.

Skill-Building Exercise¶

Name each compound with the common name, the IUPAC name, or both.

Concept Review Exercises¶

1. Name this compound with the common name and the IUPAC name.

   

2. Draw the structural formula for pentanamide.

Answers¶

1. β-bromobutyramide (3-bromobutanamide)

2. 

Exercises¶

1. Draw the structure for each compound.

   1. formamide

   2. hexanamide

2. Draw the structure for each compound.

   1. propionamide

   2. butanamide

3. Name each compound with the common name, the IUPAC name, or both.

   1. 

   2. 

4. Name the compound.

   

Answers¶

1. 1. 

   2. 

3. 1. propionamide (propanamide)

   2. α-methylbutyramide (2-methylbutanamide)

15.15 Physical Properties of Amides¶

With the exception of formamide (HCONH₂), which is a liquid, all simple amides are solids (Table 15.6 "Physical Constants of Some Unsubstituted Amides"). The lower members of the series are soluble in water, with borderline solubility occurring in those that have five or six carbon atoms. Like the esters, solutions of amides in water usually are neutral---neither acidic nor basic.

Table 15.6 Physical Constants of Some Unsubstituted Amides

The amides generally have high boiling points and melting points. These characteristics and their solubility in water result from the polar nature of the amide group and hydrogen bonding (Figure 15.7 "Hydrogen Bonding in Amides"). (Similar hydrogen bonding plays a critical role in determining the structure and properties of proteins, deoxyribonucleic acid [DNA], ribonucleic acid [RNA], and other giant molecules so important to life processes. See Chapter 19 "Nucleic Acids".)

Concept Review Exercises¶

1. Which compound has the higher boiling point---pentanamide (CH₃CH₂CH₂CH₂CONH₂) or propyl acetate (CH₃COOCH₂CH₂CH₃)? Explain.

2. Which compound is more soluble in water---propanamide (CH₃CH₂CONH₂) or 1-pentene (CH₂=CHCH₂CH₂CH₃)? Explain.

Answers¶

1. pentanamide because the nitrogen-to-hydrogen (N--H) and the carbon-to-oxygen double (C=O) bonds can engage in hydrogen bonding; propyl acetate cannot engage in hydrogen bonding

2. propanamide because the N--H and C=O bonds can engage in hydrogen bonding with water; 1-pentene cannot engage in hydrogen bonding with water

Exercises¶

1. Which compound has the higher boiling point---butyramide (CH₃CH₂CH₂CONH₂) or ethyl acetate (CH₃COOCH₂CH₃)? Explain.

2. Which compound has the higher boiling point---butyramide or dimethylacetamide [CH₃CON(CH₃)₂]? Explain.

3. Which compound is more soluble in water---acetamide (CH₃CONH₂) or 1-butene (CH₂=CHCH₂CH₃)? Explain.

4. Which compound is more soluble in water---CH₃CONHCH₃ or 2-methylbutane [CH₃CH(CH₃)CH₂CH₃)]? Explain.

Answers¶

1. butyramide because the nitrogen-to-hydrogen (N--H) and the carbon-to-oxygen double (C=O) bonds can engage in hydrogen bonding; ethyl acetate cannot engage in hydrogen bonding

3. acetamide because the N--H and C=O bonds can engage in hydrogen bonding with water; 1-butene cannot engage in hydrogen bonding with water

15.16 Formation of Amides¶

The addition of ammonia (NH₃) to a carboxylic acid forms an amide, but the reaction is very slow in the laboratory at room temperature. Water molecules are split out, and a bond is formed between the nitrogen atom and the carbonyl carbon atom.

In living cells, amide formation is catalyzed by enzymes. Proteins are polyamides; they are formed by joining amino acids into long chains. In proteins, the amide functional group is called a peptide bond. (For more information about proteins, see Chapter 18 "Amino Acids, Proteins, and Enzymes", Section 18.4 "Proteins".)

Polyamides¶

Just as the reaction of a diol and a diacid forms a polyester (Section 15.8 "Preparation of Esters"), the reaction of a diacid and a diamine yields a polyamide^[15]. The two difunctional monomers often employed are adipic acid and 1,6-hexanediamine. The monomers condense by splitting out water to form a new product, which is still difunctional and thus can react further to yield a polyamide polymer.

Some polyamides are known as nylons. Nylons are among the most widely used synthetic fibers---for example, they are used in ropes, sails, carpets, clothing, tires, brushes, and parachutes. They also can be molded into blocks for use in electrical equipment, gears, bearings, and valves.

Concept Review Exercises¶

1. Write the condensed structural formulas and give names of the two compounds from which butanamide (CH₃CH₂CH₂CONH₂) is formed.

2. Write the condensed structural formulas and names of the two compounds from which CH₃CH₂CH₂CH₂CH₂CONHCH₂CH₂CH₃ is formed.

Answers¶

1. CH₃CH₂CH₂COOH (butanoic acid) and NH₃ (ammonia)

2. CH₃CH₂CH₂CH₂CH₂COOH (hexanoic acid) and CH₃CH₂CH₂NH₂ (propylamine)

Exercises¶

1. Write the condensed structural formulas and names of the two compounds from which pentanamide (CH₃CH₂CH₂CH₂CONH₂) is formed.

2. Write the condensed structural formulas and names of the two compounds from which CH₃CONHCH₃ is formed.

Answer¶

1. CH₃CH₂CH₂CH₂COOH (pentanoic acid) and NH₃ (ammonia)

15.17 Chemical Properties of Amides: Hydrolysis¶

Generally, amides resist hydrolysis in plain water, even after prolonged heating. In the presence of added acid or base, however, hydrolysis proceeds at a moderate rate. In living cells, amide hydrolysis is catalyzed by enzymes. Amide hydrolysis is illustrated in the following example:

Example 17¶

Write the equation for the hydrolysis of each compound.

1. butyramide

2. benzamide

Solution

1. The hydrolysis of a simple amide produces an organic acid and ammonia. Butyramide thus yields butyric acid and ammonia.

   

2. The hydrolysis of an amide produces an organic acid and ammonia. Benzamide thus yields benzoic acid and ammonia.

   

Skill-Building Exercise¶

Write the equation for the hydrolysis of each compound.

1. propionamide (propanamide)

2. hexanamide

Concept Review Exercises¶

1. What are the products of the hydrolysis of an amide?

2. When the amide CH₃CH₂CH₂CH₂CONH₂ is hydrolyzed in an NaOH solution, the products are CH₃CH₂CH₂CH₂COO⁻Na⁺ and NH₃. What products are obtained when CH₃CH₂CH₂CH₂CONH₂ is hydrolyzed in a hydrochloric acid solution?

Answers¶

1. a carboxylic acid and ammonia or an amine

2. CH₃CH₂CH₂CH₂COOH and NH₄Cl

Exercises¶

1. Complete each equation.

   1. 

   2. 

2. Complete each equation.

   1. 

   2. 

Answer¶

1. 1. CH₃COOH + NH₃

   2. 

15.18 End-of-Chapter Material¶

Additional Exercises¶

1. Of the families of organic compounds discussed in this chapter, which are known for their typically unpleasant odors? Which for their characteristically pleasant aromas?

2. What is esterification of a carboxylic acid? How does it differ from neutralization?

3. Like alcohols, phenols form esters with carboxylic acids. The hydrocarbon group from phenol is called phenyl. Draw the structure of phenyl acetate.

4. Describe the hydrogen bonding in carboxylic acids, both acid-acid and acid-water. How does this influence their physical properties?

5. Which compound is more soluble in water---benzoic acid or sodium benzoate? Explain.

6. Dicarboxylic acids have two carboxyl groups and are named with the ending -dioic acid. Give the equation for the reaction of 1,5-pentanedioic acid (HOOCCH₂CH₂CH₂COOH; common name, glutaric acid) with each of the following:

   1. 1 mol of NaOH

   2. 2 mol of NaOH

7. Without consulting tables, arrange the following compounds in order of increasing boiling point: butyl alcohol, methyl acetate, pentane, and propionic acid.

8. From which alcohol might each acid be prepared via oxidation with acidic dichromate?

   1. CH₃CH₂COOH

   2. HCOOH

   3. HOOCH₂COOH

   4. (CH₃)₂CHCH₂COOH

9. The distinctive aroma and flavor of oranges are due in part to octyl acetate, an ester formed from 1-octanol (octyl alcohol) and acetic acid. Write the condensed structural formula for octyl acetate.

10. A lactone is a cyclic ester. What product is formed in the following reaction?

    

11. A lactam is a cyclic amide. What product is formed in the following reaction?

    

12. Draw the structures for the eight isomeric amines that have the molecular formula C₄H₁₁N. Give each a common name and classify it as primary, secondary, or tertiary.

13. Draw the structures for the five isomeric amines that have the molecular formula C₇H₉N and contain a benzene ring. Classify each compound as primary, secondary, or tertiary.

14. Cocaine is usually used in the form of the salt cocaine hydrochloride and sniffed up the nose. Some prefer to ingest their cocaine by smoking it (mixed with tobacco, for example). Before smoking, the cocaine hydrochloride must be converted back to the free base (that is, to the molecular form). Explain the choice of dosage form for each route of administration.

15. Draw the structures of all the isomeric amides that have the molecular formula C₄H₉NO.

16. An ester with the molecular formula C₆H₁₂O₂ was hydrolyzed in aqueous acid to yield an acid Y and an alcohol Z. Oxidation of the alcohol with potassium dichromate (K₂Cr₂O₇) gave the identical acid Y. What is the condensed structural formula of the ester?

17. The neutralization of 125 mL of a 0.400 M NaOH solution requires 5.10 g of a monocarboxylic acid. Draw all the possible structures for the acid.

18. If 3.00 g of acetic acid reacts with excess methanol, how many grams of methyl acetate are formed?

19. How many milliliters of a 0.100 M barium hydroxide solution are required to neutralize 0.500 g of dichloroacetic acid?

Answers¶

1. unpleasant: carboxylic acids; pleasant: esters

3. 

5. sodium benzoate because it is ionic and forms ion-dipole forces with water; benzoic acid can engage in hydrogen bonding only with water

7. pentane < methyl acetate < butyl alcohol < propionic acid

9. CH₃COOCH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₃

11. H₃N⁺CH₂CH₂COOH

13. 

15. 

17. 

19. 20.0 mL