1. Sample Problem 14.1 Identifying Aldehydes and Ketones
Identify each of the following compounds as an aldehyde or ketone:
a. b. c.
Solution
a. aldehyde b. ketone c. aldehyde
Study Check 14.1
Identify the following compound as an aldehyde or ketone:
Answer
ketone

2. Sample Problem 14.2 Naming Aldehydes
Give the IUPAC name for each of the following aldehydes:
a. b.
Solution
a. Step 1 Name the longest carbon chain by replacing the e in the alkane name with al. The longest carbon chain containing the carbonyl group has five carbon atoms. It is named by replacing the e in the alkane name with al to give pentanal.
Step 2 Name and number any substituents by counting the carbonyl group as carbon 1. The substituent, which is the —CH3 group on carbon 3, is methyl. The IUPAC name for this compound is 3-methylpentanal.

3. Sample Problem 14.2 Naming Aldehydes
Continued
b. Step 1 Name the longest carbon chain by replacing the e in the alkane name with al. The longest carbon chain consists of a benzene ring attached to a carbonyl group, which is named benzaldehyde.
Step 2 Name and number any substituents by counting the carbonyl group as carbon 1. Counting from carbon 1 of the ring where the carbonyl group is attached, the chloro group is attached to carbon 4.
Study Check 14.2
What is the IUPAC name of the following compound?
Answer
5-methylhexanal

4. Sample Problem 14.3 Naming Ketones
Give the IUPAC name for the following ketone:
Solution
Step 1 Name the longest carbon chain by replacing the e in the alkane name with one. The longest chain has five carbon atoms, which is named pentanone.
Step 2 Number the carbon chain starting from the end nearer the carbonyl group and indicate its location. Counting from the right, the carbonyl group is on carbon 2.

5. Sample Problem 14.3 Naming Ketones
Continued
Step 3 Name and number any substituents on the carbon chain. Counting from the right, the methyl group is on carbon 4. The IUPAC name is 4-methyl-2-pentanone.
Study Check 14.3
What is the IUPAC name of ethyl propyl ketone?
Answer
3-hexanone

6. Sample Problem 14.4 Boiling Point and Solubility
Identify each of the following compounds as an aldehyde or ketone:
a. Arrange pentane, 2-butanol, and butanone, which have similar molar masses, in order of increasing boiling points. Explain.
b. Why is butanone soluble in water, but 2-hexanone is not?
Solution
a. The only attractions between molecules of alkanes such as pentane are dispersion forces. With no dipole–dipole attractions or hydrogen bonds, pentane has the lowest boiling point of the three compounds. With a polar carbonyl group, butanone molecules form dipole–dipole attractions, but no hydrogen bonds. Butanone has a higher boiling point than pentane. Because molecules of 2-butanol can form hydrogen bonds with other butanol molecules, it has the highest boiling point of the three compounds. The actual boiling points are pentane (36 °C), butanone (80 °C), and 2-butanol (100 °C).
b. Butanone contains a carbonyl group with an electronegative oxygen atom that forms hydrogen bonds with water. 2-Hexanone also contains a carbonyl group, but its longer hydrocarbon chain reduces the impact on the solubility of the polar carbonyl group.
Study Check 14.4
If acetone molecules cannot hydrogen bond with each other, why is acetone soluble in water?
Answer
The oxygen atom in the carbonyl group of acetone forms hydrogen bonds with water molecules.

7. Sample Problem 14.5 Tollens’ Test
Draw the condensed structural formula for the product of oxidation, if any, when Tollens’ reagent is added to each of the following compounds:
a. propanal b. propanone c. 2-methylbutanal
Solution
Tollens’ reagent will oxidize aldehydes but not ketones.
a. b. no reaction c.
Study Check 14.5
Why does a silver mirror form when Tollens’ reagent is added to a test tube containing benzaldehyde?
Answer
The oxidation of benzaldehyde reduces Ag+ to metallic silver, which forms a silvery coating on the inside of the test tube.

8. Sample Problem 14.6 Reduction of Carbonyl Groups
Write the equation for the reduction of cyclopentanone using hydrogen in the presence of a nickel catalyst.
Solution
The reacting molecule is a cyclic ketone that has five carbon atoms. During the reduction, hydrogen atoms add to the carbon and oxygen in the carbonyl group, which reduces the ketone to the corresponding secondary alcohol.
Study Check 14.6
What is the name of the product obtained from the hydrogenation of 2-methylbutanal?
Answer
2-methyl-1-butanol

9. Sample Problem 14.7 Hemiacetals and Acetals
Identify each of the following as a hemiacetal or acetal:
a. b. c.
Solution
a. A hemiacetal has a carbon bonded to one alkoxy group and one hydroxyl group.
b. An acetal has a carbon bonded to two alkoxy groups.
c. A hemiacetal has a carbon bonded to one alkoxy group and one hydroxyl group.
Study Check 14.7
From the following descriptions, identify the compound as a hemiacetal or an acetal:
a. a molecule that contains a carbon atom attached to a hydroxyl group and an ethoxy group
b. a molecule that contains a carbon atom attached to two ethoxy groups
Answer
a. hemiacetal b. acetal

10. Sample Problem 14.8 Hemiacetals and Acetals
Draw the condensed structural formulas for the hemiacetal and acetal formed when methanol adds to propanal.
Solution
To form the hemiacetal, the hydrogen from methanol adds to the oxygen of the carbonyl group to form a new hydroxyl group. The remaining part of the methanol adds to the carbon atom in the carbonyl group. The acetal forms when a second molecule of methanol adds to the carbonyl carbon atom.
Study Check 14.8
Draw the condensed structural formula for the acetal formed when methanol adds to butanone.
Answer

11. Sample Problem 14.9 Chiral Carbons
For each of the following, indicate whether the carbon in red is chiral or achiral:
a. Glycerol, which is used to sweeten and preserve foods and as a lubricant in soaps, creams, and hair care products.
b. Monosodium glutamate (MSG), which is the salt of the amino acid glutamic acid used as a flavor enhancer in foods.
c. Ibuprofen, which is a nonsteroidal anti-inflammatory drug used to relieve fever and pain.

12. Sample Problem 14.9 Chiral Carbons
Continued
Solution
a. Achiral. Two of the substituents on the carbon in red are the same:
b. Chiral. The carbon in red is bonded to four different groups:
c. Chiral. The carbon in red is bonded to four different groups:

13. Sample Problem 14.9 Chiral Carbons
Continued
Study Check 14.9
Circle the chiral carbon of penicillamine, which is used in the treatment of rheumatoid arthritis.
Answer

14. Sample Problem 14.10 Fischer Projections
Identify each of the following as the D or L stereoisomer:
a. b. c.
Solution
a. When the —OH group is drawn to the left of the chiral carbon, it is the l stereoisomer.
b. When the —OH group is drawn to the right of the chiral carbon, it is the d stereoisomer.
c. When the —OH group is drawn to the right of the chiral carbon farthest from the top of the Fischer projection, it is the d stereoisomer.
Study Check 14.10
Draw the Fischer projections for the d and l stereoisomers of 2-hydroxypropanal and label the d and l isomers.
Answer