Solution To start, draw the carbon backbone of the straight-chain isomer.

1. Example 20.1Writing Structural Formulas for Hydrocarbons Write the structural formulas and carbon skeleton formulas for the five isomers of C6H14 (hexane). Solution To start, draw the carbon backbone of the straight-chain isomer. Next, determine the carbon backbone structure of the other isomers by arranging the carbon atoms in four other unique ways.

2. Example 20.1Writing Structural Formulas for Hydrocarbons Continued Fill in all the hydrogen atoms so that each carbon forms four bonds.

3. Example 20.1Writing Structural Formulas for Hydrocarbons Continued Write the carbon skeleton formulas by using lines to represent each carbon–carbon bond. Remember that each end or bend represents a carbon atom.

4. Example 20.1Writing Structural Formulas for Hydrocarbons Continued For Practice 20.1 Write the structural formulas and carbon skeleton formulas for the three isomers of C5H12 (pentane).

5. Example 20.2Naming Alkanes Name this alkane. Procedure For… Naming Alkanes Solution Step 1Count the number of carbon atoms in the longest continuous carbon chain to determine the base name of the compound. Locate the prefix corresponding to this number of atoms in Table 20.5 and add the ending -ane to form the base name. This compound has five carbon atoms in its longest continuous chain. The correct prefix from Table 20.5 is pent-. The base name is pentane.

6. Example 20.2Naming Alkanes Continued Step 2Consider every branch from the base chain to be a substituent. Name each substituent according to Table 20.6. This compound has one substituent named ethyl. Step 3Beginning with the end closest to the branching, number the base chain and assign a number to each substituent. (If two substituents occur at equal distances from each end, go to the next substituent to determine from which end to start numbering.) Number the base chain as follows: Assign the number 3 to the ethyl substituent.

7. Example 20.2Naming Alkanes Continued Step 4Write the name of the compound in the following format: (substituent number)-(substituent name)(base name) If there are two or more substituents, give each one a number and list them alphabetically with hyphens between words and numbers. The name of the compound is: 3-ethylpentane Step 5If a compound has two or more identical substituents, indicate the number of identical substituents with the prefix di- (2), tri- (3), or tetra- (4) before the substituent’s name. Separate the numbers indicating the positions of the substituents relative to each other with a comma. Do not take the prefixes into account when alphabetizing. Does not apply to this compound. For Practice 20.2 Name this alkane.

8. Example 20.3Naming Alkanes Name this alkane. Procedure For… Naming Alkanes Solution Step 1Count the number of carbon atoms in the longest continuous carbon chain to determine the base name of the compound. Locate the prefix corresponding to this number of atoms in Table 20.5 and add the ending -ane to form the base name. This compound has eight carbon atoms in its longest continuous chain. The correct prefix from Table 20.5 is oct-. The base name is octane.

9. Example 20.3Naming Alkanes Continued Step 2Consider every branch from the base chain to be a substituent. Name each substituent according to Table 20.6. This compound has one substituent named ethyl and two named methyl. Step 3Beginning with the end closest to the branching, number the base chain and assign a number to each substituent. (If two substituents occur at equal distances from each end, go to the next substituent to determine from which end to start numbering.) Number the base chain as follows: Assign the number 4 to the ethyl substituent and the numbers 2 and 7 to the two methyl substituents.

10. Example 20.3Naming Alkanes Continued Step 4Write the name of the compound in the following format: (substituent number)-(substituent name)(base name) If there are two or more substituents, give each one a number and list them alphabetically with hyphens between words and numbers. The basic form of the name of the compound is: 4-ethyl-2,7-methyloctane List ethyl before methyl because substituents are listed in alphabetical order. Step 5If a compound has two or more identical substituents, indicate the number of identical substituents with the prefix di- (2), tri- (3), or tetra- (4) before the substituent’s name. Separate the numbers indicating the positions of the substituents relative to each other with a comma. Do not take the prefixes into account when alphabetizing. This compound has two methyl substituents; therefore, the name of the compound is: 4-ethyl-2,7-dimethyloctane

11. Example 20.3Naming Alkanes Continued For Practice 20.3 Name this alkane.

12. Example 20.4Naming Alkanes Name this alkane. Solution Step 1The longest continuous carbon chain has five atoms. Therefore, the base name is pentane. Step 2This compound has two substituents, both of which are named methyl. Step 3Since both substituents are equidistant from the ends, it does not matter from which end you start numbering.

13. Example 20.4Naming Alkanes Continued Step 4, 5Use the general form for the name: (substituent number)-(substituent name)(base name) Because this compound contains two identical substituents, step 5 from the naming procedure applies and you use the prefix di-. Indicate the position of each substituent with a number separated by a comma. 2,4-dimethylpentane For Practice 20.4 Name this alkane.

14. Example 20.5Naming Alkenes and Alkynes Name each compound. a. b. Solution a. Step 1The longest continuous carbon chain containing the double bond has six carbon atoms. The base name is therefore hexene.

15. Example 20.5Naming Alkenes and Alkynes Continued Step 2The two substituents are both methyl. Step 3One of the exceptions listed previously states that, in naming alkenes, you should number the chain so that the double bond has the lowest number. In this case, the double bond is equidistant from the ends. Assign the double bond the number 3. The two methyl groups are then at positions 3 and 4.

16. Example 20.5Naming Alkenes and Alkynes Continued Step 4, 5Use the general form for the name: (substituent number)-(substituent name)(base name) Because this compound contains two identical substituents, step 5 of the naming procedure applies. Use the prefix di-. In addition, indicate the position of each substituent with a number separated by a comma. Because this compound is an alkene, specify the position of the double bond, isolated by hyphens, just before the base name. 3,4-dimethyl-3-hexene b. Step 1The longest continuous carbon chain containing the triple bond is five carbon atoms long; therefore, the base name is pentyne.

17. Example 20.5Naming Alkenes and Alkynes Continued Step 2There are two substituents; one is a methyl group and the other an isopropyl group. Step 3Number the base chain, giving the triple bond the lowest number (1). Assign the isopropyl and methyl groups the numbers 3 and 4, respectively.

18. Example 20.5Naming Alkenes and Alkynes Continued Step 4Use the general form for the name: (substituent number)-(substituent name)(base name) Since there are two substituents, list both of them in alphabetical order. Since this compound is an alkyne, specify the position of the triple bond with a number isolated by hyphens just before the base name. 3-isopropyl-4-methyl-1-pentyne For Practice 20.5 Name each compound. a. b.

19. Example 20.6Alkene Addition Reactions Determine the products of the reactions. a. b. Solution a. The reaction of 1-butene with bromine is an example of a symmetric addition. The bromine adds across the double bond and each carbon forms a single bond to a bromine atom.

20. Example 20.6Alkene Addition Reactions Continued b. The reaction of 1-butene with hydrogen bromide is an example of an unsymmetrical addition. Apply Markovnikov’s rule to determine which carbon the hydrogen bonds with and which carbon the bromine atom bonds with. Markovnikov’s rule predicts that the hydrogen bonds to the end carbon in this case. For Practice 20.6 Determine the products of the reactions. a. b.

21. Example 20.7Alcohol Reactions Determine the type of reaction (substitution, dehydration, oxidation, or reaction with an active metal) that occurs in each case, and write formulas for the products. a. b. Solution a. An alcohol reacting with an acid is an example of a substitution reaction. The product of the substitution reaction is a halogenated hydrocarbon and water.

22. Example 20.7Alcohol Reactions Continued Solution b. An alcohol in solution with sodium dichromate and acid undergoes an oxidation reaction. The product of the oxidation reaction is a carboxylic acid functional group. (We discuss carboxylic acid functional groups in detail in Section 20.11.) For Practice 20.7 Determine the type of reaction (substitution, dehydration, oxidation, or reaction with an active metal) that occurs in each case, and write formulas for the products. a. b.