Question 21.5: Naming Alkenes and Alkynes Name each compound. CH3−CH3∣CH...
Naming Alkenes and Alkynes
Name each compound.
(a) CH _3- \underset{\underset{\overset{CH _2}{\overset{|}{CH _3} } }{| } }{C} = \overset{\overset{CH _3}{|} }{C}-CH_2 – CH_3 (b) CH _3- \underset{\underset{CH_3}{|} }{CH} – \overset{\overset{CH_3- \overset{\overset{CH_3}{|} }{CH}}{|} }{CH}-C\equiv CH
| (a) 1. The longest continuous carbon chain containing the double bond has six carbon atoms. The base name is therefore hexene. | CH _3- \underset{\underset{\overset{CH _2}{\overset{|}{CH _3} } }{| } }{C} = \overset{\overset{CH _3}{|} }{C} – CH_2 – CH_3 |
| 2. The two substituents are both methyl. | \begin{matrix} \overset{\text{methyl}\xrightarrow {\hspace{1cm} }CH_3} {\downarrow\quad\quad\quad\quad| }\quad\quad\quad\quad\quad\quad\\CH _3-C=C-CH_2 – CH_3\\|\quad\quad\quad\quad\quad\\CH _2\quad\quad\quad\quad\quad\\|\quad\quad\quad\quad\quad\\CH _3\quad\quad\quad\quad\quad \end{matrix} |
| 3. One 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. | CH _3- \underset{\underset{\overset{2 CH _2}{\overset{|}{\underset{1}{CH _3} } } }{3| } }{C} = \overset{\overset{CH _3}{|} }{\underset{4}{C} }-\underset{5}{CH_2} – \underset{6}{CH_3} |
| 4, 5. Use 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) 1. The longest continuous carbon chain containing the triple bond is five carbon atoms long; therefore, the base name is pentyne. | CH _3- \underset{\underset{CH_3}{|} }{CH} – \overset{\overset{CH_3- \overset{\overset{CH_3}{|} }{CH}}{|} }{CH}-C\equiv CH |
| 2. There are two substituents; one is a methyl group and the other an isopropyl group. |  |
| 3. Number the base chain, giving the triple bond the lowest number (1). Assign the isopropyl and methyl groups the numbers 3 and 4, respectively. |
\underset{5}{CH _3} – \underset{\underset{CH_3}{4|} }{CH} – \overset{\overset{CH_3- \overset{\overset{CH_3}{|} }{CH}}{|} }{\underset{3}{CH} }-\underset{2}{C} \equiv \underset{1}{CH} |
| 4. Use the general form for the name:
(substituent number)-(substituent name)(base name) Since there are two substituents, list both of them in alphabetical order. Because 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 |