Question 11.4: Structural Isomers Identify each pair of formulas as structu...
Structural Isomers
Identify each pair of formulas as structural isomers or the same molecule.
a. \begin{array}{r c}\begin{matrix} \underset{|}{{CH_{3}}} \quad {\underset{|}{{CH_{3}}}} \\ \quad \quad {CH_{2}}-CH_{2} \quad \quad \end{matrix}\end{array} and \begin{array}{r c}\begin{matrix} \underset{|}{CH_{2}} -CH_{2}-CH_{3} \\ CH_{3} \quad \quad \quad \quad \quad \quad\end{matrix}\end{array}
b. \begin{array}{r c}\begin{matrix} {\underset{|}{{CH_{3}}}} \quad \quad \quad \quad \\ \quad \quad {CH_{3}}-CH-CH_{2}-CH_{2}-CH_{3} \end{matrix}\end{array} and \begin{array}{r c}\begin{matrix} \quad \quad \quad \quad \quad {\underset{|}{{CH_{3}}}} \quad {\underset{|}{{CH_{3}}}} \quad \quad \quad \quad \\ {CH_{3}}-CH-CH-CH_{3} \end{matrix}\end{array}
a. When we add up the number of C atoms and H atoms, they give the same molecular formula C_{4}H_{10}. The structural formula on the left consists of a chain of four C atoms. Even though the —CH_{3} ends are drawn up, they are not branches on the chain but part of the four-carbon chain. The structural formula on the right also consists of a fourcarbon chain even though one —CH_{3} end is drawn down. Thus both condensed structural formulas represent the same molecule and are not structural isomers.
b. When we add up the number of C atoms and H atoms, they give the same molecular formula C_{6}H_{14}. The structural formula on the left consists of a five-carbon chain with a —CH_{3} substituent on the second carbon of the chain. The structural formula on the right consists of a four-carbon chain with two —CH_{3} substituents, one bonded to the second carbon and one bonded to the third carbon. Thus there is a different order of bonding of atoms in these two structural formulas, which represent structural isomers.