Question 26.1: Identifying Isomers. Write structural formulas for all the c......

Analyze
First we write the longest chain of C atoms and add an appropriate number of H atoms (12, in this case) to give each C atom four bonds. Next, we look for isomers that have fewer carbon atoms in the longest chain.

Solve
The longest chain of carbon atoms that we can draw has five carbon atoms in it. When we add hydrogen atoms to give each carbon atom four bonds, we obtain structure (1).

(1)           H—\overset{\begin{matrix} H \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ H \end{matrix}}{C}}—\overset{\begin{matrix} H \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ H \end{matrix}}{C}}—\overset{\begin{matrix} H \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ H \end{matrix}}{C}}—\overset{\begin{matrix} H \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ H \end{matrix}}{C}}—\overset{\begin{matrix} H \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ H \end{matrix}}{C}}—H

Now, let’s look for isomers with four C atoms in the longest chain and one C atom as a branch (five C atoms in all). There is only one possibility. Notice that if structure (2) is flipped from left to right, the identical structure (2′) is obtained.

(2)  C—\overset{\begin{matrix} C \\ \mid \end{matrix} }{C}—C—C        (2′)  C—C—\overset{\begin{matrix} C \\ \mid \end{matrix} }{C}—C

Again, we complete the structure of this isomer by adding H atoms.

Finally, let’s consider a three-carbon chain with two one carbon branches. Again, there is only one possibility.

The number of isomers with the formula C_{5}H_{12} is three.

C—\overset{\begin{matrix} C \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ C \end{matrix}}{C}}—C        that is,            CH_{3}—\overset{\begin{matrix} CH_{3} \\ \mid \end{matrix} }{\underset{\begin{matrix} \mid \\ CH_{3} \end{matrix}}{C}}—CH_{3}

Assess
We must be careful to recognize when two possible structures are actually the same structure, as was the case for structures (2) and (2′).