Question

Draw all the possible noncyclic structural isomers of \(\mathrm{C}_{5} \mathrm{H}_{10}\). Name each compound.

Short answer

The 3 possible noncyclic structural isomers of \(\mathrm{C}_{5}\mathrm{H}_{10}\) are: 1. 1-pentene: \(\mathrm{CH}_{3} - \mathrm{CH}_{2} - \mathrm{CH}_{2} - \mathrm{CH} = \mathrm{CH}_{2}\) 2. 2-pentene: \(\mathrm{CH}_{3} - \mathrm{CH}_{2} - \mathrm{CH} = \mathrm{CH} - \mathrm{CH}_{3}\) 3. 2-methyl-1-butene: \(\mathrm{CH}_{3} - \mathrm{CH} = \mathrm{CH} - \mathrm{CH}_{2} - \mathrm{CH}_{3}\)

Step-by-step solution

Calculate Degree of Unsaturation (DOU)

The formula for DOU is given by: \[DOU = \frac{(2n+2) - x}{2}\] Where n is the number of carbon atoms, and x is the number of hydrogen atoms. In our case, n=5 and x=10. \[DOU = \frac{(2(5)+2) - 10}{2} = \frac{12 - 10}{2} = 1\] DOU is equal to 1, which means there is one double bond present in each structural isomer.

Identify the possible isomers

Now, we will start creating the isomers by adding a double bond to the carbon atoms. When adding a double bond, we will have hydrogen atoms attached to the carbons making sure that each carbon is bonded to four atoms in total. Here are the possible 3 isomers: 1. \(\mathrm{CH}_{3} - \mathrm{CH}_{2} - \mathrm{CH}_{2} - \mathrm{CH} = \mathrm{CH}_{2}\) 2. \(\mathrm{CH}_{3} - \mathrm{CH}_{2} - \mathrm{CH} = \mathrm{CH} - \mathrm{CH}_{3}\) 3. \(\mathrm{CH}_{3} - \mathrm{CH} = \mathrm{CH} - \mathrm{CH}_{2} - \mathrm{CH}_{3}\)

Name each compound

Now, we'll use IUPAC nomenclature to name each compound: 1. \(\mathrm{CH}_{3} - \mathrm{CH}_{2} - \mathrm{CH}_{2} - \mathrm{CH} = \mathrm{CH}_{2}\): The double bond starts on the fourth carbon of the longest carbon chain, so we name this compound 1-pentene. 2. \(\mathrm{CH}_{3} - \mathrm{CH}_{2} - \mathrm{CH} = \mathrm{CH} - \mathrm{CH}_{3}\): The double bond starts on the third carbon of the longest carbon chain, so the name of this compound is 2-pentene. 3. \(\mathrm{CH}_{3} - \mathrm{CH} = \mathrm{CH} - \mathrm{CH}_{2} - \mathrm{CH}_{3}\): The double bond starts on the second carbon of the longest carbon chain, so the name of this compound is 2-methyl-1-butene. Now we have drawn and named all the possible noncyclic structural isomers of \(\mathrm{C}_{5}\mathrm{H}_{10}\).