Draw all structural and geometrical (cis-trans) isomers of $\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{F}$ . Ignore any cyclic isomers.

To start, let's find all the possible ways to arrange our four carbon atoms. The different arrangements of carbon atoms can be: 1. As a straight chain: \(\mathrm{CCCC}\) 2. With a branch: \(\mathrm{CC(C)C}\)

Now, let's place the fluorine atom and hydrogen atoms onto the carbon skeletons that we found in step 1. Arrangement 1: Straight chain \(\mathrm{CCCC}\) - 1-Fluoropropane: \(\mathrm{F\!-\!C_1 C_2 C_3 C_4}\) - 2-Fluoropropane: \(\mathrm{C_1 C_2\!-\!F C_3 C_4}\) Arrangement 2: With a branch \(\mathrm{CC(C)C}\) - 1-Fluoro-2-methylpropane: \(\mathrm{F\!-\!C_1(C_2)C_3 C_4}\)

Now, let's examine each of the structures we found in step 2 for geometrical (cis-trans) isomers. Since cis-trans isomerism occurs when there is restricted rotation around a double bond or when there is stereogenic center with a specific arrangement, we'll look for these features in each structure. - 1-Fluoropropane and 2-Fluoropropane: The structures have only single bonds, so there are no cis-trans isomers for these structures. - 1-Fluoro-2-methylpropane: This structure also has only single bonds, so there are no cis-trans isomers for this structure. As a result, since there are no structures with double bonds or stereogenic centers, we conclude that there are no cis-trans (geometrical) isomers for the given chemical formula.

In this exercise, we found the following three structural isomers of \(\mathrm{C}_{4}\mathrm{H}_{7}\mathrm{F}\): 1. 1-Fluoropropane: \(\mathrm{F\!-\!C_1 C_2 C_3 C_4}\) 2. 2-Fluoropropane: \(\mathrm{C_1 C_2\!-\!F C_3 C_4}\) 3. 1-Fluoro-2-methylpropane: \(\mathrm{F\!-\!C_1(C_2)C_3 C_4}\) We've also determined that none of these structures have cis-trans (geometrical) isomers.