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

To create different structural isomers, we need to vary the position of the fluorine atom and the carbon chain. For a 4-carbon chain, we can have the fluorine atom on either the second or third carbon atom, and the carbon chain can be in a straight line or have a branching. We can generate the following isomers: 1. Fluorine atom on C2, straight chain: \(\mathrm{H}_{3}\mathrm{C}-\mathrm{CHF}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (let's call it A) 2. Fluorine atom on C2, branching: \(\mathrm{H}_{3}\mathrm{C}-\mathrm{CHF}-\mathrm{C}( \mathrm{H}_{3})-\mathrm{CH}_{3}\) (let's call it B) 3. Fluorine atom on C3, straight chain: \(\mathrm{H}_{3}\mathrm{C}-\mathrm{CH}_{2}-\mathrm{CHF}-\mathrm{CH}_{3}\) (let's call it C) Next, we will examine each of these structural isomers for any possible cis-trans isomers.

In this structural isomer, the fluorine atom is connected to a carbon that has at least two hydrogen atoms, which makes cis-trans isomerism impossible. So, A only has one geometrical isomer.

In this structural isomer, the fluorine atom is connected to a carbon that has at least two hydrogen atoms, which makes cis-trans isomerism impossible. So, B only has one geometrical isomer.

In this structural isomer, the fluorine atom is connected to a carbon that has only one hydrogen atom, and the neighboring carbons also have only one hydrogen atom. This makes cis-trans isomerism possible. We can have the following geometrical isomers of C: 1. Cis-isomer: \(\mathrm{H}_{3}\mathrm{C} - \mathrm{CH}_{2} - \mathrm{C}(\mathrm{H})(\mathrm{F}) - \mathrm{CH}_{3}\) 2. Trans-isomer: \(\mathrm{H}_{3}\mathrm{C} - \mathrm{CH}_{2} - \mathrm{C}(\mathrm{F})(\mathrm{H}) - \mathrm{CH}_{3}\) In conclusion, there are 5 isomers in total: 1 isomer of A, 1 isomer of B, and 2 geometrical isomers of C (cis and trans), plus the original structure of C as the 5th isomer.