Draw all the structural isomers of \(\mathrm{C}_{5} \mathrm{H}_{10}\). Ignore any cyclic isomers.

To find the linear isomer, we need to arrange five carbon atoms in a straight chain with single bonds. Attach hydrogen atoms to each carbon atom while maintaining the molecular formula C5H10. The linear isomer should look like this: \(CH_3-CH_2-CH_2-CH_2-CH_3\)

To find branched isomers, we need to rearrange the carbon atoms in different ways so that the molecular formula remains the same (C5H10) but the structure is different. The only way to create a branched structure in this case is to move one of the carbon atoms off the linear chain. Let's start with moving one carbon atom off the C2 position (2nd carbon atom from one end) on the linear chain, and form a new isomer: \(CH_3-CH(CH_3)-CH_2-CH_2-CH_3\) Next, move one carbon atom off the C3 position (middle carbon atom) on the linear chain: \(CH_3-CH_2-CH(CH_3)-CH_2-CH_3\) Finally, move one carbon atom off the C4 position (second carbon atom from another end) on the linear chain: \(CH_3-CH_2-CH_2-CH(CH_3)-CH_3\) Note that adding branches to the C1 or C5 (end) positions does not generate new isomers, as the structures would be identical to those already drawn.

Now we have found all four non-cyclic structural isomers of C5H10, let's list them below: 1. Linear isomer: \(CH_3-CH_2-CH_2-CH_2-CH_3\) 2. First branched isomer: \(CH_3-CH(CH_3)-CH_2-CH_2-CH_3\) 3. Second branched isomer: \(CH_3-CH_2-CH(CH_3)-CH_2-CH_3\) 4. Third branched isomer: \(CH_3-CH_2-CH_2-CH(CH_3)-CH_3\) These are all the non-cyclic structural isomers of C5H10.