Question

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

Short answer

The structural isomers of \(C_{5}H_{10}\) without any cyclic structures are: 1. Pentene: \(CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}\) 2. 2-methylbutene: \(CH_{3}CH(CH_{3})CH_{2}CH_{2}\) 3. 3-methylbutene: \(CH_{3}CH_{2}CH(CH_{3})CH_{2}\)

Step-by-step solution

Identify the Basic Structure of C5H10

The molecular formula of C5H10 consists of 5 carbon atoms and 10 hydrogen atoms. Since the molecule isn't cyclic, we can start with a straight-chain structure, and then create branches at different positions, ensuring that every carbon atom has 4 bonds and every hydrogen atom has 1 bond.

Draw the Straight-Chain Structure

Draw a straight chain of 5 carbon atoms. Each carbon atom should have 2 hydrogen atoms attached to it, except for the first and the last carbon atoms in the chain, which have 3 hydrogen atoms. This straight-chain structure is called pentene and has the following structure: \(CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}\).

Create Branches

Now, we will create branches on the carbon atoms (except for the first and the last carbon atoms, since creating a branch there would result in the same structure). While creating a branch, we need to remove one hydrogen atom from the carbon atom we are branching from and then add a new carbon atom to maintain the molecular formula of C5H10. 1. Create a branch at the second carbon atom, which results in a structure called 2-methylbutene. The structure will be: \(CH_{3}CH(CH_{3})CH_{2}CH_{2}\). 2. Create a branch at the third carbon atom, which results in a structure called 3-methylbutene. The structure will be: \(CH_{3}CH_{2}CH(CH_{3})CH_{2}\). These are all the possible structural isomers of C5H10 without any cyclic structures.