Problem 37 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

The structural isomers of C$_{5}$H$_{10}$, excluding cyclic isomers, are: 1. \(H_{2}C=CH-CH_{2}-CH_{2}-CH_{3}\) 2. \(CH_{3}-CH=CH-CH_{2}-CH_{3}\) 3. \((CH_{3})_{2}C=CH-CH_{3}\) 4. \(CH_{3}-CH_{2}-CH=CH-CH_{3}\)

Step 1: Identify the parent chain

The parent chain consists of 5 carbon atoms in a straight chain. The simplest structural isomer would have the double bond between the first and second carbon. Let's draw the isomer. H H H H H | | | | | H-C-C=C-H | | | | | H H H H H

Step 2: Move the double bond along the parent chain

Now, we need to move the double bond along the parent chain. We can place the double bond between the second and third carbon atoms. H H H H H | | | | | H-C-C=C-C-H | | | | | H H H H H At this point, we cannot move the double bond further down the chain, since doing so would result in a duplicate structure with mirror image.

Step 3: Create a branched chain

We can now create a branched chain by adding a methyl group to the parent chain. The methyl group can be added to the second carbon atom in the chain. H H H | | | H-C-C-H | C=C-H | H

Step 4: Move the double bond along the branched chain

We can now move the double bond along the branched chain to create another isomer. The double bond can be moved between the second and third carbon atoms in the branched chain. H H H H | | | | H-C-C=C | | H H | C-H | H We have now found all the structural isomers of C5H10, excluding cyclic isomers: 1. H₂C=CH-CH₂-CH₂-CH₃ 2. CH₃-CH=CH-CH₂-CH₃ 3. (CH₃)₂C=CH-CH₃ 4. CH₃-CH₂-CH=CH-CH₃