The number of isomers that can be obtained from C$_4$H$_{10}$ is

Analysis

The question asks for the number of structural isomers for the alkane with the molecular formula C$_4$H$_{10}$ (butane). Isomers are compounds that have the same molecular formula but different structural arrangements of atoms.

Concept

For alkanes, isomers are formed by changing the carbon skeleton (chain isomerism). We start with the straight-chain structure and then try to create branched structures by moving carbon atoms to different positions.

Step-by-Step Solution

1. Identify the straight-chain isomer: The first structure is the continuous chain of four carbon atoms. This is called n-butane. Structure: $CH_3 - CH_2 - CH_2 - CH_3$

2. Attempt to create a branched isomer: To create a branch, we take one carbon atom from the main chain and attach it to an internal carbon atom. If we take one carbon from the end of the 4-carbon chain, we are left with a 3-carbon chain (propane) as the parent chain. We then attach the remaining carbon to the middle carbon of the propane chain. This is called 2-methylpropane (or isobutane). Structure: $CH_3 - CH(CH_3) - CH_3$

3. Check for further isomers: If we try to move the methyl group to the end carbon of the 3-carbon chain, we simply recreate the straight-chain n-butane. Therefore, no other unique structural isomers exist for C$_4$H$_{10}$.

Summary of isomers:

1. n-butane
2. 2-methylpropane

Total number of isomers = 2.

Conclusion

Comparing our result with the options provided, we can see that there are exactly 2 isomers.

Exam Tip

For alkanes, you can use a quick mental check:

• C$_1$ to C$_3$: Only 1 isomer each.
• C$_4$: 2 isomers.
• C$_5$: 3 isomers.
• C$_6$: 5 isomers. Remembering this sequence (1, 1, 1, 2, 3, 5) can save you valuable time during your JAMB exam!