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  4. The number of possible open chain (acyclic) isomeric compounds for molecular formula C5H10 would be

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Q. The number of possible open chain (acyclic) isomeric compounds for molecular formula $C_5H_{10}$ would be

Organic Chemistry – Some Basic Principles and Techniques

Solution:

The total number of acyclic or open chain isomeric compounds for molecular formula $C _{5} H _{10}$ are as follows:
1. $CH _{2}= CH - CH _{2}- CH _{2}- CH _{3}$
2. $CH _{3}- CH = CH - CH _{2}- CH _{3}$ (cis)
3. $CH _{3}- CH = CH - CH _{2}- CH _{3}$ (trans)
4. $CH _{2}= CH - CH \left( CH _{3}\right)- CH _{3}$
5. $CH _{2}= C \left( CH _{3}\right)- CH _{2}- CH _{3}$
6. $CH _{3}- C \left( CH _{3}\right)= CH - CH _{3}$
Hence, there are total of $6$ isomers.