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Question
The number of possible open chain (acyclic) isomeric compounds for molecular formula C5H10 would be
The number of possible open chain (acyclic) isomeric compounds for molecular formula C5H10 would be
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Solution
The correct option is C 6
The total number of acyclic or open chain isomeric compounds for molecular formula C5H10 are as follows:
1. CH2=CH−CH2−CH2−CH3
2. CH3−CH=CH−CH2−CH3 (cis)
3. CH3−CH=CH−CH2−CH3 (trans)
4. CH2=CH−CH(CH3)−CH3
5. CH2=C(CH3)−CH2−CH3
6. CH3−C(CH3)=CH−CH3
Hence, there are total of 6 isomers.
The total number of acyclic or open chain isomeric compounds for molecular formula C5H10 are as follows:
1. CH2=CH−CH2−CH2−CH3
2. CH3−CH=CH−CH2−CH3 (cis)
3. CH3−CH=CH−CH2−CH3 (trans)
4. CH2=CH−CH(CH3)−CH3
5. CH2=C(CH3)−CH2−CH3
6. CH3−C(CH3)=CH−CH3
Hence, there are total of 6 isomers.
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