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Question: How many isomeric pentynes \(\left( {{C_5}{H_8}} \right)\) are possible: A. \(3\) B. \(4\) C. ...

How many isomeric pentynes (C5H8)\left( {{C_5}{H_8}} \right) are possible:
A. 33
B. 44
C. 55
D. 66
Given: The formula of the organic compound, pentynes is: (C5H8)\left( {{C_5}{H_8}} \right)

Explanation

Solution

We can write the structures of different isomers by using different carbon skeletons keeping the same formula.

Complete step by step solution:
We know that isomers are those compounds that have the same molecular formula but different properties. In organic compounds, we have basically two types of isomerism: structural isomers and stereo-isomers. Here, we will discuss the structural isomerism that arises with difference in the structure. We can further classify structural isomers as follows:
- Chain isomers: these have different carbon skeletons
- Position isomers: these have functional group/substituent positioned differently
- Functional group isomers: these have different functional groups
- Metamers: these have different alkyl chains at the sides of a functional group

We are given the organic compound, pentyne whose chemical formula is C5H8{C_5}{H_8}. let’s try to draw structural isomers with formula C5H8{C_5}{H_8} as follows:

As we can see that in pentyne, all the 55 carbons are attached in a straight chain and the triple bond is between the first and second carbon. This can also be called n-pentyne or pent−1−ynepent - 1 - yne. In the second isomer, pent−2−ynepent - 2 - yne, all the 55 carbons are attached in a straight chain but the triple bond is between second and third carbon. In the last isomer, 3−methylbut−1−yne{\rm{3 - methylbut - 1 - yne}} , only 44 carbons are attached in a straight chain, triple bond is between first and second carbon and there is one methyl group present as a substituent on the third carbon. So, all the three isomers have different structures but the molecular formula is the same for all of them.

**Hence, the correct option is A.

Note:**
We can see that positioning the triple bond between third and fourth or fourth and fifth carbons would have the same structure as that of pent−2−ynepent - 2 - yne and pent−1−ynepent - 1 - yne respectively.