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Answer 1 · 2016-08-25T23:28:55

Here are several possibilities.

The corresponding alkane would have the formula #"C"_5"H"_12#.

Each ring or double bond in the molecule removes 2 #"H"# atoms.

Our formula is missing 6 #"H"# atoms, so

∴ #"Rings + double bonds" = "12 - 6"/2 = 6/2 = 3#

There are six possible combinations.

I will give one example of each type.

3 Double bonds

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Penta-1,2,4-triene has 6 #"C-H"# σ bonds, 4 #"C-C"# σ bonds, and 3 #"C-C"# π bonds (10 σ + 3 π).

1 Triple bond + 1 double bond

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Pent-3-en-1-yne has 6 #"C-H"# σ bonds, 4 #"C-C"# σ bonds, and 3 #"C-C"# π bonds (10 σ + 3 π).

1 Ring + 1 triple bond

Ethynylcyclopropane

Ethynylcyclopropane has 6 #"C-H"# σ bonds, 5 #"C-C"# σ bonds, and 2 #"C-C"# π bonds (11 σ + 2 π).

1 Ring + 2 double bonds

Cyclopenta-1,3-diene

Cyclopenta-1,3-diene has 6 #"C-H"# σ bonds, 5 #"C-C"# σ bonds, and 2 #"C-C"# π bonds (11 σ + 2 π).

2 Rings + 1 double bond

Bicyclo[2.1.0]pent-2-ene

Bicyclo[2.1.0]pent-2-ene has 6 #"C-H"# σ bonds, 6 #"C-C"# σ bonds, and 1 #"C-C"# π bond (12 σ + 1 π).

Three rings

Tricyclo[1.1.1.0]pentane

Tricyclo[1.1.1.0]pentane has 6 #"C-H"# σ bonds, 7 #"C-C"# σ bonds, and 0 #"C-C"# π bonds (13 σ + 0 π).