Question

Question #e4dc9

Answer 1

Here are several possibilities.

Explanation:

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

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

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 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 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 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 has 6 `"C-H"` σ bonds, 7 `"C-C"` σ bonds, and 0 `"C-C"` π bonds (13 σ + 0 π).