How do you calculate the number of sigma pi bonds in $H_{3} C-CH=CH-C\equivCH$ $"H"_3"C-CH=CH-C≡CH"$ ?

Question

13576 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-08T22:07:06

Here's how you do it.

Every bond between two atoms contains 1 σ bond.

Every bond after the first bond is a π bond.

The expanded structural formula of $H_{3} C-CH=CH-C\equivCH$ $"H"_3"C-CH=CH-C≡CH"$ is

pent-3-en-1-yne

We count $6 C-H$ $"6 C-H"$ bonds and $4 C-C$ $"4 C-C"$ bonds, for a total of $10 σ$ $"10 σ"$ bonds.

The bond between $C1$ $"C1"$ and $C2$ $"C2"$ is a triple bond, so the two "extra" bonds are π bonds.

The bond between $C3$ $"C3"$ and $C4$ $"C4"$ is a double bond, so the "extra" bond is a π bond.

Hence, pent-3-en-1-yne contains $10 σ bonds$ $"10 σ bonds"$ and $3 π bonds$ $"3 π bonds"$ .

How do you calculate the number of sigma pi bonds in H3C-CH=CH-C≡CH"H"_3"C-CH=CH-C≡CH"?