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

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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 $"H"_3"C-CH=CH-C≡CH"$ is

pent-3-en-1-yne

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

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

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

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

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