Count the sigma and pi bonds in propene, CH₂=CH-CH₃, and in propyne, HC≡C-CH₃.
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Count the sigma and pi bonds in propene, , and in propyne, . Then do the same for but-1-en-3-yne, .
Need a reminder? The key ideaShow
A covalent bond forms when two half-filled orbitals overlap. Head-on overlap along the line of the nuclei gives a sigma bond. Sideways overlap of parallel p orbitals gives a pi bond, which is weaker. A single bond is one sigma bond, a double bond one sigma and one pi, and a triple bond one sigma and two pi. The bond enthalpy is the energy needed to break one mole of bonds into atoms.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Propene. Six C-H bonds and one C-C single bond, plus one C=C. Sigma: . Pi: .
Step 2. Propyne. Four C-H bonds and one C-C single bond, plus one triple bond. Sigma: . Pi: .
Step 3. But-1-en-3-yne. Four C-H bonds and one C-C single bond, one C=C and one triple bond. Sigma: . Pi: .
Answer: propene , propyne , but-1-en-3-yne .
Discussion (0)
Log in to join the discussion.
Loading…