Question

Draw the delocalized molecular orbitals for the following molecule. Are both \(\pi\) bonds of the triple bond involved in the delocalized orbitals? $$ \mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} $$

Short answer

Answer: No, both π bonds of the triple bond are not involved in the delocalized orbitals of the given molecule. Only one of the π bonds from the triple bond and the π bond from the double bond participate in the delocalized orbitals.

Step-by-step solution

Identify hybridization of the atoms

First, let's consider the molecule given: $$ \mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} $$ We will identify the hybridization of the central atoms in the molecule. For Carbon atom 1, we have: - 3 single bonds (to 3 Hydrogen atoms) The hybridization is \(\mathrm{sp^2}\) since there are 3 electron groups around the Carbon atom. For Carbon atom 2, we have: - 1 single bond (to Carbon atom 1) - 1 triple bond (to Carbon atom 3) The hybridization is \(\mathrm{sp}\) since there are 2 electron groups around the Carbon atom. For Carbon atom 3, we have: - 1 triple bond (to Carbon atom 2) - 1 single bond (to Carbon atom 4) The hybridization is \(\mathrm{sp}\) since there are 2 electron groups around the Carbon atom. For Carbon atom 4, we have: - 2 single bonds (to Carbon atom 3 and one Hydrogen atom) - 1 double bond (to Carbon atom 5) The hybridization is \(\mathrm{sp^2}\) since there are 3 electron groups around the Carbon atom. For Carbon atom 5, we have: - 1 double bond (to Carbon atom 4) - 2 single bonds (to two Hydrogen atoms) The hybridization is \(\mathrm{sp^2}\) since there are 3 electron groups around the Carbon atom.

Identify the molecular orbitals involved in delocalization

Now we will identify the molecular orbitals involved in the delocalization. This molecule contains conjugated double and triple bonds, meaning there are continuous overlapping p-orbitals. The conjugation of p-orbitals over the entire molecule forms delocalized molecular orbitals. The \(\pi\)-bonds of the double and triple bonds participate in the delocalization. The triple bond between Carbon atom 2 and Carbon atom 3 has two \(\pi\) bonds. The double bond between Carbon atom 4 and Carbon atom 5 has one \(\pi\) bond. However, one of the \(\pi\) bonds in the triple bond between Carbon atom 2 and Carbon atom 3 is required to maintain the triple bond and it cannot participate in the delocalization, leaving only one \(\pi\) bond from the triple bond and the \(\pi\) bond from the double bond to participate in the delocalization.

Draw the delocalized molecular orbitals

Drawing the delocalized molecular orbitals: - The continuous overlapping p-orbitals form delocalized molecular orbitals involving Carbon atoms 2, 3, 4, and 5. $$ \mathrm{C}_{2}-\mathrm{C}_{3}-\mathrm{C}_{4}=\mathrm{C}_{5}. $$ - Carbon atom 1 maintains its \(\mathrm{sp^2}\) hybridization and is not part of the conjugation. - Carbon atom 2 and Carbon atom 3 are both \(\mathrm{sp}\) hybridized and are involved in the delocalization. - Carbon atom 4 and Carbon atom 5 are both \(\mathrm{sp^2}\) hybridized and are involved in the delocalization. In conclusion, both π bonds of the triple bond are not involved in the delocalized orbitals, as one of them is required to maintain the triple bond. Only one π bond from the triple bond and the π bond from the double bond participate in the delocalized orbitals.