Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Which one of the following does not have ${\text{sp}}^2$ hybridised carbon?

Have you ever wondered how chemists can look at a complex molecule and instantly know its 3D shape? The secret lies in a concept called hybridization. In this problem, we are on a mission to find the odd one out: a molecule that completely lacks an ${\text{sp}}^2$ hybridized carbon atom.

To solve this, we don't need to memorize complex tables. We just need to count! Specifically, we need to count the number of sigma ($\sigma$) bonds each carbon atom forms.

Carbon is a versatile element that loves to form four bonds. Depending on how it distributes these bonds, its hybridization changes:

4 Single Bonds: If carbon forms four single ($\sigma$) bonds, it needs four hybrid orbitals. This is ${\text{sp}}^3$ hybridization (tetrahedral geometry). 1 Double Bond: A double bond consists of one $\sigma$ bond and one $\pi$ bond. If carbon has one double bond and two single bonds, it forms a total of three $\sigma$ bonds. This requires three hybrid orbitals, leading to ${\text{sp}}^2$ hybridization (trigonal planar geometry). 1 Triple Bond (or 2 Double Bonds): A triple bond has one $\sigma$ bond and two $\pi$ bonds. If carbon has one triple bond and one single bond, it forms only two $\sigma$ bonds. This means it's $\text{sp}$ hybridized* (linear geometry).

Let's apply this golden rule to our suspects.

1. Acetone (${\text{CH}}_3-\text{C}(=\text{O})-{\text{CH}}_3$) Look at the central carbon. It is double-bonded to an oxygen atom. This means it forms three $\sigma$ bonds (two with the methyl groups, one with oxygen) and one $\pi$ bond. Three $\sigma$ bonds mean it is ${\text{sp}}^2$ hybridized.

2. Acetic Acid (${\text{CH}}_3-\text{C}(=\text{O})-\text{OH}$) Similar to acetone, the central carbon is part of a carbonyl group ($\text{C}=\text{O}$). It forms three $\sigma$ bonds and one $\pi$ bond. Therefore, this carbon is also ${\text{sp}}^2$ hybridized.

3. Acetamide (${\text{CH}}_3-\text{C}(=\text{O})-{\text{NH}}_2$) Once again, we spot the familiar carbonyl group. The carbon is double-bonded to oxygen, forming three $\sigma$ bonds in total. It is firmly ${\text{sp}}^2$ hybridized.

4. Acetonitrile (${\text{CH}}_3-\text{C}\equiv \text{N}$) Now, let's look at acetonitrile. The central carbon is bonded to a nitrogen atom via a triple bond. A triple bond consists of one $\sigma$ bond and two $\pi$ bonds. The carbon also has a single $\sigma$ bond connecting it to the methyl group.

In total, this carbon forms only two $\sigma$ bonds. According to our golden rule, two $\sigma$ bonds mean the carbon is $\text{sp}$ hybridized. The other carbon in the methyl group has four single bonds, making it ${\text{sp}}^3$ hybridized.

We have found our culprit! Acetonitrile contains one ${\text{sp}}^3$ carbon and one $\text{sp}$ carbon. It is the only molecule in the list that does not have an ${\text{sp}}^2$ hybridized carbon.

This problem teaches us a valuable shortcut: functional groups are dead giveaways for hybridization. Carbonyls ($\text{C}=\text{O}$) are always ${\text{sp}}^2$, while nitriles ($\text{C}\equiv \text{N}$) are always $\text{sp}$. Keep this in your mental toolkit, and you'll be identifying hybridizations in seconds!