Animated Solution for Chemistry - Organic Chemistry: Consider the following chemical reaction. $\text{CH}\equiv \text{CH}\underset{{\text{(2) CO, HCl, AlCl}}_3}{\overset{\text{(1) Red hot Fe tube, 873 K}}{\to }}\text{Product}$ The number of $s{p}^2$ hybridised carbon atom(s) present in the product is ....

Organic chemistry is often like a beautiful puzzle where simple building blocks are stitched together to form complex, elegant architectures. In this problem, we are given a two-step reaction sequence starting from a very simple molecule—acetylene—and we need to determine the hybridization state of the carbon atoms in the final product. Let's break down this journey step by step.

The Cyclic Trimerization of Acetylene

The first step of our reaction sequence involves passing acetylene gas ($HC\equiv CH$) through a red-hot iron tube at a high temperature of $873\text{ K}$.

What happens under these extreme conditions? The high thermal energy, coupled with the catalytic surface of the iron tube, causes the $\pi$ bonds of the acetylene molecules to break and rearrange. Three molecules of acetylene come together in a concerted mechanism known as cyclic trimerization.

Instead of remaining as isolated linear molecules, they join hands to form a six-membered ring with alternating single and double bonds. This newly formed molecule is Benzene ($C_6{H}_6$). The driving force for this reaction is the immense thermodynamic stability gained through aromaticity. The delocalized $\pi$ electron cloud in benzene makes it exceptionally stable compared to the highly reactive alkyne starting material.

The Gattermann-Koch Formylation

Now that we have our stable benzene ring, we move to the second step. The reagents provided are carbon monoxide ($CO$), hydrogen chloride ($HCl$), and anhydrous aluminum chloride ($AlC{l}_3$).

This specific combination of reagents is the hallmark of the Gattermann-Koch formylation. Benzene is an electron-rich aromatic ring, making it susceptible to electrophilic aromatic substitution. However, to substitute a formyl group ($-CHO$) onto the ring, we need a strong electrophile.

Here is where the magic happens: $CO$ and $HCl$ react in the presence of the Lewis acid catalyst, $AlC{l}_3$, to generate the highly reactive formyl cation ($[H-C=O{]}^{+}$). This cation acts as a potent electrophile. It attacks the $\pi$ electron cloud of the benzene ring, temporarily breaking its aromaticity to form a sigma complex. The complex then quickly loses a proton to restore its aromatic stability, resulting in the substitution of a hydrogen atom with the formyl group.

The final product of this sequence is Benzaldehyde ($C_6{H}_{5}CHO$).

The core question asks for the total number of $s{p}^2$ hybridized carbon atoms in our final product, Benzaldehyde. To find this, we need to look at the steric number of each carbon atom. The steric number is the sum of the number of sigma ($\sigma$) bonds and the number of lone pairs on the atom. For carbon, which rarely has lone pairs in stable neutral molecules, we simply count the sigma bonds.

1. The Benzene Ring Carbons: Let's examine the six carbon atoms that make up the aromatic ring. Each carbon atom in the benzene ring is bonded to three other atoms (two adjacent carbons and one hydrogen, or the formyl carbon in one case). Because each ring carbon forms one double bond (which consists of one $\sigma$ and one $\pi$ bond) and two single bonds (both $\sigma$ bonds), it has a total of three $\sigma$ bonds.

A steric number of 3 corresponds to $s{p}^2$ hybridization. Therefore, all 6 carbon atoms in the benzene ring are $s{p}^2$ hybridized.

2. The Aldehyde Carbon: Now, we must not forget the carbon atom in the newly attached formyl group ($-CHO$). This carbon is bonded to the benzene ring via a single bond, to a hydrogen atom via a single bond, and to an oxygen atom via a double bond.

Counting the sigma bonds: one to the ring, one to the hydrogen, and one to the oxygen (the second bond to oxygen is a $\pi$ bond). This gives the aldehyde carbon a total of three $\sigma$ bonds. Just like the ring carbons, a steric number of 3 means this carbon is also $s{p}^2$ hybridized.

Bringing it all together, we have: - 6 $s{p}^2$ hybridized carbons from the benzene ring. - 1 $s{p}^2$ hybridized carbon from the aldehyde group.

Adding them up, $6+1=7$.

There are exactly 7 $s{p}^2$ hybridized carbon atoms in Benzaldehyde. This problem beautifully tests both your knowledge of classic organic name reactions and your fundamental understanding of chemical bonding and hybridization.