Total Number of Pi Electrons in Major Product Z

To determine the total number of $\pi$ electrons in the major product $z$, we analyze the sequence of organic reactions step-by-step:

Step 1: Formation of Product $x$

Benzene undergoes the Gattermann-Koch reaction when treated with carbon monoxide ($\text{CO}$) and hydrogen chloride ($\text{HCl}$) in the presence of cuprous chloride ($\text{CuCl}$): $\text{C}_6\text{H}_6 + \text{CO} + \text{HCl} \xrightarrow{\text{CuCl}} \text{C}_6\text{H}_5\text{CHO} + \text{HCl}$

Thus, product $x$ is Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$).

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Step 2: Formation of Product $y$

Benzene undergoes Friedel-Crafts acylation when reacted with ethanoyl chloride ($\text{CH}_3\text{COCl}$) in the presence of anhydrous $\text{AlCl}_3$: $\text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow{\text{Anhydrous }\text{AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl}$

Thus, product $y$ is Acetophenone ($\text{C}_6\text{H}_5\text{COCH}_3$).

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Step 3: Formation of Major Product $z$

When benzaldehyde ($x$) and acetophenone ($y$) are heated in the presence of alkali ($\text{OH}^-$), they undergo a Claisen-Schmidt condensation (a type of crossed aldol condensation followed by dehydration):

• Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$) has no $\alpha$-hydrogens.
• Acetophenone ($\text{C}_6\text{H}_5\text{COCH}_3$) has three $\alpha$-hydrogens.

The reaction proceeds as follows: $\text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{COC}_6\text{H}_5 \xrightarrow[\Delta]{\text{OH}^-} \text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{C}(=\text{O})-\text{C}_6\text{H}_5 + \text{H}_2\text{O}$

Thus, the major product $z$ is 1,3-diphenylprop-2-en-1-one (Chalcone).

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Step 4: Counting the Total Number of $\pi$ Electrons in Product $z$

The molecular structure of chalcone ($z$) is: $\text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{C}(=\text{O})-\text{C}_6\text{H}_5$

Let's count the number of $\pi$ bonds in product $z$:

1. First benzene ring ($\text{C}_6\text{H}_5$): Contains $3$ $\pi$ bonds $\rightarrow 6$ $\pi$ electrons.
2. Second benzene ring ($\text{C}_6\text{H}_5$): Contains $3$ $\pi$ bonds $\rightarrow 6$ $\pi$ electrons.
3. Alkene double bond ($\text{C}=\text{C}$): Contains $1$ $\pi$ bond $\rightarrow 2$ $\pi$ electrons.
4. Carbonyl group ($\text{C}=\text{O}$): Contains $1$ $\pi$ bond $\rightarrow 2$ $\pi$ electrons.

$\text{Total number of }\pi\text{ bonds} = 3 + 3 + 1 + 1 = 8$

Since each $\pi$ bond consists of $2$ $\pi$ electrons: $\text{Total number of }\pi\text{ electrons} = 8 \times 2 = 16$