The ratio of number of \(sp^{3}\) hybrid orbitals to number of \(sp^{2}\) hybrid orbitals in the major product (Z) of the given reaction sequence is
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An \(sp^3\) hybridized carbon atom always provides 4 hybrid orbitals, while an \(sp^2\) hybridized carbon atom provides 3 hybrid orbitals; always verify the number of atoms of each hybridization type in your final structure.
Step 1: Follow the reaction chain. Start with calcium carbide. Water gives acetylene: $CaC_2+2H_2O\rightarrow C_2H_2$. Step 2: Make the ring. Acetylene passed through a red hot iron tube joins three molecules into benzene (this is product Y). Step 3: Add the side group. Friedel-Crafts alkylation with anhydrous $AlCl_3$ attaches an alkyl group to the ring, giving the final product Z. Step 4: Sort the carbons by hybridisation. The six ring carbons are flat and $sp^2$. The carbons of the added alkyl chain are $sp^3$. Step 5: Count the hybrid orbitals. Counting the $sp^3$ orbitals and the $sp^2$ orbitals in product Z and simplifying gives the ratio. Step 6: State the ratio. \[ \boxed{sp^3:sp^2=2:3} \]
