Question

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

• A. 3:5
• B. 3:2
• C. 2:3
• D. 3:4

Hint:

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.

Answer 1

The Correct Option is C

Concept: The reaction sequence involves the synthesis of aromatic compounds starting from Calcium Carbide (\(CaC_2\)).

Step 1: Identify the reaction path.
\(CaC_2 + H_2O \rightarrow C_2H_2\) (acetylene), which is then passed through a red-hot iron tube to form Benzene (\(Y\)). Alkylation in the presence of Anhydrous \(AlCl_3\) yields the final product (\(Z\)).

Step 2: Analyze hybridization.
In the product Benzene derivative, the ring carbons are \(sp^2\) hybridized (each providing 3 \(sp^2\) hybrid orbitals), and the alkyl group carbons are \(sp^3\) hybridized (each providing 4 \(sp^3\) hybrid orbitals).

Step 3: Count the orbitals and find the ratio.
By calculating the specific total number of \(sp^3\) hybrid orbitals relative to the total number of \(sp^2\) hybrid orbitals present in the major product (\(Z\)), the simplified ratio is determined to be 2:3.