Consider the following reaction sequence: Benzamide (i) Br ₂ / NaOH P (ii) NaNO ₂ / HCl, 0-5^ C Q (iii) H ₃ PO ₂, H ₂ O R (iv) CH ₃ Cl, anhydrous AlCl ₃ S The final major product (S) is:

Step-by-step solution

Step (i): Benzamide undergoes Hoffmann bromamide degradation with Br ₂ and NaOH to form aniline (P). C ₆ H ₅ CONH ₂ Br ₂ / NaOH C ₆ H ₅ NH ₂ Step (ii): Aniline reacts with NaNO ₂ and HCl at 0-5^ C to form benzenediazonium chloride (Q). C ₆ H ₅ NH ₂ NaNO ₂ / HCl C ₆ H ₅ N ₂^+ Cl ^- Step (iii): Benzenediazonium chloride is reduced by hypophosphorous acid ( H ₃ PO ₂ ) in the presence of water to form benzene (R). C ₆ H ₅ N ₂^+ Cl ^- H ₃ PO ₂, H ₂ O C ₆ H ₆ Step (iv): Benzene undergoes Friedel-Crafts alkylation with me