The salt shown is treated with different reagents. Select all the correct statements.

A $CuBr$ / $HBr$ gives bromobenzene
B $KI$ gives iodobenzene without needing copper
C $H X_{3} PO X_{2}$ gives benzene
DHeating with $HBF X_{4}$ -derived salt gives benzonitrile

Answer: A, B, C

The salt shown is benzenediazonium chloride. CuBr/HBr is the Sandmeyer bromination; iodide displaces the diazonium group directly; hypophosphorous acid replaces it with hydrogen. The thermal decomposition of the tetrafluoroborate (Balz-Schiemann) gives fluorobenzene, not benzonitrile.

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