MCQ
Conc. $H_2SO_4$ cannot be used to prepare $HBr$ from $NaBr$ because it
  • A
    reacts slowly with $NaBr$
  • B
    oxidises $HBr$
  • C
    reduces $HBr$
  • disproportionates $HBr$

Answer

Correct option: D.
disproportionates $HBr$
d
$\pi$ Bond formation through pure orbital

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free