MCQ
Concentrated ${H_2}S{O_4}$ cannot be used to prepare $HBr$ from $NaBr$, because it
- AReduces $HBr$
- ✓Oxidises $HBr$
- CDisproportionates $HBr$
- DReacts slowly with $NaBr$
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$C{H_3}C{H_2}C{H_2}Br$ $\xrightarrow{{KOH(alc.)}}\,(A)\,\xrightarrow{{HBr}}(B)\,$ $\xrightarrow{{KOH(aq.)}}(C)$ The product $(C)$ is