MCQ
Conc. $H_2SO_4$ cannot be used to prepare $HBr$ from $NaBr$ because it
- Areacts slowly with $NaBr$
- Boxidises $HBr$
- Creduces $HBr$
- ✓disproportionates $HBr$
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$A.\,\,F -CH_2\,\,CH_2\,\,COOH$
$\begin{array}{*{20}{c}}
{B.\,\,Cl - CH - C{H_2} - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$C.\,\,F -CH_2 -COOH$
$D.\,\,Br -CH_2-CH_2 -COOH$
Correct answer is
Given :
Molar mass $N =14\,g\,mol ^{-1} ; O =16\,g\,mol ^{-1} ; C =12\,g\,mol ^{-1} ; H =1\,g\,mol ^{-1}$;

