A) \[Cr{{O}_{3}}\] in \[{{(C{{H}_{3}}CO)}_{2}}O\]
B) \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] + cone. \[{{H}_{2}}S{{O}_{4}}\]
C) Hot alkaline \[KMn{{O}_{4}}\]
D) Conc. \[HN{{O}_{3}}\]
Correct Answer: A
Solution :
\[{{k}_{2}}C{{r}_{2}}{{O}_{7}}+conc.\,{{H}_{2}}S{{O}_{4}},\]hot alkaline \[KMn{{O}_{4}}\] and cone. \[HN{{O}_{3}}\] will all oxidise toluene to benzoic acid while \[Cr{{O}_{3}}\] in \[{{(C{{H}_{3}}CO)}_{2}}O\] will oxidise toluene to only benzaldehyde.
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