Toluene is oxidised to benzoic acid by:
A. Acidified \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]
B. Acidified \[{{\rm{K}}_{\rm{2}}}{\rm{C}}{{\rm{r}}_{\rm{2}}}{{\rm{O}}_{\rm{7}}}\]
C. \[{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\]
D. Both A and B
Answer
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Hint: The oxidation reaction defines the addition of oxygen atoms to a compound. The organic compounds that help in the oxidation reaction are termed oxidising agents. Some of the oxidising agents are \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\] , \[{{\rm{K}}_{\rm{2}}}{\rm{C}}{{\rm{r}}_{\rm{2}}}{{\rm{O}}_{\rm{7}}}\]etc.
Complete step by step answer:
Let's discuss toluene first. The chemical formula for toluene is \[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}}\] . Now, we have to find out the compound which oxidize toluene (\[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}}\]) to benzoic acid (\[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}} - {\rm{COOH}}\]). Let's discuss all the options one by one.
We know that, acidified potassium permanganate (\[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]) is an oxidising agent of strong nature. So, it gives benzoic acid when toluene undergoes a reaction with it. So, option A is right.
Also, acidified potassium dichromate (\[{{\rm{K}}_{\rm{2}}}{\rm{C}}{{\rm{r}}_{\rm{2}}}{{\rm{O}}_{\rm{7}}}\]) is also a strong oxidising agent. That means, it also gives the product of benzoic acid when it undergoes a reaction with toluene. So, option B is also right.
When toluene undergoes a reaction with sulphuric acid (\[{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\]), the formation of ortho and para toluene sulphuric acid takes place. So, it does not give benzoic acid. So, option C is wrong.
Therefore, both acidified \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]and \[{{\rm{K}}_{\rm{2}}}{\rm{C}}{{\rm{r}}_{\rm{2}}}{{\rm{O}}_{\rm{7}}}\]gives product of benzoic acid on reaction with toluene.
Hence, option D is correct, that is, Both A and B.
Note: Toluene is a liquid of no colour. At room temperature, it turns vapour on, exposing it to air. The odour of the vapour of toluene is very sharp. Its use in the chemical industry is in the form of solvents and paint pigments.
Complete step by step answer:
Let's discuss toluene first. The chemical formula for toluene is \[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}}\] . Now, we have to find out the compound which oxidize toluene (\[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}}\]) to benzoic acid (\[{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}} - {\rm{COOH}}\]). Let's discuss all the options one by one.
We know that, acidified potassium permanganate (\[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]) is an oxidising agent of strong nature. So, it gives benzoic acid when toluene undergoes a reaction with it. So, option A is right.
Also, acidified potassium dichromate (\[{{\rm{K}}_{\rm{2}}}{\rm{C}}{{\rm{r}}_{\rm{2}}}{{\rm{O}}_{\rm{7}}}\]) is also a strong oxidising agent. That means, it also gives the product of benzoic acid when it undergoes a reaction with toluene. So, option B is also right.
When toluene undergoes a reaction with sulphuric acid (\[{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\]), the formation of ortho and para toluene sulphuric acid takes place. So, it does not give benzoic acid. So, option C is wrong.
Therefore, both acidified \[{\rm{KMn}}{{\rm{O}}_{\rm{4}}}\]and \[{{\rm{K}}_{\rm{2}}}{\rm{C}}{{\rm{r}}_{\rm{2}}}{{\rm{O}}_{\rm{7}}}\]gives product of benzoic acid on reaction with toluene.
Hence, option D is correct, that is, Both A and B.
Note: Toluene is a liquid of no colour. At room temperature, it turns vapour on, exposing it to air. The odour of the vapour of toluene is very sharp. Its use in the chemical industry is in the form of solvents and paint pigments.
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