The product of the below reaction is here
$ {{\text{C}}_{\text{11}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{+ conc}\text{.}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\to $
(A) $ \text{C}{{\text{O}}_{\text{2}}}\text{+ }{{\text{H}}_{\text{2}}}\text{O + glucose} $
(B) Carbon + water
(C) $ \text{fructose + }{{\text{H}}_{\text{2}}}\text{O + C}{{\text{O}}_{\text{2}}} $
(D) None of the above
Answer
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Hint: Concentrated sulphuric acid is a very strong oxidising agent and can extract water from carbohydrate molecules leaving the carbon behind in the form of charcoal, while the water molecules is/are taken up by the concentrated sulphuric acid.
Complete Step by Step Answer
When sucrose of sugar molecules reacts with concentrated sulphuric acid, then eleven water molecules are abstracted from the sucrose molecules leaving the carbon behind which is also called “sugar charcoal”.
The reaction can be shown as below:
$ {{\text{C}}_{\text{11}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{+ conc}\text{. }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\to \text{12 C + 11 }{{\text{H}}_{\text{2}}}\text{O + dilute }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} $
This reaction is highly exothermic in nature and the reaction releases a high amount of heat which vaporizes the water molecules formed in the medium which is trapped in the carbon molecules resulting in it to have a foamy structure. Along with the formation of carbon and water, certain amounts of carbon dioxide and sulphur dioxide gases are also liberated from the medium.
The experimental demonstration of the reaction involves the formation of a “carbon snake” which is a black snake like structure formed when the concentrated acid is poured over sucrose or sugar and the gases formed in the medium are trapped in the carbon structure.
Hence the correct answer is option B, carbon and water.
Note
The Pettenkofer’s Test utilizes the above reaction for the determination of Bile salt in urine. The urine sample to be tested is taken in a test tube and sucrose is added to it. Then a certain measured amount of sulphuric acid is also added to the sucrose containing urine solution. If the solution turns red then it indicates the presence of the Bile salt.
Complete Step by Step Answer
When sucrose of sugar molecules reacts with concentrated sulphuric acid, then eleven water molecules are abstracted from the sucrose molecules leaving the carbon behind which is also called “sugar charcoal”.
The reaction can be shown as below:
$ {{\text{C}}_{\text{11}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\text{+ conc}\text{. }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\to \text{12 C + 11 }{{\text{H}}_{\text{2}}}\text{O + dilute }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} $
This reaction is highly exothermic in nature and the reaction releases a high amount of heat which vaporizes the water molecules formed in the medium which is trapped in the carbon molecules resulting in it to have a foamy structure. Along with the formation of carbon and water, certain amounts of carbon dioxide and sulphur dioxide gases are also liberated from the medium.
The experimental demonstration of the reaction involves the formation of a “carbon snake” which is a black snake like structure formed when the concentrated acid is poured over sucrose or sugar and the gases formed in the medium are trapped in the carbon structure.
Hence the correct answer is option B, carbon and water.
Note
The Pettenkofer’s Test utilizes the above reaction for the determination of Bile salt in urine. The urine sample to be tested is taken in a test tube and sucrose is added to it. Then a certain measured amount of sulphuric acid is also added to the sucrose containing urine solution. If the solution turns red then it indicates the presence of the Bile salt.
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