If glucose is heated with concentrated tetraoxosulphate(vi)acid, it will be dehydrated to?

Analysis

The question asks for the product formed when glucose ($C_6H_{12}O_6$) undergoes a dehydration reaction in the presence of concentrated tetraoxosulphate(VI) acid ($H_2SO_4$).

Concept Explanation

Concentrated tetraoxosulphate(VI) acid is a powerful dehydrating agent. In chemistry, dehydration refers to the removal of water molecules ($H_2O$) from a compound.

Glucose is a carbohydrate, which literally means "hydrated carbon" ($C_n(H_2O)_m$). When concentrated $H_2SO_4$ is added to glucose, it strips away the hydrogen and oxygen atoms in the ratio of 2:1 (the ratio found in water), leaving behind the carbon atoms.

Step-by-Step Solution

1. Identify the chemical formula of glucose: $C_6H_{12}O_6$.
2. Understand the role of the reagent: Concentrated $H_2SO_4$ acts as a dehydrating agent, meaning it removes water ($H_2O$) from the molecule.
3. Write the chemical equation: $$C_6H_{12}O_6 \xrightarrow{\text{conc. } H_2SO_4} 6C + 6H_2O$$
4. Observe the products: The glucose molecule is broken down into 6 atoms of carbon (a black, porous solid) and 6 molecules of water.

Conclusion

Because the acid removes all the hydrogen and oxygen as water, only the carbon remains. Therefore, glucose is dehydrated to carbon.

Exam Tip

Whenever you see "concentrated tetraoxosulphate(VI) acid" acting on carbohydrates like sugar or glucose, always associate it with the production of carbon (which appears as a black charred mass). This is a classic dehydration reaction frequently tested in JAMB.