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Chapter 22: Problem 106

Glucose can occur in three forms: two cyclic forms and one open-chain structure. In aqueous solution, only a tiny fraction of the glucose is in the open-chain form. Yet tests for the presence of glucose depend on reaction with the aldehyde group, which is found only in the open-chain form. Explain why these tests work.

Short Answer

Expert verified
In summary, glucose tests work because they depend on reactions with the aldehyde group in the open-chain form of glucose. Despite the open-chain form comprising a nominal fraction in an aqueous solution, the equilibrium between glucose forms ensures enough of the molecule is in the open-chain form to be detected by these tests.

Step by step solution

01

Introduction to glucose forms

Glucose is a monosaccharide sugar that has the molecular formula C6H12O6. It can exist in two cyclic forms (alpha and beta-D-glucose) and one open-chain structure. These forms are in equilibrium with each other in an aqueous solution, but the cyclic forms are dominant because they are more stable.
02

Open-chain glucose and the aldehyde group

In the open-chain form of glucose, one end of the molecule has an aldehyde group (-CHO). This aldehyde group is absent in the cyclic forms of glucose. The open-chain form of glucose comprises a marginal fraction in aqueous solutions.
03

Glucose tests and aldehyde group reaction

Glucose tests typically rely on the reaction of the aldehyde group. Examples of these tests include Fehling's test, Benedict's test, and Tollens' test. The positive result of these tests confirms the presence of the aldehyde group, indicating glucose in the open-chain form.
04

Equilibrium between glucose forms

Though the open-chain glucose form is a tiny fraction in an aqueous solution, the cyclic and open-chain glucose forms are in equilibrium. When some open-chain glucose reacts with the test reagents, more cyclic glucose gets converted into the open-chain form to maintain the equilibrium. As a result, a sufficient amount of the open-chain form of glucose is available to react with the test reagents, proving the presence of glucose in the sample.
05

Conclusion

In summary, glucose tests work because they depend on reactions with the aldehyde group in the open-chain form of glucose. Despite the open-chain form comprising a nominal fraction in an aqueous solution, the equilibrium between glucose forms ensures enough of the molecule is in the open-chain form to be detected by these tests.

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Most popular questions from this chapter

Alcohols are very useful starting materials for the production of many different compounds. The following conversions, starting with 1-butanol, can be carried out in two or more steps. Show the steps (reactants/catalysts) you would follow to carry out the conversions, drawing the formula for the organic product in each step. For each step, a major product must be produced. (See Exercise \(68 . )\) (Hint: In the presence of \(\mathrm{H}^{+},\) an alcohol is converted into an alkene and water. This is the exact reverse of the reaction of adding water to an alkene to form an alcohol.) $$ \begin{array}{l}{\text { a. } 1 \text { -butanol } \longrightarrow \text { butane }} \\ {\text { b. } 1 \text { -butanol } \longrightarrow 2 \text { -butanone }}\end{array} $$

If one hydrogen in a hydrocarbon is replaced by a halogen atom, the number of isomers that exist for the substituted compound depends on the number of types of hydrogen in the original hydrocarbon. Thus there is only one form of chloroethane (all hydrogens in ethane are equivalent), but there are two isomers of propane that arise from the substitution of a methyl hydrogen or a methylene hydrogen. How many isomers can be obtained when one hydrogen in each of the compounds named below is replaced by a chlorine atom? $$ \begin{array}{ll}{\text { a. } n \text { -pentane }} & {\text { c. } 2,4 \text { -dimethylpentane }} \\ {\text { b. } 2 \text { -methylbutane }} & {\text { d. methylcyclobutane }}\end{array} $$

Is the primary, secondary, or tertiary structure of a protein changed by denaturation?

Why is it preferable to produce chloroethane by the reaction of \(\mathrm{HCl}(g)\) with ethene than by the reaction of \(\mathrm{Cl}_{2}(g)\) with ethane? (See Exercise \(68 . )\)

Estimate \(\Delta H\) for the following reactions using bond energies given in Table \(8.5 .\) $$ 3 \mathrm{CH}_{2}=\mathrm{CH}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 3 \mathrm{CH}_{3}-\mathrm{CH}_{3}(g) $$ The enthalpies of formation for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) and \(\mathrm{C}_{6} \mathrm{H}_{12}(g)\) are 82.9 and $-90.3 \mathrm{kJ} / \mathrm{mol}\( , respectively. Calculate \)\Delta H^{\circ}$ for the two reactions using standard enthalpies of formation from Appendix \( 4 .\) Account for any differences between the results obtained from the two methods.
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