Problem 102 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant of mutarotation of \(\alpha-D\)-glucose to \(\beta-D\)-glucose is \(1.8 .\) What per cent of the \(\alpha\)-form remains under equilibrium? (a) \(35.7\) (b) \(64.3\) (c) \(55.6\) (d) \(44.4\)

Step-by-Step Solution

Answer

The percentage of \(\alpha-D\)-glucose that remains at equilibrium is \(35.7\%\).

Step 1: Understand the Reaction

Mutarotation is the process where \(\alpha-D\)-glucose converts to \(\beta-D\)-glucose and vice versa. At equilibrium, both forms exist in fixed proportions dependent on the equilibrium constant \(K\). Given \(K = 1.8\), the equilibrium expression is: \(K = \frac{[\beta-D\text{-glucose}]}{[\alpha-D\text{-glucose}]}\).

Step 2: Express in Terms of Initial Concentrations

Assume the initial concentration of \(\alpha-D\)-glucose is \(C\), and \(x\) is the concentration of \(\alpha-D\)-glucose that converts to \(\beta-D\)-glucose. At equilibrium, \([\beta-D\text{-glucose}] = x\) and \([\alpha-D\text{-glucose}] = C - x\). Inserting into the equilibrium expression gives: \(1.8 = \frac{x}{C-x}\).

Step 3: Solve the Equilibrium Equation

Solve the equation \(1.8(C - x) = x\) for \(x\). Distribute and simplify to get \(1.8C - 1.8x = x\). Rearranging terms gives \(1.8C = 2.8x\), or \(x = \frac{1.8}{2.8}C\). Simplify \(x = \frac{9}{14}C\).

Step 4: Find Remaining \(\alpha-D\)-glucose

Find the concentration of \(\alpha-D\)-glucose at equilibrium, \(C - x = C - \frac{9}{14}C = \frac{5}{14}C\).

Step 5: Calculate Percentage Remaining

The percentage of \(\alpha-D\)-glucose remaining is \(\frac{5}{14} \times 100\% \). Calculating gives approximately \(35.7\%\).

Key Concepts

Mutarotation

Mutarotation refers to the spontaneous change in the optical rotation that occurs when alpha-glucose (\(\alpha-D\)-glucose) is dissolved in water and eventually reaches equilibrium with beta-glucose (\(\beta-D\)-glucose). This process is characterized by the interconversion between different anomers of glucose.
At the molecular level, mutarotation is explained by the ring-opening of the cyclic sugar, followed by the reformation of a ring that can close in either the \(\alpha\) or \(\beta\) configuration. The optical rotation changes as the ratio between these two forms changes until equilibrium is achieved.

• The initial rotation of \(\alpha\)-glucose gradually decreases while \(\beta\)-glucose gradually increases.
• Equilibrium is reached when the rates of conversion in both directions are equal and the optical rotation remains constant.

Understanding mutarotation is essential to comprehend how glucose solutions behave and why it is important in fields like biochemistry, where enzyme-substrate interactions might be sensitive to the specific form of glucose present.

Alpha and Beta Glucose

Alpha and Beta Glucose are two anomers of the molecule glucose. Anomers are isomers that differ in configuration only at the anomeric carbon.
In the case of glucose, the anomeric carbon is the carbon atom that becomes a stereocenter during ring closure.

• In \(\alpha-D\)-glucose, the hydroxyl group at the anomeric carbon is on the opposite side of the ring from the \(\text{CH}_2\text{OH}\) group.
• In \(\beta-D\)-glucose, the hydroxyl group at the anomeric carbon is on the same side as the \(\text{CH}_2\text{OH}\) group.

These two forms are conformational isomers and interconvert through the process of mutarotation as explained earlier. Knowing which form is predominant in a solution is crucial since, for instance, \(\beta-D\)-glucose is more stable in solution due to less steric hindrance between the atoms.

Equilibrium Constant

The equilibrium constant (\(K\)) is a fundamental concept in chemistry that quantifies the ratio of concentration of products and reactants at equilibrium. In the context of mutarotation, \(K\) is described as the ratio of the concentration of \(\beta-D\)-glucose to \(\alpha-D\)-glucose at equilibrium.
Using the formula: \[K = \frac{[\beta-D\text{-glucose}]}{[\alpha-D\text{-glucose}]}\]

• The equilibrium constant illustrates the extent to which reaction proceeds.
• A \(K\) value greater than 1 indicates that \(\beta-D\)-glucose is favored at equilibrium.

For example, in the given problem, \(K = 1.8\) shows that at equilibrium, more \(\beta\)-glucose is present than \(\alpha\)-glucose, which helps calculate the precise concentrations of each anomer in any given situation.

Percent Composition

Percent Composition is a way of expressing concentrations which indicates the relative amounts or proportions of substances. In chemical equilibrium, it refers to the proportion of a specific form (like \(\alpha-D\)-glucose) present compared to the total amount at equilibrium.
To compute the percentage of \(\alpha-D\)-glucose remaining, the formula used is:\[\% = \left(\frac{[\text{Remaining } \alpha-D\text{-glucose}]}{[\text{Initial }]\alpha-D\text{-glucose}}\right) \times 100\]

• The percent composition helps in understanding the dominance of one isomer over the other.
• In the exercise, the calculation ended with \(35.7\%\) \((\text{rounded})\) of \(\alpha\)-glucose present in equilibrium, indicating a significant conversion to \(\beta\)-glucose.

In practical terms, such calculations are useful in biochemistry to predict how glucose acts in reactions or in solution.

Glucose Isomers

Glucose occurs in several isomeric forms because of its ability to cyclize. These forms include different arrangements such as \(\alpha\) and \(\beta\) configurations or even linear vs. cyclic structures.
Isomeric diversity arises from the different ways atoms can be arranged in molecules while maintaining the same molecular formula.

• These variations can lead to differences in chemical behavior, reactivity, and biological roles.
• \(\alpha\)-glucose and \(\beta\)-glucose, for instance, are crucial substrates in cellular metabolism.

Recognizing the importance of isomers in glucose helps in fields such as food chemistry, pharmacology, and biochemistry. Identifying and understanding the behavior of these isomers allow scientists to manipulate and understand their role in larger biological systems, which is vital for developing treatments and studies related to energy use and storage in organisms.