Question

Gluconolactone and Glucose Oxidation States The cyclic glucose derivative 6-phosphogluconolactone is an intermediate in the pentose phosphate pathway (discussed in Chapter 14). Compare the oxidation state of \(C-1\) for the cyclic form of both gluconolactone and \(\beta\)-D-glucose.

Short answer

C-1 in glucose has an oxidation state of 0, and in gluconolactone, it is +2.

Step-by-step solution

Identify the Functional Group

For glucose and gluconolactone, identify the functional groups at \( C-1 \). In \( \beta \)-D-glucose, \( C-1 \) has an alcohol group (\(-OH\)). In gluconolactone, \( C-1 \) is part of a lactone (cyclic ester) with a carbonyl group (\(C=O\)).

Determine the Oxidation State Rules

Recall that for organic molecules, the oxidation state can be estimated by assigning \(-1\) for each bond to hydrogen, \(+1\) for each bond to oxygen or more electronegative atoms, and \(0\) for bonds to carbon.

Calculate the Oxidation State of C-1 in Glucose

In \( \beta \)-D-glucose, \(C-1\) is bonded to one oxygen (hydroxyl group) and two carbon atoms, plus another hydrogen. The oxidation state is \(-1 \) for hydrogen and \(+1 \) for the hydroxyl bond, making the oxidation state \(0\).

Calculate the Oxidation State of C-1 in Gluconolactone

In gluconolactone, \(C-1\) is doubly bonded to oxygen (carbonyl group), bonded to one other oxygen through the ester linkage, and to two carbons. The oxidation state is \(+2 \) for the carbonyl and the ester bond each contributing \(+1\) to the oxidation state, resulting in an overall oxidation state of \(+2\).

Compare the Results

\(C-1\) in \( \beta \)-D-glucose has an oxidation state of \(0\), while in gluconolactone, \(C-1\) has an oxidation state of \(+2\). This shows that gluconolactone is in a more oxidized form compared to glucose.

Key concepts

Oxidation States

Oxidation states help us understand the electron distribution around a specific carbon atom in a molecule, such as glucose or gluconolactone. To determine the oxidation state, we use a simple set of rules:

• A bond to hydrogen adds -1.
• A bond to oxygen or a more electronegative atom adds +1.
• Bonds to carbon don’t change the oxidation state.

For the cyclic glucose derivative, gluconolactone, understanding these oxidation states is crucial because they indicate how oxidized or reduced a carbon atom is.
When we compare the oxidation states of the carbon atoms involved in the pentose phosphate pathway, particularly at position C-1, we see a transformation from alcohol to carbonyl. This shift reflects the molecule's oxidative status, where glucose is less oxidized than its derivative, gluconolactone. Glucose's C-1 involvement in a hydroxyl group results in an oxidation state calculated as 0, whereas gluconolactone's C-1, dominated by carbonyl features, results in an oxidation state of +2. This is a sophisticated yet understandable way of saying that gluconolactone is more oxidized than glucose.

Gluconolactone

Gluconolactone is a fascinating intermediate found in the pentose phosphate pathway. It forms when glucose-6-phosphate undergoes oxidative decarboxylation. This reaction captures essential oxidation-reduction steps, facilitating cellular biochemical processes.
In its structure, gluconolactone features a lactone ring—more specifically, a cyclic ester. Here, C-1 participates in forming this ring by creating a bond with the oxygen atom, resulting in an ester linkage.

• This ester linkage involves a carbonyl group, a key player in the oxidation state determination.
• It is essential for the structural stability of the molecule in cellular contexts.

Gluconolactone's higher oxidation state relative to glucose showcases how its formation is an oxidative step in the pentose phosphate pathway. This pathway plays a significant role in producing NADPH, a reducing agent necessary for various biosynthetic reactions, and ribose-5-phosphate, critical for nucleotide synthesis.

Cyclic Glucose Derivatives

Cyclic glucose derivatives form when the linear glucose molecule rearranges and cyclizes. These derivatives, such as gluconolactone, play significant roles in metabolic pathways.

• Cyclization involves forming a new bond, typically between the carbonyl carbon and a hydroxyl oxygen. This action creates a ring structure.
• These compounds serve as key intermediates in various biochemical pathways.

For example, 6-phosphogluconolactone is one such cyclic derivative, integral to the pentose phosphate pathway. These cyclic forms allow crucial metabolic intermediates to be transported and modified more efficiently within cells.
The cyclic structure is not just structural; it influences reactivity and the type of biochemical pathways the molecule can enter. It also ensures that certain metabolic processes occur seamlessly, emphasizing the importance of cyclic derivatives in life's complex chemistry.