Biomolecules · Lesson 2 of 12
Glucose: Deducing the Open-Chain Structure
“Use reactions, conservation of atoms and stereochemical conventions to build and evaluate the open-chain structure of D-glucose.”
• Infer glucose functional groups from specific chemical evidence. • Distinguish gluconic acid from glucaric acid. • Read a Fischer projection of D-glucose. • Separate D/L configuration from measured optical rotation. • Use hydrolysis stoichiometry in glucose preparation.
A structure must explain several independent observations
A molecular formula tells us which atoms are present but not how they connect. To build a structure for glucose, combine evidence from reactions that probe different parts of the molecule. A proposed formula must satisfy all observations together. One positive carbonyl test does not by itself prove whether the group is an aldehyde or ketone, and a carbon count does not specify stereochemistry.
Glucose occurs in fruits and honey and as units in starch and cellulose. The name dextrose refers to the rightward optical rotation of ordinary D-(+)-glucose. In the open-chain representation, number the aldehyde carbon C1 and the terminal primary alcohol carbon C6. The condensed connectivity is CHO–CH(OH)–CH(OH)–CH(OH)–CH(OH)–CH₂OH, often abbreviated CHO–(CHOH)₄–CH₂OH.
Preparing glucose by hydrolysis
Heating sucrose with dilute acid produces equimolar glucose and fructose. The acid catalyses cleavage and is not consumed in the net equation. This route gives a mixture, so merely hydrolysing sucrose does not produce pure glucose without separation. Industrial hydrolysis of starch provides glucose-rich solutions; one acid-hydrolysis route uses dilute sulfuric acid at 393 K under pressure, approximately 2–3 atm. Enzymatic routes also exist, but the chemistry here is cleavage of glycosidic bonds by water.
For the repeat-unit equation, each residue has effectively lost H₂O relative to free glucose. Adding water restores those atoms. For an exact isolated linear chain made from n glucose molecules, formation of n−1 links releases n−1 water molecules; hydrolysis consumes the same number. Keep these two bookkeeping conventions distinct.
Problem
What mass of glucose can be obtained from 34.2 g sucrose on complete hydrolysis? Use molar masses 342 g mol⁻¹ for sucrose and 180 g mol⁻¹ for glucose.
- 1.Convert sucrose mass to amount: 34.2/342 = 0.100 mol.
- 2.The balanced equation gives one mole of glucose per mole of sucrose, so glucose amount is 0.100 mol.
- 3.Mass of glucose = 0.100 × 180 = 18.0 g. Fructose also contributes 18.0 g. The total 36.0 g is larger than 34.2 g because 1.80 g water is incorporated, not because mass is created.
Six observations establish the carbon skeleton and functional groups
| Observation | Structural conclusion | Limit of the evidence |
|---|---|---|
| Molecular formula C₆H₁₂O₆ | Six carbons and six oxygens | Connectivity still unknown |
| Prolonged heating with HI gives n-hexane | Consistent with an unbranched six-carbon skeleton | Does not give OH stereochemistry |
| Hydroxylamine gives an oxime; HCN gives a cyanohydrin | An accessible carbonyl group exists | Aldehydes and ketones can both do this |
| Bromine water gives gluconic acid | The accessible carbonyl is aldehydic | Mild oxidation affects C1 here |
| Acetic anhydride gives a pentaacetate | Five alcohol-type OH groups can be acetylated | Does not alone specify the ring arrangement |
| Nitric acid gives glucaric acid | Both terminal groups become COOH | Includes oxidation of the terminal CH₂OH |
Hydroxylamine converts a carbonyl to C=N–OH, releasing water. HCN adds across C=O to form a carbon bearing OH and CN. In glucose these reactions are possible through access to its open-chain form. They demonstrate carbonyl reactivity but cannot distinguish aldehyde from ketone by themselves. Bromine water provides the more selective clue: oxidation at C1 gives HOOC–(CHOH)₄–CH₂OH, called gluconic acid.
With nitric acid under oxidising conditions, both the aldehyde terminus and the primary alcohol terminus become carboxyl groups. The product HOOC–(CHOH)₄–COOH is glucaric acid, also called saccharic acid, with formula C₆H₁₀O₈. Gluconic acid can also be oxidised further to this same dicarboxylic acid. In oxidation bookkeeping, glucose + 3[O] → glucaric acid + H₂O is balanced; [O] denotes oxygen equivalents and is not a full nitric-acid equation.
Acetylation with acetic anhydride replaces the hydrogen of each of five OH groups with an acetyl group, –COCH₃. Each ester formation releases acetic acid. This probes hydroxyl groups, whereas bromine water probes aldehyde reactivity. The pentaacetate has a protected cyclic anomeric centre and does not behave as though it contained a free aldehyde. That distinction requires the cyclic model developed next.
Problem
A six-carbon sugar produces a monocarboxylic acid with Br₂ water and a dicarboxylic acid with HNO₃. Which terminal groups are indicated?
- 1.Mild oxidation converts CHO to COOH, indicating an aldehyde at one end.
- 2.Strong oxidation also converts the other end into COOH; a terminal CH₂OH group accounts for this transformation.
- 3.A skeleton CHO–(CHOH)₄–CH₂OH fits both observations. The observations do not independently determine whether each interior OH points left or right in a Fischer drawing.
Fischer projections and the meaning of D
A Fischer projection is a stereochemical convention. Vertical bonds point away from the viewer and horizontal bonds toward the viewer. For open-chain D-glucose, write CHO at the top and CH₂OH at the bottom. The OH positions from C2 through C5 are right, left, right, right. This spatial pattern is information that the condensed formula alone omits.
D or L is assigned by comparison with glyceraldehyde. In the standard carbohydrate Fischer orientation with the carbonyl near the top, inspect the stereocentre farthest from the carbonyl. OH on the right places the sugar in the D series; OH on the left places it in the L series. For glucose that reference centre is C5. D does not mean that every OH is on the right, and it does not specify the sign of rotation.
The symbols (+) and (−) report measured optical rotation under specified conditions: dextrorotatory and laevorotatory. D-(+)-glucose and D-(−)-fructose belong to the same configurational series but rotate light in opposite directions. D/L, α/β and (+)/(−) describe different properties. Keeping these labels separate prevents errors when interpreting rings, hydrolysis products and optical measurements.
Quick Quiz
Quiz
Which reagent gives gluconic acid from glucose?
Explanation: Mild bromine-water oxidation converts the aldehyde terminus to COOH.
Which observation most directly supports five acetylatable OH groups?
Explanation: Five acetylations indicate five OH sites; the other observations probe different properties.
What distinguishes glucaric acid from gluconic acid?
Explanation: Stronger oxidation also converts C6–CH₂OH into COOH.
What does D in D-glucose specify?
Explanation: D/L is a configurational relationship, independent of optical rotation sign.
Complete hydrolysis of 0.20 mol sucrose produces how much glucose?
Explanation: One sucrose molecule yields one glucose and one fructose molecule.
Practice Problems
- Predict the main carbon skeleton or functional-group change when glucose reacts with prolonged hot HI, bromine water and nitric acid.
- Why do oxime and cyanohydrin formation not alone prove an aldehyde group?
- Describe the Fischer projection of D-glucose and identify the D/L reference centre.
- Calculate glucose yield from 68.4 g sucrose, using 342 and 180 g mol⁻¹, and account for total product mass.
- A student says D-fructose must be dextrorotatory. Correct the reasoning.
- Write an oxidation-equivalent equation for glucose to glucaric acid and check the atoms.
Prolonged hot HI gives n-hexane, supporting the straight six-carbon skeleton. Bromine water oxidises CHO at C1 to give gluconic acid, HOOC–(CHOH)₄–CH₂OH. Nitric acid oxidises both terminal groups to give glucaric acid, HOOC–(CHOH)₄–COOH. These are different probes, not interchangeable reagents.
Key Takeaways
• Glucose evidence supports an unbranched aldohexose with five OH groups. • Bromine water gives gluconic acid; nitric acid gives glucaric acid. • Oxime and cyanohydrin formation probe carbonyl reactivity. • Acetylation probes alcohol-type OH groups. • A Fischer projection records spatial arrangement as well as connectivity. • D/L configuration and optical rotation sign are independent. • Hydrolysis yields follow balanced mole ratios and incorporate water.