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Structure of Glucose

Chemical Evidence for Open Chain and Cyclic Structures.

By chemca Team • Updated Jan 2026

Glucose is an aldohexose ($C_6H_{12}O_6$). Its structure was elucidated using a series of specific chemical reactions.

1. Evidence for Open Chain Structure

A. Straight Carbon Chain

On prolonged heating with HI (Red P), glucose forms n-Hexane.

$$ \text{Glucose} \xrightarrow{HI, \Delta} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3 \text{ (n-Hexane)} $$

Conclusion: All 6 carbons are linked in a straight chain.

B. Presence of Carbonyl Group (>C=O)

Glucose reacts with Hydroxylamine ($NH_2OH$) to form an Oxime and with HCN to form a Cyanohydrin.

$$ \text{Glucose} + NH_2OH \rightarrow \text{Glucoxime} $$ $$ \text{Glucose} + HCN \rightarrow \text{Cyanohydrin} $$

C. Presence of Aldehyde Group (-CHO)

Glucose gets oxidized to Gluconic Acid (6C acid) by mild oxidizing agents like Bromine water.

$$ \text{Glucose} \xrightarrow{Br_2/H_2O} \text{Gluconic Acid} $$

Conclusion: The carbonyl group is an Aldehyde.

D. Presence of 5 -OH Groups

Acetylation with Acetic Anhydride gives Glucose Pentaacetate.

$$ \text{Glucose} + 5(CH_3CO)_2O \rightarrow \text{Glucose Pentaacetate} $$

E. Presence of Primary Alcohol

Oxidation with Conc. $HNO_3$ gives Saccharic Acid (Dicarboxylic acid).

$$ \text{Glucose} \xrightarrow{HNO_3} \text{Saccharic Acid} $$

2. Cyclic Structure of Glucose

Limitations of Open Chain

Despite the aldehyde group, Glucose does not give the Schiff’s test and does not form the hydrogensulphite addition product with $NaHSO_3$.

Hemiacetal Formation

The -OH group at $C_5$ reacts with the $-CHO$ group at $C_1$ to form a cyclic hemiacetal (Pyranose structure).

Anomers ($\alpha$ and $\beta$)

This cyclization creates a new chiral center at $C_1$ (Anomeric carbon).

• $\alpha$-D-Glucose: -OH at $C_1$ is on the right (in Fischer projection).
• $\beta$-D-Glucose: -OH at $C_1$ is on the left.

Mutarotation: The spontaneous change in specific rotation of an optically active compound when dissolved in water ($\alpha \rightleftharpoons \text{Equilibrium} \rightleftharpoons \beta$).

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