Glucose: Structure, Evidence & Tests

CHECarbohydrates

Glucose is the benchmark aldohexose; this subtopic tests its open-chain structure, the evidence for individual groups, and why its cyclic (hemiacetal) form modifies the expected aldehyde tests. Anomers, mutarotation and epimers are recurring definition questions.

Open-chain glucose

CH_2OH\text{-}(CHOH)_4\text{-}CHO \qquad (C_6H_{12}O_6,\ \text{4 chiral C})

Aldohexose with one

\text{-}CHO

and five

\text{-}OH

groups.

Reduction of chain (HI / red P)

C_6H_{12}O_6 \xrightarrow{HI,\ \text{red P},\ \Delta} CH_3(CH_2)_4CH_3 \ (\text{n-hexane})

Proves an unbranched six-carbon chain.

Oxidation to saccharic acid

CH_2OH\text{-}(CHOH)_4\text{-}CHO \xrightarrow{HNO_3} HOOC\text{-}(CHOH)_4\text{-}COOH

Strong

HNO_3

oxidises both ends to give the dicarboxylic glucaric (saccharic) acid.

Equilibrium specific rotation (mutarotation)

[\alpha]_D:\ \alpha\text{-form } +112.2^\circ \rightleftharpoons \text{equilibrium } +52.7^\circ \rightleftharpoons \beta\text{-form } +18.7^\circ

Either pure anomer drifts to the same equilibrium mixture.

Open-chain glucose is $CH_2OH\text{-}(CHOH)_4\text{-}CHO$ , an aldohexose with five $\text{-}OH$ groups and one $\text{-}CHO$ ; D-(+)-glucose has four chiral carbons.
Evidence for structure: forms a pentaacetate (5 $\text{-}OH$ groups); $HI$ with red P gives n-hexane (straight 6-C chain); addition of $HCN$ confirms a carbonyl; oxidation with $HNO_3$ gives the dicarboxylic saccharic (glucaric) acid.
Glucose gives a negative Schiff's test and does not form the hydrogensulphite ( $NaHSO_3$ ) addition product because it exists almost entirely in the cyclic hemiacetal (pyranose) form, so a free $\text{-}CHO$ is rarely available.
The pentaacetate of glucose does not react with hydroxylamine, confirming that the carbonyl is masked in the cyclic hemiacetal form.
Anomers: the $\alpha$ - and $\beta$ - cyclic forms differing only in configuration at the anomeric (hemiacetal C1) carbon.
Mutarotation is the gradual change in optical rotation when either pure anomer is dissolved, until the equilibrium specific rotation $+52.7$ degrees is reached (a mixture of $\alpha$ , $+112.2$ degrees, and $\beta$ , $+18.7$ degrees).
Epimers differ in configuration at only one chiral carbon (other than the anomeric one), e.g. glucose and galactose (at C4), or glucose and mannose (at C2).
Mild oxidation by bromine water (a mild oxidant) attacks only the $\text{-}CHO$ to give the monocarboxylic gluconic acid, confirming the aldehyde group is the more easily oxidised end.
Reaction with hydroxylamine gives an oxime and with $HCN$ gives a cyanohydrin — both prove the presence of a carbonyl ( $\text{-}CHO$ ) group in the open-chain form.
The cyclic form is a six-membered ring (pyranose) formed when the C5 $\text{-}OH$ adds across the C1 aldehyde, creating the new anomeric stereocentre; this is depicted in Haworth projection.
Reaction with acetic anhydride (or acetyl chloride) gives glucose pentaacetate, confirming five hydroxyl groups.
The open-chain (Fischer) reactivity (oxime, cyanohydrin, oxidation) coexists with hemiacetal behaviour because a small equilibrium amount of open chain is always present, replenished as it reacts.

Where the marks go

Confusing the products of mild vs strong oxidation: bromine water gives gluconic acid (mono-acid) at the $\text{-}CHO$ only, whereas $HNO_3$ gives saccharic/glucaric acid (di-acid) at both ends.
Saying glucose is fully aldehyde and so gives a positive $NaHSO_3$ /Schiff's test — the dominant cyclic hemiacetal masks the carbonyl, so these tests are negative.
Confusing anomers with epimers: anomers differ at the anomeric C1 (hemiacetal carbon), epimers differ at any one other chiral carbon.
Misquoting the equilibrium specific rotation — it is $+52.7^\circ$ , between $\alpha$ ( $+112.2^\circ$ ) and $\beta$ ( $+18.7^\circ$ ), and reflects the unequal anomer ratio (about 36% $\alpha$ , 64% $\beta$ ), not a simple average.
Drawing the cyclic form as a five-membered ring for glucose — glucose's stable ring is the six-membered pyranose (C5-OH onto C1).

How the board asks it

Give reasonsnegative schiff's and nahso3 tests; cyclic hemiacetal masking
Account for the following: glucose does not give the Schiff's test and does not form the hydrogensulphite addition product with $NaHSO_3$ , even though its open-chain structure contains a $\text{-}CHO$ group.
Give reasonsevidence reactions for each functional group
Give the structural evidence, naming the reagent in each case, that establishes the presence of (i) five $\text{-}OH$ groups, (ii) a straight six-carbon chain and (iii) a carbonyl group in the open-chain structure of glucose.
Predict the productmild oxidation (bromine water) vs strong oxidation ( $HNO_3$ )
Name and write the structure of the product formed when glucose is treated with (i) bromine water and (ii) concentrated $HNO_3$ , and account for the difference between the two products.
Define / stateanomers, epimers, mutarotation
Define the terms anomers, epimers and mutarotation, giving one example each from the chemistry of glucose.
Give reasonsmutarotation via cyclic-open chain equilibrium
Explain why a freshly prepared solution of pure $\alpha$ -D-glucose shows a gradual change in its specific rotation until it reaches $+52.7$ degrees.
Structure / namingpyranose hemiacetal ring formation in haworth projection
Draw the Haworth projection of $\alpha$ -D-glucopyranose and explain how the six-membered hemiacetal ring is formed from the open-chain form of glucose.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.