(a) Reducing sugars
Definition : Carbohydrates that contain a free aldehyde or free ketone group (or a potential one at a free anomeric carbon) and are therefore able to reduce Tollens' reagent and Fehling's solution are called reducing sugars.
The sugar itself is oxidised while it reduces $Ag^+$ to metallic silver (a silver mirror) or $Cu^{2+}$ to $Cu_2O$ (a red-brown precipitate).
All monosaccharides are reducing sugars.
Example of a non-reducing sugar : Sucrose
Why sucrose is non-reducing : both its anomeric carbons — C-1 of glucose and C-2 of fructose — are locked into the glycosidic linkage, so neither ring can open to expose a carbonyl group.
(b) Glucose with bromine water
Bromine water is a mild oxidising agent. It attacks only the $-CHO$ group, leaving the primary alcoholic $-CH_2OH$ untouched :
$$\underset{\text{glucose}}{CHO-(CHOH)_4-CH_2OH} \xrightarrow{Br_2\ \text{water}} \underset{\text{gluconic acid}}{COOH-(CHOH)_4-CH_2OH}$$
Product : gluconic acid
What it proves : the carbonyl group in glucose is specifically an aldehyde, since a ketone would not be oxidised by so mild a reagent.
(c) Two reactions not explained by the open chain structure
1. Glucose does not give Schiff's test and does not form a bisulphite addition product with $NaHSO_3$.
If a free $-CHO$ group were present, both of these characteristic aldehyde reactions should occur. Their failure shows that the aldehyde group is not freely available.
2. Glucose pentaacetate does not react with hydroxylamine $(H_2N-OH)$.
Once all five $-OH$ groups are acetylated — including the one at C-1 — the ring is locked shut and can never reopen. No free $-CHO$ group can be produced, so no oxime forms.
(A third acceptable answer : glucose exists in two crystalline forms, $\alpha$ and $\beta$, differing in melting point and specific rotation — something the open chain formula, with its fixed arrangement, cannot account for. A fourth : glucose shows mutarotation.)
What these observations led to : the cyclic hemiacetal structure, in which the $-OH$ at C-5 attacks the $-CHO$ at C-1, making C-1 a new chiral centre — the anomeric carbon.
OR
(c) Significance of 'D' and '(+)' in D-(+)-glucose
The prefix 'D' — a matter of configuration
'D' refers to the spatial arrangement (configuration) of the groups at the highest-numbered chiral carbon — C-5 in glucose.
The molecule is compared with the reference compound glyceraldehyde :
- if the $-OH$ at that carbon lies on the right in the Fischer projection, the sugar is D
- if on the left, it is L
Crucially, 'D' says nothing about the direction of rotation of light.
The prefix '(+)' — a matter of observed rotation
'(+)' means the compound is dextrorotatory — it rotates plane-polarised light to the right (clockwise), as measured experimentally with a polarimeter.
$$[\alpha]_D = +52 \cdot 5^\circ \text{ for D-glucose}$$
A '(−)' prefix would mean laevorotatory, rotating light to the left.
Why both are needed
The two prefixes are independent of each other. A D sugar may be either dextro- or laevorotatory.
The clearest illustration : D-(−)-fructose. It has the D configuration at C-5, yet it rotates light to the left $([\alpha]_D = -92 \cdot 4^\circ)$. Configuration and optical rotation must therefore be specified separately.
