Q.Ketones are normally not reducing. Fructose however is reducing towards Fehling's solution and Tollen's reagent in spite of the fact that it has ketonic group. Give reason.

Sugars with a free aldehydic or ketonic group reduce Fehling's and Tollens' reagents and are called reducing sugars.

The distinction is whether a free carbonyl (or a hemiacetal that can open to reveal one) is available to reduce Fehling's/Tollens' reagent.

Reducing sugars

A sugar whose anomeric carbon retains a free $-OH$ in equilibrium with an open-chain form bearing a free aldehyde ($-CHO$), or — in special cases — a free ketone group capable of tautomerising to give one, is a reducing sugar. Because a free carbonyl is available, it can reduce mild oxidising agents — Fehling's solution (giving a red $Cu_2O$ precipitate) and Tollens' reagent (giving a silver mirror). Examples: glucose, fructose, and disaccharides like maltose and lactose (which still have one free anomeric OH).

Non-reducing sugars

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False - household sugar is sucrose, not glucose.

The most common sugar used in our homes (cane/table sugar) is sucrose, a disaccharide made of one glucose unit joined to one fructose unit. Glucose is a monosaccharide and is not what we normal …

All monosaccharides (whether aldose or ketose) have a free, reactive carbonyl group and therefore act as reducing sugars.

A carbohydrate is classed as a 'reducing sugar' if it has a free aldehyde or (via tautomerisation) a free ketone group that can reduce mild oxidising agents such as Fehling's or Tollens' reagent. Every monosaccharide (glucose, fructose, ribose, etc.) has exactly this kind of free, un-combined carbonyl group in its open-chain form, so all monosaccharides are reducing …

Glucose gives a positive Fehling's test because it has a free aldehyde group that acts as a reducing agent.

Fehling's solution (alkaline Cu²⁺ tartrate complex) is reduced by free aldehyde groups to give a brick-red precipitate of Cu2O. Glucose, an aldose sugar with a free –CHO group, is a reducing sugar and gives this positive test. Sucrose (a non-reducing disaccharide, glycosidic bo …

Whether a sugar reduces Tollens'/Fehling's reagent depends on whether it retains a free, reactive anomeric carbon; when both monosaccharide units' anomeric carbons are locked into the glycosidic bond, no such group remains, and the sugar is non-reducing.

A sugar shows reducing behaviour (turning Tollens' reagent into a silver mirror, or Fehling's solution into a brick-red $Cu_2O$ precipitate) if it possesses a free aldehyde or (via tautomerisation) an accessible ketone group, generally arising from a free anomeric carbon (hemiacetal $-OH$) that can open to the reactive carbonyl form.

In a non-reducing sugar, the glycosidic bond joining two monosaccharide units is formed using the anomeric carbons of both units, so neither anomeric carbon is left free to open into an aldehyde/ketone form — hence there is no group available to reduce Tollens'/Fehling's reagent.

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A reducing sugar has a free -CHO/-C=O group and reduces Tollens'/Fehling's reagent; sucrose is the main non-reducing exception.

Carbohydrates that contain a free aldehydic group (in aldoses) or a free ketonic group that can tautomerise to give a reactive carbonyl species (in ketoses, via enediol formation) are called reducing sugars, because this free carbonyl group can reduce mild oxidising agents:

• Tollens' reagent (ammoniacal AgNO$_3$) $\rightarrow$ silver mirror (Ag$^+ \rightarrow$ Ag)
• Fehling's solution (alkaline Cu$^{2+}$ tartrate complex) $\rightarrow$ red precipitate of Cu$_2$O …

A sugar is 'reducing' when at least one of its anomeric carbons is left free (not tied up in a glycosidic bond), allowing the ring to open and expose a free aldehyde/ketone group that can reduce mild oxidising agents like Tollens' or Fehling's reagent.

A carbohydrate is classified as a reducing sugar if it possesses a free aldehyde group ($-CHO$) or a free ketonic group that is able to reduce mild oxidising agents such as Tollens' reagent (ammoniacal $AgNO_3$, giving a silver mirror) or Fehling's solution (giving a brick-red $Cu_2O$ precipitate).

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