Sugar which does not give reddish brown precipitate with Fehling's reagent, is :

Solution

Chemistry ma’am · 3 steps

Fehling's test needs a free hemiacetal group that can open up into an aldehyde, and sucrose has none.

1. What Fehling's detects. Fehling's reagent contains Cu²⁺ in alkaline solution. A sugar that can open into a free aldehyde reduces it to Cu₂O, the reddish brown precipitate. Such a sugar is called a reducing sugar.
2. Free anomeric carbon. A sugar is reducing only if at least one anomeric carbon (C1 of an aldose, C2 of a ketose) is not tied up in a glycosidic bond. That free hemiacetal can open into the chain form.

   Sugar	Linkage	Free anomeric carbon?	Fehling's test
   Glucose	monosaccharide	yes, C1	positive
   Sucrose	α-glucose C1 to β-fructose C2	no, both are used	negative
   Lactose	β-galactose C1 to glucose C4	yes, C1 of glucose	positive
   Maltose	α-glucose C1 to glucose C4	yes, C1 of the second glucose	positive

3. So. In sucrose the glycosidic bond joins the two anomeric carbons to each other, so the molecule cannot open into an aldehyde or ketone form. Sucrose is non-reducing and gives no reddish brown precipitate, so the answer is (b).

The idea to keep. For any disaccharide, look at the glycosidic bond. If it joins anomeric carbon to anomeric carbon, as in sucrose, the sugar is non-reducing. If one anomeric carbon stays free, it reduces Fehling's.

Answer from the NTA final answer key. Solution by PrepWiser.