Which one of the following is a reducing sugar?

Solution

Chemistry ma’am · 3 steps

Look only at the anomeric carbon: a free OH there, a hemiacetal, means the sugar is reducing.

1. What reduces. Tollens' and Fehling's reagents are basic. A sugar reduces them only if its anomeric carbon can open up into a free -CHO (or C=O) group. A hemiacetal carbon (ring O plus a free -OH) opens easily. An acetal carbon (two -OR groups) stays locked in base.

   Option	What is on the anomeric carbon	Type	Reducing?
   (a) pentose in pyranose ring	ring O, H and a free -OH	hemiacetal	yes
   (b) furanose ring	ring O, H and -OCH₃	acetal (methyl glycoside)	no
   (c) two furanose rings	each anomeric C bonded to the bridging O	both acetals	no
   (d) open-chain hexose	C1 is CH(OCH₃)₂, not CHO	dimethyl acetal	no

2. Why (a) works. In (a), C1 carries the ring oxygen and a free -OH. In solution the ring opens to the open-chain aldehyde, which is oxidised by Tollens' or Fehling's reagent.
3. So. Only (a) has a free hemiacetal -OH at the anomeric carbon, so it is the reducing sugar.

The idea to keep. Find the carbon bonded to the ring oxygen (or C1 in open chain). Free OH there means hemiacetal, which is reducing. OR or a glycosidic link there means acetal, which is non-reducing.

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