How many stereocentres (chiral carbon atoms) are present in the open-chain structure and in the cyclic (pyranose) structure of glucose, respectively?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4 and 5
A stereocentre (chiral carbon) is a carbon atom bonded to four different groups. Count them in each structure of glucose (carbons numbered from the –CHO end):
Open chain, C1: the –CHO carbon has only three groups (H, =O, C2): not a stereocentre.
Open chain, C2 to C5: each carries H, OH and two different carbon groups: 4 stereocentres.
Open chain, C6: the \( \ce{CH2OH} \) carbon has two H atoms: not a stereocentre.
Cyclic (pyranose) form, C1: the –OH at C5 adds to the –CHO group, so C1 now carries H, OH, the ring O and C2 — four different groups. This is the anomeric carbon, a new stereocentre.
Cyclic form, C2 to C5: still 4 stereocentres; C6 is still not one.
Total: 4 in the open-chain structure and 5 in the cyclic structure.
NCERT (2026-27 edition):
10.1.2.1 Glucose – “… It was proposed that one of the —OH groups may add to the —CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a six-membered ring in which —OH at C-5 is involved in ring formation. … The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group at C1, called anomeric carbon (the aldehyde carbon before cyclisation).”
