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Number of chiral carbons in $\beta$–D–(+)–glucose is: -
A
Six
B
Three
C
Four
D
Five
Detailed Solution
$\beta$-D-(+)-glucose is the cyclic (pyranose) form of glucose, in which the C-5 –OH has added to the C-1 aldehyde group to form a six-membered ring containing oxygen.
Check each carbon:
C-1 (anomeric carbon): attached to H, OH, ring O and C-2, i.e. four different groups, so it is chiral.
C-2, C-3 and C-4: each carries H, OH and two different ring carbons, so each is chiral.
C-5: attached to H, ring O, C-4 and $CH_2OH$, so it is chiral.
C-6 ($CH_2OH$): it has two H atoms, so it is not chiral.
So $\beta$-D-glucose has five chiral carbons (C-1 to C-5). The open-chain form has only four (C-2 to C-5); ring formation makes C-1 a new chiral centre.
Check each carbon:
C-1 (anomeric carbon): attached to H, OH, ring O and C-2, i.e. four different groups, so it is chiral.
C-2, C-3 and C-4: each carries H, OH and two different ring carbons, so each is chiral.
C-5: attached to H, ring O, C-4 and $CH_2OH$, so it is chiral.
C-6 ($CH_2OH$): it has two H atoms, so it is not chiral.
So $\beta$-D-glucose has five chiral carbons (C-1 to C-5). The open-chain form has only four (C-2 to C-5); ring formation makes C-1 a new chiral centre.