Q.Open chain formula of glucose does not contain
a. Formyl group
b. Anomeric hydroxyl group
c. Primary hydroxyl group
d. Secondary hydroxyl group

Glucose's penta-acetate formation with acetic anhydride confirms it has 5 –OH groups.

When glucose is treated with excess acetic anhydride, all five hydroxyl groups get acetylated to give glucose pentaacetate:

$C_6H_7O(OH)_5 + 5(CH_3CO)_2O \rightarrow C_6H_7O(OCOCH_3)_5 + 5CH_3COOH$

Since exactly five acetate groups are incorporated, this reaction is used as chemical proof that glucose contains five –OH groups. (Bromine water tests for the –CHO group by oxidising it to –COOH; hydroxylamine tests for the carbonyl group forming an oxime; dilute nitric acid oxidises both terminal groups to a dicarboxylic acid.)

The pyranose (cyclic, six-membered) form of glucose carries 5 free –OH groups: on C1 (anomeric), C2, C3, C4, and the primary –OH on C6.

In the open-chain form, glucose ($C_6H_{12}O_6$) has –OH groups on C2, C3, C4, C5 and C6 (C1 is the aldehyde carbon, with no –OH). On cyclisation, the C5–OH attacks the C1 aldehyde carbon to form the six-membered pyranose ring (hemiacetal); C5's oxygen becomes the ring oxygen, but a new –OH (the anomeric hydroxyl) appears at C1. So the total count of free hydroxyl groups is unchanged: C1 (anomeric –OH, $\alpha$ or $\beta$), C2, C3, C4 (secondary, ring –OH's) and C6 (primary –OH of the $CH_2OH$ group) — 5 in all, arranged with the characteristic trans (alternating axial/equatorial in the chair form) disposition drawn for $\alpha$-D-(+)-glucopyranose.