(a) Two fat-soluble vitamins
Vitamin A and Vitamin D
(Vitamins E and K are equally acceptable — the fat-soluble group consists of exactly these four : A, D, E and K.)
Why the classification matters : being fat soluble, these vitamins are stored in the liver and adipose tissue, so they need not be supplied every day. An excess can, however, accumulate and become harmful.
The water-soluble vitamins — the B-complex group and vitamin C — cannot be stored and must be taken regularly in the diet.
(b) Confirming five $-OH$ groups on different carbon atoms
Experiment : acetylation with acetic anhydride
$$C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow \underset{\text{glucose pentaacetate}}{C_6H_7O(OCOCH_3)_5} + 5CH_3COOH$$
Two separate conclusions follow :
How many $-OH$ groups —
Acetylation converts each $-OH$ group into an $-OCOCH_3$ group, one acetyl unit per hydroxyl. Since exactly five acetyl groups are taken up, glucose must contain five hydroxyl groups.
Why they must be on different carbons —
Two $-OH$ groups on the same carbon would constitute a gem-diol, which is unstable and immediately loses water to give a carbonyl group :
$$>C(OH)_2 \rightleftharpoons \ >C=O + H_2O$$
Since all five hydroxyl groups survive and are successfully acetylated, each must sit on a separate carbon atom.
This experiment forms one of the key pieces of evidence for the open-chain structure of glucose :
$$CHO-(CHOH)_4-CH_2OH$$
Interestingly, the same reaction gives further evidence for the cyclic structure too — glucose pentaacetate does not react with hydroxylamine, showing that once C-1 is acetylated the ring can no longer open to release a free $-CHO$ group.
