Write the reactions of $D-Glucose$ which cannot be explained by its open-chain structure. How can the cyclic structure of glucose explain these reactions?

(N/A) The open-chain structure of $D-Glucose$ fails to explain the following reactions:
$1$. $D-Glucose$ does not form the hydrogen sulfite addition product with $NaHSO_3$.
$2$. $D-Glucose$ does not give a positive Schiff's test.
$3$. The pentaacetate of $D-Glucose$ does not react with hydroxylamine $(NH_2OH)$,indicating the absence of a free $-CHO$ group.
$4$. $D-Glucose$ exists in two anomeric forms,$\alpha-D-Glucose$ (m.p. $419 \ K$) and $\beta-D-Glucose$ (m.p. $423 \ K$),which cannot be explained by a single open-chain structure.
Explanation by cyclic structure:
These observations are explained by the formation of a cyclic hemiacetal structure. The $-CHO$ group at $C-1$ reacts with the $-OH$ group at $C-5$ to form a six-membered ring (pyranose structure). In this cyclic form,the aldehyde group is involved in hemiacetal formation and is not free to react with $NaHSO_3$,Schiff's reagent,or $NH_2OH$. The existence of two anomers arises due to the formation of a new chiral center at $C-1$ during cyclization.