Step 1: What the open chain picture actually promises.
The open-chain model draws glucose as a straight six-carbon chain with one free aldehyde group at C-1 and five $-OH$ groups spread over the remaining carbons. Any behaviour that comes purely from this aldehyde group or from the plain six-carbon backbone should be explained by this picture without trouble.
Step 2: Testing the chain-length and carbonyl-based reactions.
Heating glucose with HI for a long time strips away every oxygen and leaves a plain chain of six carbons, n-hexane, which only tells us the skeleton is six carbons long, nothing about rings. Reaction with hydroxylamine to give an oxime, and oxidation by bromine water to gluconic acid, both use the free $-CHO$ group directly, so the open-chain aldehyde explains both of these comfortably too.
Step 3: The anomer problem.
Glucose is isolated as two separate solids, $\alpha$-glucose and $\beta$-glucose, which slowly interconvert in solution, a behaviour called mutarotation. A single open chain with one fixed aldehyde carbon gives only one structure, so it cannot produce two distinct crystalline forms on its own. This only makes sense once the $-OH$ on C-5 folds back and attacks the $-CHO$ carbon, closing a six-membered ring and creating a brand new stereocentre at C-1 that can sit in two different ways.
Step 4: Picking the answer.
Since it is only the two-anomer behaviour that needs the ring form to be explained, the open-chain structure falls short specifically there. \[ \boxed{\text{Existence of } \alpha \text{ and } \beta \text{ crystalline forms}} \]