Haworth's research of glucose

Question

Firstly haworth used CH3I/Ag2O with glucose which replaced 5 H by CH3 forming a acetal which locked the anomer and stopped the equilibirium of ring chain tautomerism. Under dilute acidic conditions acetal group hydrolyze and ethers do not.

So my text (Organic chemistry by L.G. Wade 8th edition) says through these reactions haworth suggested that C5 contains free hydroxyl group and hence proved the pyranose(6 membered cyclic ring) form of glucose. But I fail to understand how did he find that the hydroxyl group is present on the 5th carbon in the chain after hydrolyzation by dilute acid.

Answer 1

I cannot guarantee that the following explanation for the proof of the ring size of the principal cyclic form of glucose is due to Haworth. The link I provided in the comment above explains one method.

In "Methylation of Pyranoses" methods are illustrated for the permethylation of the hemiacetal form of glucose and its subsequent hydrolysis to the tetramethoxy hemiacetal. Although the 6-membered ring is shown, a priori the ring size is unknown. This is the issue to be determined. continued below

In "Ring Size of Pyranosides" mild HNO₃ oxidation of the tetramethoxy hemiacetal affords a keto carboxylic acid (upper left). Continued oxidation of either enol of the keto group leads to oxidative degradation to an optically-active dimethoxysuccinic acid (red path) and a meso trimethoxyglutaric acid (green path). These two dicarboxylic acids are only derived from a 6-membered ring of glucose.

Source: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/carbohydrates44.pdf