Question

Ribose, a carbohydrate with the formula shown, forms a cyclic hemiacetal, which, in principle, could contain either a four-membered, five-membered, or six-membered ring. When D-ribose is treated with methanol in the presence of an acid catalyst, two cyclic acetals, \(\mathrm{A}\) and \(\mathrm{B}\), are formed, both with molecular formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{5}\). These are separated, and each is treated with sodium periodate (Section 10.8C) followed by dilute aqueous acid. Both \(\mathrm{A}\) and \(\mathrm{B}\) yield the same three products in the same ratios.

Short answer

Answer: Cyclic acetal A is a furanose with a five-membered ring, and cyclic acetal B is a pyranose with a six-membered ring. After cleavage by sodium periodate and subsequent aqueous acid treatment, both A and B yield the same three products: two molecules with the formula C2H4O2 and one molecule with the formula C2H4O1.

Step-by-step solution

Identify the cyclic acetals A and B.

Since ribose can form four-membered, five-membered, or six-membered rings as cyclic acetals, and A and B have the same molecular formula, we can conclude that one of them is a five-membered ring (furanose) and the other is a six-membered ring (pyranose). Let's assume A is the furanose and B is the pyranose.

Determine the structure of the two cyclic acetals.

Since ribose is an aldopentose, the carbon (C1) at the aldehyde group will react with an alcohol group (OH) at C4 in furanose and C5 in pyranose to form the cyclic acetals. The structures of cyclic acetals A (furanose) and B (pyranose) can be drawn as follows: A (furanose): ``` O O \ / C1---C2 `---' C4 ``` B (pyranose): ``` O O \ / C1---C2 | C5 ```

Identify the glycosidic linkage by adding methanol group (OCH3) to ribose

In order to understand the glycosidic linkage, we need to add a methanol group (OCH3) to ribose (which exists after adding an acid catalyst). This methanol group replaces the -OH on ribose's C1, and the structure would be: A (furanose): ``` O O \ / C1---C2 `---' C4 | OCH3 ``` B (pyranose): ``` O O \ / C1---C2 | C5 OCH3 ```

Analyze the cleavage of A and B by Sodium Periodate and subsequent aqueous acid treatment.

Sodium periodate cleaves C-C bonds adjacent to the acetals. Since both acetals have the same reaction products, we need to find which C-C bonds are cleaved. In both cases (A and B), the common bonds are the C1-C2 and C4-C5 bonds. After the cleavage, we can deduce the three products formed as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\), \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) and \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{1}\). The same three products are formed from both A and B after cleavage with sodium periodate and subsequent treatment with dilute aqueous acid.