Question

Ribose, a carbohydrate with the formula shown, forms a cyclic hemiacetal, which, in principle, could contain either a four-membered, a five-membered, or a sixmembered ring. When D-ribose is treated with methanol in the presence of an acid catalyst, two cyclic acetals, A and 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 \(10.8 \mathrm{C}\) ) followed by dilute aqueous acid. Both \(\mathrm{A}\) and \(\mathrm{B}\) yield the same three products in the same ratios.

Short answer

Question: Determine the structures of acetals A and B formed from the reaction of D-ribose with methanol, considering that they have a molecular formula of C6H12O5 and differ in the position of a methoxy group in the ring. Answer: Acetal A has a four-membered ring with the structure: $$\left(\begin{array}{c}\mathrm{O}-\mathrm{CHO}\\ |\mathrm{CH}_{2} \mathrm{OCH}_{3}\end{array}\right)-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CH}_{2} \mathrm{OH}$$ Acetal B has a five-membered ring with the structure: $$\left(\begin{array}{c}\mathrm{O}-\mathrm{CHO}\\ |\mathrm{CHOH}\end{array}\right)-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CH}_{2} \mathrm{OCH}_{3}$$

Step-by-step solution

Ribose structure and cyclic hemiacetal formation

To start, let's first understand the structure of Ribose. Ribose is an aldopentose sugar with the formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\). Its structure is as follows: $$\mathrm{HOCH}_{2} \left( \begin{array}{c}-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CHOH}\\ |\text{(}\mathrm{C} H_{2} \mathrm{OH}\text{)}\end{array} \right.$$ When treated with methanol in the presence of an acid catalyst, a cyclic hemiacetal is formed from the oxygen attached to the carbonyl carbon. The cyclic hemiacetal can have a four, five, or six-membered ring.

Formation of Acetals A and B

During the reaction with methanol, two cyclic acetals A and B are formed. The molecular formula for these acetals is \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{5}\). In this step, we are not yet able to determine the structures of A and B, but we know that during this reaction, a methoxy group (\(\mathrm{-OCH}_{3}\)) is added to the cyclic hemiacetal.

Reaction of Acetals A and B with Sodium Periodate

Acetals A and B are treated with sodium periodate, which cleaves the 1,2-diols, followed by dilute aqueous acid. It is important to note that both A and B yield the same three products in the same ratios. This indicates that they have similar structure and may differ in the position of the methoxy group in the five or six-membered ring.

Identifying the Products of the Reaction

Since A and B yield the same three products, we do not need to worry about the identity of the acetals yet. We can first determine the products formed after cleaving the 1,2-diols with sodium periodate and then look for possible structures of A and B. Suppose the products formed after cleavage are: 1. Formaldehyde (\(\mathrm{CH}_{2}\mathrm{O}\)) 2. A dialdehyde with formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{3}\) 3. A triol with formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{2}\)

Constructing the Structures of A and B

Based on the possible products formed, we can deduce the possible structures of A and B. Considering that they have a molecular formula of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{5}\) and differ in the position of a methoxy group in the ring, we can construct the following structures: - Acetal A: $$\left(\begin{array}{c}\mathrm{O}-\mathrm{CHO}\\ |\mathrm{CH}_{2} \mathrm{OCH}_{3}\end{array}\right)-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CH}_{2} \mathrm{OH}$$ - Acetal B: $$\left(\begin{array}{c}\mathrm{O}-\mathrm{CHO}\\ |\mathrm{CHOH}\end{array}\right)-\mathrm{CHOH}-\mathrm{CHOH}-\mathrm{CH}_{2} \mathrm{OCH}_{3}$$ arsimp In summary, by understanding the reaction mechanism and identifying the products formed after cleaving 1,2-diols, we were able to determine the likely structures of cyclic acetals A and B formed from the reaction of D-ribose with methanol. Acetal A has a four-membered ring, while Acetal B has a five-membered ring. They differ in the position of the methoxy group in the ring.