Question

Product(s) formed in the given reaction is(are): 

 

Hint:

Vicinal \(\ce{RO^-}\) adjacent to \(\ce{R\text{-}OTs}\) prefers intramolecular \(\mathrm{S_N2}\) \(⇒\) \(\ce{epoxide}\) (anhydrosugar) formation. Under basic conditions, epoxide formation (intramolecular) usually outcompetes intermolecular substitution by external nucleophiles. In carbohydrate scaffolds, both \(2,3\)- and \(1,2\)-anhydrosugars can arise via neighboring-group participation from \(\ce{O3^-}\) and the ring oxygen, respectively.

Answer 1

Step 1: Base generates a neighboring alkoxide.
\( \mathrm{MeO}^- \) deprotonates the free vicinal \( 3\text{-OH} \) to give the C-3 alkoxide. This alkoxide then performs an intramolecular attack on the adjacent tosylate-bearing C-2. The mechanism proceeds via an \( \mathrm{S_N2} \) pathway, which leads to inversion at C-2, expelling \( \mathrm{TsO}^- \). As a result, a \( 2,3 \)-anhydrosugar is formed, where an epoxide is created across the C-2/C-3 positions. It is important to note that depending on the chair conformer adopted at the moment of attack, two equivalent drawings of the same \( 2,3 \)-epoxide connectivity are obtained. These are depicted in structures (A) and (B). Thus, this step produces the \( 2,3 \)-epoxide, and the outcome is highly dependent on the conformational preference during the reaction.

Step 2: Competing neighboring-group participation by the ring oxygen.
Under strongly basic conditions, which favor intramolecular interactions, the endocyclic ring oxygen can also act as a neighboring nucleophile. This oxygen attacks C-2, forming a \( 1,2 \)-epoxide fused to the ring (also called a \( 1,2 \)-anhydrosugar). This reaction pathway leads to structure (C). The oxygen’s nucleophilic attack forms an epoxide at the C-2 position, resulting in a fused cyclic structure. The intramolecular participation of the ring oxygen leads to a different epoxide product than the one formed in Step 1, specifically the \( 1,2 \)-epoxide. This alternative pathway competes with the \( 2,3 \)-epoxide formation and is observed in the reaction.

Step 3: Why (D) is not chosen.
Direct intermolecular substitution by \( \mathrm{MeO}^- \) at C-2 to furnish a simple \( 2 \)-\( \mathrm{OMe} \) product (D) is not favored. This is because the intramolecular cyclizations that form epoxides are much faster than the direct substitution. In basic methanol, the epoxide-opening reaction by \( \mathrm{MeO}^- \) is slower and requires more specific conditions. Consequently, the intermolecular substitution leading to the simple \( 2 \)-\( \mathrm{OMe} \) product is not the main pathway. Instead, the reaction predominantly proceeds through intramolecular cyclization steps that form the epoxides. Therefore, the products observed are the epoxides (A), (B), and (C), rather than the solvolysis product (D).

Hence, the reaction yields the epoxides in (A), (B), and (C), not the solvolysis product (D).