Question:

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

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Vicinal \(\ce{RO^-}\) adjacent to \(\ce{R\text{-}OTs}\) prefers intramolecular \(\mathrm{S_N2}\) \(\Rightarrow\) \(\ce{epoxide}\) (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.
Updated On: Aug 28, 2025
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The Correct Option is A, B, C

Solution and Explanation

Step 1: Base generates a neighboring alkoxide.
\(\ce{MeO^-}\) deprotonates the free vicinal \(\ce{3\text{-}OH}\) to give the C-3 alkoxide. Intramolecular attack of this alkoxide on the adjacent tosylate-bearing C-2 proceeds by \(\mathrm{S_N2}\) with inversion at C-2, expelling \(\ce{TsO^-}\). This produces a \(2,3\)-\emph{anhydrosugar} (epoxide across C-2/C-3). Depending on the chair conformer adopted at the moment of attack, two equivalent drawings of the same \(2,3\)-epoxide connectivity are obtained (depicted as (A) and (B)).
Step 2: Competing neighboring-group participation by the ring oxygen.
Under strongly basic, intramolecularly favored conditions, the endocyclic ring oxygen can also act as a neighboring nucleophile and attack C-2, giving a \(1,2\)-epoxide fused to the ring (a \(1,2\)-anhydrosugar). This pathway corresponds to structure (C).
Step 3: Why (D) is not chosen.
Direct intermolecular substitution by \(\ce{MeO^-}\) at C-2 to furnish a simple \(2\)-\(\ce{OMe}\) product (D) is disfavored relative to the much faster \emph{intramolecular} cyclizations that give epoxides. In basic methanol, the epoxide-opening by \(\ce{MeO^-}\) is slower and requires additional conditions; thus the epoxide products (A/B/C) are the observed outcomes.
Hence, the reaction yields the epoxides in (A), (B), and (C), not the solvolysis product (D).
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