See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Trimethyl orthoformate (HC(OMe)3) reacts with ketones under acid catalysis and aqueous conditions via a transketalization/orthoacid exchange mechanism. In this reaction, the orthoformate acts as a source of a methoxycarbocation equivalent (MeO+), effectively converting the ketone into a ketal (dimethyl ketal), while releasing methyl formate (HCOOMe) as a byproduct. Step 1: Under acid catalysis (H+), trimethyl orthoformate (HC(OMe)3) is activated. One OMe is protonated and leaves as MeOH, generating the electrophilic species [HC(OMe)2]+. Step 2: The ketone oxygen of tetrahydropyran-4-one attacks the electrophilic carbon of [HC(OMe)2]+, forming a mixed orthoester intermediate. Step 3: The intermediate collapses with loss of methyl formate (H-C(=O)-OMe), which is explicitly shown as the co-product in the reaction equation. Step 4: The net result is that both OMe groups from the orthoformate are transferred to the ketone carbon (C-4 of the tetrahydropyran ring), converting the C=O into C(OMe)2 — a dimethyl ketal. This gives tetrahydropyran-4,4-dimethyl ketal, i.e., 4,4-dimethoxytetrahydropyran. Verification against options: - Option (a): has OH and OMe at C-4 — this would be a hemiketal, not the full ketal; inconsistent with loss of water and methyl formate stoichiometry. - Option (b): has two OMe groups at C-4 — this is the dimethyl ketal, consistent with the mechanism and the methyl formate byproduct. CORRECT. - Option (c): has OMe and Me at C-4 — Me groups are not introduced in this reaction; incorrect. - Option (d): has two Me groups at C-4 — no source of methyl groups; incorrect. Therefore, the correct answer is B.