See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The hydration of alkynes under Markovnikov conditions using H2SO4 and HgSO4 (mercury(II) sulfate) proceeds via an enol intermediate. The mercury catalyst facilitates anti-Markovnikov-free, Markovnikov addition of water across the triple bond, forming a vinyl alcohol (enol) as the intermediate, which then tautomerizes to a ketone. Step 1 - Identify the reaction mechanism: When an internal or terminal alkyne is treated with H2SO4/HgSO4 and water, the first step produces an enol (vinyl alcohol) intermediate. The enol then undergoes keto-enol tautomerism to yield the final ketone product. Step 2 - Identify the correct enol structure: The intermediate must be a vinyl alcohol (enol), meaning it has an OH group directly attached to a carbon that is part of a C=C double bond. This is the defining feature of the enol intermediate. Step 3 - Evaluate the options: - Option (a): Shows an allylic alcohol where OH is on a saturated carbon adjacent to a double bond (allyl alcohol type, -CH2-OH next to C=C). This is NOT an enol; it is an allylic alcohol. This is not the correct intermediate. - Option (b): Shows an enol structure where OH is on an internal sp2 carbon with a terminal =CH2 group (i.e., a 1,1-disubstituted vinyl alcohol with OH on the substituted carbon). While this is an enol, the position of OH relative to the double bond and substituents does not match the expected Markovnikov product correctly for the given alkyne. - Option (c): Shows an enol structure similar to (b) but with OH on a different carbon. Again an enol, but with substituent arrangement that does not represent the correct intermediate. - Option (d): Shows a vinyl alcohol (enol) where the OH is on the terminal =CH- carbon (i.e., HO attached directly to the sp2 carbon of a terminal double bond), with the molecule having a branched chain. This represents the correct enol intermediate: the OH is on the vinylic carbon (sp2), making it a true enol that will tautomerize to a ketone. The structure in (d) has OH on the terminal carbon of the double bond (=C(OH)-) consistent with enol formation from Markovnikov addition of water to an alkyne. Step 4 - Why (d) is correct: Option (d) depicts a vinyl alcohol (enol) with the hydroxyl group directly on the sp2 carbon at the end of the double bond, which is the hallmark of the enol intermediate formed during acid/mercury-catalyzed hydration of an alkyne. Upon tautomerization, the C=C-OH system converts to C-C=O, yielding a ketone. The structure in (d) correctly represents this enol intermediate with OH on the vinylic (sp2) carbon and an adjacent branched alkyl group. Therefore, the correct answer is D.