See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is hex-5-yn-1-ol (a terminal alkyne with a hydroxyl group at the other end, HC≡C-(CH2)4-OH). Step 2 - Hydrogenation step (A): Treatment with H2 (with a suitable catalyst, implied to be a full/syn hydrogenation catalyst) reduces the terminal alkyne to a terminal alkene or fully to an alkane. Given the product is 2-methyltetrahydropyran, the alkyne is hydrogenated to give hex-5-en-1-ol (the alkene intermediate) or more precisely the terminal alkyne is reduced to give CH3-CH=... wait - let me reconsider. With H2 (full hydrogenation), HC≡C-(CH2)4-OH becomes CH3-CH2-(CH2)4-OH = hexan-1-ol. But that cannot cyclize to give a methylated ring. Step 3 - Re-analysis: With H2/Lindlar catalyst (partial hydrogenation), the terminal alkyne HC≡C-(CH2)4-OH becomes CH2=CH-(CH2)4-OH (hex-5-en-1-ol). Then acid-catalyzed intramolecular cyclization (H+) of the alcohol onto the alkene gives 2-methyltetrahydropyran. The OH attacks the double bond (Markovnikov), forming a 6-membered ring. The carbon bearing the methyl group (C2 of the ring) becomes a new stereocenter. Step 4 - Stereochemistry of product: The acid-catalyzed cyclization proceeds via a protonated alkene (carbocation or protonated intermediate). The nucleophilic attack of the hydroxyl oxygen on the carbocation can occur from either face of the planar carbocation with equal probability. This generates a new chiral center at C2 of the tetrahydropyran ring (the carbon bearing CH3 and O). Since both faces are equally accessible, equal amounts of (R) and (S) enantiomers are formed. Step 5 - Nature of the product: Equal amounts of R and S enantiomers constitute a racemic mixture. The product has one chiral center (C2 with CH3, O, H, and the ring carbons giving four different substituents), so it cannot be meso. Both enantiomers are formed in equal amounts, making it a racemic mixture, which is optically inactive as a mixture but each component is optically active. Step 6 - Why other options fail: (a) Optically active - incorrect because equal R and S are produced, canceling optical rotation. (c) Meso - impossible with only one stereocenter. (d) Diastereomers - requires at least two stereocenters; only one exists here. Therefore, the correct answer is B.