See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: Coumarin (2H-chromen-2-one) is a bicyclic lactone with a benzene ring fused to a pyranone ring. The carbonyl (lactone C=O) is at C2, oxygen at O1, and there is a C3=C4 double bond. Step 2 - Reaction with 2 equivalents of PhMgBr (Grignard reagent): The first equivalent of PhMgBr adds to the carbonyl (C2) of the lactone in a 1,2-addition. This opens the lactone ring to give an alkoxide intermediate. However, since 2 equivalents are used, the second equivalent also adds. Actually, with a lactone and 2 equivalents of Grignard: the first addition opens the lactone to give a ketone intermediate (hemi-ketal collapses to ketone after ring opening), and the second Grignard adds to that ketone. The net result (after H2O workup) is addition of two phenyl groups to C2, giving a diol or after ring closure a tertiary alcohol at C2 with two phenyl groups. More precisely: coumarin reacts with 2 PhMgBr - the lactone carbonyl undergoes double Grignard addition. After hydrolysis (H2O workup), the product A is 2,2-diphenylchroman-2-ol or equivalently the ring-opened diol that recyclizes. The actual product A is 2,2-diphenyl-2H-chromen... more accurately, A = 2,2-diphenylchroman with an OH, i.e., the tertiary alcohol: the lactone C2 now bears two Ph groups and an OH (tertiary alcohol), with the ring oxygen still intact - giving 2,2-diphenylchroman-2-ol (a hemiketal-type structure that is a stable tertiary alcohol in the chromanol ring). Step 3 - Reaction of A with H2SO4/heat: The tertiary alcohol at C2 (bearing two Ph groups) undergoes acid-catalyzed dehydration. Elimination of water from the tertiary alcohol gives a carbocation at C2 stabilized by two phenyl groups and the adjacent oxygen (oxocarbenium). Loss of a proton from C3 gives the alkene, forming the C3=C4 double bond... wait, dehydration would form C2=C3 or C3=C4. Since C2 bears the OH and C3 is adjacent, dehydration gives C2=C3 double bond - but that would place the double bond between C2 and C3 in the chromene ring. However, looking at option D: it shows 2,2-diphenyl substitution at C2 (the oxygen carbon) with a double bond at C3=C4. This means the carbocation forms at C3 after loss of OH from C2 as water, and then proton loss from C4 gives C3=C4 alkene, retaining both Ph at C2. This gives 2,2-diphenyl-2H-chromene with the endocyclic double bond between C3 and C4. Step 4 - Why option D: Option D shows the chromene ring system with both phenyl groups on C2 (the carbon bearing the ring oxygen) and a double bond between C3 and C4 (the endocyclic position), which is consistent with dehydration of the tertiary alcohol at C2 where the C-OH bond breaks, carbocation at C2 is stabilized by oxygen lone pairs forming an oxocarbenium, then elimination gives C3=C4 alkene. This is 2,2-diphenyl-2H-chromene. Step 5 - Why other options fail: (a) and (b) show only one Ph group at C2 - inconsistent with 2 equivalents of PhMgBr adding two Ph groups. (c) shows geminal Ph groups at C3 rather than C2, which is not the product of addition to the lactone carbonyl at C2. Therefore, the correct answer is D.