Determine Major Compound Y Obtained from Grignard Reaction and Elimination

To determine the structure of the major compound $\mathbf{Y}$, we analyze the reaction sequence step-by-step:

Step 1: Identification of the starting material $\text{C}_8\text{H}_6$

The molecular formula $\text{C}_8\text{H}_6$ has a degree of unsaturation (DU) calculated as: $\text{DU} = C + 1 - \frac{H}{2} = 8 + 1 - \frac{6}{2} = 6$ Since it undergoes Oxymercuration-Demercuration/Hydration (using $\text{HgSO}_4 / \text{H}_2\text{SO}_4$) and partial hydrogenation (using $\text{Pd-BaSO}_4 / \text{H}_2$), $\text{C}_8\text{H}_6$ contains a benzene ring ($\text{DU} = 4$) and a terminal alkyne group ($\text{DU} = 2$). Thus, $\text{C}_8\text{H}_6$ is phenylacetylene: $\text{Ph}-\text{C}\equiv\text{CH}$

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Step 2: Hydration of $\text{C}_8\text{H}_6$ to yield $\text{C}_8\text{H}_8\text{O}$

Hydration of phenylacetylene using $\text{H}_2\text{O}/\text{HgSO}_4/\text{H}_2\text{SO}_4$ follows Markovnikov's addition: $\text{Ph}-\text{C}\equiv\text{CH} + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{2+}/\text{H}^+} \text{Ph}-\text{C}(\text{OH})=\text{CH}_2 \xrightarrow{\text{Tautomerization}} \text{Ph}-\overset{\underset{||}{\text{O}}}{\text{C}}-\text{CH}_3$ The intermediate $\text{C}_8\text{H}_8\text{O}$ is acetophenone.

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Step 3: Reaction of acetophenone with Grignard reagent ($\text{EtMgBr}$) followed by hydrolysis

Nucleophilic addition of ethylmagnesium bromide ($\text{CH}_3\text{CH}_2\text{MgBr}$) to acetophenone occurs as follows:

1. The nucleophilic ethyl group ($\text{CH}_3\text{CH}_2^-$) attacks the electrophilic carbonyl carbon of acetophenone: $\text{Ph}-\overset{\underset{||}{\text{O}}}{\text{C}}-\text{CH}_3 + \text{CH}_3\text{CH}_2\text{MgBr} \longrightarrow \text{Ph}-\underset{\underset{\text{CH}_2\text{CH}_3}{|}}{\overset{\underset{|}{\text{O}^-\text{MgBr}^+}}{\text{C}}}-\text{CH}_3$
2. Acidic hydrolysis ($\text{H}_2\text{O}$) yields the tertiary alcohol, 2-phenylbutan-2-ol: $\text{Ph}-\underset{\underset{\text{CH}_2\text{CH}_3}{|}}{\overset{\underset{|}{\text{OH}}}{\text{C}}}-\text{CH}_3$

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Step 4: Dehydration (Elimination) using $\text{H}^+$, heat

Acid-catalyzed dehydration of 2-phenylbutan-2-ol proceeds via an $\text{E1}$ mechanism:

1. Protonation and loss of water: $\text{Ph}-\text{C}(\text{OH})(\text{CH}_3)\text{CH}_2\text{CH}_3 \xrightarrow{\text{H}^+} \text{Ph}-\text{C}(\text{OH}_2^+)(\text{CH}_3)\text{CH}_2\text{CH}_3 \xrightarrow{-\text{H}_2\text{O}} \text{Ph}-\overset{+}{\text{C}}(\text{CH}_3)\text{CH}_2\text{CH}_3$ This forms a highly stable benzylic carbocation.

2. Deprotonation to form the alkene: Deprotonation can occur from either the $-\text{CH}_3$ group or the adjacent $-\text{CH}_2-$ group:

   • Deprotonation from the methylene ($-\text{CH}_2-$) group yields the more substituted, highly conjugated alkene (Saytzeff product): $\text{Ph}-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \quad (\text{2-phenylbut-2-ene})$

Because 2-phenylbut-2-ene has extended $\pi$-conjugation with the phenyl ring along with greater hyperconjugative stabilization, it is formed as the major product $\mathbf{Y}$.

This structure corresponds to option (D).