Synthesis of Tertiary Alcohol from Carbonyl and Grignard

To determine the structure of tertiary alcohol $\mathbf{H}$ and the reaction that forms it, we analyze the reaction sequence given in the passage:

1. Identification of Compound $\mathbf{K}$ and $\mathbf{M}$:
   Compound $\mathbf{K}$ reacts with $\mathrm{KOH}$ to give solely $\mathbf{M}$.
   From the provided structure, $\mathbf{M}$ is dypnone:
   $\mathbf{M} = \mathrm{Ph}-\mathrm{C}(\mathrm{CH}_3)=\mathrm{CH}-\mathrm{CO}-\mathrm{Ph}$
   Dypnone is the aldol condensation product formed by the self-condensation of acetophenone ($\mathrm{PhCOCH}_3$) in the presence of a base:
   $2\,\mathrm{PhCOCH}_3 \xrightarrow{\mathrm{KOH}} \mathrm{Ph}-\mathrm{C}(\mathrm{CH}_3)=\mathrm{CH}-\mathrm{CO}-\mathrm{Ph} + \mathrm{H}_2\mathrm{O}$
   Therefore, compound $\mathbf{K}$ is acetophenone, $\mathrm{PhCOCH}_3$.

2. Identification of Compound $\mathbf{J}$ and $\mathbf{L}$:
   Compound $\mathbf{J}$ undergoes reaction with $\mathrm{KOH}$ to give benzyl alcohol ($\mathrm{PhCH}_2\mathrm{OH}$) and compound $\mathbf{L}$. This is the classic disproportionation via the Cannizzaro reaction:
   $2\,\mathrm{PhCHO} \xrightarrow{\mathrm{KOH}} \mathrm{PhCH}_2\mathrm{OH} + \mathrm{PhCOO^-K^+}$
   Thus:

   • $\mathbf{J}$ is benzaldehyde, $\mathrm{PhCHO}$.
   • $\mathbf{L}$ is potassium benzoate, $\mathrm{PhCOO^-K^+}$.

3. Identification of Alkene $\mathbf{I}$:
   The passage states that ozonolysis of alkene $\mathbf{I}$ yields $\mathbf{J}$ ($\mathrm{PhCHO}$) and $\mathbf{K}$ ($\mathrm{PhCOCH}_3$).
   Recombining the two carbonyl fragments reveals the structure of alkene $\mathbf{I}$:
   $\mathbf{I} = \mathrm{PhCH}=\mathrm{C}(\mathrm{CH}_3)\mathrm{Ph}$

4. Identification of Tertiary Alcohol $\mathbf{H}$:
   Acid-catalyzed dehydration of $\mathbf{H}$ produces the conjugated alkene $\mathbf{I}$.
   Since $\mathbf{H}$ is a tertiary alcohol, the hydroxyl group must reside on the tertiary carbon atom:
   $\mathbf{H} = \mathrm{PhCH}_2-\mathrm{C}(\mathrm{OH})(\mathrm{CH}_3)\mathrm{Ph} \quad (\text{1,2-diphenylpropan-2-ol})$

5. Synthesis of Compound $\mathbf{H}$:
   The tertiary alcohol $\mathrm{PhCH}_2-\mathrm{C}(\mathrm{OH})(\mathrm{CH}_3)\mathrm{Ph}$ is formed by the nucleophilic addition of benzylmagnesium bromide ($\mathrm{PhCH}_2\mathrm{MgBr}$) to acetophenone ($\mathrm{PhCOCH}_3$), followed by acidic workup:
   $\mathrm{PhCOCH}_3 + \mathrm{PhCH}_2\mathrm{MgBr} \xrightarrow{\mathrm{H_3O^+}} \mathrm{PhCH}_2-\mathrm{C}(\mathrm{OH})(\mathrm{CH}_3)\mathrm{Ph}$

This corresponds to Option (B).

Correct Answer: B