Identification of Cleavage Products J K and L

Based on the reaction sequence in the passage, compound $\mathbf{I}$ is $(E)$-1,2-diphenylprop-1-ene:

$\text{PhCH}=\text{C}(\text{CH}_3)\text{Ph}$

1. Ozonolysis / Cleavage of Alkene $\mathbf{I}$:
   Reductive cleavage (such as ozonolysis in the presence of $\text{Zn}/\text{H}_2\text{O}$) of the double bond in $\mathbf{I}$ yields benzaldehyde and acetophenone:
   $\text{PhCH}=\text{C}(\text{CH}_3)\text{Ph} \xrightarrow{\text{O}_3 / \text{Zn}, \text{H}_2\text{O}} \text{PhCHO} \, (\mathbf{J}) + \text{PhCOCH}_3 \, (\mathbf{K})$

2. Formation of Compound $\mathbf{L}$:

   • Benzaldehyde ($\text{PhCHO}$) on treatment with concentrated aqueous $\text{KOH}$ undergoes the Cannizzaro reaction:
     $2\,\text{PhCHO} \xrightarrow{\text{KOH}} \text{PhCOO}^-\text{K}^+ \, (\mathbf{L}) + \text{PhCH}_2\text{OH}$
   • Alternatively, acetophenone ($\text{PhCOCH}_3$) undergoes the haloform reaction with $\text{I}_2/\text{KOH}$ to yield potassium benzoate ($\text{PhCOO}^-\text{K}^+$) along with iodoform ($\text{CHI}_3$).

Thus, the structures of compounds $\mathbf{J}$, $\mathbf{K}$, and $\mathbf{L}$ are:

• $\mathbf{J} = \text{PhCHO}$
• $\mathbf{K} = \text{PhCOCH}_3$
• $\mathbf{L} = \text{PhCOO}^-\text{K}^+$

Therefore, the correct option is D.