Animated Solution for Chemistry - Aldehydes and Ketones: The compound, A in the following :

The beauty of organic chemistry lies in its logical flow. Sometimes, the best way to move forward is to take a step back. This problem is a perfect example of retrosynthetic analysis, where we start from the final products and deduce our way back to the starting material. Let's break down this fascinating journey step by step.

The Final Clues

Our first major clues lie in the reactions of compounds $C$ and $D$.

Compound $C$ reacts with concentrated $\text{KOH}$ and heat to yield potassium benzoate (${\text{Ph-COO}}^{-}{\text{K}}^{+}$) and benzyl alcohol (${\text{Ph-CH}}_2\text{OH}$). This is the hallmark of the Cannizzaro reaction, a disproportionation reaction typical of aldehydes lacking alpha-hydrogens. Since the products are derived from a benzene ring, compound $C$ must unequivocally be benzaldehyde ($\text{Ph-CHO}$).

Next, we examine compound $D$. It reacts with barium hydroxide (${\text{Ba(OH)}}_2$) and heat to form mesityl oxide, an $\alpha ,\beta$-unsaturated ketone with the structure ${\text{H}}_3\text{C}-\text{C}({\text{CH}}_3)=\text{CH}-\text{CO}-{\text{CH}}_3$. This reaction is a classic Aldol condensation. By mentally breaking the double bond and adding water across it, we can deduce that the precursor was two molecules of acetone. Thus, compound $D$ is acetone (${\text{CH}}_3-\text{CO}-{\text{CH}}_3$).

Reverse Ozonolysis

Now that we have identified $C$ (benzaldehyde) and $D$ (acetone), we can determine the structure of $B$. We know that $B$ undergoes ozonolysis (${\text{O}}_3$ followed by $\text{Zn}/{\text{H}}_2\text{O}$) to yield $C$ and $D$.

To find $B$, we perform a mental reverse ozonolysis. We take benzaldehyde and acetone, remove their carbonyl oxygen atoms, and stitch the remaining carbon atoms together with a double bond.

Joining $\text{Ph-CHO}$ and ${\text{CH}}_3-\text{CO}-{\text{CH}}_3$ gives us the alkene $B$: $\text{Ph}-\text{CH}=\text{C}({\text{CH}}_3{)}_2$.

The Grignard Connection

We are now at the final stage of our retrosynthetic journey. Compound $A$ reacts with methylmagnesium bromide (${\text{CH}}_3\text{MgBr}$) followed by acidic dehydration (${\text{Conc. H}}_2{\text{SO}}_4/\Delta$) to form the alkene $B$.

The double bond in $B$ ($\text{Ph}-\text{CH}=\text{C}({\text{CH}}_3{)}_2$) is formed by the dehydration of a tertiary alcohol. The most stable alcohol precursor that would yield this specific highly conjugated alkene is $\text{Ph}-{\text{CH}}_2-\text{C}(\text{OH})({\text{CH}}_3{)}_2$.

This tertiary alcohol is the product of the Grignard reaction. Since the Grignard reagent provided a methyl group (${\text{CH}}_3^{-}$), we must remove one methyl group from the alcohol and convert the hydroxyl group back into a carbonyl to find $A$.

Doing this reveals that compound $A$ is 1-phenylpropan-2-one, with the structure $\text{Ph}-{\text{CH}}_2-\text{CO}-{\text{CH}}_3$.

By systematically working backwards, we have successfully unraveled the entire reaction sequence. The starting material $A$ is indeed 1-phenylpropan-2-one, which perfectly matches option (a). This problem beautifully illustrates how mastering fundamental reactions like Cannizzaro, Aldol, and Ozonolysis allows us to solve complex multi-step syntheses with confidence.