Animated Solution for Chemistry - Carbonyl Compounds: Compound P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is $C_8{H}_{8}O$. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction. The option(s) with suitable combination of P and R, respectively, is(are)

To solve this organic puzzle, we must first play detective and deduce the exact structures of compounds Q and S using the chemical tests provided.

Let's start with compound Q. The problem states that Q gives a positive Cannizzaro reaction. This is a massive clue! The Cannizzaro reaction is strictly given by aldehydes that lack $\alpha$-hydrogens. However, Q fails the haloform test, which means it does not contain a methyl ketone group ($-CO-C{H}_3$). Given its molecular formula of $C_8{H}_{8}O$ and its high degree of unsaturation, Q must be an aromatic aldehyde. An aldehyde group attached directly to a benzene ring has no $\alpha$-hydrogens, fitting perfectly. To satisfy the 8 carbon atoms, it must be a methylbenzaldehyde, commonly known as tolualdehyde.

Now, let's decode compound S. It gives a positive haloform test, meaning it definitely contains a methyl ketone group. It does not give a Cannizzaro reaction, confirming the presence of $\alpha$-hydrogens. With the formula $C_8{H}_{8}O$, S can only be acetophenone ($Ph-CO-C{H}_3$).

We are told that reductive ozonolysis ($O_3$ followed by $Zn/H_{2}O$) of compound P yields Q, and ozonolysis of compound R yields S. Reductive ozonolysis acts like a pair of molecular scissors, completely cleaving the carbon-carbon double bond and capping both ends with oxygen atoms to form aldehydes or ketones.

Our goal is to find the correct pair of alkenes (P and R) that will yield tolualdehyde and acetophenone, respectively.

Let's evaluate the options one by one.

Option A: Compound P is p-methylstyrene. When we cleave its double bond via ozonolysis, we get p-tolualdehyde and formaldehyde. p-tolualdehyde matches all the conditions for Q! Compound R is $\alpha$-methylstyrene. Cleaving its double bond yields acetophenone and formaldehyde. Acetophenone perfectly matches S. Therefore, Option A is a correct combination.

Option B: Here, compound R is an ortho-methyl derivative (o-methyl-$\alpha$-methylstyrene). Its ozonolysis would yield o-methylacetophenone. But wait, count the carbons! That product has 9 carbon atoms ($C_9{H}_{10}O$), whereas our compound S strictly has 8. Therefore, Option B is incorrect.

Option C: Compound P is m-methyl-$\beta$-methylstyrene. Ozonolysis breaks the double bond, producing m-tolualdehyde and acetaldehyde. m-tolualdehyde is $C_8{H}_{8}O$ and fits Q perfectly. For compound R, which is 2-phenyl-2-butene, cleaving the double bond gives acetophenone and acetaldehyde. Acetophenone is exactly S. Thus, Option C is also a correct combination.

Option D: Similar to B, compound R here has an extra methyl group on the aromatic ring (p-methyl-$\alpha$-methylstyrene), meaning its ozonolysis product will have 9 carbons instead of 8. So, Option D is incorrect.

By carefully analyzing the chemical tests and performing mental ozonolysis, we have successfully identified the correct starting materials. The correct options are A and C.