Question

Acarbonyl compound \(\mathbf{P}\), which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of \(\mathbf{Q}\) leads to a dicarbonyl compound \(\mathbf{R}\), which undergoes intramolecular aldol reaction to give predominantly S.

Short answer

Compound \( P \) is a methyl ketone, which reacts with \( \text{MeMgBr} \) to produce compound \( Q \). Ozonolysis of \( Q \) yields compound \( R \), followed by intramolecular aldol reaction to form \( S \).

Step-by-step solution

- Identify Compound \( P \)

Compound \( P \) gives a positive iodoform test, which indicates that it is a methyl ketone. Therefore, \( P \) could be \( \text{CH}_3CO- \) group with an \( R \) group, giving the general structure \( \text{CH}_3COR \).

- Reaction with \( \text{MeMgBr} \)

When methyl magnesium bromide (MeMgBr) reacts with the carbonyl compound \( P \), it results in the formation of a tertiary alcohol after the addition of water. The Grignard reaction is as follows: \[ \text{CH}_3COR + \text{MeMgBr} \rightarrow \text{CH}_3C(OMgBr)R \xrightarrow{H_2O} \text{CH}_3C(OH)R\text{Me} \].

- Dehydration to form \( Q \)

The tertiary alcohol then undergoes dehydration (loss of water) to form the corresponding olefin (alkene). The dehydration reaction can be represented as: \[ \text{CH}_3C(OH)R\text{Me} \xrightarrow{-H_2O} \text{CH}_3C(R)=\text{CMe}_2 \]. This olefin is compound \( Q \).

- Ozonolysis of \( Q \)

Ozonolysis of the olefin \( Q \) cleaves the double bond and forms two carbonyl compounds. The reaction for ozonolysis is: \[ \text{CH}_3C(R)=\text{CMe}_2 \xrightarrow{O_3} \text{CH}_3COR + \text{CH}_3CO \]. Thus, compound \( R \) is an acetone and another carbonyl compound depending on \( R \).

- Intramolecular Aldol Reaction of \( R \)

Intramolecular aldol reaction suggests that compound \( R \) must have reactive sites within the same molecule to form a cyclic structure. Depending on the structure, the compound \( R \) cyclizes to form compound \( S \), which is a predominant product.

Key concepts

Carbonyl Compound

Carbonyl compounds are organic molecules containing a carbon-oxygen double bond (C=O). These include aldehydes and ketones.
As per the problem, compound **P** is a carbonyl compound that gives a positive iodoform test, indicating that it is a methyl ketone.
The general structure of such a compound is \(\text{CH}_3COR\). This means it has a methyl group (\(\text{CH}_3\)) attached to the carbon of the carbonyl group (C=O), along with another group (R).

Iodoform Test

The iodoform test is a chemical reaction used to identify methyl ketones. If a compound has the structure \(\text{CH}_3CO-\), it will react with iodine and a base to form iodoform (CHI3), which is a yellow solid.
This test is very specific and effective in identifying the presence of methyl ketone groups within a molecule like compound **P** in the exercise.
Compounds like alcohols having an extra methyl group on the carbon connected to the hydroxyl group also test positive.

Grignard Reaction

Grignard reaction involves the addition of a Grignard reagent (R-Mg-Br) to a carbonyl compound, leading to the formation of an alcohol.
In this case, we used methyl magnesium bromide (MeMgBr) with compound **P**. The reaction transforms \(\text{CH}_3COR \rightarrow \text{CH}_3C(OH)R\text{Me} \).
This tertiary alcohol undergoes dehydration (removal of water) to form the olefin **Q**, an alkene.

Ozonolysis

Ozonolysis is a reaction where an alkene (olefin) is cleaved using ozone (O3) to form carbonyl compounds.
In the reaction, **Q** (an olefin) reacts with ozone to break its double bond. Here, \(\text{CH}_3C(R)=\text{CMe}_2 \rightarrow \text{CH}_3COR + \text{CH}_3CO \).
This results in compound **R**, which consists of two carbonyl-containing products, one of which is acetone.

Aldol Reaction

An aldol reaction involves the combining of two carbonyl compounds to form a β-hydroxy carbonyl compound (aldol).
In an intramolecular aldol reaction, compounds within the same molecule react to form a cyclic structure.
In compound **R**, this reaction leads to a predominant cyclic product **S**, which completes the process started with the original carbonyl compound **P**.