Compound 'P' `(C_10H_12O)` evolves `H_2` gas with Na metal.It reacts with `Br_2//C Cl_4` to give 'Q' `(C_10H_12Br_2O)`.With `I_2//NaOH` it forms iodoform and an acid 'R' `(C_9H_8O_2)` .'P' has geometrical and optical isomers. The struture of 'P' and 'R' should be

Options

A.
B. PH-CH=CH-COOH
C.
D.

Explanation

To solve the problem, we need to identify the structures of compounds P and R based on the information provided. Let's break down the information step by step.

### Step 1: Analyze the information about compound P
1. **Molecular formula**: Compound P has the formula \( C_{10}H_{12}O \).
2. **Reaction with Na metal**: It evolves hydrogen gas, indicating the presence of an alcohol functional group (–OH).
3. **Reaction with Br2/CCl4**: This suggests that P likely contains a double bond (alkene) that can undergo electrophilic addition.
4. **Presence of isomers**: P has geometrical (cis/trans) and optical isomers, indicating that it has a chiral center and/or a double bond that can exhibit cis/trans isomerism.

### Step 2: Propose a structure for compound P
Based on the above information, we can propose a structure for P. A suitable candidate is:
- **Structure of P**:
- A phenolic compound with a double bond and an alcohol group.
- One possible structure is:

\[ \text{C}_6\text{H}_5\text{C}(CH_3)=CH\text{CH}_2\text{OH} \]

This structure has:
- A benzene ring (C6H5)
- A double bond (C=CH)
- An alcohol group (–OH)

### Step 3: Analyze the reaction of compound P to form compound Q
1. **Reaction with Br2/CCl4**: The double bond in compound P will react with bromine to form compound Q, which will have two bromine atoms added across the double bond.
2. **Structure of Q**:
- The addition of bromine will yield:

\[ \text{C}_6\text{H}_5\text{C}(Br)=C(Br)\text{CH}_2\text{OH} \]

### Step 4: Analyze the reaction of compound P with I2/NaOH
1. **Iodoform reaction**: The presence of a methyl group adjacent to the alcohol indicates that P can undergo the iodoform reaction, leading to the formation of iodoform (CHI3) and an acid.
2. **Formation of compound R**:
- The iodoform reaction will oxidize the alcohol to a carboxylic acid.
- The resulting acid R will have the structure:

\[ \text{C}_6\text{H}_5\text{C}(CH_3)=C(COOH) \]

### Step 5: Final structures
- **Compound P**:
\[ \text{C}_6\text{H}_5\text{C}(CH_3)=CH\text{CH}_2\text{OH} \]

- **Compound R**:
\[ \text{C}_6\text{H}_5\text{C}(CH_3)=C(COOH) \]

### Summary
- The structure of compound P is a phenolic compound with a double bond and an alcohol group.
- The structure of compound R is the corresponding carboxylic acid formed after the iodoform reaction.