The reaction of compounf P with `CH_(3)MgBr` (excess) in `(C_(2)H_(5)_(2)O` followed by addition of `H_(2)O` give Q the compound Q on treatment with `H_(2)SO_(4)` at `0^(@)` C gives R. the reaaction of R with `CH_(3) COCl` in the presence of anhydrous `AlCl_(3)` in `CH_(2)Cl_(2)` followed by treatment with `H_(2)O` "produces compound S`. [Et in compound P is ethyl group]
The reaction, Q to R and R TO S, are

Options

A. Aromatic sulfnation and Friedel-Crafts acylation
B. Friedel-Crafts alkylation and Friedel-Crafts acylation
C. Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation
D. Dehydration and Friedel acylation

Answer

To solve the problem step by step, we will analyze the reactions described in the question, identifying the compounds and the reactions taking place.

### Step 1: Identify Compound P
Compound P is described as having an ethyl group (Et). For the sake of this solution, let's assume compound P is ethylbenzene (C6H5C2H5).

### Step 2: Reaction of Compound P with CH3MgBr
When ethylbenzene (P) reacts with excess methylmagnesium bromide (CH3MgBr) in diethyl ether (C2H5)2O, it undergoes a nucleophilic attack. The ethyl group (C2H5) is replaced by a methyl group (CH3) leading to the formation of a new compound, which we will call compound Q. The reaction can be summarized as follows:
- Ethylbenzene + CH3MgBr → Q (which is 1-phenylpropane, C6H5C3H7)

### Step 3: Addition of Water
After the reaction with CH3MgBr, the addition of water (H2O) will protonate the alkoxide formed during the Grignard reaction, yielding compound Q:
- Q (1-phenylpropane) is formed.

### Step 4: Treatment of Compound Q with H2SO4
Compound Q is then treated with sulfuric acid (H2SO4) at 0°C. This leads to dehydration, forming a carbocation intermediate. The reaction can be summarized as:
- Q + H2SO4 → R (where R is likely to be a more stable alkene, such as styrene, C6H5C=CH2)

### Step 5: Reaction of Compound R with CH3COCl
Next, compound R (the alkene) is treated with acetyl chloride (CH3COCl) in the presence of anhydrous AlCl3 in dichloromethane (CH2Cl2). This is a Friedel-Crafts acylation reaction, where the acetyl group (COCH3) is introduced to the aromatic ring:
- R + CH3COCl → S (where S is acetophenone, C6H5COCH3)

### Step 6: Treatment of Compound S with Water
Finally, the product S is treated with water, but since S is already a stable ketone (acetophenone), no further reaction occurs.

### Summary of Reactions
1. **P to Q**: Ethylbenzene reacts with CH3MgBr → Q (1-phenylpropane) via nucleophilic substitution.
2. **Q to R**: Q treated with H2SO4 → R (styrene) via dehydration.
3. **R to S**: R treated with CH3COCl and AlCl3 → S (acetophenone) via Friedel-Crafts acylation.

### Final Answer
- The reactions are:
- From Q to R: **Dehydration**
- From R to S: **Friedel-Crafts Acylation**