In the following reaction C is, `CH_(3) - CH = CH_(2) `+ HCl
`underset("Peroxide")overset("Organic")(rarr) A underset("Moist")overset(Ag2O)(rarr) B overset("Et MgBr")(rarr) `C

Options

A. n –propanol
B. Isopropanol
C. Propane
D. Ethane

Answer

To solve the given reaction step by step, we will analyze each part of the reaction and identify the products formed at each stage.

### Step 1: Reaction of Propene with HCl in the presence of Peroxide
The starting compound is propene, which is represented as CH₃-CH=CH₂. When propene reacts with HCl in the presence of organic peroxide, the addition follows Markovnikov's rule.

- **Markovnikov's Rule** states that in the addition of HX to an alkene, the hydrogen atom (H) will attach to the carbon with the greater number of hydrogen atoms, while the halide (Cl in this case) will attach to the carbon with fewer hydrogen atoms.

In propene:
- The double bond is between the second and third carbon atoms.
- The first carbon (C1) has 3 hydrogen atoms, the second carbon (C2) has 1 hydrogen atom, and the third carbon (C3) has 2 hydrogen atoms.

According to Markovnikov's rule:
- The Cl will attach to C2 (the carbon with fewer hydrogens), and H will attach to C3.

Thus, the product A formed will be:
\[ A = CH_3-CH(Cl)-CH_3 \]
This can also be written as:
\[ A = CH_3-CHCl-CH_3 \]

### Step 2: Reaction of Product A with Moist Ag₂O
Next, we react product A with moist silver oxide (Ag₂O). The presence of moisture means that water (H₂O) is involved, leading to the formation of silver hydroxide (AgOH).

In this reaction:
- The nucleophilic part is the hydroxide ion (OH⁻) from AgOH.
- The Cl atom in product A will be substituted by the OH group.

The reaction can be represented as:
\[ CH_3-CH(Cl)-CH_3 + OH^- \rightarrow CH_3-CH(OH)-CH_3 + Cl^- \]
Thus, the product B formed will be:
\[ B = CH_3-CH(OH)-CH_3 \]
This is an alcohol.

### Step 3: Reaction of Product B with Ethyl Magnesium Bromide (C₂H₅MgBr)
Finally, we react product B with ethyl magnesium bromide (C₂H₅MgBr), which is a Grignard reagent.

In this reaction:
- The Grignard reagent acts as a nucleophile and will attack the carbon atom bonded to the hydroxyl group (OH) in product B.
- The OH group will be converted into an alkoxide (O⁻) and the ethyl group (C₂H₅) will be added to the carbon.

The reaction can be represented as:
\[ CH_3-CH(OH)-CH_3 + C_2H_5MgBr \rightarrow CH_3-CH(C_2H_5)-CH_3 + Mg(OH)Br \]
Thus, the final product C will be:
\[ C = CH_3-CH(C_2H_5)-CH_3 \]
This is an alkane.

### Final Summary of Products
- Product A: \( CH_3-CH(Cl)-CH_3 \)
- Product B: \( CH_3-CH(OH)-CH_3 \)
- Final Product C: \( CH_3-CH(C_2H_5)-CH_3 \)