Identify `Z` in the following series
`CH_(2)=CH_(2) overset(HBr)rarr X overset(Hydrolysis)rarr Y overset(Na_(2) CO_(3))underset(I_(2)excess)rarrZ`

Options

A. `C_(2)H_(5)I`
B. `C_(2)H_(5)OH`
C. `CHI_(3)`
D. `CH_(3)CHO`

Explanation

To identify compound `Z` in the given reaction series, we will follow the steps outlined in the question.

### Step-by-Step Solution:

1. **Identify the Starting Compound**:
The starting compound is `CH₂=CH₂`, which is ethylene (ethene).

2. **Reaction with HBr**:
When ethylene reacts with hydrogen bromide (HBr), an electrophilic addition reaction occurs. The bromine (Br) adds to one carbon atom, and a hydrogen (H) adds to the other carbon atom. According to Markovnikov's rule, since both carbon atoms are equivalent in this case, we can add Br to either carbon. The product formed is `C2H5Br`, which is bromoethane (ethyl bromide).
- **Product X**: `C2H5Br` (Bromoethane)

3. **Hydrolysis of Bromoethane**:
The next step involves hydrolysis of bromoethane. In this reaction, the bromine atom is replaced by a hydroxyl group (OH) from water. Thus, bromoethane reacts with water to form ethanol.
- **Product Y**: `C2H5OH` (Ethanol)

4. **Reaction with Sodium Carbonate and Iodine**:
The final step involves the reaction of ethanol with sodium carbonate (Na2CO3) and excess iodine (I2). This reaction is known as the hydroform reaction. In this context, ethanol reacts to form a methyl ketone, specifically acetaldehyde (CH3CHO) and sodium formate (HCOONa) as a byproduct.
- **Product Z**: `CH3CHO` (Acetaldehyde)

### Final Answer:
Thus, the compound `Z` in the reaction series is **acetaldehyde (CH3CHO)**.

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