`CH_(2)=CH_(2)overset(HBr)rarrXoverset(Ag.KOH)rarrYunderset(I_(2))overset(Na_(2)CO_(3))rarrZ`. In the above reaction sequence, Z is `:`

Options

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

Explanation

To solve the given reaction sequence, we will identify the compounds X, Y, and Z step by step.

### Step 1: Identify Compound X
We start with ethene (CH₂=CH₂) and treat it with HBr. This is an electrophilic addition reaction where HBr adds across the double bond.

- **Reaction**:
\[ CH_2=CH_2 + HBr \rightarrow CH_3-CH_2Br \]
- **Compound X**: The product of this reaction is bromoethane (CH₃CH₂Br).

### Step 2: Identify Compound Y
Next, we treat compound X (bromoethane) with Ag in KOH. The presence of Ag in KOH facilitates the substitution of the bromine atom (Br) with a hydroxyl group (OH).

- **Reaction**:
\[ CH_3-CH_2Br + Ag/KOH \rightarrow CH_3-CH_2OH \]
- **Compound Y**: The product of this reaction is ethanol (CH₃CH₂OH).

### Step 3: Identify Compound Z
Now, we treat compound Y (ethanol) with Na₂CO₃ and I₂. This is indicative of the iodoform reaction, which occurs with alcohols that have a methyl group adjacent to the hydroxyl group.

- **Reaction**:
\[ CH_3-CH_2OH + I_2 + Na_2CO_3 \rightarrow CHI_3 + HCOONa \]
- **Compound Z**: The products of this reaction are iodoform (CHI₃) and sodium formate (HCOONa). Therefore, Z is the combination of these products.

### Final Answer
Z is: Iodoform (CHI₃) and Sodium Formate (HCOONa).

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