The product `C` is
`CH_(3).CH_(2).C-=CH+HCl rarrBoverset(HI)rarrC`

Options

A. `CH_(3)CH_(2)-underset(Cl)underset(|)overset(I)overset(|)(C)-CH_(3)`
B. `CH_(3)underset(Cl)underset(|)CH-CH_(2)-CH_(2)I`
C. `CH_(3)-CH_(2)-CH_(2)-underset(Cl)underset(|)overset(I)overset(|)(C)-H`
D. `CH_(3)-CH_(2)-overset(I)overset(|)CH-CH_(2)Cl`

Answer

To determine the product C from the given reaction sequence, we will break down the steps involved in the reaction of 1-butyne (CH₃CH₂C≡CH) with HCl followed by HI.

### Step-by-Step Solution:

1. **Identify the Reactant**:
The starting compound is 1-butyne, which has the structure CH₃CH₂C≡CH.

2. **Reaction with HCl**:
When 1-butyne reacts with hydrochloric acid (HCl), the reaction proceeds via an electrophilic addition mechanism. The H⁺ from HCl will add to the carbon atom of the triple bond that is more substituted (the terminal carbon), leading to the formation of a carbocation.

- The reaction can be represented as:
\[ CH₃CH₂C≡CH + HCl \rightarrow CH₃CH₂C^+(CH₂Cl) \]
- Here, a double bond is formed, and the product is 2-chlorobutene (B).

3. **Formation of Product B**:
The structure of product B is:
\[ CH₃CH₂C(Cl)=CH₂ \]
This is 2-chlorobutene.

4. **Reaction with HI**:
Next, product B reacts with hydrogen iodide (HI). Similar to the previous step, the H⁺ from HI will add to the carbon with the double bond, and the I⁻ will attack the other carbon.

- The reaction can be represented as:
\[ CH₃CH₂C(Cl)=CH₂ + HI \rightarrow CH₃CH₂C(I)Cl \]
- The iodine will replace the chlorine because it is a better nucleophile.

5. **Final Product C**:
The final product C is:
\[ CH₃CH₂C(I)Cl \]
This compound can be named as 2-chloro-2-iodobutane.

### Final Answer:
The product C is **2-chloro-2-iodobutane**, represented structurally as:
\[ CH₃CH₂C(I)Cl \]