Identify the product for the following reaction:
`CH-=CH+HOCl to`

Options

A. `Cl_(2)CHCHO`
B. `CH(OH)=CHCl`
C. `ClCH_(2)CH_(2)OH`
D. `CH_(3)COCl`

Explanation

To identify the product of the reaction between 1-alkyne (ethyne) and HOCl, we will follow these steps:

### Step 1: Understand the Reactants
The reactants are ethyne (CH≡CH) and hypochlorous acid (HOCl). Ethyne is an alkyne with a triple bond between two carbon atoms.

### Step 2: Identify the Type of Reaction
The reaction between an alkyne and HOCl is an addition reaction. In this case, the HOCl will add across the triple bond of the alkyne.

### Step 3: Determine the Electrophilic Attack
In HOCl, chlorine (Cl) is less electronegative than oxygen (O). Therefore, the O in HOCl will act as the nucleophile and attack one of the carbon atoms in the triple bond, while Cl will act as the electrophile.

### Step 4: Apply Markovnikov's Rule
According to Markovnikov's rule, when HX (where X is a halogen) adds to an unsymmetrical alkene or alkyne, the hydrogen atom will attach to the carbon with more hydrogen atoms already attached. In this case, both carbons in the triple bond have the same number of hydrogen atoms, so we can expect the addition to occur at either carbon.

### Step 5: Form the Intermediate
When HOCl adds to the alkyne, we form an enol intermediate. The oxygen from HOCl will bond with one carbon, and the chlorine will bond with the other carbon. The intermediate can be represented as:
- C1: CH (with OH)
- C2: CH (with Cl)

### Step 6: Tautomerization to Form the Final Product
The enol form is typically unstable and will undergo tautomerization to form a more stable keto form. In this case, the enol will convert to an aldehyde:
- The enol (C1=CH-OH) will rearrange to form the aldehyde (C=O) and a dichloromethane derivative (CHCl2).

### Step 7: Final Product
The final product of the reaction is:
- CHCl2 (dichloroethene) and CHO (aldehyde).

### Conclusion
The product of the reaction between ethyne (CH≡CH) and HOCl is:
- **CHCl2CHO** (2-chloroacetaldehyde).