`CH_(3)D+Cl_(2)underset( "(1 mole)")overset("hv")rarr` Product
The product is

Options

A. `CH_(2)DCl`
B. `CH_(3)Cl`
C. `CDCl_(3)`
D. `C Cl_(4)`

Answer

To solve the question regarding the reaction of CH₃D with Cl₂ in the presence of light, we can follow these steps:

### Step 1: Initiation of the Reaction
In the presence of light (hv), chlorine (Cl₂) undergoes homolytic cleavage to form two chlorine radicals (Cl·):
\[ \text{Cl}_2 \xrightarrow{hv} 2 \text{Cl} \cdot \]

**Hint:** Remember that light can provide the energy needed to break the Cl-Cl bond, creating radicals.

### Step 2: Formation of Alkyl Radical
Next, we have CH₃D (methane with one deuterium atom). The chlorine radical can abstract a hydrogen atom from CH₃D. Since the C-H bond is weaker than the C-D bond, the chlorine radical will preferentially abstract a hydrogen atom (H) rather than deuterium (D). This results in the formation of a methyl radical (CH₂D·) and HCl:
\[ \text{CH}_3\text{D} + \text{Cl} \cdot \rightarrow \text{CH}_2\text{D} \cdot + \text{HCl} \]

**Hint:** Consider the bond strengths; the weaker bond will break more easily.

### Step 3: Termination of the Reaction
The methyl radical (CH₂D·) can then react with another chlorine radical to form the final product, which is CH₂DCl:
\[ \text{CH}_2\text{D} \cdot + \text{Cl} \cdot \rightarrow \text{CH}_2\text{DCl} \]

**Hint:** Look for the combination of radicals to form stable products.

### Final Product
The final product of the reaction is CH₂DCl, which contains one deuterium atom and one chlorine atom.

### Summary of the Reaction
The overall reaction can be summarized as:
\[ \text{CH}_3\text{D} + \text{Cl}_2 \xrightarrow{hv} \text{CH}_2\text{DCl} + \text{HCl} \]

### Conclusion
The product formed from the reaction of CH₃D with Cl₂ in the presence of light is CH₂DCl.