The presence of carbon in an organic compound is detected by heating it with

Options

A. sodium metal to convert it into NaCN
B. CaO to convert into CO which burns with a blue flame
C. CuO to convert it into `CO_(2)` which turns lime water milky
D. Cu wire to give a bluish -green flame

Answer

To determine the presence of carbon in an organic compound, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Question**: We need to identify a method to detect carbon in an organic compound by heating it with a specific reagent.

2. **Evaluating Options**:
- **Option A: Heating with Sodium**: When an organic compound is heated with sodium, it can produce sodium cyanide (NaCN) if nitrogen is present. However, this does not indicate the presence of carbon. Thus, this option is incorrect.

- **Option B: Heating with Calcium Oxide (CaO)**: Heating an organic compound with CaO does not specifically indicate the presence of carbon. Although it can produce carbon monoxide (CO), this method is not used for carbon detection. Therefore, this option is also incorrect.

- **Option C: Heating with Copper(II) Oxide (CuO)**: When an organic compound is heated with CuO, it oxidizes the carbon present in the compound to carbon dioxide (CO2). This CO2 can then be bubbled through lime water (Ca(OH)2), resulting in a milky appearance due to the formation of calcium carbonate (CaCO3). This is a well-known test for detecting carbon, making this option correct.

- **Option D: Heating with Copper Wire**: Heating an organic compound with copper wire can produce a bluish-green flame, which indicates the presence of halogens, not carbon. Thus, this option is incorrect.

3. **Conclusion**: The correct method for detecting carbon in an organic compound is by heating it with copper(II) oxide (CuO), as it leads to the formation of carbon dioxide, which can be confirmed by the milky appearance of lime water.

### Final Answer:
The presence of carbon in an organic compound is detected by heating it with **copper(II) oxide (CuO)**.