Reduction of Aldehydes and Ketones to hydrocarbon take place in the presence of

Options

A. Zn amalgam and HCl acid
B. `Pd//BaSO_(4)`
C. Anhydrous `AlCl_(3)`
D. `Ni//Pt`

Answer

To solve the question regarding the reduction of aldehydes and ketones to hydrocarbons, we need to identify the correct reagent from the given options.

### Step-by-Step Solution:

1. **Understanding the Reduction Process**:
Aldehydes and ketones can be reduced to hydrocarbons through specific reduction reactions. The two common methods are:
- **Clemensen Reduction**: Involves the use of zinc amalgam in the presence of hydrochloric acid (HCl).
- **Wolff-Kishner Reduction**: Involves the use of hydrazine (NH2-NH2) and a strong base like KOH.

2. **Analyzing the Given Options**:
- **Option A**: Zinc amalgam in the presence of HCl (Clemensen Reduction).
- **Option B**: Palladium in the presence of BaSO4 (not typically used for this reduction).
- **Option C**: Anhydrous AlCl3 (not a reducing agent for aldehydes/ketones).
- **Option D**: Nickel in the presence of platinum (not a standard method for this reduction).

3. **Identifying the Correct Option**:
- Among the options, **Option A** (Zinc amalgam + HCl) is the correct choice as it corresponds to the Clemensen reduction method, which effectively reduces aldehydes and ketones to hydrocarbons.

4. **Writing the Reaction**:
- For example, if we take a generic aldehyde (RCHO) or ketone (R2C=O), the reaction with zinc amalgam and HCl can be represented as:
\[ RCHO \xrightarrow{Zn/HCl} RCH_2 + H_2O \]
- This shows the conversion of the carbonyl group (C=O) into a methylene group (CH2) while releasing water.

### Final Answer:
The reduction of aldehydes and ketones to hydrocarbons takes place in the presence of **Zinc amalgam and HCl (Clemensen Reduction)**.