In the given reaction, product formed is `CH_3CHO+HCN⟶CH_3CH(OH)CN`

Options

A. Laevorotatory
B. Racemic nixture
C. Meso compound
D. Dextrorotatory

Answer

To solve the reaction \( \text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN} \), we will break down the process step by step.

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Acetaldehyde (\( \text{CH}_3\text{CHO} \))
- Hydrogen cyanide (\( \text{HCN} \))

### Step 2: Understand the Reaction Mechanism
The reaction involves the nucleophilic addition of cyanide ion (\( \text{CN}^- \)) to the carbonyl carbon of acetaldehyde. The carbonyl group (C=O) is planar, which allows the nucleophile to attack from either side.

### Step 3: Nucleophilic Attack
1. The nucleophile (\( \text{CN}^- \)) attacks the carbonyl carbon from one side (let's say the top side).
2. This leads to the formation of a tetrahedral intermediate. The double bond between carbon and oxygen breaks, and the oxygen becomes negatively charged.

### Step 4: Protonation of the Intermediate
1. The negatively charged oxygen atom can then be protonated by a proton (\( \text{H}^+ \)) from the solution.
2. This results in the formation of the product \( \text{CH}_3\text{CH(OH)CN} \).

### Step 5: Consider the Other Possible Attack
1. The nucleophile can also attack from the opposite side (the bottom side), leading to another tetrahedral intermediate.
2. This will also be protonated to give a different stereoisomer of the same product.

### Step 6: Formation of a Racemic Mixture
Since the nucleophile can attack from both sides of the planar carbonyl group, two different stereoisomers are formed:
- One product from the top attack (let's say it has the configuration R)
- One product from the bottom attack (let's say it has the configuration S)

These two products are enantiomers, resulting in a racemic mixture.

### Final Product
The final product of the reaction is \( \text{CH}_3\text{CH(OH)CN} \), which exists as a racemic mixture of two enantiomers.