Observe the following reaction `CH_(3)-underset(O)underset(||)(C)-CH_(2)-underset(O)underset(||)(C) -CH_3 overset(HCN "(excess)")to overset(H_(3)overset(o+)(O)//Delta)to` products . The correct statements is

Options

A. the products is a mixture of two compounds
B. the product is optically inactive
C. the product is a mixture of two chiral and one achiral stereoisomer
D. The product is a mixture of three stereoisomers.

Answer

To solve the question regarding the reaction of acetylacetone with excess HCN followed by hydrolysis and heating, we will analyze the steps involved in the reaction and the resulting products.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is acetylacetone, which has the structure:
\[ CH_3C(=O)CH_2C(=O)CH_3 \]
This molecule contains two carbonyl (C=O) groups.

2. **Reaction with HCN**:
When acetylacetone reacts with excess HCN, the cyanide ion (CN⁻) acts as a nucleophile and attacks the electrophilic carbonyl carbon. This results in the formation of cyanohydrin intermediates:
\[ CH_3C(OH)(CN)CH_2C(OH)(CN)CH_3 \]
After the reaction with HCN, we have two cyanohydrin groups formed.

3. **Hydrolysis**:
The next step involves hydrolysis of the cyanohydrin in the presence of H₃O⁺ (acidic conditions). This converts the cyanohydrin into carboxylic acids:
\[ CH_3C(OH)COOHCH_2C(OH)COOHCH_3 \]
This results in the formation of two carboxylic acid groups.

4. **Heating**:
Upon heating, decarboxylation occurs, where carbon dioxide (CO₂) is eliminated. This leads to the formation of a product that retains one carboxylic acid group and a ketone:
\[ CH_3C(CH_3)COOH \]
The final product is a compound with one carboxylic acid group and one ketone group.

5. **Analyze the Product**:
The product formed has a center of symmetry, indicating that it is a meso compound. Therefore, it is optically inactive.

6. **Chirality Assessment**:
In the product, we can identify two chiral centers (the carbon atoms attached to four different groups), while one carbon is not chiral (it has two identical groups). Thus, we have two chiral centers and one achiral center.

7. **Stereoisomer Count**:
The product can exist in multiple stereoisomeric forms due to the presence of chiral centers. Specifically, we can have a mixture of stereoisomers (D and L forms).

### Final Statements:
- The product is a mixture of two components: **Incorrect** (only one major product is formed).
- The product is optically active: **Incorrect** (the product is optically inactive due to symmetry).
- The product is a mixture of two chiral and one achiral derivatives: **Correct** (two chiral centers and one achiral center).
- The product is a mixture of three stereoisomers: **Correct** (due to the presence of chiral centers).

### Conclusion:
The correct statements are that the product is a mixture of two chiral and one achiral derivatives and that it can exist as a mixture of three stereoisomers.