In the given reaction , the number of isomeric products formed is -
`underset(("Racemic mixture"))(CH_3CHClCOOH)+underset(("Optically pure d-isomer"))(CH_3CHDOH)overset(H^(+))to`Ester

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To determine the number of isomeric products formed in the given reaction, we can follow these steps:

### Step 1: Identify the Reactants
The reactants are:
1. A racemic mixture of `CH3CHClCOOH` (which contains both D and L forms).
2. An optically pure D-isomer of `CH3CHDOH`.

### Step 2: Understand the Reaction Mechanism
The reaction occurs in the presence of H⁺, which facilitates the formation of a carbocation intermediate. The alcohol (`CH3CHDOH`) will then react with this carbocation to form an ester.

### Step 3: Analyze the Chiral Centers
- The racemic mixture `CH3CHClCOOH` has a chiral carbon, which can exist in two configurations: D and L.
- The alcohol `CH3CHDOH` is optically pure and only exists in the D form.

### Step 4: Determine Possible Isomeric Products
Since the first reactant can exist in both D and L forms, while the second reactant is fixed as D, we can analyze the combinations:
1. **D from the first reactant + D from the second reactant** = D-D configuration.
2. **L from the first reactant + D from the second reactant** = L-D configuration.

### Step 5: Count the Isomeric Products
From the analysis:
- We have two distinct configurations based on the combinations of the chiral centers:
1. D-D isomer
2. L-D isomer

Thus, the total number of isomeric products formed is **2**.

### Final Answer
The number of isomeric products formed in the reaction is **2**.

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