? `underset(ETOH_(2)H_(2)O)overset(Delta, CN^(-))(rarr)` Benzoin.
The reactant is obtained by dry distillation of the calcium salts of the following pairs –

Options

A. `C_(6)H_(5)CH_(2)`COOH,HCOOH
B. `C_(6)H_(5)`COOH, HCOOH
C. `C_(6)H_(4)` (OH)COOH, HCOOH
D. `C_(6)H_(4) (NH_(2))`COOH, HCOOH

Answer

To solve the question regarding the reactant obtained from the dry distillation of calcium salts that leads to the formation of benzoin, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The question involves the reaction of a compound with cyanide ion (CN⁻) and ethanol-water mixture (EtOH/H₂O) to yield benzoin. This reaction is known as benzoin condensation.

2. **Identify the Reactant**: In benzoin condensation, the reactants are typically aromatic aldehydes. The most common example is benzaldehyde (C₆H₅CHO).

3. **Dry Distillation of Calcium Salts**: The reactant (benzaldehyde) can be obtained from the dry distillation of calcium salts of specific carboxylic acids. We need to identify which pairs of carboxylic acids can yield benzaldehyde upon dry distillation.

4. **Pairs of Carboxylic Acids**:
- The first pair is **C₆H₅CH₂COOH (benzoic acid)** and **HCOOH (formic acid)**.
- The second pair is **C₆H₅COOH (benzoic acid)** and **HCOOH (formic acid)**.

5. **Formation of Calcium Salts**:
- For the first pair: The calcium salt would be C₆H₅CH₂COO⁻Ca²⁺ and HCOO⁻Ca²⁺.
- For the second pair: The calcium salt would be C₆H₅COO⁻Ca²⁺ and HCOO⁻Ca²⁺.

6. **Dry Distillation Process**: When these calcium salts are subjected to dry distillation, they decompose to form benzaldehyde (C₆H₅CHO) and calcium carbonate (CaCO₃) as a byproduct.

7. **Conclusion**: The correct pairs of carboxylic acids that can yield benzaldehyde upon dry distillation are:
- **C₆H₅CH₂COOH (benzoic acid) and HCOOH (formic acid)**
- **C₆H₅COOH (benzoic acid) and HCOOH (formic acid)**

### Final Answer:
The reactant is obtained by dry distillation of the calcium salts of the pairs:
- C₆H₅CH₂COOH and HCOOH (or)
- C₆H₅COOH and HCOOH.

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