Match the compounds given in List I with List II and select the suitable option using the code given below
`{:(List I,List II),((A)"Benzaldehyde",(i)"Phenolphthalein"),((B)"Phthalic anhydride",(ii)"Benzoin condensation"),((C)"Phenyl benzoate",(iii)"Oil of wintergreen"),((D)"Methyl salicylate",(iv)"Fire rearrangement"):}`
Code

Options

A. `{:((A),(B),(C),(D).),((ii),(iii),(iv),(i).):}`
B. `{:((A),(B),(C),(D).),((ii),(i),(iv),(iii).):}`
C. `{:((A),(B),(C),(D).),((iv),(i),(iii),(ii).):}`
D. `{:((A),(B),(C),(D).),((iv),(ii),(iii),(i).):}`

Explanation

To solve the matching question, we will analyze each compound in List I and determine which compound from List II corresponds to it.

### Step-by-Step Solution:

1. **Identify Compound A: Benzaldehyde**
- Benzaldehyde is known to undergo a reaction called **Benzoin condensation**. In this reaction, benzaldehyde reacts with cyanide ions (KCN) to form benzoin.
- **Match:** A (Benzaldehyde) → (ii) Benzoin condensation

2. **Identify Compound B: Phthalic Anhydride**
- Phthalic anhydride is used in the synthesis of **Phenolphthalein**. This compound is formed through the reaction of phthalic anhydride with phenol in the presence of an acid catalyst.
- **Match:** B (Phthalic anhydride) → (i) Phenolphthalein

3. **Identify Compound C: Phenyl Benzoate**
- Phenyl benzoate can undergo a reaction known as the **Fries rearrangement**. In this reaction, phenyl benzoate is treated with a Lewis acid, leading to the formation of ortho- and para-hydroxybenzophenone.
- **Match:** C (Phenyl benzoate) → (iv) Fries rearrangement

4. **Identify Compound D: Methyl Salicylate**
- Methyl salicylate is commonly referred to as **Oil of wintergreen**. It is a compound that is used for its aromatic properties and is derived from salicylic acid.
- **Match:** D (Methyl salicylate) → (iii) Oil of wintergreen

### Final Matches:
- A → (ii) Benzoin condensation
- B → (i) Phenolphthalein
- C → (iv) Fries rearrangement
- D → (iii) Oil of wintergreen

### Code:
The final matching can be represented as:
- A → (ii)
- B → (i)
- C → (iv)
- D → (iii)