Correct options for the given below reaction:
`Ph-OH overset((i) CHCl_(3)+OH^(-))underset((ii) H^(+)) to` Products

Options

A.
B.
C.
D.

Answer

To solve the given reaction of phenol with chloroform in the presence of a base and then with an acid, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reaction involves phenol (Ph-OH), chloroform (CHCl3), and hydroxide ion (OH^-). The reaction will also involve a subsequent step with H^+ (acid).

2. **Initial Reaction**: The first step is the reaction of phenol with chloroform in the presence of a strong base (like NaOH). This reaction is known as the Riemann-Tiemann reaction. The base deprotonates phenol, making it a stronger nucleophile.

3. **Formation of Intermediate**: The deprotonated phenol (Ph-O^-) reacts with chloroform (CHCl3) to form an intermediate. The reaction leads to the formation of a compound where the chloroform contributes a carbon atom and chlorine atoms are eliminated, resulting in a compound with a structure that includes a carbon atom bonded to the phenol oxygen.

4. **Elimination of Water**: The intermediate undergoes dehydration (removal of water) to form a compound with a carbonyl group (C=O) adjacent to the phenolic group. This results in the formation of salicylaldehyde (ortho-hydroxybenzaldehyde).

5. **Final Step with Acid**: The addition of H^+ (acid) leads to the protonation of the oxygen in the intermediate, stabilizing the product. This step may also lead to the formation of two products: the major product (salicylaldehyde) and a minor product (para-hydroxybenzaldehyde).

6. **Identify Major and Minor Products**: The major product formed is salicylaldehyde (with the aldehyde group at the ortho position relative to the hydroxyl group), while the minor product is para-hydroxybenzaldehyde.

### Conclusion:
The correct options for the products formed in this reaction are:
- **Option A**: Salicylaldehyde (major product)
- **Option C**: Para-hydroxybenzaldehyde (minor product)