Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional group ?

Options

A. (a) `-CHCl_(2)`
B. (b) `-CHO`
C. (c) `-CH_(2)Cl`
D. (d) `-COOH`

Explanation

To solve the question regarding the reaction of phenol with chloroform in the presence of dilute sodium hydroxide, we will follow these steps:

### Step 1: Understand the Reactants
We start with phenol (C6H5OH) and chloroform (CHCl3), and we are using dilute sodium hydroxide (NaOH) as a base.

### Step 2: Identify the Reaction Type
This reaction is known as the Riemer-Tiemann reaction, which involves the introduction of an aldehyde group (CHO) at the ortho position of the phenol ring.

### Step 3: Formation of the Intermediate
When phenol reacts with chloroform in the presence of dilute NaOH, an intermediate is formed. The hydroxyl (OH) group of phenol is deprotonated by NaOH, resulting in the formation of a phenoxide ion (C6H5O^-Na^+).

At the ortho position of the phenoxide ion, chloroform reacts to introduce a dichloromethyl group (CHCl2), leading to the formation of an intermediate compound.

### Step 4: Further Reaction of the Intermediate
The intermediate undergoes hydrolysis. The Na^+ ion is replaced by a hydrogen ion (H^+) from the surrounding medium, leading to the formation of a hydroxyl group (OH) at the ortho position and the conversion of the CHCl2 group into an aldehyde group (CHO).

### Step 5: Final Product
The final product of this reaction is salicylaldehyde (C6H4(OH)(CHO)), which contains the aldehyde functional group (CHO) at the ortho position relative to the hydroxyl group.

### Conclusion
Thus, the functional group introduced in the reaction of phenol with chloroform in the presence of dilute sodium hydroxide is the aldehyde group (CHO).

### Answer
**The correct answer is: B) CHO.**

---