The given reaction involves the conversion of phenol to 2-Hydroxybenzaldehyde (also known as salicylaldehyde). This reaction is known as the Reimer-Tiemann reaction. Let's break down the process step-by-step:
Thus, the treatment of phenol with chloroform and sodium hydroxide followed by hydrolysis results in the formation of 2-Hydroxybenzaldehyde.
Here's why the other options are incorrect:
The correct answer is 2-Hydroxybenzaldehyde.
The reaction of phenol with chloroform in the presence of sodium hydroxide, followed by hydrolysis in an acidic medium, is known as the Reimer-Tiemann reaction. This reaction results in the formation of 2-hydroxybenzaldehyde (commonly known as salicylaldehyde). The reaction can be represented as:
\(\text{Phenol} + \text{CHCl}_3 + \text{NaOH} \rightarrow 2\text{-Hydroxybenzaldehyde (salicylaldehyde)}\)
The reaction involves the introduction of a formyl group (-CHO) at the ortho position relative to the hydroxyl group (-OH), producing 2-hydroxybenzaldehyde.
The Correct Answer is: 2-Hydroxybenzaldehyde
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,