Question

When phenol is treated with chloroform and aqueous sodium hydroxide, the major product formed is:

• A. Salicylic acid
• B. Salicylaldehyde
• C. Benzoic acid
• D. Benzyl chloride

Hint:

Remember the reagent distinction: Phenol + \(\text{CHCl}_3\) / NaOH yields Salicylaldehyde (Reimer-Tiemann formylation), whereas Phenol + \(\text{CO}_2\) / NaOH yields Salicylic acid (Kolbe's reaction).

Answer 1

The Correct Option is B

Step 1: Identify the reagents.
Phenol is treated with chloroform $\text{CHCl}_3$ and aqueous $\text{NaOH}$. This exact combination is the signature of a famous named reaction.
Step 2: Name the reaction.
This is the Reimer-Tiemann reaction, which installs an aldehyde group onto a phenol ring.
Step 3: Generate the active species.
The base $\text{NaOH}$ strips a proton from $\text{CHCl}_3$ to make dichlorocarbene $:\text{CCl}_2$, a hungry electron-poor intermediate.
Step 4: Attack the ring.
Meanwhile $\text{NaOH}$ turns phenol into the electron-rich phenoxide ion. The dichlorocarbene attacks this ring, favouring the ortho position next to the oxygen.
Step 5: Hydrolyse to the product.
Workup with water converts the resulting intermediate into a $-\text{CHO}$ group, giving ortho-hydroxybenzaldehyde, better known as salicylaldehyde.
Step 6: Eliminate the other options.
Salicylic acid would form if $\text{CCl}_4$ were used instead of $\text{CHCl}_3$; benzoic acid and benzyl chloride do not arise from this pathway. So the answer is salicylaldehyde, option (B).
\[ \boxed{\text{Salicylaldehyde}} \]