The intermediate X, in the reaction :

Solution

Chemistry ma’am · 4 steps

This is Reimer–Tiemann: carbene adds at ortho as CHCl₂, and in NaOH the phenol stays as phenoxide.

1. Make the electrophile. OH⁻ removes the H from CHCl₃ to give CCl₃⁻, which loses Cl⁻ to form dichlorocarbene, :CCl₂. This carbene is the electrophile.
2. Phenoxide attacks. In aqueous NaOH, phenol is present as sodium phenoxide. Its ring attacks :CCl₂ at the ortho position. A proton shift then turns the CCl₂⁻ into a CHCl₂ group. The medium is still basic, so the oxygen stays as O⁻ Na⁺.

   C₆H₄(O⁻Na⁺)CHCl₂ →(NaOH) C₆H₄(O⁻Na⁺)CHO →(H⁺) C₆H₄(OH)CHO

3. Check against step 2. NaOH hydrolyses the CHCl₂ group to CH(OH)₂, which loses water to give CHO. The final H⁺ is needed to protonate the phenoxide back to OH, giving salicylaldehyde. This fits only if X is still the phenoxide.
4. So. X is sodium 2-(dichloromethyl)phenoxide, option (c).

The idea to keep. In Reimer–Tiemann, CHCl₃ gives CHCl₂, which becomes CHO. CCl₄ gives CCl₃, which becomes COOH. In base, always draw the phenol as phenoxide until the acid step.

Answer from the NTA final answer key. Solution by PrepWiser.