Question

Write the products of the following reactions:
(i) Phenol (C₆H₅OH) + CHCl₃ + aq. NaOH → ?
(ii) Phenol (C₆H₅OH) + Zn / Δ → ?
OR
(i) Explain hydrogen bonding in o-nitrophenol and p-nitrophenol.
(ii) What is formed in the nitration of phenol with concentrated HNO₃?

Hint:

(i) is Reimer-Tiemann (dichlorocarbene) giving salicylaldehyde; Zn/Δ removes –OH to give benzene. For the alternative, compare intramolecular chelation in o-nitrophenol with intermolecular association in p-nitrophenol, and remember conc. HNO₃ gives picric acid.

Answer 1

Option 1 (products of reactions)

Step 1: Reaction (i) is the Reimer-Tiemann reaction. When phenol is treated with chloroform (CHCl₃) in the presence of aqueous NaOH, an aldehyde group (–CHO) is introduced at the ortho position of the ring.
Step 2: How it works. Aqueous NaOH first reacts with CHCl₃ to generate the electrophile dichlorocarbene, \( :CCl_2 \). NaOH also converts phenol into the phenoxide ion, which is highly activated at the ortho and para positions. The electron-rich ortho carbon attacks the dichlorocarbene, and on subsequent hydrolysis the \( -CHCl_2 \) group is converted to \( -CHO \).
Step 3: Product of (i). The main product is salicylaldehyde (2-hydroxybenzaldehyde, ortho-hydroxybenzaldehyde).
\[ C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{\Delta} o\text{-}HO\text{-}C_6H_4\text{-}CHO + 3NaCl + 2H_2O \]
Step 4: Reaction (ii) is zinc-dust distillation. When phenol vapour is passed over heated zinc dust, the \( -OH \) group is removed (reduced) and replaced by \( -H \); zinc takes up the oxygen as ZnO.
Step 5: Product of (ii). The product is benzene.
\[ C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO \]

Option 2 (hydrogen bonding and nitration)

Step 1: Nature of H-bonding in o-nitrophenol. In o-nitrophenol the \( -OH \) group and the \( -NO_2 \) group are on adjacent (ortho) carbons and lie close in space. The \( -OH \) hydrogen bonds to an oxygen of the neighbouring \( -NO_2 \) group within the same molecule, forming a stable six-membered chelate ring. This is intramolecular hydrogen bonding.
Step 2: Consequence for o-nitrophenol. Because the \( -OH \) is locked in an internal H-bond, the molecules do not associate strongly with one another. Hence o-nitrophenol has a lower boiling point, is steam-volatile, and is less soluble in water.
Step 3: Nature of H-bonding in p-nitrophenol. In p-nitrophenol the \( -OH \) and \( -NO_2 \) groups are far apart (para positions), so an intramolecular bond is not possible. Instead the \( -OH \) of one molecule bonds to the \( -NO_2 \)/\( -OH \) of another molecule. This is intermolecular hydrogen bonding, which links many molecules together.
Step 4: Consequence for p-nitrophenol. The strong intermolecular association gives p-nitrophenol a higher boiling point, makes it non-steam-volatile, and more soluble in water than the ortho isomer.
Step 5: Nitration of phenol with concentrated HNO₃. Phenol is very strongly activated by the \( -OH \) group, so concentrated nitric acid nitrates it at all three activated positions (two ortho and one para). The product is 2,4,6-trinitrophenol, commonly called picric acid.
\[ C_6H_5OH + 3HNO_3 \xrightarrow{\text{conc.}} 2,4,6\text{-}(NO_2)_3C_6H_2OH + 3H_2O \]
\[\boxed{\text{Products: salicylaldehyde; benzene; and picric acid (2,4,6-trinitrophenol)}}\]