Consider the following reaction chart and identify the correct products of the following reaction sequence:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) = Phenol (B) = o-Hydroxybenzaldehyde (C) = 2,4,6-tribromophenol
The chart combines three reactions of benzenediazonium chloride and phenol. The related named reactions of phenol are best remembered together:
Reimer–Tiemann reaction: phenol + CHCl3 and aqueous NaOH, then acid, puts a –CHO group ortho to –OH, giving salicylaldehyde.
Kolbe’s reaction: sodium phenoxide + CO2 (400 K, 4–7 atm), then acid, puts a –COOH group ortho to –OH, giving salicylic acid.
Bromination: with bromine water, phenol gives a white precipitate of 2,4,6-tribromophenol; with \( \ce{Br2} \) in a non-polar solvent (\( \ce{CHCl3} \) or \( \ce{CS2} \)) at low temperature, it gives monobromophenols.
Working through the chart:
(A) Warming the diazonium salt with water replaces –\( \ce{N2^{+}} \) by –OH, so A is phenol.
\( \displaystyle \ce{C6H5N2^+Cl^- + H2O} \xrightarrow{\text{warm}} \ce{C6H5OH + N2 + HCl} \)
(B) \( \ce{CHCl3} \) + aq NaOH, then NaOH and \( \ce{H^{+}} \), is the Reimer–Tiemann reaction. The –\( \ce{CHCl2} \) group formed at the ortho position is hydrolysed to –CHO, so B is o-hydroxybenzaldehyde (salicylaldehyde).
\( \displaystyle \ce{C6H5OH} \xrightarrow[\text{(ii) }\ce{NaOH}\text{ (iii) }\ce{H+}]{\text{(i) }\ce{CHCl3}\text{ + aq }\ce{NaOH}} \text{o-}\ce{HO-C6H4-CHO} \)
(C) In bromine water the polar solvent ionises phenol, the ring becomes very strongly activated, and all three ortho and para positions are brominated, so C is 2,4,6-tribromophenol.
\( \displaystyle \ce{C6H5OH + 3Br2} \xrightarrow{\ce{H2O}} \text{2,4,6-tribromophenol (white ppt)} + \ce{3HBr} \)
Shortcut: The \( \ce{CHCl3} \)/NaOH step gives an aldehyde, not an acid, which rules out Options (3) and (4). Bromine water gives the tribromo product, which leaves Option (1).