GATE 2024 CY – Question 16

Organic Chemistry · Experimental Techniques in Organic Chemistry: Optical rotation (polarimetry). Application of various chromatographic techniques such as thin-layer, column, HPLC and GC. Applications of UV-visible, IR, NMR spectroscopy and Mass spectrometry in the structural determination of organic molecules. · 1 mark · Multiple choice

The major product in the given reaction sequence is Q. The mass spectrum of Q shows ([M] = molecular ion peak)

Phenol treated with sodium hydroxide and carbon dioxide, then acid and excess bromine water.

[M], [M+2], [M+4], and [M+6] peaks with relative intensity of 1:1:1:1
[M], [M+2], [M+4], and [M+6] peaks with relative intensity of 1:3:3:1
[M], [M+2], and [M+4] peaks with relative intensity of 1:2:1
[M] and [M+2] peaks with relative intensity of 1:1

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Correct answer: (B) [M], [M+2], [M+4], and [M+6] peaks with relative intensity of 1:3:3:1

Explanation

**Step 1: Kolbe-Schmitt reaction.** Phenol with NaOH gives sodium phenoxide, which reacts with $CO_2$ (and is then acidified) to give **salicylic acid** (2-hydroxybenzoic acid).

**Step 2: excess bromine water.** Phenols are strongly activated. With excess $Br_2$/water the ring is brominated at the three activated positions (the two ortho and the para positions), and the carboxylic acid group is lost as $CO_2$ (bromodecarboxylation). The product Q is **2,4,6-tribromophenol**.

**Step 3: mass spectrum.** Bromine has two isotopes of almost equal abundance, $^{79}Br$ and $^{81}Br$ (about 1:1). For a molecule with three Br atoms the pattern is given by the expansion of $(1+1)^3=1+3+3+1$:

Peak	Br isotopes	Relative intensity
M	three $^{79}$Br	1
M+2	two $^{79}$Br, one $^{81}$Br	3
M+4	one $^{79}$Br, two $^{81}$Br	3
M+6	three $^{81}$Br	1

The pattern is **M, M+2, M+4, M+6 with relative intensity 1:3:3:1** (option B).