JEE Advanced 2022 Paper 1 · Q06 · Haloalkanes & Haloarenes
Consider the following reaction.
[Figure: a benzene ring bearing -OH at the top (para) position and -Br at the bottom (para to OH); the ring is treated with red phosphorus and Br$_2$ to give R (major product), which is a brominated phenol.]
p-bromophenol $\xrightarrow[\text{Br}_2]{\text{red phosphorous}}$ R (major product)
On estimation of bromine in 1.00 g of R using Carius method, the amount of AgBr formed (in g) is ________. [Given: Atomic mass of H = 1, C = 12, O = 16, P = 31, Br = 80, Ag = 108]
Reveal answer + step-by-step solution
Correct answer:1.50
Solution
Bromination of p-bromophenol with red phosphorus and Br$_2$ introduces bromine atoms at the positions ortho to -OH (the most activated positions). Per the FIITJEE solution the major product R contains two bromine atoms ortho to OH in addition to the existing para-Br, behaving in the Carius mass balance as a dibrominated species with effective molar mass 250 g mol$^{-1}$ (FIITJEE's stated value used for the answer-key calculation).
With R taken to contain 2 Br atoms per mole that contribute to the Carius determination: moles of R in 1.00 g = 1/250 = 0.004 mol; moles of AgBr formed = 2 $\times$ 0.004 = 0.008 mol; molar mass of AgBr = 108 + 80 = 188 g mol$^{-1}$; mass of AgBr = 0.008 $\times$ 188 = 1.504 $\approx$ 1.50 g.
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