The given reaction can occur in the presence of: (a) Bromine water (b) \(Br_2\) in \(CS_2\), 273 K (c) \(Br_2/FeBr_3\) (d) \(Br_2\) in \(CHCl_3\), 273 K Choose the correct answer from the options given below:
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To stop the substitution of phenol at the mono-stage, avoid polar protic solvents. Use \(CS_2\), \(CCl_4\), or \(CHCl_3\) at \(0^\circ C\).
Step 1: Understanding the Concept:
Bromination of phenol is highly sensitive to the solvent and conditions. Phenol is extremely reactive towards electrophilic aromatic substitution due to the strong activating effect of the \(-OH\) group. Step 2: Key Formula or Approach:
1. In highly polar solvents (like water), phenol ionizes to phenoxide ion, which is even more reactive, leading to polybromination.
2. In non-polar solvents (like \(CS_2\) or \(CHCl_3\)) at low temperatures, the reactivity is controlled, favoring monobromination. Step 3: Detailed Explanation:
- (a) Bromine water: Reaction with \(Br_2/H_2O\) yields a white precipitate of 2,4,6-tribromophenol. This is not the desired monobromo product.
- (b) \(Br_2\) in \(CS_2\), 273 K: The low polarity of \(CS_2\) and low temperature results in monobromination, giving p-bromophenol as the major product.
- (c) \(Br_2/FeBr_3\): This is the standard reagent for brominating benzene. While it works for phenol, it is unnecessarily strong and often leads to mixtures or over-bromination unless carefully controlled. It is not the "typical" laboratory condition for this specific transformation.
- (d) \(Br_2\) in \(CHCl_3\), 273 K: Similar to \(CS_2\), \(CHCl_3\) is a solvent of low polarity that prevents the formation of phenoxide ion, thus yielding the monobromo product. Step 4: Final Answer:
The monobromination of phenol to p-bromophenol is best achieved using \(Br_2\) in non-polar solvents like \(CS_2\) or \(CHCl_3\) at low temperatures. Thus, (b) and (d) are the correct conditions.
