Problem 214
Question
Phenol is heated with a solution of mixture of \(\mathrm{KBr}\) and \(\mathrm{KBrO}_{3} .\) The major product obtained in the above reaction is (a) 3-Bromophenol (b) 4-Bromophenol (c) \(2,4,6\)-Tribromophenol (d) 2-Bromophenol
Step-by-Step Solution
Verified Answer
The major product is (c) \(2,4,6\)-Tribromophenol.
1Step 1: Understand the Reagents
Phenol, when treated with a combination of \(\text{KBr}\) and \(\text{KBrO}_3\), undergoes bromination. The presence of \(\text{KBrO}_3\) acts as an oxidizing agent, which facilitates the production of bromine from \(\text{KBr}\). This setup leads to the formation of bromine molecules in situ, which can then react with phenol.
2Step 2: Analyze the Reaction Pathway
Phenol is known for its high reactivity towards electrophilic aromatic substitution reactions due to the -OH group's activating effect. When bromine is introduced, it preferentially substitutes at ortho and para positions relative to the -OH group. However, the reaction with such a powerful brominating system will result in multiple substitutions on the aromatic ring.
3Step 3: Determine the Bromination Outcome
Given that the \(\text{KBr}/\text{KBrO}_3\) mixture is a strong brominating agent, phenol undergoes bromination at all available ortho and para positions. Therefore, the major product of this reaction will be \( 2,4,6\)-tribromophenol, as the significant activation by the hydroxyl group leads to full bromination at all positions that are ortho and para to the hydroxyl group.
Key Concepts
BrominationPhenolAromatic Compounds
Bromination
Bromination is a chemical reaction where bromine atoms are introduced into a compound. In the context of aromatic compounds like phenol, bromination typically involves an electrophilic aromatic substitution. In this type of reaction, the compound reacts with an electrophile, in this case, bromine. The bromine molecule
initiates the reaction by interacting with the electron-rich aryl ring.
An essential aspect to consider is the brominating agents. A mixture of
KBr and KBrO
3 serves this purpose effectively. KBrO
3 acts as an oxidizing agent converting KBr into bromine molecules. This creates bromine in situ, which is highly reactive and enhances the bromination process. The reactivity of this mixture allows for the multiple bromination of phenol, resulting in tribromophenol under the right conditions.
Phenol
Phenol is a type of aromatic compound which features a hydroxyl group (-OH) directly attached to a benzene ring. This structural characteristic makes phenol considerably more reactive than benzene in electrophilic aromatic substitution reactions. The oxygen in the hydroxyl group donates electron density
into the benzene ring through resonance and inductive effects.
This electron-donating effect activates the benzene ring, making it more susceptible to bromination.
When bromine is present, the increased electron density makes the ortho and para positions relative to the hydroxyl group more
prone to substitution. In strong bromination reactions, phenol can undergo bromination at all available ortho and para positions.
In the reaction with KBr and KBrO
3, phenol's reactivity leads to the formation of 2,4,6-tribromophenol as multiple substitutions occur.
Aromatic Compounds
Aromatic compounds are a class of organic compounds characterized by the presence of one or more planar ring structures with delocalized
electrons. These compounds form stable arrangements due to resonance.
In electrophilic aromatic substitution reactions, such as bromination, the aromatic ring serves as a nucleophile. Its electron-rich pi
system interacts with electrophiles, like bromine, allowing substitutions to occur.
The degree of substitution on an aromatic ring can be influenced by substituents already attached to the ring. For example, in phenol, the
hydroxyl group increases the ring's nucleophilicity, influencing the reaction to favor ortho and para substitutions.
Through appropriate activation, aromatic compounds can undergo multiple substitutions, which is evident in the case of phenol being
transformed into 2,4,6-tribromophenol when reacting with a potent brominating agent.
Other exercises in this chapter
Problem 212
From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous \(\mathrm{ZnCl}_{2}\), is (a) 2-Butanol (b) 2 -Methylpropan \(
View solution Problem 213
Which of the following reagents may be used to distinguish between phenol and benzoic acid? (a) Tollen's reagent (b) Molisch reagent (c) Neutral \(\mathrm{FeCl}
View solution Problem 215
Iodoform can be prepared from all except (a) 3 - methyl \(-2\)-butanone (b) Isopropyl alcohol (c) Ethyl methyl ketone (d) Isobutyl alcohol
View solution Problem 216
Ortho \(-\) Nitrophenol is less soluble in water than pand \(\mathrm{m}-\) Nitrophenols because (a) Melting point of o- Nitrophenol is lower than those of \(\ma
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