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Phenol reacts with bromine in carbon disulphide at low temperature to give (a) \(m\)-bromophenol (b) \(o\) - and \(p\)-bromophenol (c) \(p\)-bromophenol (d) \(2,4,6\)-tribromophenol

Short Answer

Expert verified
The reaction produces (b) o- and p-bromophenol.

Step by step solution

01

Understanding the Reaction Context

Phenol, which is a hydroxyl group attached to a benzene ring, acts as an activating group for electrophilic aromatic substitution reactions due to the resonance stabilization it provides. This makes it highly susceptible to react with bromine to form brominated derivatives.
02

Analyzing Reaction Conditions

The reaction takes place at low temperatures and in carbon disulfide as a solvent. These conditions favor controlled substitution reactions, typically leading to less substituted products compared to reactions occurring in water where more substitution may occur.
03

Identifying Bromination Pattern

The hydroxyl group on phenol is an ortho-para directing group. This means that electrophiles, like bromine, are more likely to add to the ortho (adjacent) and para (opposite) positions relative to the hydroxyl group.
04

Predicting the Major Product

Given the ortho-para directing nature of the hydroxyl group and the mild conditions (low temperature, non-polar solvent), the reaction is more controlled, primarily producing mono-substituted bromophenol. Thus, the para position, which is less sterically hindered compared to ortho positions (owing to one of them being blocked by another bromine), is most favored.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phenol Bromination
Phenol bromination is an interesting process used in organic chemistry, particularly in electrophilic aromatic substitution reactions. Phenol, a benzene ring with a hydroxyl group, makes it more reactive than benzene alone due to the activation from the hydroxyl group. This increased reactivity allows phenol to easily undergo a substitution reaction with bromine.

When bromine is reacted with phenol, the hydroxyl group's influence makes the benzene ring more attractive to bromine, which acts as an electrophile. This causes the bromine to attach to the benzene ring.
  • The process is called electrophilic because bromine acts as an electron-seeking species.
  • It's aromatic because the reactions occur on a benzene ring.
  • It's a substitution as the bromine replaces a hydrogen atom on the ring.
The mild reaction conditions and the specific chemical nature of phenol favor certain positions on the benzene ring for bromination. This is where the directing nature of the hydroxyl group plays a pivotal role.
Ortho-Para Directing Groups
In organic chemistry, understanding directing groups is crucial to predicting how substitution reactions will progress. The hydroxyl group in phenol is a classic example of an ortho-para directing group, meaning it encourages substitution at the ortho (adjacent) and para (opposite) positions relative to itself on the aromatic ring. This happens due to resonance structures where the oxygen atom donates electron density through the benzene ring, substantially increasing electron density at ortho and para positions.

Ortho-para directing groups are generally activating groups. Here are some features to recognize them:
  • They enhance the reactivity of the benzene ring towards electrophiles.
  • They stabilize the intermediate carbocation formed during the reaction by resonance.
  • They lead to higher electron density at ortho and para positions, making these sites more attractive for electrophilic attack.
For phenol, this means that bromine will likely attach to these positions, resulting in ortho-bromophenol and para-bromophenol as major products.
Reaction Conditions
The reaction conditions play a significant role in determining the outcome of a chemical reaction. For phenol bromination, specific conditions are required to control the reaction and obtain desired products. The reaction is carried out in carbon disulfide ( ext{CS}_2) at low temperatures. These conditions are critical for a few reasons:

  • Low temperatures help minimize over-bromination, which could lead to more bromine atoms attaching to the benzene ring than desired, such as in the production of 2,4,6-tribromophenol.
  • ext{CS}_2 is a non-polar solvent, which aids in facilitating a more controlled reaction environment compared to polar solvents like water.
  • In a controlled non-polar environment, there is a tendency for the electrophile to selectively target the less hindered para position more than the ortho position.
By understanding and controlling these conditions, chemists can effectively steer the reaction towards the production of the preferred mono-brominated product, particularly targeting the para position in the case of phenol.

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Most popular questions from this chapter

When \(t\)-butanol and \(n\)-butanol are separately treated with a few drops of dilute \(\mathrm{KMnO}_{4}\), in one case only the purple colour disappears and a brown precipitate is formed. Which of the two alcohols gives the above reaction and what is the brown precipitate?

Match the following, choosing one item from column \(\mathbf{X}\) and one from column Y. \(\mathbf{X}\) \(\mathbf{Y}\) (i) pyrolysis of alkanes (a) elimination reaction (ii) benzene+chloroethane (b) saponification (+anhydrous \(\mathrm{AlCl}_{3}\) ) (iii) \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\) (c) Wurtz reaction \(+\mathrm{NaOH}\) (iv) preparation of alkanes (d) Friedel-Crafts reaction (v) phenol (e) Reimer-Tiemann \(+\mathrm{CHCl}_{3}(\mathrm{NaOH})\) reaction (vi) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+\mathrm{alc} . \mathrm{KOH}\) (f) cracking

The reaction products of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OCH}_{3}+\mathrm{HI} \stackrel{\Delta}{\longrightarrow}\) is : (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{I}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}+\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}+\mathrm{HOI}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3} \mathrm{OH}\)

Allyl phenyl ether can be prepared by heating: [Main Online April 9, 2014] (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{ONa}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{ONa}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{Br}+\mathrm{CH}_{3}-\mathrm{ONa}\) (d) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Br}+\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{ONa}\)

A compound of molecular formula \(\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}\) is insoluble in water and dilute sodium bicarbonate but dissolves in dilute aqueous sodium hydroxide. On treatment with bromine water, it readily gives a precipitate of \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{OBr}_{3} .\) Write down the structure of the compound.

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