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Which of the following is/are correct? (a) Phenol gives paraquinol with \(\mathrm{S}_{2} \mathrm{O}_{8}^{-2} / \mathrm{OH}^{-}\)as a major product. (b) Phenol can give electrophilic aromatic substitution reactions in absence of lewis acid. (c) Salicylic acid can give white/yellow solution with \(\mathrm{Br}_{2} / \mathrm{CS}_{2}\) (d) Ipso substitution is possible in phenol derivative if it contains \(-\mathrm{COOH}\) or \(-\mathrm{SO}_{3} \mathrm{H}\) group at ortho or para position with respect to OH group.

Short Answer

Expert verified
All options (a), (b), (c), and (d) are correct.

Step by step solution

01

Evaluating Option (a)

Phenol reacts with S_{2}O_{8}^{-2} / OH^{-} to form para-quinol as the major product. This reaction is correct as the strong oxidizing agent S_{2}O_{8}^{-2} facilitates this transformation. Therefore, option (a) is correct.
02

Evaluating Option (b)

Phenol is known to undergo electrophilic aromatic substitution reactions even in the absence of a Lewis acid. This is due to the activating effect of the hydroxyl group on the benzene ring, which makes the ring more reactive towards electrophiles. Hence, option (b) is correct.
03

Evaluating Option (c)

Salicylic acid, when treated with Br_{2}/CS_{2} , can form a white/yellow solution due to the formation of brominated products. This indicates that electrophilic substitution can occur, potentially at the ortho position relative to the carboxylic acid group. Thus, option (c) is correct.
04

Evaluating Option (d)

Ipso substitution refers to substitution at the same position as a leaving group on the aromatic ring. In phenol derivatives containing substituents like -COOH or -SO_{3}H at ortho or para positions relative to the OH group, ipso substitution can occur due to the acidic nature and increased resonance stabilization. Therefore, option (d) is also correct.

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Key Concepts

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

Electrophilic Aromatic Substitution
Electrophilic aromatic substitution is a fundamental reaction for aromatic compounds, like benzene and its derivatives. This process involves the replacement of a hydrogen atom on the aromatic ring by an electrophile, which is a species attracted to electrons. A fascinating aspect of this reaction is that phenol and its compounds can undergo this substitution even without the aid of a Lewis acid.

Here's why: the hydroxyl group (-OH) present in phenol is an incredibly powerful activating group. It increases the electron density on the benzene ring through resonance and inductive effects, making the ring more reactive. This heightened reactivity allows electrophiles to attack more easily, facilitating the substitution even in the absence of additional catalysts. Understanding this process helps us appreciate how phenol can naturally undergo these substitutions efficiently.
Ipso Substitution
Ipso substitution is a specific type of substitution that occurs directly at the position of a pre-existing substituent on an aromatic ring. In derivatives of phenol, when substituents such as -COOH or -SO3H are present at ortho or para positions relative to the OH group, ipso substitution becomes particularly feasible.

This potential arises because these groups enhance the stability of potential intermediates via resonance. Acids like -COOH and -SO3H can increase acidity and create better leaving conditions for the substituent. The aromatic ring, after such substitution, continues to demonstrate its extraordinary stabilization through resonance, accommodating the transiently disrupted electronic framework that arises during substitution.
Oxidation of Phenol
In chemistry, oxidation often involves the loss of electrons or an increase in oxidation state of a molecule. For phenol, oxidation has a specific transformative effect. When phenol is treated with strong oxidizing agents like persulfate ion ( S2O8^{-2} ), it converts primarily into para-quinol.

This reaction highlights phenol’s susceptibility to undergo oxidation due to its activated aromatic ring. The hydroxyl group (-OH) not only enhances reactivity but also directs these transformations towards specific products like para-quinol. Such oxidation reactions depict the intricate relationship between structure and reactivity in aromatic compounds.
Bromination of Salicylic Acid
The bromination of salicylic acid is a notable example of electrophilic aromatic substitution. When salicylic acid reacts with bromine ( Br2 ) in carbon disulfide (CS2), it forms brominated products that often appear as a white or yellow-colored solution.

This reaction emphasizes the ortho-directing influence of carboxylic acids, particularly when bromine specifically targets the position adjacent to the carboxylic acid group. Additionally, the reaction medium can have a significant role in driving the formation of specific brominated structures. Through this process, students can appreciate the nuances of how substituents on the aromatic ring influence the course of substitution reactions.

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Most popular questions from this chapter

Among the following, the most stable compound is (a) cis-1, 2-cyclohexanediol (b) trans- 1,2 -cyclohexanediol (c) cis-1, 3-cyclohexenediol (d) trans- 1,3 -cyclohexanediol

\(\mathrm{A} \frac{\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{\mathrm{H}_{2} \mathrm{SO}_{4}}-\mathrm{B} \underset{\text { vigrous oxidation }}{[\mathrm{O}]} \mathrm{CH}_{3} \mathrm{COOH}\) Here, B is propanone so A will be (a) Ethyl alcohol (b) Isopropyl alcohol (c) n-propyl alcohol (d) tertiary butyl alcohol

Iodoform can be prepared from all except (a) 3 - methyl \(-2\)-butanone (b) Isopropyl alcohol (c) Ethyl methyl ketone (d) Isobutyl alcohol

Identify the correct statements. (a) \(\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{O}-\mathrm{CH}_{3}\) on cleavage with \(\mathrm{PCl}_{5}\) gives equimolar mixture of ethyl chloride and methyl chloride (b) \(\mathrm{ClCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) is more reactive than \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) to prepare ether using Williamson's synthesis (c) Anisole can be prepared by the reaction of 1 \(\mathrm{CH}_{3} \mathrm{ONa}\) on chlorobenzene (d) Sodium metal can be used to dry diethyl ether, benzene and ethyl alcohol.

The compound which gives the most stable carbonium ion on dehydration is (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{OH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\)

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