Phenols MCQ Quiz - Objective Question with Answer for Phenols - Download Free PDF

Option (2) is correct. 

Explanation :

In the first step, the -OH of -CH₂-OH group is substituted by Br atom.

In this reaction, the reaction of ether takes place with HBr in which the deavage of ether taken place in such a way that aromatic alcohol is formed

Concept:

The Kolbe-Schmitt reaction is a carboxylation reaction in which sodium phenoxide reacts with carbon dioxide to form sodium salicylate, which is then converted to salicylic acid upon acidification. This reaction is used in the industrial synthesis of salicylic acid, a precursor to aspirin.

• Electrophile: In this reaction, the electrophile is carbon dioxide (CO₂), which reacts with sodium phenoxide to form the salicylate ion.

• Reaction Conditions: The reaction occurs at high temperature and pressure to drive the carboxylation step.

• Product: The product formed is sodium salicylate, which is converted to salicylic acid upon treatment with an acid (H⁺).

• Reaction Mechanism:

  • 

Explanation: 

• Statement 1 is correct: CO₂ acts as the electrophile in the reaction, reacting with sodium phenoxide.

• Statement 2 is correct: The product formed is sodium salicylate.

• Statement 3 is correct: The reaction occurs under high pressure to facilitate the carboxylation of phenoxide.

• Statement 4 is incorrect: Sodium salicylate is converted to salicylic acid only after acidification, not directly in the reaction.

Conclusion:

The correct answer is: The product 'B' formed in the above reaction is salicylic acid upon acidification, not directly after the reaction.

CONCEPT:

Reaction of Compounds with Dilute NaOH

• Compounds that exhibit acidic properties can react with dilute NaOH to form a salt and water.
• Phenol (C₆H₅OH) is more acidic than alcohols like ethanol (C₂H₅OH) due to resonance stabilization of its conjugate base (phenoxide ion).
• Other alcohols like ethanol, benzyl alcohol (C₆H₅CH₂OH), and tertiary butyl alcohol (CH₃)₃COH do not react with NaOH readily as they are less acidic than phenol.

EXPLANATION:

• Phenol (C₆H₅OH) has an -OH group directly attached to an aromatic ring (benzene), which increases its acidity due to resonance stabilization of the phenoxide ion after deprotonation.
• When phenol reacts with dilute NaOH, it forms sodium phenoxide (C₆H₅O⁻Na⁺) and water, indicating that it can readily react with a base.
• Other alcohols such as ethanol, benzyl alcohol, and tertiary butyl alcohol are less acidic and do not form stable anions in the presence of NaOH, so they do not react as readily.

REACTION:

C₆H₅OH + NaOH → C₆H₅O⁻Na⁺ + H₂O

CONCEPT:

Nucleophilic Substitution and Aromatic Amine Formation

• The reaction involves the nucleophilic substitution of a chlorine atom on a methyl group-substituted phenyl ring by an amine group (NH₂) in the presence of sodium amide (NaNH₂), followed by the addition of ammonia (NH₃).
• The nucleophilic substitution proceeds through an S_NAr mechanism due to the presence of an electron-withdrawing group (-OCH₃) on the aromatic ring, which makes the ring more susceptible to attack by the nucleophile (NH₂).
• In the reaction, the chlorine atom is replaced by the NH₂ group at the position ortho to the methoxy group (-OCH₃) due to the electron-withdrawing nature of the methoxy group, which favors the attack at the ortho position.

EXPLANATION:

• Step 1: The NaNH₂ base deprotonates the amine, generating a strong nucleophile (NH₂⁻), which attacks the aromatic ring where the chlorine atom is present.
• Step 2: The methoxy group (-OCH₃) directs the nucleophilic substitution to the ortho position, resulting in the displacement of chlorine and the attachment of the NH₂ group.
• Step 3: The addition of ammonia (NH₃) in the final step ensures the amine (-NH₂) remains in the product at the ortho position relative to the methoxy group.

Therefore, the correct product is the compound shown in option A,

CONCEPT:

pKa and Acidity

• The pKa value of a compound is a measure of its acidity. A lower pKa value indicates a stronger acid, while a higher pKa value indicates a weaker acid.
• The presence of electron-withdrawing groups (EWGs) such as NO₂ can stabilize the negative charge on the conjugate base, making the compound more acidic and thus lowering its pKa value.
• The position of the EWG relative to the hydroxyl group (OH) also affects the acidity. Ortho (o) and para (p) positions have a stronger influence due to resonance and inductive effects compared to the meta (m) position.

EXPLANATION:

• Phenol (C₆H₅OH) has no electron-withdrawing groups attached, so it will have a higher pKa value compared to the nitro-substituted phenols.
• o-O₂N-C₆H₄-OH (ortho-nitrophenol) — The nitro group is at the ortho position, and it has a strong electron-withdrawing effect, making this compound more acidic.
• p-O₂N-C₆H₄-OH (para-nitrophenol) — The nitro group is at the para position, and it also has a strong electron-withdrawing effect, making this compound more acidic.
• m-O₂N-C₆H₄-OH (meta-nitrophenol) — The nitro group is at the meta position, and its electron-withdrawing effect is weaker than in the ortho and para positions, making this compound less acidic compared to ortho and para.
• C₆H₅OH (phenol) — Without any electron-withdrawing groups, phenol is the least acidic among the options.
• Therefore, phenol (C₆H₅OH) has the highest pKa value among the given compounds.

Therefore, the compound having the maximum value of pKa is phenol (C₆H₅OH).

CONCEPT:

Oxidative Cleavage of Cycloalkenes

• Oxidative cleavage involves breaking the carbon-carbon double bond of a cycloalkene using strong oxidizing agents like KMnO₄ in acidic conditions.
• This reaction typically leads to the formation of dicarboxylic acids, depending on the structure of the cycloalkene.
• The reaction mechanism involves the addition of oxygen atoms to the double bond, followed by cleavage and oxidation of the resulting fragments.

EXPLANATION:

• In the given reaction, cyclohexene is oxidized by KMnO₄ in the presence of H₂SO₄ (acidic medium) under heat.
• This leads to the oxidative cleavage of the double bond and formation of adipic acid, which is a six-carbon dicarboxylic acid.
• Thus, the product "A" formed in the reaction is adipic acid, corresponding to option (4).

The correct answer is option (4): Adipic acid.

The correct answer is Anthocyanins.Key Points

• Anthocyanins is a water-soluble pigment that is present in pomegranate, eggplant, black carrots, and red cabbage.
• It belongs to a phenolic group.
• These are responsible for the red, purple, and blue colors in many fruits, vegetables, and flowers.
• They have antioxidant and anti-inflammatory properties and are being studied for their potential health benefits.

Additional Information

• Phycoerythrin is a water-soluble pigment found in red algae and cyanobacteria.
  • It is used in biotechnology as a fluorescent label and in medical research as a diagnostic tool.
• Phycobilins are water-soluble pigments found in cyanobacteria and some red algae.
  • They are involved in photosynthesis and act as accessory pigments that capture light energy.
• Betalains are water-soluble pigments found in beets and cacti.
  • They are responsible for the red and yellow colors in these plants and have antioxidant properties.

Option (2) is correct. 

Explanation :

In the first step, the -OH of -CH₂-OH group is substituted by Br atom.

In this reaction, the reaction of ether takes place with HBr in which the deavage of ether taken place in such a way that aromatic alcohol is formed

Concept:

The Kolbe-Schmitt reaction is a carboxylation reaction in which sodium phenoxide reacts with carbon dioxide to form sodium salicylate, which is then converted to salicylic acid upon acidification. This reaction is used in the industrial synthesis of salicylic acid, a precursor to aspirin.

• Electrophile: In this reaction, the electrophile is carbon dioxide (CO₂), which reacts with sodium phenoxide to form the salicylate ion.

• Reaction Conditions: The reaction occurs at high temperature and pressure to drive the carboxylation step.

• Product: The product formed is sodium salicylate, which is converted to salicylic acid upon treatment with an acid (H⁺).

• Reaction Mechanism:

  • 

Explanation: 

• Statement 1 is correct: CO₂ acts as the electrophile in the reaction, reacting with sodium phenoxide.

• Statement 2 is correct: The product formed is sodium salicylate.

• Statement 3 is correct: The reaction occurs under high pressure to facilitate the carboxylation of phenoxide.

• Statement 4 is incorrect: Sodium salicylate is converted to salicylic acid only after acidification, not directly in the reaction.

Conclusion:

The correct answer is: The product 'B' formed in the above reaction is salicylic acid upon acidification, not directly after the reaction.

CONCEPT:

Electrophilic Substitution and Reaction Feasibility

• Electrophilic substitution reactions often occur on aromatic rings, but the feasibility depends on the stability of the intermediate formed during the reaction.
• In some cases, certain reactions are not possible due to the lack of suitable conditions or the inability of the reagent to react with the given substrate.
• Alcohols generally do not react with HCl directly to form aryl halides due to the instability of the transition state or intermediate.

EXPLANATION:

• Option 2 is not possible because the direct reaction of phenol (an -OH group attached to a benzene ring) with HCl does not result in the formation of chlorobenzene.
• The hydroxyl group in phenol is not easily replaced by a halogen in the absence of a suitable activating reagent or catalyst.
• Thus, option (2) is the correct choice as this reaction is not feasible under normal conditions.

The correct answer is option (2).

CONCEPT:

Formation of Picric Acid from Phenol

• In this reaction sequence, phenol first reacts with concentrated sulfuric acid (H2SO4) to form phenol-2,4-disulfonic acid (Product A).
• Subsequently, Product A undergoes nitration with concentrated nitric acid (HNO3) to form 2,4,6-trinitrophenol (picric acid), which is Product B.
• Picric acid has three nitro groups and one hydroxyl group attached to the benzene ring.

Explanation:-

• Product A (Phenol-2,4-disulfonic acid): Contains 8 oxygen atoms (2 SO3H groups, each contributing 3 oxygen atoms, plus 1 OH group contributing 1 oxygen atom).
• Product B (Picric acid, 2,4,6-trinitrophenol): Contains 6 oxygen atoms (3 NO2 groups, each contributing 2 oxygen atoms, plus 1 OH group contributing 1 oxygen atom).
• Total sum of oxygen atoms in Product A and Product B: 8 (Product A) + 6 (Product B) = 14 oxygen atoms.
• 

Picric acid is prepared by treating phenol first with concentrated sulphuric acid which converts it to phenol-2, 4-disulphonic acid and then with concentrated nitric acid to get 2, 4, 6 trinitrophenol. 

Conclusion:-

• The total sum of oxygen atoms in Product A and Product B is 14.

Concept:

Decarboxylation of acids - The process of removing CO₂ from carboxylic acid is known as decarboxylation.

Decarboxylation is carried out heating the acid with sadalime i.e. a mixture of NaOH + CaO at 630K.

Thus, the sodium salt of carboxylic acid on distillation gives hydrocarbons.

\(CH_3COONa + NaOH → CH_4 + Na_2CO_3\)

Explanation:

→ Decarboxylation of salicylic acid will give phenol.

• This is an important method for preparation of phenol.
• When salicylic acid is heated with a mixture of NaOH +CaO phenol and CO₂ is produced.

The reaction is shown as - 

→ Malonic acid on decarboxylation gives carboxylic acid - 

→ Benzoic acid on decarboxylation gives  benzene -

→ Phthalic acid on decarboxylation gives benzoic acid -

Conclusion:

Thus, only salicylic acid give Phenol as the product when its Sodium salt is heated in presence of Sodalime.

Correct answer: 4)

Concept:

• The Friedel–Crafts acylation is the reaction of an arene with acyl chlorides or anhydrides using a strong Lewis acid catalyst.
• This reaction proceeds via electrophilic aromatic substitution to form monoacetylated products.
• The Friedel–Crafts alkylation is the reaction of an arene with alkyl halides using a strong Lewis acid catalyst.
• This reaction proceeds via electrophilic aromatic substitution to form monoacetylated products such as toluene.

Explanation:

• When phenol is treated with chloromethane in presence of AlCl₃, we get a mixture of o & p-cresol (o-hydroxy toluene and p-hydroxy toluene).
• It is an example of Friedel Craft's alkylation reaction.

Conclusion:

Thus, the product of the given reaction is o-hydroxy toluene and p-hydroxy toluene.

CONCEPT:

Formation of Methyl Groups in Organic Synthesis

• This sequence of reactions involves the conversion of compound P through intermediate stages to compound S, with specific transformations adding methyl (-CH₃) groups at each step.
• Reactions with reagents like NH₂OH, KOH, and CH₃MgBr are critical in adding or rearranging functional groups to increase the number of methyl groups in the final structure.

EXPLANATION:

• Step-by-step transformation:
  • Compound P reacts with hydroxylamine (NH₂OH) to form compound Q with multiple oxime groups.
  • Subsequent reactions with KOH and methylation reagents yield compound R, which sets up the structure for further methyl additions.
  • In the final step, excess CH₃MgBr reacts with R, resulting in compound S, where additional methyl groups are introduced around the central structure.

• 

  

• Counting the methyl groups in compound S gives a total of 12 -CH₃ groups.

Answer: 12

Concept:

Electrophilic substitution reactions are a key class of reactions in organic chemistry, particularly important for aromatic compounds like phenols. Phenols are highly reactive towards electrophilic substitution compared to benzene due to the presence of the hydroxyl group (-OH), which is an activating and ortho/para-directing group. Here are 3-4 important points about the electrophilic substitution reactions of phenols:

• Activating Effect of Hydroxyl Group: The -OH group in phenol increases the electron density on the benzene ring, particularly at the ortho (2, 6) and para (4) positions. This makes phenol more reactive towards electrophiles compared to benzene.
• Ortho/Para Directing: Due to resonance, the electron-donating effect of the hydroxyl group activates the ortho and para positions, directing incoming electrophiles to these sites. This results in major products being ortho and para isomers.
• Common Electrophilic Substitution Reactions: Phenols commonly undergo halogenation (e.g., bromination, chlorination), nitration, sulfonation, and Friedel-Crafts alkylation/acylation. For example, in bromination, phenol reacts with bromine water to give 2,4,6-tribromophenol.
• Factors Affecting Electrophilic Substitution: The reactivity and orientation of electrophilic substitution on phenol can be influenced by the nature of the electrophile, the specific conditions (e.g., solvent, temperature), and the presence of additional substituents on the aromatic ring.

Explanation:

The OH group of Phenol is ortho and Para directing whereas SO₃H is meta directing.

The 3rd equivalent of Bromine attack at para of OH group. Since, SO₃H is a good leaving group, The final product has 3 bromine on ortho and para position of OH.

Reaction: 

Conclusion:

The correct product of the given reaction is: