Phenols MCQ Quiz in मराठी - Objective Question with Answer for Phenols - मोफत PDF डाउनलोड करा

CONCEPT:

Reaction Sequence for the Formation of Ca-soap (Compound R)

• In this reaction sequence, the main product P is first formed by the transformation of the given starting material.
• P then reacts with glycerol in the presence of an acid catalyst to form compound Q.
• Compound Q undergoes saponification with excess NaOH, followed by treatment with CaCl₂, to yield the Ca-soap, R.

EXPLANATION:

• 

  1 mole of Q will give 1.5 mole of R.

  So, mass of R produced = 606 g × 1.5

  = 909 g

Conclusion:

• The amount of R produced starting with one mole of Q is 909 g.

CONCEPT:

Reaction of Acetaldehyde with Excess Formaldehyde and Subsequent Reactions

• When acetaldehyde reacts with excess formaldehyde in the presence of NaOH, it undergoes an aldol condensation and Cannizzaro reaction to produce compounds P and Q.
• P does not give Tollens’ test, indicating it is not an aldehyde, while Q gives a positive Tollens’ test, suggesting it contains an aldehyde group.
• Treatment of compound P with excess cyclohexanone in the presence of p-toluenesulfonic acid (PTSA) produces a cyclic compound R.
• To find the answer, we need to determine the number of methylene (–CH₂–) groups and oxygen atoms in R.

EXPLANATION:

• The formation of compound P and Q from acetaldehyde and formaldehyde is shown in the reaction mechanism.
  • Compound P is the product of the aldol reaction, while Q is the Cannizzaro product (formic acid).
  • P is then treated with cyclohexanone in the presence of PTSA to form a large cyclic compound R.
• Counting the number of –CH₂– groups and oxygen atoms in R:
  • Number of –CH₂– groups = 14
  • Number of oxygen atoms = 4
• The sum of the methylene groups and oxygen atoms in R is:

  14 + 4 = 18

Conclusion:

• The sum of the number of methylene (–CH₂–) groups and oxygen atoms in compound R is 18.

Concept:

In the given reaction sequence, we analyze the transformation of a phenolic compound through various steps to determine the correct structures for intermediates and the final product.

Explanation: 

• Step 1: The reaction starts with a phenol derivative with two -OH groups in an ortho position. Upon treatment with NaOH, one of the hydroxyl groups is deprotonated, forming a phenoxide ion, which acts as a nucleophile.

• Step 2: The phenoxide ion reacts with CH₂=CH–COCl (an acid chloride), attacking the carbonyl carbon and forming an ester linkage.

  The product Q matches the structure in Option 3, which has an ester group attached to the phenolic ring.

• Step 3: Treatment with NaOH leads to hydrolysis of the ester, converting it into a carboxylate intermediate.

• Step 4: The final step involves reduction with LiAlH₄, which reduces the carboxylate group to a -CH₂CH₂OH group attached to the benzene ring.

Conclusion:

The correct structures for intermediates and products are:

• For Q, Option 3 is correct, showing the ester formation.

• For S, Option 1 is correct, indicating the final reduced product with a -CH₂CH₂OH group.

The correct answer is Options A and C.

CONCEPT:

Ozonolysis of Alkenes

• Ozonolysis is a reaction in which an alkene reacts with ozone (O₃) to form an ozonide, which is then cleaved by reductive work-up (e.g., Zn/H₂O) to give carbonyl compounds.
• The reaction splits the double bond of the alkene, and each carbon of the double bond becomes part of a carbonyl group (C=O).

EXPLANATION:

• Given Reaction:

  CH₃CH=CHCH₂CH₃ + O₃ → [Ozonide] → CH₃CHO + CH₃CH₂CHO (on reductive work-up with Zn/H₂O)

• 
• Product Analysis:
  • The alkene splits at the double bond, forming two products: acetaldehyde (CH₃CHO) and propanal (CH₃CH₂CHO).
  • Each product contains one carbonyl group (C=O), which has one oxygen atom.
• Total Oxygen Atoms:
  • Since the product molecule contains only one carbonyl group, the total number of oxygen atoms present per molecule is 1.

Therefore, the total number of oxygen atoms present per molecule of the product (P) is 1.

Correct answer:1)

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.

Correct answer: 2)

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.

Correct answer: 2)

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:

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.