Heterocyclic Compounds MCQ Quiz in मल्याळम - Objective Question with Answer for Heterocyclic Compounds - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Hexane-2,5-dione Reaction with P₂S₅ and Raney Nickel

Let's break down the two-step transformation of hexane-2,5-dione to hexane:

1. Step 1: Reaction with P₂S₅ (Phosphorus Pentasulfide)
   • P₂S₅ is a reagent that converts carbonyl groups (C=O) to thio-carbonyl groups (C=S).
   • When hexane-2,5-dione reacts with P₂S₅, the two carbonyl groups (C=O) in the molecule are converted to thiocarbonyl groups (C=S), forming hexane-2,5-dithione.
2. Step 2: Reduction with Raney Nickel
   • Raney Nickel is a finely divided, highly active form of nickel used as a catalyst for hydrogenation reactions.
   • In the presence of Raney Nickel and hydrogen (H₂), the C-S or C=S are reduced to methylene groups (CH₂), forming the corresponding alkane.

Explanation:

• Mechanism with P₂S₅:
  • P₂S₅ reacts with the ketones in hexane-2,5-dione by replacing the oxygen atoms in the C=O bonds with sulfur atoms, forming the corresponding dithione with C=S bonds.
  • This transformation is crucial because the subsequent reduction step is more efficient on thiocarbonyl groups.
• Reduction with Raney Nickel:
  • Raney Nickel catalyzes the reduction of both C-S and C=C bond to convert it to alkane.

Conclusion:

The product of the reaction is:

The correct answer is Option 2.

Concept:

Indole Synthesis using 1-Methylnitrobenzene and Diethyl Oxalate

• This process involves the formation of an indole ring from 1-methylnitrobenzene and diethyloxalate.
• The reaction sequence includes the formation of an intermediate, followed by cyclization and subsequent reduction.

Explanation:

Step 1: The base OEt^- abstracts the acidic proton which attacks on the carbonyl group of the ester.

Step 2: H₂/Pt reduces the nitro group to amine group, which in the presence of carbonyl group near undergoes cyclization through condensation process.

Conclusion:

The product of the reaction is:

Concept:

Nucleophilic Aromatic Substitution (NAS)

• Nucleophilic Aromatic Substitution (NAS) occurs when a nucleophile displaces a leaving group on an aromatic ring.
• The presence of electron-withdrawing groups (EWGs) such as nitrogen in the pyridine ring tends to stabilize the negatively charged intermediate, facilitating the reaction.
• The position of the substituent (chlorine) on the pyridine ring affects the rate of substitution due to resonance and inductive effects.
• Resonating structure of Pyridne:
  • ​

​

Explanation:

The reactivity of chloro-substituted pyridine derivatives towards nucleophilic aromatic substitution follows the stability of the intermediate Meisenheimer complex formed during the substitution process. The order of ease of substitution is influenced by the electron-withdrawing effect of the nitrogen and the position of the chlorine atom:

• p-Chloropyridine (I): The chlorine is in the para position relative to the nitrogen. The para position is the most reactive as the nitrogen significantly stabilizes the negatively charged intermediate by resonance.
• o-Chloropyridine (II): The chlorine is in the ortho position relative to the nitrogen. The ortho position allows for significant stabilization by the nitrogen, albeit slightly less effective compared to the para position due to steric hindrance.
• m-Chloropyridine (III): The chlorine is in the meta position. This position experiences the least stabilization from the nitrogen, making the substitution reaction less favorable.
  • 

Conclusion:

The correct order of displacement of chloride with MeO⁻ is p-chloropyridine (I) > o-chloropyridine (II) > m-chloropyridine (III), Therefore, the reactivity order is I > II > III.

CONCEPT:

Sulfur as a Heteroatom

• Heteroatoms are atoms other than carbon or hydrogen that are part of a compound's structure, such as nitrogen, oxygen, sulfur, etc.
• Compounds containing sulfur as a heteroatom may include sulfur in rings or functional groups like thiol (-SH) or thioether (-S-).

EXPLANATION:

• Given compounds: Furan, Thiophene, Pyridine, Pyrrole, Cysteine, Tyrosine
• 
• 
• Analysis:
  • Furan: Contains oxygen as the heteroatom, no sulfur.
  • Thiophene: Contains sulfur in its five-membered aromatic ring.
  • Pyridine: Contains nitrogen as the heteroatom, no sulfur.
  • Pyrrole: Contains nitrogen as the heteroatom, no sulfur.
  • Cysteine: Contains a thiol (-SH) group, so it has sulfur as a heteroatom.
  • Tyrosine: Contains a phenolic (-OH) group, no sulfur.

• Total compounds with sulfur:
  • Thiophene
  • Cysteine

  Total: 2 compounds

The number of compounds containing sulfur as a heteroatom is 2.

Concept:-

Chichibabin Reaction:

• Chichibabin amination reaction is the amination of pyridines, quinoline, and other N-heterocycles with alkali-metal amides (NaNH₂) in a solvent.
• The nucleophilic attack at C-2 or C-6 takes place.
• If both the ortho-positions are blocked then amination takes place at the 4-position.
• Amination of pyridine takes place at the C-2 position by heating it with sodamide in dry toluene at 110^oC.
• The mechanism of the reaction is shown below:

Explanation:-

Zieglar Alkylation:

• When organolithium compounds are used in the amination of pyridines, quinoline, and other N-heterocycles then it is known as Zieglar Alkylation Reaction.
• The reaction pathway is shown below:

Conclusion:- 

• Hence, option 1 is the correct answer.

CONCEPT:

Reaction Mechanism and Intermediate Formation:

• This reaction involves the formation of an intermediate iminium ion followed by nucleophilic substitution.
• The initial step uses formaldehyde and dimethylamine to form the iminium intermediate.
• Methyl iodide (MeI) acts as the alkylating agent to form an N-methyl iminium salt.
• Finally, sodium cyanide (NaCN) replaces the methyl group with a cyanide group through a nucleophilic attack.

EXPLANATION:

• Step 1: Formaldehyde reacts with dimethylamine to form the iminium ion intermediate.

  HCHO + Me₂NH → H₂C=NMe₂

• Step 2: The iminium intermediate reacts with methyl iodide (MeI), resulting in the formation of an N-methyl iminium salt:

  H₂C=NMe₂ + MeI → Me₃N⁺−CH₂

• Step 3: Sodium cyanide replaces the methyl group with a cyanide group:

  Me₃N⁺−CH₂ + NaCN → Me₂N−CH₂CN

The major structures of A and B are as follows A: (CH₃)₂N−CH₂ and B: (CH₃)₂N−CH₂−CN

Therefore, the correct option is: (b)

Explanation:-

• The reaction pathway is shown below:

Conclusion:-

Hence, the major product formed in the following reaction is

Explanation:-

The reaction pathway is shown below:

Conclusion:-

• Hence, the product of the above reaction is​

Concept:

​Reduction of carbon-carbon double with H₂/Pd:

The reduction of the carbon-carbon double with H₂/Pd involves the following steps:

• H₂ dissociatively adsorbed onto the metal surface.
• Alkene π-bonds coordinated to the catalyst surface.
• Alkene π-bonds adsorbed onto the catalyst surface.
• A hydrogen atom is added sequentially onto both carbons.
• The reduced product can dissociate from the catalyst surface.

Explanation:

• The reaction pathway is shown below :
  • 
• In the first step of the reaction, H₂/Pd reduces the aromatic heterocycle to give a saturated heterocycle.
• The next step of the reaction involves the oxidation of the saturated heterocycle to give the final product.
• The oxidation of the saturated heterocycle involves both the lone pairs of the S atom, that become involved in bonds to oxygen.

Therefore, the correct option is 1.

Explanation:-

• The reaction pathway is shown below:

Conclusion:-

• Hence, the major product of the following reaction is