Heterocyclic Compounds MCQ Quiz in मल्याळम - Objective Question with Answer for Heterocyclic Compounds - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 15, 2025
Latest Heterocyclic Compounds MCQ Objective Questions
Top Heterocyclic Compounds MCQ Objective Questions
Heterocyclic Compounds Question 1:
Match List I with List II
LIST I |
LIST II |
||
A. |
|
I. |
Oxepane |
B. |
|
II. |
Oxolane |
C. |
|
III. |
Oxetane |
D. |
|
IV. |
Oxirane |
Choose the correct answer from the options given below:
Answer (Detailed Solution Below)
Heterocyclic Compounds Question 1 Detailed Solution
Concept:
Classification of Cyclic Ethers
- Cyclic ethers are compounds containing oxygen within a ring structure.
- They are classified based on the number of atoms in the ring:
- Oxirane: A three-membered cyclic ether (e.g., ethylene oxide).
- Oxetane: A four-membered cyclic ether.
- Oxolane: A five-membered cyclic ether (e.g., tetrahydrofuran).
- Oxepane: A seven-membered cyclic ether.
Explanation:
Matching:
- A → I
- B → III
- C → IV
- D → II
Therefore, the correct option is 4.
Heterocyclic Compounds Question 2:
The major product of the reaction is:
Answer (Detailed Solution Below)
Heterocyclic Compounds Question 2 Detailed Solution
The correct answer is Option 4.
Explanation:
Step 1: Conjugated Addition of enamine followed by the cyclization to form the new six member ring. The enol must attack five member ring in a cis fashion.
Step 2: The reaction of hydroxylamine in acid solution. Formation of oxime of ketone followed by the hydrolysis of acetal and pyridine synthesis is completed by the cyclization and double dehydration.
Conclusion:
The major product of the reaction is:
Heterocyclic Compounds Question 3:
The product of the reaction is:
Answer (Detailed Solution Below)
Heterocyclic Compounds Question 3 Detailed Solution
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:
- 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.
- 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:
Heterocyclic Compounds Question 4:
The product of the reaction is
Answer (Detailed Solution Below)
Heterocyclic Compounds Question 4 Detailed Solution
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:
Heterocyclic Compounds Question 5:
The correct order of displacement of chloride with MeO- is
Answer (Detailed Solution Below)
Heterocyclic Compounds Question 5 Detailed Solution
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.
Heterocyclic Compounds Question 6:
Number of compounds among the following which contain sulphur as heteroatom is____.
Furan, Thiophene, Pyridine, Pyrrole, Cysteine, Tyrosine
Answer (Detailed Solution Below) 2
Heterocyclic Compounds Question 6 Detailed Solution
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.
Heterocyclic Compounds Question 7:
The major product formed in the following reaction is
Answer (Detailed Solution Below)
Heterocyclic Compounds Question 7 Detailed Solution
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 110oC.
- 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.
Heterocyclic Compounds Question 8:
The major products A and B in the following reaction sequence are
Answer (Detailed Solution Below)
Heterocyclic Compounds Question 8 Detailed Solution
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)
Heterocyclic Compounds Question 9:
The major product formed in the following reaction is
Answer (Detailed Solution Below)
Heterocyclic Compounds Question 9 Detailed Solution
Explanation:-
- The reaction pathway is shown below:
Conclusion:-
Hence, the major product formed in the following reaction is
Heterocyclic Compounds Question 10:
The product of the above reaction is
Answer (Detailed Solution Below)
Heterocyclic Compounds Question 10 Detailed Solution
Explanation:-
The reaction pathway is shown below:
Conclusion:-
- Hence, the product of the above reaction is