Organic Reactive Intermediates MCQ Quiz in தமிழ் - Objective Question with Answer for Organic Reactive Intermediates - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Reaction of Amines with Nitrous Acid (HONO)

• Primary aliphatic amines react with nitrous acid (HONO) to produce unstable diazonium salts, which immediately decompose to form alcohols.
• The reaction mechanism involves the generation of a nitrosating agent, which reacts with the amine group to form the diazonium intermediate.
• The diazonium intermediate gives the corresponding product.

Explanation:

Therefore, the correct answer is 2.

The correct answer is Option 1.

Explanation:

The given reaction is an example of one electron reduction of ketone using SmI₂.

• Reaction proceeds via the generation of carbon radical of carbonyl group followed by the cyclization with the alkene.

• 5 member ring is favoured over 6 member ring in radical reaction because most of the radical reaction are kinetic in nature. The rate of formation is greater for 5 member ring.

• In the next step, The electrophile Ac₂O also generates radical MeCOO^. and ^.COMe. SmI2 attached to more reactive MeCOO^. radical whereas other radical ^.COMe gets attached to the reactant.

Conclusion:

The major product of the transformation is:

CONCEPT:

Hunsdiecker Reaction

Hunsdiecker Reaction, also known as Hunsdiecker–Borodin reaction or Borodin reaction, is a chemical reaction in which silver salts of carboxylic acids react with a halogen to produce an unstable intermediate. The intermediate further undergoes a decarboxylation reaction thermally, leading to the creation of a final product known as alkyl halides. This reaction is an example of both halogenation and decarboxylation reactions.

The Hunsdiecker reaction is a radical reaction that converts silver salts of carboxylic acids into alkyl halides. Here’s a brief outline of the mechanism:

1. Formation of Silver Carboxylate: A carboxylic acid reacts with silver oxide or silver nitrate to form a silver carboxylate salt.

2. Halogenation: The silver carboxylate is treated with a halogen (like Br₂), forming a radical intermediate.

3. Homolytic Cleavage: The carbon-carbon bond of the carboxylate radical breaks, generating an alkyl radical.

4. Recombination: The alkyl radical reacts with a halogen atom to form the alkyl halide.

5. Byproducts: Silver halide (AgX) and carbon dioxide (CO₂) are produced.

EXPLANATION:

CONCLUSION:

So, the correct option is 1.

CONCEPT:

Stability of Carbocation:

The stability of a carbocation depends on several factors, including:

1. Inductive Effect:
   - Alkyl groups (e.g., methyl, ethyl) donate electron density through sigma bonds, helping stabilize the positively charged carbon.
   - More alkyl groups around the carbocation increase its stability (tertiary > secondary > primary).

2. Hyperconjugation:
   - The ability of adjacent C-H or C-C bonds to delocalize electron density into the empty p-orbital of the carbocation stabilizes it.
   - More neighboring C-H bonds (as in tertiary carbocations) contribute to greater stability.

3. Resonance Effect:
   - Carbocations that can stabilize the positive charge through resonance (delocalization of the charge) are more stable.
   - For example, allyl or benzyl carbocations are highly stabilized due to resonance delocalization.

4. Nature of Substituents:
   - Electronegative atoms or groups (e.g., halogens, nitro groups) can destabilize a carbocation by withdrawing electron density via the inductive or resonance effects, making the carbocation less stable.

5. Aromaticity:
   - Carbocations that contribute to an aromatic structure (e.g., cyclopropylmethyl carbocation) are more stable due to the stabilization of the positive charge by aromatic resonance.

6. Solvent:
   - Polar solvents, particularly polar protic solvents, can stabilize carbocations by solvation, where the positive charge is stabilized through interactions with solvent molecules.

EXPLANATION:

Analyze the given compounds:

Compound I: This compound is stabilized via resonance.

Compound II: This compound is anti-aromatic.

Compound III: This compound is aromatic. 

Stability order: Aromatic>Resonance>Anti-Aromatic

So the order is III>I>II

CONCLUSION:

So, the correct option is 4.

Concept:

The reaction involves the stereospecific opening of epoxides under both acidic and basic conditions using DMSO as a solvent. The stereochemistry of the reaction depends on the type of catalyst used (acid or base). Under acidic conditions, the epoxide undergoes ring opening with inversion of configuration at the benzylic position, leading to an anti product. Under basic conditions, the epoxide opens via a trans-diaxial mechanism, where the base attacks the less hindered carbon, preserving the stereochemistry.

• Acid-Catalyzed Epoxide Opening: In acidic conditions (DMSO, H⁺), the epoxide opens at the benzylic carbon via an anti mechanism. This results in the inversion of configuration at the benzylic carbon, producing an anti diol as the product (Q).

• Base-Catalyzed Epoxide Opening: In basic conditions (DMSO, OH^-), the base attacks the epoxide, leading to a trans-diaxial opening. This attack occurs at the less hindered carbon, and the reaction proceeds without inversion, preserving the stereochemistry, producing the diol product (R).

Explanation:

• Step 1: Acid-Catalyzed Opening – The acid protonates the oxygen of the epoxide, making it more electrophilic. The ring opens at the benzylic carbon, leading to inversion of configuration and formation of the anti diol product (Q).
• Step 2: Base-Catalyzed Opening – The hydroxide ion (OH^-) attacks the epoxide at the less hindered carbon, leading to a trans-diaxial opening. This results in a diol product (R) without inversion of configuration.
• 

Conclusion:

The correct answer is Option 3. In acidic conditions (H⁺), the epoxide opens with inversion at the benzylic position to form product Q, while under basic conditions (OH^-), the epoxide opens via a trans-diaxial mechanism to form product R, preserving the stereochemistry.

​Concept:

The elimination-addition mechanism occurs during nucleophilic aromatic substitution (SNAr) when using strong bases like NH₂^-. This mechanism is facilitated by the removal of a proton from the carbon adjacent to the halide, leading to the formation of a highly reactive benzyne intermediate. The relative reactivity of different halides (X = F, Cl, Br, I) depends on their leaving group ability and bond strengths.

• Benzyne Formation: The elimination step involves the removal of a halide and an adjacent proton, leading to the formation of benzyne. This is influenced by the leaving group ability of the halide. The weaker the bond between the halide and the benzene ring, the easier it is to eliminate, favoring benzyne formation.

• Reactivity Order: The reactivity order reflects the balance between bond strengths and the ability of the halide to leave. For halides like Br and I, the bond strength is weaker, leading to easier elimination, whereas F has a very strong bond with carbon, making elimination harder.

Explanation:

• Step 1: Deprotonation – The strong base NH₂^- abstracts a proton from the carbon adjacent to the halide (X), forming a carbanion.
• Step 2: Benzyne formation – The halide leaves as a halide ion (X^-), leading to the formation of a triple bond between the two carbons, creating a strained benzyne intermediate.
• Step 3: Nucleophilic attack – The nucleophile (NH₂^-) attacks the benzyne intermediate, leading to substitution at the carbon previously occupied by the halide.
• 

Conclusion:

The correct order of reactivity for halides in the elimination-addition mechanism with NH₂^- is Br > I > Cl > F, as shown in Option 3. This order is primarily due to the leaving group ability, with bromine and iodine being the best leaving groups, followed by chlorine and finally fluorine, which has a very strong bond with carbon and is the least reactive in this reaction.

Concept:

The given reaction proceeds via enamine formation, which is a key method in organic chemistry to perform alkylation or acylation at the alpha position of a carbonyl compound. Enamines are formed by the condensation of a secondary amine with a ketone or aldehyde. They act as nucleophiles, particularly effective for alkylation reactions.

• Enamine Formation: Enamines are formed when a ketone (cyclopentanone in this case) reacts with a secondary amine (such as pyrrolidine) under mildly acidic conditions. The enamine has a nucleophilic carbon at the alpha position, which is highly reactive towards electrophiles.

• Reactivity of Enamines: Enamines are more nucleophilic than enolates in some cases because the lone pair of electrons on nitrogen stabilizes the system. They react readily with alkyl halides in an SN2 fashion, leading to alpha-alkylation of the parent ketone after hydrolysis.

Explanation:

• Step 1: Enamine formation – Cyclopentanone reacts with a secondary amine (shown as pyrrolidine) to form the enamine, where the lone pair on nitrogen delocalizes, creating a nucleophilic alpha carbon.
• Step 2: Alkylation – The enamine, acting as a nucleophile, attacks the alkyl halide (i-PrBr) in an SN2 manner, resulting in alkylation at the alpha position of the ketone.
• Step 3: Hydrolysis – After alkylation, the enamine is hydrolyzed with aqueous acid (H₃O⁺), regenerating the ketone with the new alkyl group attached at the alpha position.
•  
  • Formation of α,β unsaturated ketone is more favorable than the formation of terminal alkene.

Conclusion:

The correct major product is option 1.

Concept:

Cation-pi cyclization is a type of cyclization reaction where a carbocation is stabilized by interaction with a pi system, such as a double bond or an aromatic ring. This stabilizing interaction makes the cation more reactive toward cyclization, leading to the formation of a cyclic structure. In this reaction, a carbocation is generated, and the pi bond of an alkene helps stabilize the transition state, facilitating the cyclization.

• Formation of Carbocation: The reaction starts with the formation of a carbocation at a suitable position, typically through protonation or loss of a leaving group.

• Pi Interaction: The carbocation interacts with the pi electrons of the double bond (or aromatic ring), allowing the cyclization to occur, leading to the formation of a more stable cyclic product.

Explanation:

• Step 1: Protonation of the Alkene – The alkene in the substrate is protonated by H₂SO₄, leading to the formation of a carbocation at the adjacent carbon.
• Step 2: Cation-Pi Cyclization – The carbocation interacts with the pi bond of the neighboring alkene, facilitating the cyclization to form a new ring structure. This step is driven by the stability provided by the pi-electron cloud.
• Step 3: Formation of the Product Q – After cyclization, a stable product (Q) is formed through loss of a proton, completing the cyclization reaction.
• Step 4: Formation of Product R – In the second reaction, a similar cation-pi cyclization occurs at a different position, leading to the formation of product R through the same mechanism.
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  • Mechanism:
    • ​

Conclusion:

The correct products Q and R are given in option 3.

Concept:

The reaction shown is an example of a **titanium-mediated pinacol coupling reaction**. This reaction involves the reductive coupling of two aldehyde groups to form a 1,2-diol (pinacol) using titanium-based reagents. The following key points explain the concept:

• Titanium Mediated Reduction: Titanium trichloride (TiCl₃) in the presence of zinc acts as a reducing agent, facilitating the conversion of aldehydes to radicals, which then couple to form a pinacol.
• Radical Mechanism: The reaction proceeds via a radical mechanism where two carbonyl (aldehyde) groups are reduced to radical intermediates, which dimerize to form a C–C bond.
• Formation of 1,2-Diols: The coupling of two aldehydes results in the formation of a vicinal diol (1,2-diol), also known as a pinacol.
• Stereoselectivity: The reaction can exhibit stereoselectivity, depending on the conditions, leading to the formation of specific stereoisomers of the diol product.

Explanation:

A mechanistic scheme for the titanium-mediated coupling of two carbonyl groups involves an initial one-electron transfer from titanium metal to the aldehyde or ketone carbonyl group to produce radical anions (ketyl species), which then dimerize to a Ti-pinacolate. Hydrolysis of the pinacolate generates the vicinal diol.

• Step 1: Reduction of Aldehydes – The aldehyde groups are reduced by the titanium trichloride-zinc system, generating radical anions (ketyl species).
• Step 2: Radical Coupling – The two radical intermediates dimerize to form a C–C bond, resulting in the formation of a Ti-pinacolate intermediate.
• Step 3: Hydrolysis – Hydrolysis of the Ti-pinacolate intermediate with aqueous K₂CO₃ yields the final 1,2-diol (pinacol) product.
• Reaction:
  • ​
• Mechanism:
  • ​

Conclusion:

The correct major product is option 2, which is the 1,2-diol formed through the titanium-mediated pinacol coupling reaction.

Concept:

The reaction shown proceeds via the formation of a nitrene intermediate, which is highly reactive and can participate in various types of reactions. Key points about nitrene formation and its reactivity include:

• Nitrene Formation: Nitrenes are nitrogen-centered species with a lone pair of electrons and an empty p-orbital. They are typically generated from azides or imines in the presence of heat, light, or chemical reagents, such as triphenylphosphine (PPh₃)/DEAD in this reaction.
• Reactivity of Nitrenes: Nitrenes are highly reactive intermediates that can undergo various transformations, such as C-H insertion, cycloaddition, and rearrangement. Due to their electron deficiency, they are electrophilic and can insert into carbon-hydrogen bonds to form amines.
• Nitrene Insertion: In this reaction, the nitrene inserts into a C-H bond, leading to the formation of the product. This type of insertion typically occurs with high regioselectivity and stereoselectivity.

Explanation:

The reaction mechanism proceeds as follows:

• Step 1: Nitrene Formation – The azide group (-N₃) is converted into a nitrene intermediate through a Staudinger reaction using PPh₃/DEAD, which cleaves the N-N bonds and generates the reactive nitrene species.
• Step 2: Nitrene Insertion – The nitrene intermediate inserts into a nearby C-H bond, forming a new C-N bond. This insertion is highly regioselective and results in the formation of the amine group (-NH₂) in the product.
• Step 3: Cyclization – The newly formed amine group participates in a cyclization reaction, forming a new ring structure by interacting with the ester functionality, leading to the final product.
• Step 4: Hydrolysis – The final step involves the hydrolysis of any intermediate ester groups under aqueous conditions, resulting in the formation of the amine product.
• 

Conclusion:

The correct major product is option 1, where the nitrene inserts into the C-H bond, followed by cyclization and hydrolysis to yield the final product.