Generation, Stability and Reactivity MCQ Quiz in తెలుగు - Objective Question with Answer for Generation, Stability and Reactivity - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:-

Stability of Carbocation:

• The stability of a carbocation depends on several factors like hyperconjugation, resonance, and inductive effect. 
• Out of these factors, the resonance effect is the most important factor for the stability of a carbocation.
• The observed order of stability for carbocations is as follows:

tertiary>secondary>primary>methyl.

Explanation:-

• The carbocations I and III are bridgehead carbocations. A carbocation is sp² hybridized. Hence it has a planar geometry.

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• The C atom at the bridgehead position of a  bridgehead carbocation is not planar. This is because the C atom experiences a sufficient angle strain at the bridgehead position due to sp² hybridization. Hence carbocations I and III are unstable carbocations.
• Among the bridgehead carbocations, the carbocation with more number of C atoms in the ring will have less steric strains than the other carbocations and hence will be stable to some extent.\
• Thus, Carbocation II with a higher number of C atoms in the ring will be more stable than Carbocation I.
• Now, carbocation II is a tertiary carbocation. The tertiary Carbocation III is the most stable carbocation among all the three carbocations as it is a tertiary carbocation, which is stabilized by both the inductive and hyperconjugation effect.
• Thus the stability order will be II>III>I

Conclusion:-

• Hence, the correct order of stability of the following carbocation is II>III>I.

The correct answer is option 1

Concept:-

• Nucleophilic Carbon: The deprotonated 1,3-dithiane acts as a nucleophile due to its carbanionic character.
• SN2 Reaction: This reaction mechanism involves the backside attack of a nucleophile and is typically used for alkylation steps.
• Steric Considerations: The intramolecular SN2 reaction is favored due to the proximity of the reacting centers within the same molecule, leading to cyclization.
• Oxidative Desulfurization: The conversion of sulfur-containing groups to carbonyl groups using oxidative conditions.

Explanation:-

• Deprotonation and Alkylation:
• Step 1a: The 1,3-dithiane is treated with n-butyllithium (n-BuLi) in tetrahydrofuran (THF). n-BuLi serves as a strong base to deprotonate the 1,3-dithiane, generating a carbanion at the position between the sulfur atoms.
• Step 1b: The resulting carbanion is then alkylated with diiodomethane ((CH2)2I), where the carbanion displaces one of the iodide ions through an SN2 reaction, leading to a new carbon-sulfur bond formation.

 

Intramolecular Cyclization:
Step 2: The compound is treated again with n-BuLi in THF at -78 °C, inducing the formation of another carbanion, which then intramolecularly attacks the adjacent carbon with the iodide leaving group, resulting in the formation of a five-membered ring via an intramolecular SN2 displacement of the iodide.
Oxidative Desulfurization:
Step 3: Mercury(II) oxide (HgO) and mercury(II) chloride (HgCl₂) in methanol (MeOH) and water (H₂O) are used to oxidize the thioacetal to a carbonyl group, removing the sulfur atoms and yielding a cyclic ketone. This step is a desulfurization process that involves the oxidative removal of sulfur from the molecule.

Conclusion:-

So, The major product will be option1.

- Without the specific reaction details or the images of options 1 to 4, a precise explanation for the chemical reaction and its major product cannot be provided. However, I can guide you on how to approach the analysis of a reaction's major product and explain why a certain product (in a hypothetical scenario) might be favored over others.

- Understanding the Reaction Type: The first step is to identify the type of reaction (e.g., addition, elimination, substitution, rearrangement). Each reaction type has its own set of rules that govern the outcome.

- Stability of the Product: Generally, the major product is the one that is most stable. For example, in elimination reactions, Zaitsev's rule often applies, predicting that the alkene with the most substituents (and thus the most stable) will be the major product.

- Steric and Electronic Factors: These can greatly influence the outcome of a reaction. Sterically hindered reactants or intermediates might favor pathways that are less crowded. Electron-withdrawing or electron-donating groups can stabilize or destabilize intermediates, affecting the product distribution.

- Mechanism Pathways: Understanding the mechanism can provide insights into which product will be favored. For instance, in a substitution reaction, whether the mechanism follows SN1 or SN2 can change the outcome based on factors like the solvent, substrate, and nucleophile.

- Incorrect Options Analysis:
- Option 2, 3, and 4 might represent products that are less stable than the one in Option 1 due to reasons such as less favorable substitution patterns, formation of less stable intermediates, or being products of minor pathways due to kinetic or thermodynamic control issues.

- Why Other Options Are Incorrect:
- They could involve forming less stable carbocations (if it's a reaction that involves carbocations) or less substituted alkenes, which are generally less stable.
- They might not adhere to the stereoselectivity or regioselectivity rules of the reaction.
- They could be products of side reactions that are less favored under the reaction conditions given.

- Conclusion: Without the specifics, the explanation remains general. However, the key to identifying the major product lies in understanding the reaction mechanism, the stability of potential products, and the influence of steric and electronic factors.

The correct answer is option 3

Concept:-

• Sodium Borohydride (NaBH₄) : NaBH4 is a selective reducing agent that is commonly used to reduce carbonyl compounds to alcohols. It is particularly effective for the reduction of aldehydes and ketones to primary and secondary alcohols, respectively. However, under standard conditions, NaBH₄ is generally not reactive enough to reduce esters or amides.
• Methanol (MeOH): Methanol can act as a solvent for NaBH4 reductions. It's a common choice for several reasons. Methanol is a polar solvent that can solubilize a wide range of organic compounds, including many aldehydes and ketones. Additionally, methanol can participate in the reaction by donating a hydrogen atom to the alkoxide intermediate, effectively protonating it to form the final alcohol product.

Explanation:-

The mechanism is as follows

Conclusion:-

So, the final product is option 3.

The above reaction is known as mercuration and demercuration, detailed mechanism is given below:

Solvolysis ∝ stability of carbocation

Carbocation and their stabilities after removal of -Br group is given as

So, II > IV > I > III

At bridge head carbovation is very much unstable

In (a) and (b) reaction centre is sp² hybrid carbon, thus it is not feasible for S_N2 reaction, while in case of (c), cation or reaction centre is generated at bridgehead position which is highly unstable due to strain. Thus none of the given options will undergo SN2 reaction.