Organic Reaction Mechanisms MCQ Quiz in मराठी - Objective Question with Answer for Organic Reaction Mechanisms - मोफत PDF डाउनलोड करा

Concept:-

Nucleophilic substitution reactions-

Substitution reactions are the types of reactions where a nucleophile is an attacking reagent.

• There are three types of substitution reactions depending on the nature of the substrate.
  • Nucleophilic substitution at saturated carbon. 
  • Nucleophilic acyl substitution
  • Nucleophilic aromatic substitution.

Nucleophilic substitution reaction is mainly of two types. These are 

• SN1 or Unimolecular nucleophilic substitution and SN2 or bimolecular nucleophilic substitution

1. SN1 or Unimolecular nucleophilic substitution:

• Depends upon the concentration of the substrate.
• Is independent of the concentration of the nucleophile.
• Follows first-order kinetics.

2. SN2 or bimolecular nucleophilic substitution:

• The rate depends on the concentration of both the reactant and the substrate.
• It follows second-order kinetics.

Given below are the examples of SN2 and SN1 nucleophilic substitution reactions:

This reaction is a nucleophilic substitution reaction.

Here, Dimethyl formamide (DMF) is a polar aprotic solvent that stabilizes cation i.e. Na⁺ and gives the desired nucleophile PhS^- to which the reaction proceeds through the S_N2 mechanism with an inversion of configuration.

Polar aprotic solvents like DMF, DMSO, Acetone, etc. don't form a Hydrogen bond increasing the reactivity nucleophile.

As it occurs in a single step so the nucleophile attacks from the back side (the front side is blocked due to leaving the group). So it follows inversion stereochemistry.

Here is the diagram of the transition state. 

Below is an example of an S_N2  reaction.

Explanation:-

The Tosyl group is an excellent leaving group in substitution reactions. So the nucleophile phenyl sulfide (PhS^-) will attack on the back side as the tosyl group is bulky. It will follow the inversion stereochemistry. 

This reaction won't follow the S_N1 mechanism because

(i) The solvent should be polar protic like H₂O, ROH, etc. which increases the stability of carbocation formed as the intermediate in a SN1 mechanism.

Conclusion:-

• Hence, the major product formed in the following reaction is 
  

Concept:

• Methyl amine (MeNH₂) can act as a base as well as the nucleophile.  
• if substrate has acidic proton, it will preferably act as the base and abstracts the most acidic proton.
• the H is considered acidic if the negative charge formed after  its abstraction can be stabilized by conjugation or presence of some electronegative atom.

Explanation:

• In the first step, Methyl amine will abstract the most acidic proton attached to N in 5-membered ring.

          The generated negative charge is resonance stablized.

• In the next step, negative charge will move to C and will substitute Br^- to form 3 membered ring (SN²).

• Next step will follow up with  nucleophillic attack of another methylamine molecule at electrophillic carbon centre of Carbonyl bond.

• Finally, the back conjugation of negative change on O, will facilitate the breaking of  3-membered ring (which is unstable) in such a way that the aromaticity of 5-membered ring is regained.

Conclusion:

The final product of the reaction is :

Explanation:- 

• Diethyl aluminium chloride is used as a catalyst in  Ziegler-Natta catalysis.
• The organic compound is also used as a Lewis acid in various organic synthesis reactions.
• They are highly reactive and pyrophoric in nature.

Thus, the final product is an alpha-beta unsaturated carbonyl compound.

Reaction:-

Concept: 

NUCLEOPHILIC SUBSTITUTION REACTION

• A chemical reaction in which the displacement of a leaving group is taking place by a nucleophile.
• Nucleophilic substitution reactions mainly take place through two reaction mechanisms, S_N1 and S_N2.

Comparing the SN1 and the S_N2 reaction

Explanation:

Consider the reaction of 2-bromo-propanoic acid with OH^-.

• The reaction proceeds through S_N2 mechanism. 
• Backside attack of OH^- take place and product formed with inversion of configuration. 

So here the product P formed has an inversion of configuration. 

Consider the other reaction of 2-bromo-propanoic acid with OH^- in presence of Ag₂O/H₂O.

• The presence of Ag⁺ as a Lewis acid helps the ionization of substrate molecules. 
• The reaction proceeds through the intermolecular S_N2 mechanism with neighbouring group participation. 
• The result of the neighbouring group participation is the formation of substituted product with retention of configuration. 

So here the product Q formed has retention of configuration.

Concept:

Reaction of Alcohols with Thionyl Chloride (SOCl₂)

• Thionyl chloride (SOCl₂) is commonly used to convert alcohols into alkyl chlorides.

Explanation:

• When converting alcohols to alkyl chlorides using thionyl chloride (SOCl₂), the reaction generally proceeds with retention of configuration meaning the stereochemistry at the chiral carbon is maintained.
• The reaction of thionyl chloride with chiral 2º-alcohols has been observed to proceed with either inversion or retention. In the presence of a base such as a pyridine, the intermediate chlorosulfite ester reacts to form an "pyridinium" salt, which undergoes a relatively clean S_N2 reaction to the inverted chloride.

​Therefore, the correct major product is the chlorinated compound shown in option 3.

Concept:

Ozonolysis

Definition and Purpose: Ozonolysis is a reaction where ozone (O₃) cleaves the carbon-carbon double bonds (alkenes) to form carbonyl compounds. It is a widely-used method to break down alkenes into smaller, more functionalized fragments such as aldehydes or ketones.

Reaction Mechanism:

• Ozonide Formation: The reaction begins with the addition of ozone to the carbon-carbon double bond, forming an unstable intermediate called a molozonide, which rearranges to a more stable ozonide.
• Hydrolysis Step: The ozonide is then typically subjected to hydrolytic workup, which can be either reductive or oxidative.
• Reductive Workup: The ozonide can be cleaved under reductive conditions, often using zinc and acetic acid (Zn/HOAc) or dimethyl sulfide (DMS) as the reducing agents. This leads to the formation of aldehydes or ketones.
  • Example: RCH=CH₂ → RCHO (if hydrolysis conditions are reductive).
• Oxidative Workup: Using hydrogen peroxide (H₂O₂) under oxidative conditions results in the formation of carboxylic acids.
  • Example: RCH=CHR' → RCOOH + R'COOH (if hydrolysis conditions are oxidative).

Explanation:

Reaction Mechanism:

Conclusion:

Number of carbonyl groups present in the final product of the following reaction sequence is: 4

Concept:

The reactions P and Q involve dehydrohalogenation using i-PrOH (isopropanol) and a base at 43°C, leading to the formation of alkyne products. The reactions can proceed through different elimination mechanisms, such as E2 (bimolecular elimination) and E1cB (unimolecular elimination via a conjugate base intermediate). The pathway depends on the nature of the substrate and the reaction conditions.

Key Points on Elimination Mechanisms (E2 vs. E1cB):

• E2 Mechanism: A one-step mechanism where the base abstracts a proton as the leaving group departs. This reaction occurs in a concerted fashion and is favored by strong bases and good leaving groups.
• E1cB Mechanism: A two-step elimination where the base abstracts a proton, forming a carbanion intermediate. The leaving group is eliminated in a second step. This mechanism is favored when the leaving group is poor or when a strong base is used with a poor electrophile.
• The reaction rate constants k_P and k_Q reflect the relative reactivity of the substrates in these two elimination pathways.

Explanation: 

• P proceeds via the E2 mechanism: The structure of P allows for a concerted E2 elimination because the base can easily abstract a proton, leading to the rapid formation of the alkyne product. This is why the rate constant k_P is higher.
  • ​
• Q proceeds via the E1cB mechanism: In contrast, the structure of Q leads to an E1cB pathway due to a more difficult proton abstraction, forming a carbanion intermediate before the bromide leaves. This makes the reaction slower, reflected by k_Q being lower.
  • ​
  •  

Conclusion:

Thus, the correct statement is: k_P > k_Q; P proceeds via an E2 mechanism, and Q proceeds via an E1cB mechanism.

Concept:

The dissociation constant (K_D) for hydrates of carbonyl compounds indicates the ease with which the hydrate form dissociates back into the carbonyl compound and water. The stability of the hydrate form is influenced by electronic effects, steric effects, and resonance, with electron-withdrawing groups stabilizing the hydrate and bulky groups destabilizing it.

• Electron-Withdrawing Groups (EWG): Carbonyl compounds with strong electron-withdrawing groups (e.g., CCl₃) stabilize the hydrate form by increasing the electrophilicity of the carbonyl carbon, leading to a lower dissociation constant.

• Steric Hindrance: Bulky groups, like tert-butyl (C(CH₃)₃), hinder the formation of stable hydrates due to steric crowding, leading to a higher dissociation constant (K_D).

• Resonance and Aromatic Rings: Phenyl (C₆H₅) groups provide some resonance stabilization but also contribute steric effects, making their impact intermediate between the other two groups.

Explanation: 

• For compound R, the phenyl group (C₆H₅) is bulky and introduces steric hindrance, which makes the hydrate less stable compared to compound Q. However, it provides more stability than P due to less steric hindrance than the trichloromethyl group (CCl₃).

• For compound P, the methyl group (CH₃) is the least bulky, allowing for higher stability in the hydrate form, leading to a lower dissociation constant than Q or R.

• For compound Q, the trichloromethyl group (CCl₃) is a strong electron-withdrawing group that stabilizes the hydrate form the most, giving it the lowest dissociation constant (K_D) among the compounds.

Conclusion:

The correct order of dissociation constants is R > P > Q.

The correct answer is option 1

Explanation:-

• Grignard Reagent Formation: Formation of the Grignard reagent using Mg and alkyl chloride.
• Gilman Reagent Formation: Conversion of the Grignard reagent to a Gilman reagent and subsequent reaction with acid chloride to form a ketone.
• Nucleophilic Addition: Nucleophilic addition of MeLi to the ketone to form a tertiary alcohol.
• Dehydration Reactions: Dehydration of the tertiary alcohol under basic (KH) and acidic (TsOH) conditions, yielding different ratios of alkenes.

Conclusion:-

So, The major product will be option 1

The correct answer is option 1.

Explanation:-

β-Halocarbonyl compounds can be rather unstable: the combination of a good leaving group and an acidic proton means that E_1cB elimination is extremely easy. This mixture of diastereoisomers is fi rst of all lactonized in acid, and then undergoes E_1cB elimination with triethylamine to give a product known as a butenolide. Butenolides are common structures in naturally occurring compounds.

Conclusion:-

So, the major product of given reaction and mechanism it follows is option 1 and E1cB.