Named Reactions MCQ Quiz in বাংলা - Objective Question with Answer for Named Reactions - বিনামূল্যে ডাউনলোড করুন [PDF]

CONCEPT:

Reformatsky Reaction Mechanism

• The Reformatsky reaction involves the formation of a zinc enolate from an α-halo ester and an aldehyde or ketone in the presence of zinc, which then adds to the carbonyl compound.
• This reaction is used to form β-hydroxy esters, which upon further hydrolysis, yield β-hydroxy acids.
• Here, the α-bromo ester (BrCH₂COOEt) reacts with the ketone in the presence of zinc to form a β-hydroxy ester intermediate. Acidic hydrolysis of this intermediate leads to the formation of a carboxylic acid.

REACTION:

The reaction proceeds as follows:

• Step 1: Formation of Zinc Enolate
  • The α-bromo ester, BrCH₂COOEt, reacts with Zn to form a zinc enolate.
• Step 2: Addition to the Ketone
  • The zinc enolate then undergoes nucleophilic addition to the carbonyl group of the ketone, forming a β-hydroxy ester intermediate.
• Step 3: Hydrolysis and Decarboxylation
  • Under acidic conditions (H₃O⁺), the β-hydroxy ester undergoes hydrolysis to form a β-hydroxy acid, which can further undergo dehydration to yield a carboxylic acid product.
• 

CONCLUSION:

The correct option is: Option 2

Concept:

Intramolecular Nucleophilic Substitution (S_N2) Reaction Leading to Cyclization

• This reaction involves an alkyl halide with a nitro group (-NO₂) on the same chain, with K₂CO₃ as the base.
• In the presence of a base like K₂CO₃, the nitro group acts as a nucleophile, attacking the carbon attached to the bromine in an intramolecular S_N2 reaction.
• This intramolecular substitution leads to the formation of a five-membered ring due to the favorable formation of a stable cyclopentane ring structure.

Explanation:

Step 1: Generation of the Nucleophile

The base K₂CO₃ deprotonates the molecule, activating the nitro group as a nucleophile.

Step 2: Intramolecular Nucleophilic Substitution (S_N2)

The nitro group attacks the carbon attached to the bromine, displacing the bromine in an S_N2 fashion. This forms a cyclopentane ring with the nitro group attached.

The correct option is 3.

Solution (3)

The correct answer is option 1.

Explanation:-

Deprotonation to form Enolate: The starting compound, which is an ester with a tosylate (OTs) leaving group, is treated with Lithium Diisopropylamide (LDA) in Tetrahydrofuran (THF) at -78°C.
LDA is a strong, non-nucleophilic base that selectively deprotonates the less hindered alpha-proton of the ester to form an enolate. The low temperature helps to stabilize the enolate ion formed.
SN2 Reaction: The enolate ion then undergoes an S_N2 reaction with an alkyl halide (shown as Iodine with a methyl group attached), where the enolate oxygen acts as a nucleophile and attacks the electrophilic carbon of the alkyl iodide.
This leads to the formation of the substituted ester product, with inversion of configuration at the carbon where the substitution took place.
Formation of a Directed Enolate: A different part of the molecule, now bearing the ester and tosylate, undergoes a similar deprotonation with LDA. However, due to steric reasons and the directionality of the previous reaction, the resulting enolate is formed with high regioselectivity, meaning that the enolate forms preferentially at a specific position of the molecule.
This is referred to as a "directive enolate" because the structure of the molecule directs the formation of the enolate to a specific carbon.
Diastereoselective SN₂ Reaction: This directed enolate then reacts with another molecule of methyl iodide, but this time the reaction is diastereoselective, with one diastereomer being formed in a much larger proportion than the other (94% of one, 6% of the other).
The high diastereoselectivity is indicated to be 88%, suggesting that the reaction favors the formation of one diastereomer significantly over the other.
Formation of a Silyl Enol Ether: In a separate reaction pathway, the ester is treated with (Hexamethyl)disilazide and chlorotrimethylsilane (TMSCl) to form a silyl enol ether.
The disilazide serves as a base to deprotonate the alpha-proton next to the ester, while the TMSCl reacts with the resulting enolate to protect it as a silyl enol ether.
SN2 Reaction of Silyl Enol Ether: The silyl enol ether then undergoes an SN2 reaction with a methyl group transfer reagent (shown as MeLi, which is likely methyl lithium, although typically MeLi would not be used for transmetallation in this type of chemistry).
This leads to the formation of a new product where the silyl group has been replaced with a methyl group.

Conclusion:-

So, the major product will be option 1.

The correct answer is option 1

Explanation:- 

The depicted steps are:

• Formation of tosylate (leaving group) from alcohol by reaction with Py/Me-I.
• Elimination of the tosylate by E2 mechanism using t-BuOK, with formation of the alkene.
• Nucleophilic substitution of the remaining tosylate group by a nucleophile.

These reactions are a simplified depiction and in a real-world scenario, may involve additional steps or intermediate structures. The bulky base is used to abstract a proton leading to the formation of the more stable (and thus major) alkene product through the E2 elimination mechanism. The S_N2 reaction leads to the substitution at the less hindered site (the primary carbon in this case), which is a characteristic of S_N2 reactions.

Conclusion:-

So the product is option 1.

The correct answer is option 2

Explanation:-

The mechanism is as follows:

• Deprotonation: The first step shows a ketone being deprotonated by a strong base, potassium hydride (KH), to form an enolate ion. The electron pair on the oxygen of the enolate then delocalizes to form a resonance-stabilized enolate with the negative charge on the carbon atom.
• Nucleophilic Attack: In the second step, this enolate ion acts as a nucleophile and attacks a nitro alkene (presumably formed in a separate reaction not shown in the image), resulting in the addition of the enolate to the beta position of the nitro alkene. This forms a new carbon-carbon bond.
• Protonation: Following the nucleophilic attack, the intermediate is protonated to form a ketone with an adjacent nitro group.
• Nucleophilic Substitution (S_N2): The nitro group positioned next to the ketone is then replaced by a nucleophile (Nu) through an S_N2 reaction. This is typically characterized by a backside attack of the nucleophile, leading to the inversion of configuration at the carbon atom, and the departure of the leaving group.
• Reduction: The final step is the reduction of the nitro group to an amine. This is achieved by treatment with zinc/ammonium chloride (Zn/Ammonium chloride), which reduces the nitro group first to a nitroso group (N=O), and subsequently to an oxime (N-OH), and finally to an amine (NH₂). The oxime may also isomerize to an imine before the final reduction to the amine.

Conclusion:-

So, the final product will be option 2.

Concept:-

Diels–Alder reaction:

• Diels–Alder reactions occur between a conjugated diene and an alkene, usually called the dienophile.
• This reaction goes in a single step simply on heating.
• Here are some examples: first an open-chain diene with a simple unsaturated aldehyde as the dienophile

Diene:

• The diene component in the Diels–Alder reaction can be open-chain or cyclic and can have many different substituents.
• There is only one limitation: The diene must have the s-cis conformation.
• Butadiene typically prefers the s-trans conformation with the two double bonds as far away from each other as possible for steric reasons. The barrier to rotation about the central σ bond is small (about 30 kJ.mol−1 at room temperature) and rotation to the less favorable but reactive s-cis conformation is rapid.

• Cyclic dienes that are permanently in the s-cis conformation are exceptionally good at Diels–Alder reactions—cyclopentadiene is a classic example—but cyclic dienes that are permanently in the s-trans conformation and cannot adopt the s-cis conformation will not do the
  Diels–Alder reaction at all.
• Dienes permanently in the s-cis conformation are excellent for Diels–Alder reactions.

• The Approach of dienophile from the below and the above of the diene plane is shown below:

Explanation:-

• The reaction pathway is shown below:

Conclusion:-

• Hence, the major product of the reaction is

Explanation:-

The Chugaev Reaction

• The reaction pathway is shown below:

• The loss of C(O)S and MeSH occur at lower ternperatures than the carboxylic ester pyrolysis and thus is rnore suitable for the preparation of sensitive alkenes.
• Heating the xanthate derived by sequential treatment of cis-2-benzylcyclopentan with sodium in toluene followed by addition of excess carbon disulfide and methyl iodide furnishes, via a syn-elimination, 3-benzylcyclopentene.

Conclusion:-

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

Concept:

• In an aldol reaction, the enolate of an aldehyde or a ketone reacts at the α-carbon atom with the carbonyl of another molecule under basic or acidic conditions to obtain a β-hydroxy carbonyl compound.
• The mechanism of the aldol reaction is shown below:

• The intramolecular aldol reaction is the condensation reaction of two aldehyde groups or ketone groups in the same molecule in presence of a base.
• An example is shown below:

Explanation:-

• The reaction pathway is shown below:

Conclusion:-

Hence, the major product formed in the following reaction is

Concept:-

Simmons-Smith Cyclopropanation Reaction:

• Simmons-Smith Cyclopropanation Reaction involves the treatment of a zinc-copper couple with di-iodomethane in ether, which produces a reagent that adds to alkenes to form cyclopropanes.
• Cyclopropanation reaction of simple alkenes appears to proceed via stereospecific syn-addition of a Zn-carbenoid (carbene-like species) to the double bond without the involvement of a free carbene.
• A general mechanism of the Simmons-Smith Cyclopropanation reaction with Zn-carbenoid is given below:

​

Explanation:-

• A particularly interesting aspect of the Simmons-Smith cyclopropanation reaction is the stereoelectronic control exhibited by proximal OH, OR groups, which favor cyclopropanation to occur from the same face of the double bond as the oxy substituents.
• A general mechanism of the Simmons-Smith Cyclopropanation reaction with oxy substituents IS given below:

• Thus, the reaction pathway will be as follows

​

• The cyclopropanation reaction occurs from the same face of the double bond as the oxy substituents.

Conclusion:-

• Hence, option 4 is the correct answer.