Common Named Reactions MCQ Quiz in मल्याळम - Objective Question with Answer for Common Named Reactions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

• Prins reaction is nucleophillic addition of alkene to aldehyde or ketone assisted by an acid
• intramolecular Prins reaction leads to cyclization and thus called Prins cyclization.
• Pinacole arrangement involves conversion of 1,2 diol system to carbonyl system in the presence of an acid (lewis acid or proton ).

Explanation:

• Firstly, SnCl4 (lewis acid) coordinates with oxygen whose lone pairs are  most available for donation.
• The positive charge on O coordinated to lewis acid will undergo neutralization by bond breaking and oxonium ion forms.
• In the next step, alkene adds to carbonyl carbon to form a 6-membered ring with a tertiary carbocation. This is called prins cyclization.
• Further, the formed carbocation undergoes pincaole rearrangement to give 5-membered ring

Conclusion:

The sequence of steps involved in the given chemical reaction are:

(1) formation of oxonium ion, (2) Prins cyclization and finally (3) Pinacole rearrangement.

Concept:

Electrophilic Substitution on Benzene Derivatives

• Benzene derivatives undergo electrophilic aromatic substitution reactions, where the nature of the substituent influences the reactivity and orientation of the reaction.
• The -CCl₃ group is an electron-withdrawing group due to its inductive and resonance effects, which deactivate the aromatic ring towards electrophilic substitution.
• Electron-withdrawing groups are meta-directing, meaning substitution typically occurs at the meta position relative to the substituent.
• Chlorination in the presence of FeCl₃ generates the electrophile Cl⁺, which reacts with the deactivated benzene ring.

Explanation:

• The aromatic ring has a deactivating -CCl₃ group, making the meta position the preferred site for electrophilic substitution.
• The chlorine electrophile (Cl⁺) substitutes at the meta position relative to the -CCl₃ group.
• Ortho and para substitution are disfavored due to steric hindrance and lower electron density at those positions.

The major product is a meta-chlorinated derivative of benzotrichloride.

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

Concept:-

Mitsunobu reaction

• The Mitsunobu reaction is a chemical reaction in an organic reaction that converts an alcohol into a variety of functional groups, such as an ester, by using triphenylphosphine (PPh3) and an azodicarboxylate such as diethyl azodicarboxylate (DEAD).
• The first stage of the Mitsunobu reaction involves
  attack of the phosphine (PPh3) adds to the weak N=N π bond of diethyl azodicarboxylate (DEAD) giving an anion that is stabilized by the electron-withdrawing effect of the ester group.

• In the next step, the generated anion abstracts a proton from the NuH and gives the Nu-.

• In the third step, positively charged phosphorus is now attacked by the alcohol, displacing a second nitrogen anion in an SN2 reaction at the phosphorus atom. The nitrogen anion generated in this step is stabilized by
  conjugation with the ester, but rapidly removes the proton from the alcohol to give an electrophilic
  R–O+–PPh3.

• In the last step the Nu- attacks the phosphorus derivative of the alcohol in a normal SN2 reaction at carbon with the phosphine oxide as the leaving group.

Explanation:-

• The reaction pathway is shown below:

• The first step of the reaction follows the Mitsunobu reaction, which involves the formation of esters from secondary alcohols with inversion of configuration.
• In the next step of the reaction, LiAlH4 reacts with the ester group and undergoes reduction to form alcohol with retention of configuration.

Conclusion:-

• Hence, the correct combination of reagents to effect the following reaction is A. Ph3P, DEAD, PhCO2H; B. LiAlH4
• Correct option is (c).

Explanation:-

The reaction pathway is shown below:

Conclusion:-

Hence, the major product formed in the given reaction is 

Explanation:-

The reaction pathway is shown below:

This is an example of favorskii type rearrangement.

Conclusion:-

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