Asymmetric Synthesis MCQ Quiz in मल्याळम - Objective Question with Answer for Asymmetric Synthesis - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 13, 2025
Latest Asymmetric Synthesis MCQ Objective Questions
Top Asymmetric Synthesis MCQ Objective Questions
Asymmetric Synthesis Question 1:
The following reaction sequences produce the major stereoisomer
Answer (Detailed Solution Below)
Asymmetric Synthesis Question 1 Detailed Solution
Concept:-
Aldol Condensation:
- 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:
Explanation:-
- The reaction pathway is shown below:
- The stereoselectivity can be rationalized by a six-membered (Zimmerman–Traxler) transition state 71, in which the aldehyde approaches from
the less-hindered face. The product imide can be hydrolyzed to give the enantiomerically pure carboxylic acid or derivative, or reduced to give the primary alcohol product. This procedure, therefore, allows the formation of either enantiomer of the desired syn aldol product.
Conclusion:-
- Hence, the major stereoisomer will be
Asymmetric Synthesis Question 2:
Predict the major predict of the reaction
Answer (Detailed Solution Below)
Asymmetric Synthesis Question 2 Detailed Solution
Explanation:-
- The reaction pathway is shown below:
- In the first step of the reaction, reduction to the alcohol gives the C2-symmetric compound mannitol.
- In the next step, mannitol is converted to a useful aldehyde by selective protection as a bis-acetal with acetone and using a Lewis acid ZnCl2.
- In the last step of the reaction, cleavage of the remaining diol with sodium periodate gives two equivalents of a useful protected form of glyceraldehyde.
Conclusion:-
- Hence, the major product of the reaction is
Asymmetric Synthesis Question 3:
Predict the major product of the reaction below
Answer (Detailed Solution Below)
Asymmetric Synthesis Question 3 Detailed Solution
Concept:-
- Alkylation of lithiated hydrazones forms the basis of an efficient method for the asymmetric alkylation of aldehydes and ketones, using the optically active hydrazines (S)-1-amino-2-(methoxymethyl)pyrrolidine (SAMP).
- Deprotonation of the optically active hydrazones, alkylation, and removal of the chiral auxiliary under mild conditions (ozonolysis or acid hydrolysis of the N-methyl salt) gives the alkylated aldehyde or ketone with, generally, greater than 95% optical purity.
Explanation:-
- This procedure has been exploited in the asymmetric synthesis of several natural products. Thus, (S)-4-methyl-3- heptanone, the principal alarm pheromone of the leaf-cutting ant Atta texana, was prepared from 3-pentanone in very high optical purity as shown in the scheme below:
Conclusion:-
- Hence, the major product of the reaction is
Asymmetric Synthesis Question 4:
The major products A and B formed in the following reaction sequence are
Answer (Detailed Solution Below)
Asymmetric Synthesis Question 4 Detailed Solution
Correct option is (c)
Asymmetric Synthesis Question 5:
The role of BF3 as an industrial plymerization catalyst is to generate
Answer (Detailed Solution Below)
Asymmetric Synthesis Question 5 Detailed Solution
Concept:
→ BF3 is a Lewis acid, which means that it can accept a pair of electrons from a molecule with a lone pair of electrons, such as an alkene monomer. In the case of polymerization, BF3 reacts with the double bond of an alkene monomer to form a complex between BF3 and the alkene.
→ This complex is highly reactive and can undergo a series of reactions that eventually lead to the formation of a carbocation intermediate. The carbocation is a positively charged species that is highly reactive and can react with other alkene monomers to form longer polymer chains.
Explanation:
The role of BF3 as an industrial polymerization catalyst is to activate the double bond of the alkene monomer and generate a carbocation intermediate that initiates the polymerization reaction.
This mechanism allows for the formation of long polymer chains with specific properties, such as molecular weight and branching.
\(\rm H_2O + BF_3 → H^+ + HO\overset \Theta{ B}F_3 \)
cation formation ⇒ carbocation
Conclusion: The correct answer is option 2.