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

Last updated on Apr 10, 2025

നേടുക Common Named Reactions ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Common Named Reactions MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Common Named Reactions MCQ Objective Questions

Top Common Named Reactions MCQ Objective Questions

Common Named Reactions Question 1:

The major product formed in the following reaction is
F3 Vinanti Teaching 29.05.23 D8

  1. F3 Vinanti Teaching 29.05.23 D9
  2. F3 Vinanti Teaching 29.05.23 D10
  3. F3 Vinanti Teaching 29.05.23 D11
  4. F3 Vinanti Teaching 29.05.23 D12

Answer (Detailed Solution Below)

Option 2 : F3 Vinanti Teaching 29.05.23 D10

Common Named Reactions Question 1 Detailed Solution

 

Concept:

The reaction follows McMurry Mechanism  

McMurry Reaction:

  • It's the reaction in which reductive dimerization of carbonyl compounds gives alkenes in the presence of TiCl3 and a reducing agent like Zn/Cu or LiAlH4.
  • The reaction is inactive in finely divided Ti but works in a faster rate in TiCl+ AlClas it gives an electron rich Ti (0) particle.
  • The reaction proceeds in two steps, ​radical coupling followed by deoxygenation on the surface of Ti(0). 

                Example: 
F3 Vinanti Teaching 29.05.23 D13F3 Vinanti Teaching 29.05.23 D14

F3 Vinanti Teaching 29.05.23 D15
Explanation: 

The mechanism of the reaction involves radical bond cleavage followed by hydrolysis gives cyclic ketone.

F3 Vinanti Teaching 29.05.23 D16

Conclusion

Therefore option (2) is the correct one.

Common Named Reactions Question 2:

Identify the product formed in the following transformations
F1 Vinanti Teaching 30.03.23 D14

  1. F1 Vinanti Teaching 30.03.23 D15
  2. F1 Vinanti Teaching 30.03.23 D16
  3. F1 Vinanti Teaching 30.03.23 D17
  4. F1 Vinanti Teaching 30.03.23 D18

Answer (Detailed Solution Below)

Option 1 : F1 Vinanti Teaching 30.03.23 D15

Common Named Reactions Question 2 Detailed Solution

Explanation:-

  • The reaction pathway is shown below:

qImage649298fe21b36ea1b6277c35

Conclusion:-

  • Hence,  the product formed in the following transformations is

F1 Vinanti Teaching 30.03.23 D15

Common Named Reactions Question 3:

The major product formed in the following reaction

F1 Madhuri Teaching 06.02.2023 D35

  1. F1 Madhuri Teaching 06.02.2023 D36
  2. F1 Madhuri Teaching 06.02.2023 D37
  3. F1 Madhuri Teaching 06.02.2023 D38
  4. F1 Madhuri Teaching 06.02.2023 D39

Answer (Detailed Solution Below)

Option 2 : F1 Madhuri Teaching 06.02.2023 D37

Common Named Reactions Question 3 Detailed Solution

Explanation: -

The reaction will involve the following steps: -

Step 1: Oxidative-addition i.e. addition of organic compound on metal complex via sigma complex state with an increase in oxidation state on the metal ion.

F1 Madhuri Teaching 06.02.2023 D40

Step 2: Transmetalation rearrangement reaction, this reaction involves the transfer of ligands from one metal to another metal complex.

F1 Madhuri Teaching 06.02.2023 D41

Step 3: Reduction-elimination reaction, in this reaction metal, will undergo reduction by two oxidation states and two of the ligand will leave and form a single molecular species.

F1 Madhuri Teaching 06.02.2023 D42

Conclusion: -

The reaction will produce the product preserved stereochemistry as follows: -

F1 Madhuri Teaching 06.02.2023 D43

Hence, The correct option is - 2.

Common Named Reactions Question 4:

Find the major product in the following reaction

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  1. qImage6794971db58e851d164f7429
  2. qImage6794971db58e851d164f742a
  3. qImage6794971eb58e851d164f742b
  4. qImage6794971eb58e851d164f742c

Answer (Detailed Solution Below)

Option 2 : qImage6794971db58e851d164f742a

Common Named Reactions Question 4 Detailed Solution

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:

qImage67988d4e7ce28154874a8189

  • 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.

Common Named Reactions Question 5:

In the given reaction, the major product is:

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  1. qImage6769c8442aab0b4cc5aead79
  2. qImage6769c8442aab0b4cc5aead7a
  3. qImage6769c8452aab0b4cc5aead7b
  4. qImage6769c8452aab0b4cc5aead7c

Answer (Detailed Solution Below)

Option 2 : qImage6769c8442aab0b4cc5aead7a

Common Named Reactions Question 5 Detailed Solution

Concept:

Curtius Rearrangement

Curtius Rearrangement is also called Curtius degradation or Curtius reaction. Curtius rearrangement is a thermal decomposition of acyl acid to form isocyanate with a loss of nitrogen as stated by Theodor Curtius in the year 1885. It is also known as Curtius degradation or Curtius reaction. This reaction is identical to the Schmidt reaction.

Isocyanates are subjected to attack by various nucleophiles namely alcohols, water, and amines which in turn output urea derivative or carbamate and essential amines.

1280px-Curtius rearrangement rxn

Explanation:

Mechanism

Screenshot 2024-12-25 163040

Therefore, the correct option is 2.

Common Named Reactions Question 6:

What is the final product C.

F3 Teaching SSC 9-12-24 D51

  1. F3 Teaching SSC 9-12-24 D52
  2. F3 Teaching SSC 9-12-24 D53
  3. F3 Teaching SSC 9-12-24 D54
  4. F3 Teaching SSC 9-12-24 D55

Answer (Detailed Solution Below)

Option 2 : F3 Teaching SSC 9-12-24 D53

Common Named Reactions Question 6 Detailed Solution

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.
  • F3 Teaching SSC 9-12-24 D56

CONCLUSION:

The correct option is: Option 2

Common Named Reactions Question 7:

Major product formed

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  1. F3 Teaching SSC 9-12-24 D9
  2. F3 Teaching SSC 9-12-24 D10
  3. F3 Teaching SSC 9-12-24 D11
  4. F3 Teaching SSC 9-12-24 D12

Answer (Detailed Solution Below)

Option 3 : F3 Teaching SSC 9-12-24 D11

Common Named Reactions Question 7 Detailed Solution

Concept:

Intramolecular Nucleophilic Substitution (SN2) 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 SN2 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 (SN2)

The nitro group attacks the carbon attached to the bromine, displacing the bromine in an SN2 fashion. This forms a cyclopentane ring with the nitro group attached.

F3 Teaching SSC 9-12-24 D13

The correct option is 3.

Common Named Reactions Question 8:

The major product S of the following reaction is

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  1. qImage66eea3a3380a624cdb3af446
  2. qImage66eea3a3380a624cdb3af449
  3. qImage66eea3a3380a624cdb3af44b
  4. qImage66eea3a4380a624cdb3af44c

Answer (Detailed Solution Below)

Option 3 : qImage66eea3a3380a624cdb3af44b

Common Named Reactions Question 8 Detailed Solution

Solution (3)

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Common Named Reactions Question 9:

The major product of the given reaction:
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  1. F1 Savita Teaching 8-4-24 D21
  2. F1 Savita Teaching 8-4-24 D22
  3. F1 Savita Teaching 8-4-24 D23
  4. F1 Savita Teaching 8-4-24 D24

Answer (Detailed Solution Below)

Option 1 : F1 Savita Teaching 8-4-24 D21

Common Named Reactions Question 9 Detailed Solution

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.

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Conclusion:-

So, the major product will be option 1.

Common Named Reactions Question 10:

The major product for the given reaction is :

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  1. F1 Savita Teaching 8-4-24 D8
  2. F1 Savita Teaching 8-4-24 D9
  3. F1 Savita Teaching 8-4-24 D10
  4. F1 Savita Teaching 8-4-24 D11

Answer (Detailed Solution Below)

Option 1 : F1 Savita Teaching 8-4-24 D8

Common Named Reactions Question 10 Detailed Solution

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 SN2 reaction leads to the substitution at the less hindered site (the primary carbon in this case), which is a characteristic of SN2 reactions.

F1 Savita Teaching 8-4-24 D12

Conclusion:-

So the product is option 1.

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