Organic Reaction Mechanisms MCQ Quiz in தமிழ் - Objective Question with Answer for Organic Reaction Mechanisms - இலவச PDF ஐப் பதிவிறக்கவும்

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பெறு Organic Reaction Mechanisms பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Organic Reaction Mechanisms MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

Latest Organic Reaction Mechanisms MCQ Objective Questions

Top Organic Reaction Mechanisms MCQ Objective Questions

Organic Reaction Mechanisms Question 1:

The major product formed in the following reaction is

F1 Madhuri Teaching 08.02.2023 D6

  1. F1 Madhuri Teaching 08.02.2023 D7
  2. F1 Madhuri Teaching 08.02.2023 D8
  3. F1 Madhuri Teaching 08.02.2023 D9
  4. F1 Madhuri Teaching 08.02.2023 D10

Answer (Detailed Solution Below)

Option 3 : F1 Madhuri Teaching 08.02.2023 D9

Organic Reaction Mechanisms Question 1 Detailed Solution

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.

          F1 Madhuri Teaching 08.02.2023 D11

          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).

          F1 Madhuri Teaching 08.02.2023 D12

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

         F1 Madhuri Teaching 08.02.2023 D13

  • 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.

         F1 Madhuri Teaching 08.02.2023 D14

Conclusion:

The final product of the reaction is :

  F1 Madhuri Teaching 08.02.2023 D15

 

Organic Reaction Mechanisms Question 2:

The major product formed in the following reaction is

F1 Utkarsha Madhuri 18.11.2021 D7

  1. F2 Vinanti Teaching 20.06.23 D2 V2
  2. F2 Vinanti Teaching 20.06.23 D1 V2
  3. F1 Utkarsha Madhuri 18.11.2021 D10
  4. F1 Utkarsha Madhuri 18.11.2021 D11

Answer (Detailed Solution Below)

Option 2 : F2 Vinanti Teaching 20.06.23 D1 V2

Organic Reaction Mechanisms Question 2 Detailed Solution

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:-

F1 Utkarsha Madhuri 18.11.2021 D7    F1 Utkarsha Madhuri 18.11.2021 D9

F2 Vinanti UG Entrance 02.02.23 D1

Organic Reaction Mechanisms Question 3:

The CORRECT statement with respect to the stereochemistry of a-hydroxy acids P and Q formed in the following reactions is:

F1 Puja Madhuri 18.05.2021 D56

  1. Both P and Q are formed with retention of configuration.
  2. Both P and Q are formed with inversion of configuration.
  3. P is formed with retention of configuration and Q with inversion of configuration.
  4. P is formed with inversion of configuration and Q with retention of configuration.

Answer (Detailed Solution Below)

Option 4 : P is formed with inversion of configuration and Q with retention of configuration.

Organic Reaction Mechanisms Question 3 Detailed Solution

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, SN1 and SN2.

 

Comparing the SN1 and the SN2 reaction

 

SN1

N2

Rate Law

Unimolecular

(substrate only)

Bimolecular

(Substrate and nucleophile)

“Big Barrier”

Carbocation stability

Steric hindrance

Alkyl halide

(electrophile)

3° > 2° ≫ 1° (worst)

1° > 2° ≫ 3° (worst)

Nucleophile

Weak (generally neutral)

Strong (generally bearing a negative charge)

Solvent

Polar protic (e.g. alcohols)

Polar aprotic (e.g. DMSO, acetone)

Stereochemistry

Mix of retention and inversion

Inversion only

Explanation:

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

  • The reaction proceeds through SN2 mechanism. 
  • Backside attack of OH- take place and product formed with inversion of configuration. 

F1 Utkarsha Ravi 30.07.21 D18

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 SN2 mechanism with neighbouring group participation
  • The result of the neighbouring group participation is the formation of substituted product with retention of configuration

  F1 Utkarsha Ravi 30.07.21 D19.1 

So here the product Q formed has retention of configuration.

Organic Reaction Mechanisms Question 4:

Predict the correct major product in the following reaction. 

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  1. qImage679e4507835d60b28381ba7f
  2. n-Butane 
  3. qImage679e4507835d60b28381ba81
  4. CH3-CH=CH-CH3 

Answer (Detailed Solution Below)

Option 3 : qImage679e4507835d60b28381ba81

Organic Reaction Mechanisms Question 4 Detailed Solution

Concept:

Reaction of Alcohols with Thionyl Chloride (SOCl2)

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

Explanation:

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  • 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 SN2 reaction to the inverted chloride.

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

Organic Reaction Mechanisms Question 5:

Write the product A in the following reaction:

qImage679494d306f85dad8fedd684

The options are:

  1. qImage679494d406f85dad8fedd686
  2. qImage679494d506f85dad8fedd688
  3. qImage679494d506f85dad8fedd68b
  4. qImage679494d506f85dad8fedd68c

Answer (Detailed Solution Below)

Option 2 : qImage679494d506f85dad8fedd688

Organic Reaction Mechanisms Question 5 Detailed Solution

Concept:

Electrophilic Sulfonation of Aromatic Compounds

  • In the presence of concentrated H2SO4 at elevated temperatures, aromatic compounds undergo electrophilic substitution, specifically sulfonation.
  • The sulfonation reaction introduces a sulfonic acid group (-SO3H) into the aromatic ring.
  • The position of substitution depends on the stability of the intermediate carbocation, governed by the resonance effects of the existing substituents.

Explanation:

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  • The substrate in this reaction is naphthalene, a fused aromatic compound with two benzene rings.
  • Under sulfonation conditions at high temperatures (160°C), the sulfonic acid group predominantly attaches to the β-position (2-position) of naphthalene.
  • This positional preference occurs because the intermediate carbocation formed during substitution is stabilized by resonance when the substituent is in the β-position.
  • Hence, the product formed is β-naphthalenesulfonic acid (2-naphthalenesulfonic acid).

The product A is β-naphthalenesulfonic acid.

Organic Reaction Mechanisms Question 6:

Number of carbonyl groups present in the final product of the following reaction sequence is ___________.

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Answer (Detailed Solution Below) 4.0

Organic Reaction Mechanisms Question 6 Detailed Solution

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:

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Conclusion:

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

Organic Reaction Mechanisms Question 7:

The correct statement for the reactions P and Q is

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  1. kP > kQ; P goes via an E2 and Q goes via an E1cB pathway
  2. kP > kQ; both P and Q go via an E2 pathway
  3. k> kP; P goes via an E1cB and Q goes via an E2 pathway
  4. k> kP; both P and Q go via an E1cB pathway

Answer (Detailed Solution Below)

Option 1 : kP > kQ; P goes via an E2 and Q goes via an E1cB pathway

Organic Reaction Mechanisms Question 7 Detailed Solution

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.

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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 kP and kQ 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 kP is higher.
    • F1 Priya CSIR 7-10-24 D71
  • 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 kQ being lower.
    • qImage670e5edff03497d7ef7bd61c
    •  

Conclusion:

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

Organic Reaction Mechanisms Question 8:

The correct order of the dissociation constants for hydrates of the following compounds is

F1 Priya CSIR 7-10-24 D16

  1. R > Q > P
  2. R > P > Q
  3. Q > R > P
  4. P > Q > R

Answer (Detailed Solution Below)

Option 2 : R > P > Q

Organic Reaction Mechanisms Question 8 Detailed Solution

Concept:

The dissociation constant (KD) 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 (KD).

  • 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 (KD) among the compounds.

Conclusion:

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

Organic Reaction Mechanisms Question 9:

The major product will be:
F1 Priyas Teaching 28-5-24 D71   in presence of TsOH the final product wioo be

  1. F1 Priya Teaching 28-5-24 D72
  2. F1 Priya Teaching 28-5-24 D73
  3. F1 Priya Teaching 28-5-24 D74
  4. F1 Priya Teaching 28-5-24 D75

Answer (Detailed Solution Below)

Option 1 : F1 Priya Teaching 28-5-24 D72

Organic Reaction Mechanisms Question 9 Detailed Solution

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.

F1 Priya Teaching 28-5-24 D76

Conclusion:-

So, The major product will be option 1

Organic Reaction Mechanisms Question 10:

The major product of given reaction and mechanism it follows is :
F1 Teaching Savita 27-3-24 D36

  1. F1 Teaching Savita 27-3-24 D37
  2. F1 Teaching Savita 27-3-24 D38
  3. F1 Teaching Savita 27-3-24 D39
  4. F1 Teaching Savita 27-3-24 D40

Answer (Detailed Solution Below)

Option 1 : F1 Teaching Savita 27-3-24 D37

Organic Reaction Mechanisms Question 10 Detailed Solution

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 E1cB elimination is extremely easy. This mixture of diastereoisomers is fi rst of all lactonized in acid, and then undergoes E1cB elimination with triethylamine to give a product known as a butenolide. Butenolides are common structures in naturally occurring compounds.

F1 Teaching Savita 27-3-24 D41

Conclusion:-

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

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