Organic Reaction Mechanisms MCQ Quiz - Objective Question with Answer for Organic Reaction Mechanisms - Download Free PDF

Last updated on Jul 16, 2025

Latest Organic Reaction Mechanisms MCQ Objective Questions

Organic Reaction Mechanisms Question 1:

Which one of the following ylides will be the least reactive for nucleophilic addition to >C=O group?

  1. qImage684706eabc1e5f6cad37963f
  2. qImage684706eabc1e5f6cad37964e
  3. qImage684706ebbc1e5f6cad37964f
  4. qImage684706ebbc1e5f6cad379650

Answer (Detailed Solution Below)

Option 4 : qImage684706ebbc1e5f6cad379650

Organic Reaction Mechanisms Question 1 Detailed Solution

CONCEPT:

Reactivity of Phosphonium Ylides in Nucleophilic Addition to Carbonyl Compounds

  • Phosphonium ylides (Wittig reagents) react with carbonyl groups (>C=O) to form alkenes in the Wittig reaction.
  • The reactivity of the ylide depends on the stability of the ylide:
    • Unstabilized ylides (without electron-withdrawing groups) are more reactive.
    • Stabilized ylides (containing electron-withdrawing groups like esters, ketones) are less reactive because they are resonance-stabilized and less nucleophilic.

EXPLANATION:

  • Option 1: Ph3P=CH2 — Simple ylide, highly reactive, no stabilization.
  • Option 2: Ph3P=CH–CH3 — Alkyl-substituted ylide, still fairly reactive.
  • Option 3: Ph3P=CH–CH2–CH3 — Slightly bulkier alkyl group, but still reactive.
  • Option 4: Ph3P=CH–COOC2H5 — Strongly stabilized ylide due to the electron-withdrawing ester group (–COOC2H5), making it the least reactive.
  • Electron-withdrawing groups stabilize the negative charge on the ylide carbon through resonance, lowering its nucleophilicity.

Therefore, the least reactive ylide is option 4) Ph3P=CH–COOC2H5.

Organic Reaction Mechanisms Question 2:

For the following reaction, the possible product(s) is/are

qImage6842f6222b3c8dde0f93d974

  1. qImage6842f6222b3c8dde0f93d992
  2. qImage6842f6232b3c8dde0f93d993
  3. qImage6842f6232b3c8dde0f93d996
  4. qImage6842f6242b3c8dde0f93d999

Answer (Detailed Solution Below)

Option :

Organic Reaction Mechanisms Question 2 Detailed Solution

CONCEPT:

Reagent Sequence in Organic Reactions

  • The reaction sequence involves the following steps:
    • Step 1: Grignard reaction with MeMgBr in THF, followed by the addition of H2O to obtain the alcohol after protonation.
    • Step 2: Use of PCC (Pyridinium chlorochromate) to oxidize the alcohol to a ketone.
    • Step 3: Reduction of the ketone using H2 and Pd/C to achieve the desired alkane or hydrocarbon product.

EXPLANATION:

  • For the given reaction sequence:

    Me-C6H5-CH2 (pure enantiomer) undergoes a Grignard reaction.

    • MeMgBr reacts with the aldehyde to form a magnesium alkoxide intermediate.
    • Hydrolysis (with H2O) converts the alkoxide into an alcohol.

qImage687619cfd9a407f6af381745

  • In the second step, PCC is used to selectively oxidize the alcohol to a ketone.
  • In the final step, H2 and Pd/C reduce the ketone to an alkane by hydrogenating the carbonyl group.

qImage6842f6222b3c8dde0f93d992 and qImage6842f6232b3c8dde0f93d996 form the same configuration i.e. R.

Therefore, the possible products after following this sequence are the structures from option 1 and option 3.

Organic Reaction Mechanisms Question 3:

Consider the following two reactions and their corresponding Hammett plots 

qImage6842f278741e4022aa9844e2

qImage6842f279741e4022aa9844fe

Choose the option(s) that correctly match(es) the points on the graph given in Column-I with substituents X given in Column-II in accordance with their substituents constant σ

Column-I (points on the graph)  Column-II (substituent X) 
p NH2
q NO2
r OMe 
s Cl 
t Me 
u CN

  1. \(\rm \mathrm{s} \rightarrow \sigma_{(\mathrm{X}=\mathrm{Cl})} ; \mathrm{t} \rightarrow \sigma_{(\mathrm{X}=\mathrm{OMe})} ; \mathrm{u} \rightarrow \sigma_{\left(\mathrm{X}=\mathrm{NH}_{2}\right)} ; \mathrm{r} \rightarrow \sigma_{\left(\mathrm{X}=\mathrm{NO}_{2}\right)}\)\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
  2. \(\rm \mathrm{s} \rightarrow \sigma_{(\mathrm{X}=\mathrm{Me})} ; \mathrm{u} \rightarrow \sigma_{\left(\mathrm{X}=\mathrm{NH}_{2}\right)} ; \mathrm{t} \rightarrow \sigma_{(\mathrm{X}=\mathrm{OMe})} ; \mathrm{r} \rightarrow \sigma_{(\mathrm{X}=\mathrm{Br})}\)
  3. \(\rm p \rightarrow \sigma_{(X=M e)} ; q \rightarrow \sigma_{(X=C N)} ; r \rightarrow \sigma_{\left(X=N O_{2}\right)} ; t \rightarrow \sigma_{(X=O M e)}\)
  4. \(\rm \mathrm{p} \rightarrow \sigma_{(\mathrm{X}=\mathrm{Cl})} ; \mathrm{q} \rightarrow \sigma_{\left(\mathrm{X}=\mathrm{NO}_{2}\right)} ; \mathrm{r} \rightarrow \sigma_{(\mathrm{X}=\mathrm{CN})} ; \mathrm{t} \rightarrow \sigma_{(\mathrm{X}=\mathrm{Me})}\)

Answer (Detailed Solution Below)

Option :

Organic Reaction Mechanisms Question 3 Detailed Solution

CONCEPT:

Hammett Equation and σ Constants

  • The Hammett equation relates the effect of substituents on the reactivity of aromatic compounds to a substituent constant (σ).
  • σ represents the electronic nature of a substituent:
    • Electron-withdrawing groups (EWG) have positive σ values.
    • Electron-donating groups (EDG) have negative σ values.
  • Reaction rate trends:
    • Reactions involving carbocation intermediates (like SN1) are stabilized by EDGs.
    • Reactions involving negative charge development (like ester hydrolysis) are stabilized by EWGs.
  • The Hammett plot graphs log(kX/kH) vs. σX:
    • A steep slope implies strong sensitivity to substituent effects.
    • Points to the left = EDGs (negative σ), points to the right = EWGs (positive σ)

EXPLANATION:

  • Given reactions:
    • Reaction M: SN1 reaction – favors EDGs to stabilize carbocation.
    • Reaction N: Ester hydrolysis – favors EWGs to stabilize the transition state.
  • We compare substituent constants (σ) with graph coordinates:
    • NH2, OMe, Me = EDGs → appear at negative σ (left side of graph).
    • NO2, CN, Cl = EWGs → appear at positive σ (right side of graph).
  • Matching using Hammett plot:
    • Option 1: Correctly matches:
      • s → CN (strong EWG)
      • t → OMe (EDG)
      • u → NH2 (strong EDG)
      • r → NO2 (strong EWG)
    • Option 3: Also correctly matches:
      • p → Me (weak EDG)
      • q → CN (EWG)
      • r → NO2 (strong EWG)
      • t → OMe (EDG)

Therefore, the correct options are Option 1 and Option 3.

Organic Reaction Mechanisms Question 4:

In the following asymmetric transformation, the key aldol reaction involves the attack of  

qImage6842eccc441f4cb769c7a22f

  1. Si face of enolate on to the Re face of aldehyde
  2. Si face of enolate on to the Si face of aldehyde 
  3. Re face of enolate on to the Re face of aldehyde
  4. Re face of enolate on to the Si face of aldehyde 

Answer (Detailed Solution Below)

Option 4 : Re face of enolate on to the Si face of aldehyde 

Organic Reaction Mechanisms Question 4 Detailed Solution

CONCEPT:

Asymmetric Aldol Reaction – Felkin-Anh Model and Enolate Geometry

  • In asymmetric aldol reactions, the stereoselectivity is controlled by the geometry of the enolate and the face of the electrophile (usually an aldehyde) that is attacked.
  • The reaction shown uses a chiral auxiliary to generate a Z-enolate (cis-enolate) from the oxazolidinone derivative.
  • Facial selectivity depends on:
    • The face of the enolate delivering the nucleophile (Re or Si)
    • The face of the aldehyde receiving the nucleophile (Re or Si)

EXPLANATION:

qImage68752f816510fe78f97dd5ad

qImage68752f816510fe78f97dd5ae

  • From the oxazolidinone, a Z-enolate is formed due to steric hindrance and chelation control.
  • This enolate attacks the aldehyde via Felkin-Anh transition state, where:
    • The nucleophile (enolate) attacks the Si face of the aldehyde to minimize steric hindrance.
    • Based on the 3D arrangement, the enolate delivers the nucleophile from its Re face.
  • This leads to the formation of a syn-aldol product as shown in the image (pure enantiomer).

Therefore, the correct answer is: Re face of enolate on to the Si face of aldehyde (Option 4).

Organic Reaction Mechanisms Question 5:

Which of the following is an Electrophilic addition reaction?

  1. CH3CH = CH2 + HBr → CH3CH(Br)CH3
  2. \(CH_3CH=CH_2+HBr\xrightarrow{peroxide}CH_3CH2Br\)
  3. \(C_6H_6+Cl_2\xrightarrow[AlCl_{3}]{Anhydraes}C_6H_5Cl\)
  4. \(C_6H_6+Cl_2\xrightarrow[AlCl_{3}]{Anhydraes}C_6H_5Cl+HCl \)
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : CH3CH = CH2 + HBr → CH3CH(Br)CH3

Organic Reaction Mechanisms Question 5 Detailed Solution

Concept:

Addition reactions are of two types:

Electrophilic Addition reaction:

  • The addition of an electrophile to a double bond is an example of an addition reaction.
  • Alkenes react with a solution of halogen and water to form halohydrins.
  • The reaction takes place following Markownikoff's rule which states that: 'In an electrophilic addition reaction of an unsymmetrical alkene with an unsymmetrical addendum (reagent), the positive part of the addendum adds to the less substituted carbon and the negative part adds to the more substituted carbon of the alkene.

​Nucleophilic Addition:

  • In this type of reaction, the nucleophile is added to a substrate containing an electrophilic centre such as a double or a triple bond.
  • The nucleophile gets added to the electrophilic centre and the double or the triple bond is broken in the process.

Explanation:

  • Alkenes add on a molecule of halogen acids to form alkyl halides.
  • The general reaction is :

F1 Puja J Anil 15.04.21  D7

  • The addition of HBr to symmetrical alkenes yields only one product.
  • The addition of HBr to unsymmetrical alkenes yields one major and one minor product.
  • The addition of HBr to unsymmetrical alkenes yields products according to Markownikof's Rule which states that: During the addition of a polar molecule to an unsymmetrical alkene or alkyne, the negative part of the addendum is added to the more substituted Carbon atom.
  • The positive part of the addendum gets added to the less substituted carbon part of the alkyne.
  • Hence, the reaction of hydrogen bromide with propylene is an example of an addition reaction.

  • Kharash discovered that the addition of HBr to unsymmetrical alkenes in presence of organic peroxides takes place against Markownikoff's rule.
  • The more saturated carbon gets the positive part of the addendum whereas the less saturated carbon atom gets the negative part of the addendum when an addition reaction of an unsymmetrical alkane takes place in presence of peroxide.
  • The reaction takes place by a free-radical mechanism.
  • Hence,\(CH_3CH=CH_2+HBr\xrightarrow{peroxide}CH_3CH2Br\) is a free radical substitution reaction.
  • Electrophilic aromatic substitution (EAS) is where benzene acts as a nucleophile to replace a substituent with a new electrophile.
  • Benzene needs to donate electrons from inside the ring. An electrophile attacks the region of high electron density. Hydrogen is replaced by an electrophile.
  • Benzene reacts with chlorine or bromine in an electrophilic substitution reaction, but only in the presence of a catalyst. The catalyst is either aluminium chloride or iron.

Hence, \(C_6H_6+Cl_2\xrightarrow[AlCl_{3}]{Anhydraes}C_6H_6Cl\) is an example of an electrophilic substitution reaction.

Hence, the reaction CH3CH = CH2 + HBr → CH3CH(Br)CHis an addition reaction.

Top Organic Reaction Mechanisms MCQ Objective Questions

Which of the following is an Electrophilic addition reaction?

  1. CH3CH = CH2 + HBr → CH3CH(Br)CH3
  2. \(CH_3CH=CH_2+HBr\xrightarrow{peroxide}CH_3CH2Br\)
  3. \(C_6H_6+Cl_2\xrightarrow[AlCl_{3}]{Anhydraes}C_6H_5Cl\)
  4. \(C_6H_6+Cl_2\xrightarrow[AlCl_{3}]{Anhydraes}C_6H_5Cl+HCl \)

Answer (Detailed Solution Below)

Option 1 : CH3CH = CH2 + HBr → CH3CH(Br)CH3

Organic Reaction Mechanisms Question 6 Detailed Solution

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Concept:

Addition reactions are of two types:

Electrophilic Addition reaction:

  • The addition of an electrophile to a double bond is an example of an addition reaction.
  • Alkenes react with a solution of halogen and water to form halohydrins.
  • The reaction takes place following Markownikoff's rule which states that: 'In an electrophilic addition reaction of an unsymmetrical alkene with an unsymmetrical addendum (reagent), the positive part of the addendum adds to the less substituted carbon and the negative part adds to the more substituted carbon of the alkene.

​Nucleophilic Addition:

  • In this type of reaction, the nucleophile is added to a substrate containing an electrophilic centre such as a double or a triple bond.
  • The nucleophile gets added to the electrophilic centre and the double or the triple bond is broken in the process.

Explanation:

  • Alkenes add on a molecule of halogen acids to form alkyl halides.
  • The general reaction is :

F1 Puja J Anil 15.04.21  D7

  • The addition of HBr to symmetrical alkenes yields only one product.
  • The addition of HBr to unsymmetrical alkenes yields one major and one minor product.
  • The addition of HBr to unsymmetrical alkenes yields products according to Markownikof's Rule which states that: During the addition of a polar molecule to an unsymmetrical alkene or alkyne, the negative part of the addendum is added to the more substituted Carbon atom.
  • The positive part of the addendum gets added to the less substituted carbon part of the alkyne.
  • Hence, the reaction of hydrogen bromide with propylene is an example of an addition reaction.

  • Kharash discovered that the addition of HBr to unsymmetrical alkenes in presence of organic peroxides takes place against Markownikoff's rule.
  • The more saturated carbon gets the positive part of the addendum whereas the less saturated carbon atom gets the negative part of the addendum when an addition reaction of an unsymmetrical alkane takes place in presence of peroxide.
  • The reaction takes place by a free-radical mechanism.
  • Hence,\(CH_3CH=CH_2+HBr\xrightarrow{peroxide}CH_3CH2Br\) is a free radical substitution reaction.
  • Electrophilic aromatic substitution (EAS) is where benzene acts as a nucleophile to replace a substituent with a new electrophile.
  • Benzene needs to donate electrons from inside the ring. An electrophile attacks the region of high electron density. Hydrogen is replaced by an electrophile.
  • Benzene reacts with chlorine or bromine in an electrophilic substitution reaction, but only in the presence of a catalyst. The catalyst is either aluminium chloride or iron.

Hence, \(C_6H_6+Cl_2\xrightarrow[AlCl_{3}]{Anhydraes}C_6H_6Cl\) is an example of an electrophilic substitution reaction.

Hence, the reaction CH3CH = CH2 + HBr → CH3CH(Br)CHis an addition reaction.

The major product formed in the following reaction

F1 Madhuri Teaching 06.02.2023 D26

  1. F1 Madhuri Teaching 06.02.2023 D27
  2. F1 Madhuri Teaching 06.02.2023 D28
  3. F1 Madhuri Teaching 06.02.2023 D29
  4. F1 Madhuri Teaching 06.02.2023 D30

Answer (Detailed Solution Below)

Option 1 : F1 Madhuri Teaching 06.02.2023 D27

Organic Reaction Mechanisms Question 7 Detailed Solution

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Explanation: -

The reaction will occur as follows: -

Step 1: the diethyl succinate with loses an acidic α-hydrogen in reaction with base t-BuOK, to give a nucleophile.

F1 Madhuri Teaching 06.02.2023 D31

Step 2: In the second step this nucleophile will attack the electron-deficient carbonyl carbon of benzaldehyde. To give a condensation product

F1 Madhuri Teaching 06.02.2023 D32

Step 3: In the second part of the reaction acid is added, we know that in an acidic medium alcohols give a dehydration reaction as follows: 

F1 Madhuri Teaching 06.02.2023 D33

Step 4: hydrolysis of conjugated ester in the presence of acidic medium
F1 Madhuri Teaching 06.02.2023 D27

Conclusion:-

Hence, the correct option is 1.

The rates of alkaline hydrolysis of the compounds shown below

F1 Puja J 29.4.21 Pallavi D53

follow the order:

  1. I > II > III
  2. II > I > III
  3. II > III > I
  4. III > I > II

Answer (Detailed Solution Below)

Option 3 : II > III > I

Organic Reaction Mechanisms Question 8 Detailed Solution

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Concept:

  • Hydrolysis is one of the important reactions of esters.
  • The acidic hydrolysis of esters gives carboxylic acid and alcohol, whereas the basic hydrolysis of an ester gives a carboxylate ion and an alcohol.
  • Amides are not easy to hydrolyse, and hydrolysis of amides takes place when heated in aqueous acids for extended periods.

F1 Puja j Anil 18-05.21 D5

  • Amides are difficult to hydrolyse owing to the presence of a Carbon - Nitrogen double bond in the resonating structure given below:

F1 Puja j Anil 18-05.21 D6

 

  • As nitrogen is more basic than oxygen, the double bond character in amides is much more than in esters and thus amides are difficult to hydrolysed compared to esters.

Explanation:

  • As structure 1 is an amide, it will be more difficult to hydrolyse.
  • Between structures 2 and 3, both are esters, so let's see which one will be hydrolysed more easily.

F1 Puja J 18.5.21 Pallavi D9

  • Clear from the description above that structure 2 gives us a more stable intermediate as compared to structure III and thus, structure II will be hydrolysed more easily.

Hence, the rates of hydrolysis are: II > III > I

Additional Information

F1 Puja j Anil 18-05.21 D7 

The major product formed in the following reaction is

F1 Madhuri Teaching 06.02.2023 D4

  1. F1 Madhuri Teaching 06.02.2023 D5
  2. F1 Madhuri Teaching 06.02.2023 D6
  3. F1 Madhuri Teaching 06.02.2023 D7
  4. F1 Madhuri Teaching 06.02.2023 D8

Answer (Detailed Solution Below)

Option 1 : F1 Madhuri Teaching 06.02.2023 D5

Organic Reaction Mechanisms Question 9 Detailed Solution

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Concept: -

Electrophilic ring substitution reaction:

  • Benzene and other aromatic compounds show characteristic electrophilic substitution reactions.
  • In this reaction, a hydrogen atom of the aromatic ring is substituted by an electrophile.
  • The substitution takes place through an addition-elimination mechanism.
  • In the first step, the benzene ring donates pi electrons to the electrophile.
  • One of the carbon atoms forms a bond with the electrophile.
  • In the second step, the complex formed then loses a proton from the saturated carbon atom with the help of a base.
  • The aromatic ring is finally regenerated in the last step.

Nitration:

  • Nitric acid and sulfuric acid used in nitration generate nitronium NO2+ ions as electrophiles.
  • Nitronium NO2+ is the nitrating agent.
  • The electrophile then attacks the benzene ring forming a sigma complex.
  • The σ complex is resonance stabilized.
  • The complex then loses a proton to form nitrobenzene.
  • Effect of halogen on Nitration of benzene ring: 
    • Halogens show the +R effect as follows

F1 Madhuri Teaching 06.02.2023 D9

As we can see from the resonance hybrids, the ortho and para position is activated.

Thus, halogens are ortho-para directing.

​​Explanation:

As Discussed halogens are ortho para directing, So the reaction will proceed as follows:

Step 1: - Production of electrophile

F1 Madhuri Teaching 06.02.2023 D10

The nitronium ion will behave as an electrophile.

Step 2: - Attack of electrophile.

The resonance will activate the ortho-para position.

F1 Madhuri Teaching 06.02.2023 D11

Thus, on the attack following products will form:

F1 Madhuri Teaching 06.02.2023 D12

Conclusion:- 

Hence,

qImage64465d2c1cde7ac8ec553bd8

The correct option is (1).

The major product formed in the following reaction

F1 Puja J 29.4.21 Pallavi D15 is:

  1. F1 Puja J 29.4.21 Pallavi D16
  2. F1 Puja J 29.4.21 Pallavi D17
  3. F1 Puja J 29.4.21 Pallavi D18
  4. F1 Puja J 29.4.21 Pallavi D19

Answer (Detailed Solution Below)

Option 1 : F1 Puja J 29.4.21 Pallavi D16

Organic Reaction Mechanisms Question 10 Detailed Solution

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Concept:

  • Barton reaction is the reaction used to carry out the conversion of nitrous acid esters to γ nitroso alcohols.
  • The reaction occurs in presence of ultraviolet light and leads to the production of an alkoxy radical species and nitrous oxide.
  • The alkoxy radical then abstracts a hydrogen atom via a cyclic intermediate to give a carbon radical species.
  • The carbon radical species then reacts with the nitrous oxide radical to yield γ nitroso alcohols.

​Explanation:

  • The mechanism of the reaction occurring is given below:
  • The carbon-free radical abstracts the alpha hydrogen which is nearest to it and in favourable orientation to it which is the γ hydrogen present in the axial position.

 

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Hence, the major product formed is F1 Puja J 29.4.21 Pallavi D16.

The least acidic among the following compounds 

F1 Puja J 29.4.21 Pallavi D14

is:

  1. M
  2. N
  3. O
  4. P

Answer (Detailed Solution Below)

Option 1 : M

Organic Reaction Mechanisms Question 11 Detailed Solution

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Concept:

The acidity of compounds:

  • The acidity of compounds is determined by their ability to donate hydrogen ions in solution.
  • The greater the ease of donation or liberation of the hydrogen ions, the stronger is the acid.
  • The acidic proton of the compound is generally attached to an electronegative atom.
  • The strength of the acidity is greatly influenced by the substituents or groups attached.
  • The strength of an acid is measured by its pKa value. Lower the pKa, stronger is the acid.

Factors influencing the acid strength-

  • Inorganic acids are much stronger than Organic acids.
  • The stability of the conjugate base-
    • if the negative charge is resonance stabilized in the conjugate base, then the compound is more acidic compared to the compound whose conjugate base has the charge localized.
  • Electronegative substituents or groups like F, Cl, Br, I increase the acidity via inductive electron withdrawal (-I).
  • Electron donating groups such as - OR, -Me, etc. decrease the acidity via the +R and +I effect.
  • Electron withdrawing groups such as NO2,-CF3, -COOH, -CN increases the acidity via the –R effect.
  • Hydrogen attached to sp2 Carbon is more acidic than hydrogen attached to spcarbon.
  • The acidity order is sp> sp2>sp3.

Explanation:

  • The conjugate bases of the acids are given below:

F1 Puja Madhuri 06.05.2021 D13

F1 Puja Madhuri 06.05.2021 D12 (1)

F1 Puja Madhuri 06.05.2021 D14

 

F1 Puja Madhuri 06.05.2021 D15
  • From the explanation above, we see that the molecules N, O, P are resonance stabilized, whereas M is not, hence, M is the least acidic.

Major products P and Q, formed in the reactions given below, are:

 

 F1 Puja Madhuri 18.05.2021 D41F1 Puja Madhuri 18.05.2021 D42

  1.  F1 Puja Madhuri 18.05.2021 D43
  2. F1 Puja Madhuri 18.05.2021 D44
  3. F1 Puja Madhuri 18.05.2021 D45
  4.  F1 Puja Madhuri 18.05.2021 D46

Answer (Detailed Solution Below)

Option 3 : F1 Puja Madhuri 18.05.2021 D45

Organic Reaction Mechanisms Question 12 Detailed Solution

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Explanation:
Here in the first case, that is cis-2-bromo-4-methyl cyclohexanol, the hydrogen anti to the bromine will assist loss of axial bromide and following oxidation by Ag2O forms 4-methyl cyclohexanone.

So, the product P formed is 4-methyl cycloheanone. 

F1 Utkarsha Singh Anil 30-06.21 D6

In the case of  trans-2-bromo-4-methyl cyclohexanol there occur fragmentation by Ag2O forms 3-methyl cyclopentanal. So the product Q formed is 3-methyl cyclopentanal.

  F1 Utkarsha Singh Anil 30-06.21 D7

The major product formed in the following reaction is
F1 Savita Teaching 25-5-23 D15

  1. F1 Savita Teaching 25-5-23 D16
  2. F1 Savita Teaching 25-5-23 D17
  3. F1 Savita Teaching 25-5-23 D18
  4. F1 Savita Teaching 25-5-23 D19

Answer (Detailed Solution Below)

Option 2 : F1 Savita Teaching 25-5-23 D17

Organic Reaction Mechanisms Question 13 Detailed Solution

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Concept:-

Nucleophilic substitution reactions-

Substitution reactions are the types of reactions where a nucleophile is an attacking reagent.

  • There are three types of substitution reactions depending on the nature of the substrate.
    • Nucleophilic substitution at saturated carbon. 
    • Nucleophilic acyl substitution
    • Nucleophilic aromatic substitution.

 

Nucleophilic substitution reaction is mainly of two types. These are 

  • SN1 or Unimolecular nucleophilic substitution and SN2 or bimolecular nucleophilic substitution

 

1. SN1 or Unimolecular nucleophilic substitution:

  • Depends upon the concentration of the substrate.
  • Is independent of the concentration of the nucleophile.
  • Follows first-order kinetics.

2. SN2 or bimolecular nucleophilic substitution:

  • The rate depends on the concentration of both the reactant and the substrate.
  • It follows second-order kinetics.

Given below are the examples of SN2 and SN1 nucleophilic substitution reactions:

F1 Utkarsha.S 08-03-21 Savita D10

F1 Utkarsha.S 08-03-21 Savita D11

This reaction is a nucleophilic substitution reaction.

Here, Dimethyl formamide (DMF) is a polar aprotic solvent that stabilizes cation i.e. Naand gives the desired nucleophile PhS- to which the reaction proceeds through the SN2 mechanism with an inversion of configuration.

Polar aprotic solvents like DMF, DMSO, Acetone, etc. don't form a Hydrogen bond increasing the reactivity nucleophile.

As it occurs in a single step so the nucleophile attacks from the back side (the front side is blocked due to leaving the group). So it follows inversion stereochemistry.

Here is the diagram of the transition state. 
F1 Savita Teaching 25-5-23 D20
Below is an example of an SN2  reaction.
F1 Savita Teaching 25-5-23 D21
Explanation:-

The Tosyl group is an excellent leaving group in substitution reactions. So the nucleophile phenyl sulfide (PhS-) will attack on the back side as the tosyl group is bulky. It will follow the inversion stereochemistry. 

This reaction won't follow the SN1 mechanism because

(i) The solvent should be polar protic like H2O, ROH, etc. which increases the stability of carbocation formed as the intermediate in a SN1 mechanism.
F1 Savita Teaching 25-5-23 D22
Conclusion:-

  • Hence, the major product formed in the following reaction is 
    F1 Savita Teaching 25-5-23 D17

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 14 Detailed Solution

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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

 

The reaction that is expected to show a primary kinetic isotope effect for the indicated H‐atom (C‐H) is

  1. F1 Madhuri Teaching 06.02.2023 D15
  2. F1 Madhuri Teaching 06.02.2023 D16
  3. F1 Madhuri Teaching 06.02.2023 D17
  4. F1 Madhuri Teaching 06.02.2023 D18

Answer (Detailed Solution Below)

Option 3 : F1 Madhuri Teaching 06.02.2023 D17

Organic Reaction Mechanisms Question 15 Detailed Solution

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Concept:-

  • The kinetic isotope effect (KIE) describes the change in the reaction rate of a chemical reaction when one of the atoms in the reactants is replaced by one of its isotopes.
  • It is the ratio of the rate constant of reactions involving the light (kL) and the heavy (kH) isotopically substituted reactants.
  • primary kinetic isotope (PKI) effect may be found when a bond to the isotopically labeled atom is formed or broken at the rate-limiting step.
  • The difference in bond strength will be reflected in different rates of breaking of the two bonds under comparable conditions.
  • Quantum mechanical calculation suggests a maximum rate difference observed when,

\({{{{\rm{k}}_{\rm{H}}}} \over {{{\rm{k}}_{\rm{D}}}}}{\rm{ = 7}}\)

Explanation:-

  • Out of these four reactions, only the bromination of acetone in presence of acid (reaction C) involves the breaking of the C-H bond in the rate-limiting step of the reaction.
  • The mechanism of bromination of acetone in presence of acid is given by,

F1 Madhuri Teaching 06.02.2023 D19

  • The first step involves acid-catalyzed enolization followed by the electrophilic attack of the bromine molecule on the nucleophilic carbon of the enol.
  • As the rate-limiting step involves the breaking of the C-H bond,  this reaction is expected to show a primary kinetic isotope effect.

​Conclusion:-

Hence, the reaction that is expected to show a primary kinetic isotope effect for the indicated H‐atom (C‐H) is

F1 Madhuri Teaching 06.02.2023 D17 

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