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

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 :

​

• 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,$C{H}_{3}CH=C{H}_2+HBr\stackrel{peroxide}{\to }C{H}_{3}CH2Br$ 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, $C_6{H}_6+C{l}_2\underset{AlC{l}_3}{\overset{Anhydraes}{\to }}{C}_6{H}_{6}Cl$ is an example of an electrophilic substitution reaction.

Hence, the reaction CH3CH = CH2 + HBr → CH3CH(Br)CH3 is an addition reaction.

CONCEPT:

E2 Elimination followed by NaBH₄ Reduction

• E2 Reaction (NaOEt as base):
  • Anti-periplanar elimination requires a β-hydrogen and a leaving group (Br) in trans-diaxial or anti-coplanar orientation.
  • This leads to formation of a trans-alkene through ring-opening or rotation to satisfy stereoelectronic requirements.
• Carbonyl Intermediate:
  • The alkene intermediate tautomerizes to a carbonyl compound (enone or enal).
• Reduction with NaBH₄:
  • NaBH₄ selectively reduces aldehydes and ketones to corresponding alcohols.

EXPLANATION:

Bond Rotation 

• Initial rotation aligns the β-H and Br anti-periplanar, enabling smooth E2 elimination.
• This generates a trans-alkene due to stereoelectronic constraints and bulky substituent repulsion.
• Upon tautomerization and subsequent NaBH₄ reduction, the ketone intermediate is reduced to a secondary alcohol.
• Option 3 correctly shows the product as:
  • A trans-alkene backbone
  • And –OH on the carbon originally part of the carbonyl group

Hence, the major product formed in the reaction is the trans-reduced alcohol shown in Option 3.

CONCEPT:

Elimination–Addition Mechanism (Conjugate Addition Pathway)

• This transformation proceeds in two distinct mechanistic steps:
  1. 1,4-Elimination – formation of a conjugated diene system through E1cb mechanism under basic conditions.
  2. 1,4-Addition (Michael-type addition) – nucleophilic attack by CN⁻ at the β-position of the enone system.
• Why not SN1/SN2?
  • The leaving group (Cl) is not undergoing direct substitution.
  • Instead, elimination is favored under basic (Et₃N/DMSO) and heat conditions.

EXPLANATION:

• Step 1: The base abstracts a β-hydrogen → elimination occurs to form a conjugated enone (α,β-unsaturated ketone system).
• Step 2: The nucleophile (CN⁻) attacks the β-carbon of the enone via a 1,4-conjugate addition mechanism, leading to the final product.

Therefore, the correct answer is 1,4-Elimination followed by a 1,4-Addition reaction.

CONCEPT:

Hydroboration–Oxidation Using Thexylborane

• Thexylborane (R₂BH) is a bulky borane reagent used for selective hydroboration of less hindered alkenes.
• Hydroboration occurs via a syn-addition mechanism where boron and hydrogen add to the same face of the alkene.
• Oxidation with H₂O₂/NaOH converts the C–B bond to a C–OH, retaining stereochemistry at the site of addition.
• In molecules with multiple alkenes, sterics determine the regioselectivity — thexylborane prefers the less substituted (less hindered) double bond.

EXPLANATION:

• The substrate contains two double bonds: one more substituted, and one less substituted (less hindered).
• Thexylborane adds to the less hindered alkene due to steric bulk, forming a syn-addition intermediate.
• The hydroboration intermediate forms a cyclic transition state with boron and hydrogen adding to the same face (above the plane).
• Oxidation with H₂O₂/NaOH replaces the B with OH at the same position, maintaining the stereochemistry.
• The final product shows the OH and H added syn to each other, with the OH located at the terminal carbon (from anti-Markovnikov addition), matching Option 1.

Therefore, the correct answer is Option 1

CONCEPT:

Samarium(II) Iodide (SmI₂) Mediated Radical Reactions

• SmI₂ is a single electron transfer (SET) reagent that facilitates radical-type reactions.
• It is commonly used for initiating radical cyclizations, especially in molecules containing carbonyl groups and unsaturated systems (alkenes, alkynes).
• The reaction generally proceeds through:
  • Single electron reduction of the carbonyl to form a ketyl radical
  • Radical addition to a multiple bond (like an alkyne)
  • Subsequent radical rearrangement or cyclization to form a stable product

EXPLANATION:

• In the given reaction, SmI₂ reduces the ketone to form a ketyl radical (a carbon-centered radical adjacent to the oxygen).
• This radical undergoes a 5-exo-trig cyclization onto the alkyne, generating a new carbon-carbon bond and forming a radical at the terminal carbon.
• Subsequent ring closure results in the formation of a fused bicyclic system.
• The geometry and stability of this fused system direct the formation of the major product as shown in Option 1.
• This product is thermodynamically favored due to conjugation and reduced ring strain.

Therefore, the correct answer is Option 1, where the major product is a fused bicyclic compound formed via a radical cascade initiated by SmI₂.

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 :

​

• 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,$C{H}_{3}CH=C{H}_2+HBr\stackrel{peroxide}{\to }C{H}_{3}CH2Br$ 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, $C_6{H}_6+C{l}_2\underset{AlC{l}_3}{\overset{Anhydraes}{\to }}{C}_6{H}_{6}Cl$ is an example of an electrophilic substitution reaction.

Hence, the reaction CH3CH = CH2 + HBr → CH3CH(Br)CH3 is an addition reaction.

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.

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

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: 

Step 4: hydrolysis of conjugated ester in the presence of acidic medium

Conclusion:-

Hence, the correct option is 1.

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.

​

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

• 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.

• 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

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

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

The nitronium ion will behave as an electrophile.

Step 2: - Attack of electrophile.

The resonance will activate the ortho-para position.

Thus, on the attack following products will form:

Conclusion:- 

Hence,

The correct option is (1).

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.

Hence, the major product formed is .

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 sp3 carbon.
• The acidity order is sp> sp2>sp3.

Explanation:

• The conjugate bases of the acids are given below:

• 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.

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. 

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.

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:

This reaction is a nucleophilic substitution reaction.

Here, Dimethyl formamide (DMF) is a polar aprotic solvent that stabilizes cation i.e. Na⁺ and gives the desired nucleophile PhS^- to which the reaction proceeds through the S_N2 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. 

Below is an example of an S_N2  reaction.

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 S_N1 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.

Conclusion:-

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

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.

          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).

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

• 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.

Conclusion:

The final product of the reaction is :

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:-