Mechanism of Organic Reactions

The mechanism of an organic reaction describes the step-by-step sequence of elementary reactions by which overall chemical change occurs. It provides insight into the bond breaking and bond forming processes, the intermediates formed, and the transition states involved. Understanding reaction mechanisms is crucial for predicting reaction outcomes and designing new synthetic routes.

1. Fundamental Concepts

1.1. Types of Bond Fission (Bond Breaking)

• Homolytic Fission (Homolysis):
  • A covalent bond breaks in such a way that each atom retains one of the shared electrons.
  • Results in the formation of free radicals (species with unpaired electrons).
  • Indicated by half-headed arrows (⌢).
  • Favoured by: High temperatures, UV light, or presence of peroxides.
  • Example: Cl−Cl UV light​ Cl^⋅+Cl^⋅
• Heterolytic Fission (Heterolysis):
  • A covalent bond breaks in such a way that one atom retains both of the shared electrons, and the other atom gets none.
  • Results in the formation of ions (carbocations, carbanions, or other charged species).
  • Indicated by full-headed curved arrows (↷).
  • Favoured by: Polar solvents and presence of polar reagents.
  • Example: CH₃​−ClH₂​O​CH₃⁺​ + Cl⁻

1.2. Electron Displacement Effects

These effects influence the distribution of electron density within a molecule, affecting its reactivity.

• Inductive Effect (+I and -I):
  • The permanent displacement of electron density along a saturated carbon chain towards a more electronegative atom or group.
  • -I Effect: Electron-withdrawing groups (e.g., −NO₂​, −CN, −COOH, −X (halogens), −OH, −OR). They pull electron density away.
  • +I Effect: Electron-donating groups (e.g., alkyl groups −CH₃​, −C₂​H₅​). They push electron density.
  • Effect decreases rapidly with distance.
• Resonance Effect (Mesomeric Effect, +M and -M):
  • The delocalization of π-electrons within a conjugated system (alternating single and multiple bonds or multiple bonds with lone pairs/empty orbitals).
  • +M Effect: Electron-donating groups through resonance (e.g., −OH, −OR, −NH₂​, −NR₂​, −X). They push lone pairs/π-electrons into the conjugated system.
  • -M Effect: Electron-withdrawing groups through resonance (e.g., −CHO, −COR, −CN, −NO₂​, −COOH). They pull π-electrons from the conjugated system.
  • Permanent effect. Stronger than inductive effect in conjugated systems.
• Hyperconjugation (No bond resonance):
  • The delocalization of σ-electrons (from C-H bonds) of an alkyl group interacting with an adjacent empty p-orbital (in a carbocation), a half-filled p-orbital (in a free radical), or a π-bond (in an alkene).
  • Stabilizes carbocations, free radicals, and alkenes.
  • More α-hydrogens (hydrogens on carbon adjacent to the electron-deficient center/double bond) lead to greater hyperconjugation and stability.
  • Stability of Carbocations/Free Radicals: 3∘>2∘>1∘>CH₃⁺​

1.3. Types of Reagents

• Electrophiles (Electron-loving species):
  • Electron-deficient species that seek electrons.
  • Usually positively charged ions or neutral molecules with an empty orbital (Lewis acids).
  • Examples: H⁺, NO₂⁺​, R₃​C⁺, Cl⁺, BF₃​, AlCl₃​.
• Nucleophiles (Nucleus-loving species):
  • Electron-rich species that donate electrons.
  • Usually negatively charged ions or neutral molecules with at least one lone pair of electrons (Lewis bases).
  • Examples: OH−, CN−, RO−, RNH₂​, H₂​O, Cl−.

2. Reaction Intermediates

Transient species formed during a reaction that are not reactants or products but are formed and consumed within the reaction pathway.

• Carbocations (Carbonium Ions):
  • Carbon atom carrying a positive charge.
  • sp2 hybridized, trigonal planar geometry.
  • Electron deficient (sextet of electrons).
  • Stability: 3∘>2∘>1∘>CH₃⁺​ (due to +I and hyperconjugation).
• Carbanions:
  • Carbon atom carrying a negative charge.
  • sp3 hybridized, pyramidal geometry (similar to ammonia).
  • Electron rich (lone pair + octet).
  • Stability: CH₃⁻​>1∘>2∘>3∘ (opposite to carbocations, as electron-donating groups destabilize the negative charge).
• Free Radicals:
  • Species with an unpaired electron on a carbon atom.
  • sp2 hybridized (planar) or sp3 hybridized (pyramidal), often planar or near-planar.
  • Electron deficient (septet of electrons).
  • Stability: 3∘>2∘>1∘>CH₃^⋅​ (due to hyperconjugation).
• Carbenes:
  • Neutral carbon species with two unshared electrons and two bonds.
  • Singlet carbene (paired electrons, sp², bent) and Triplet carbene (unpaired electrons, sp, linear).
  • Highly reactive.
• Nitrenes:
  • Nitrogen analog of carbenes, with two unshared electrons and one bond.
  • Highly reactive, involved in reactions like Hofmann degradation.

3. Types of Organic Reactions

3.1. Substitution Reactions

An atom or group is replaced by another atom or group.

• Nucleophilic Substitution (SN​): Common for alkyl halides.
  • SN​¹ (Unimolecular Nucleophilic Substitution):
    • Two-step process.
    • Rate-determining step (RDS) involves formation of a carbocation intermediate.
    • Rate depends only on the concentration of the substrate: Rate = k[Substrate].
    • Favoured by: 3^∘ alkyl halides (stable carbocation), weak nucleophiles, polar protic solvents.
    • Racemization occurs if the reactant is chiral (due to planar carbocation).
  • SN​² (Bimolecular Nucleophilic Substitution):
    • One-step (concerted) process, no intermediate.
    • Involves a transition state where new bond forms and old bond breaks simultaneously.
    • Rate depends on concentrations of both substrate and nucleophile:

Rate = k[Substrate][Nucleophile]

• Favoured by: 1^∘ alkyl halides (less steric hindrance), strong nucleophiles, polar aprotic solvents.
• Inversion of configuration (Walden inversion) occurs if the reactant is chiral.
• Electrophilic Substitution (SE​): Common for aromatic compounds.
  • Electrophile replaces a hydrogen atom on the aromatic ring.
  • Examples: Nitration, halogenation, sulfonation, Friedel-Crafts alkylation/acylation.
• Free Radical Substitution (SR​): Common for alkanes.
  • Mechanism involves initiation, propagation, and termination steps.
  • Example: Halogenation of methane.

3.2. Addition Reactions

A molecule adds across a multiple bond (double or triple bond), breaking the π-bond(s) and forming new σ-bonds.

• Electrophilic Addition (AE​): Common for alkenes and alkynes.
  • Electrophile attacks the π-bond first, forming a carbocation intermediate.
  • Markovnikov’s rule governs regioselectivity (hydrogen adds to the carbon with more hydrogens).
  • Example: Addition of HBr to propene.
• Nucleophilic Addition (AN​): Common for aldehydes and ketones (due to polar carbonyl group).
  • Nucleophile attacks the electrophilic carbon of the carbonyl group.
  • Example: Addition of HCN to acetaldehyde.
• Free Radical Addition (AR​): Less common, occurs under specific conditions (e.g., HBr addition to alkenes in presence of peroxides, anti-Markovnikov).

3.3. Elimination Reactions

Atoms or groups are removed from adjacent carbon atoms, forming a multiple bond.

• E₁ (Unimolecular Elimination):
  • Two-step process, carbocation intermediate formed in RDS.
  • Favoured by: 3^∘ alkyl halides, weak bases, polar protic solvents.
  • Competes with SN​¹.
• E₂ (Bimolecular Elimination):
  • One-step (concerted) process.
  • Rate depends on concentrations of substrate and base:

Rate = k[Substrate][Base]

• Favored by: 1^∘ and 2^∘ alkyl halides, strong bases, polar aprotic solvents.
• Requires anti-periplanar arrangement of leaving group and β-hydrogen.
• Competes with SN​².
• Saytzeff’s Rule (Zaitsev’s Rule): In elimination reactions, the major product is the more substituted alkene (the one with fewer hydrogens on the double bond carbons).

3.4. Rearrangement Reactions

An atom or group migrates from one atom to another within the same molecule, leading to a new structural isomer. Often driven by the formation of more stable intermediates (e.g., carbocation rearrangements via hydride or alkyl shifts).

• Example: Pinacol rearrangement, Wagner-Meerwein rearrangement.

4. Important Considerations

• Solvent Effects: Polar protic solvents (H₂​O, ROH) stabilize ions (favor SN​¹, E₁). Polar aprotic solvents (DMSO, acetone, DMF) do not solvate nucleophiles strongly (Favour SN​², E₂).
• Leaving Group Ability: Good leaving groups are weak bases (e.g., I−>Br−>Cl−>F−, OTs−, H₂​O).
• Nucleophilicity/Basicity: Strong nucleophiles and strong bases favour SN​² and E₂.
• Steric Hindrance: Increased steric hindrance around the reaction centre disfavours

SN​² and favours SN​¹/ E₁.

Understanding these fundamental principles provides a powerful framework for comprehending the vast array of organic reactions.

40 Important MCQs on Mechanism of Organic Reactions with Explanations

1. Which of the following bond fissions results in the formation of ions? (a) Homolytic fission (b) Heterolytic fission (c) Both homolytic and heterolytic fission (d) Neither homolytic nor heterolytic fission

Explanation: The correct answer is (b). Heterolytic fission involves the unequal breaking of a covalent bond, where one atom takes both bonding electrons, leading to the formation of a cation and an anion (ions). Homolytic fission produces free radicals.

2. A species with an unpaired electron is called a(n): (a) Carbocation (b) Carbanion (c) Free radical (d) Electrophile

Explanation: The correct answer is (c). Free radicals are atoms or groups of atoms that possess an unpaired electron. Carbocations are positively charged, and carbanions are negatively charged. Electrophiles are electron-deficient species, which can be ions or neutral molecules.

3. Which of the following effects involves the delocalization of σ-electrons? (a) Inductive effect (b) Resonance effect (c) Hyperconjugation (d) Mesomeric effect

Explanation: The correct answer is (c). Hyperconjugation involves the delocalization of σ-electrons (typically from C-H bonds) into an adjacent empty p-orbital or π-system. Inductive and resonance (mesomeric) effects involve the displacement or delocalization of σ and π electrons, respectively, but hyperconjugation is distinct in its involvement of σ-bonds for stabilization without an explicit bond formation.

4. An electrophile is a species that is: (a) Electron-rich (b) Electron-deficient (c) Negatively charged (d) A proton donor

Explanation: The correct answer is (b). Electrophiles are “electron-loving” species, meaning they are electron-deficient and seek electrons. They are typically positively charged or have an incomplete octet (Lewis acids).

5. Which of the following is an example of a nucleophile? (a) H₂​O (b) BF₃​ (c) AlCl₃​ (d) NO₂^+​

Explanation: The correct answer is (a). Nucleophiles are electron-rich species that can donate a lone pair of electrons. Water (H₂​O) has lone pairs on oxygen and can act as a nucleophile. BF₃​, AlCl₃​, and NO₂⁺​ are electrophiles (Lewis acids).

6. The stability order of carbocations is: (a) 1^∘>2^∘>3^∘ (b) 3^∘>2^∘>1^∘ (c) CH₃⁺​>1^∘>2^∘ (d) All are equally stable

Explanation: The correct answer is (b). The stability of carbocations increases with the number of alkyl groups attached to the positively charged carbon due to the electron-donating (+I) effect of alkyl groups and hyperconjugation. Thus, tertiary (3^∘) carbocations are the most stable.

7. Which of the following organic reactions proceeds via a carbocation intermediate? (a) SN^​2 reaction (b) SN^​1 reaction (c) E₂ reaction (d) Free radical addition

Explanation: The correct answer is (b). SN​¹ (unimolecular nucleophilic substitution) reactions are two-step processes where the rate-determining step involves the formation of a carbocation intermediate. E₁ reactions also involve carbocations. SN​² and E₂ are concerted (one-step).

8. Walden inversion is associated with which reaction mechanism? (a) SN​¹ (b) SN​² (c) E₁ (d) E₂

Explanation: The correct answer is (b). Walden inversion, which is the inversion of configuration at the chiral centre, is a characteristic feature of SN​² reactions due to the backside attack of the nucleophile. SN^​1 reactions typically lead to racemization.

9. Which type of solvent favours SN​² reactions? (a) Polar protic solvents (b) non-polar solvents (c) Polar aprotic solvents (d) Any solvent

Explanation: The correct answer is (c). Polar aprotic solvents (e.g., DMSO, acetone, DMF) do not solvate the nucleophile effectively, leaving it “naked” and highly reactive, thus favouring the SN​² mechanism. Polar protic solvents solvate nucleophiles, reducing their reactivity.

10. Markovnikov’s rule is primarily applicable to: (a) Nucleophilic substitution reactions (b) Elimination reactions (c) Electrophilic addition reactions (d) Free radical substitution reactions

Explanation: The correct answer is (c). Markovnikov’s rule states that in the addition of an unsymmetrical reagent to an unsymmetrical alkene (or alkyne), the positive part of the reagent (hydrogen) adds to the carbon atom of the double bond that already has more hydrogen atoms. This is characteristic of electrophilic addition reactions where carbocation stability dictates regioselectivity.

11. Which of the following is the best leaving group? (a) OH⁻ (b) F⁻ (c) Cl⁻ (d) I⁻

Explanation: The correct answer is (d). Good leaving groups are weak bases. Basicity generally decreases down a group in the periodic table. Thus, I⁻ is the weakest base among the halides and therefore the best leaving group (I⁻>Br⁻>Cl⁻>F⁻). OH⁻ is a strong base and a poor leaving group.

12. Saytzeff’s rule (Zaitsev’s rule) predicts the major product in: (a) Addition reactions (b) Substitution reactions (c) Elimination reactions (d) Rearrangement reactions

Explanation: The correct answer is (c). Saytzeff’s rule states that in an elimination reaction, the preferred product is the more substituted alkene, i.e., the alkene that has the greater number of alkyl groups attached to the double-bonded carbon atoms.

13. The geometry of a carbocation is: (a) Tetrahedral (b) Pyramidal (c) Trigonal planar (d) Linear

Explanation: The correct answer is (c). The carbon atom in a carbocation is sp² hybridized, and the three groups attached to it lie in a plane, resulting in a trigonal planar geometry.

14. Which factor favours SN​¹ reactions over SN​2 reactions? (a) Strong nucleophile (b) Primary alkyl halide (c) Polar protic solvent (d) Low temperature

Explanation: The correct answer is (c). Polar protic solvents stabilize the carbocation intermediate formed in SN​¹ reactions, thus favouring them. SN^​2 is favored by strong nucleophiles, primary alkyl halides (less steric hindrance), and polar aprotic solvents.

15. The carbon atom in a carbanion is typically: (a) sp² hybridized (b) sp hybridized (c) sp³ hybridized (d) dsp² hybridized

Explanation: The correct answer is (c). The carbon atom in a carbanion has three bonded pairs and one lone pair of electrons. This corresponds to sp³ hybridization, and the geometry is typically pyramidal (similar to ammonia).

16. Which reaction is characteristic of alkanes? (a) Electrophilic addition (b) Nucleophilic substitution (c) Free radical substitution (d) Nucleophilic addition

Explanation: The correct answer is (c). Alkanes are relatively unreactive due to their strong C-H and C-C σ-bonds. They primarily undergo free radical substitution reactions, such as halogenation in the presence of UV light.

17. The intermediate formed during the nitration of benzene is a(n): (a) Carbanion (b) Carbocation (sigma complex) (c) Free radical (d) Nitrene

Explanation: The correct answer is (b). Nitration of benzene is an electrophilic aromatic substitution. The electrophile (NO₂⁺​) attacks the benzene ring, forming a resonance-stabilized carbocation intermediate, also known as a sigma complex or arenium ion.

18. Which of the following is a two-step reaction mechanism? (a) SN​² (b) E₂ (c) SN​¹ (d) All of the above

Explanation: The correct answer is (c). Both SN​¹ and E₁ reactions are two-step processes involving the formation of a carbocation intermediate. SN​² and E₂ are one-step (concerted) mechanisms.

19. In an SN​² reaction, the rate depends on: (a) Concentration of substrate only (b) Concentration of nucleophile only (c) Concentrations of both substrate and nucleophile (d) Concentration of leaving group only

Explanation: The correct answer is (c). SN​² is a bimolecular reaction, meaning its rate-determining step involves two molecules: the substrate and the nucleophile. Therefore, Rate = k[Substrate][Nucleophile].

20. Which effect explains the ortho/para-directing nature of electron-donating groups in electrophilic aromatic substitution? (a) Inductive effect (b) Hyperconjugation (c) Resonance effect (d) Steric effect

Explanation: The correct answer is (c). Electron-donating groups with lone pairs (e.g., -OH, -NH₂, -OR) activate the ortho and para positions of the aromatic ring by increasing electron density through the resonance effect (+M effect).

21. A molecule with an incomplete octet on a carbon atom and a positive charge is a: (a) Carbanion (b) Free radical (c) Carbocation (d) Electrophile (but specifically an ion)

Explanation: The correct answer is (c). A carbocation (carbonium ion) has a carbon atom with only six valence electrons (an incomplete octet) and a positive charge. While it acts as an electrophile, the specific term for this intermediate is carbocation.

22. Which of the following reactions involves the formation of a free radical intermediate? (a) Friedel-Crafts alkylation (b) Halogenation of methane under UV light (c) Nitration of benzene (d) Dehydration of alcohol

Explanation: The correct answer is (b). The halogenation of alkanes (like methane) with halogens (e.g., chlorine) in the presence of UV light proceeds via a free radical chain mechanism.

23. The relative stability of free radicals is: (a) 1⋅>2⋅>3⋅ (b) 3⋅>2⋅>1⋅ (c) CH₃⋅​>1⋅>2⋅ (d) All are equally stable

Explanation: The correct answer is (b). Similar to carbocations, the stability of free radicals increases with increasing alkyl substitution due to hyperconjugation.

24. Which type of reaction mechanism is dominant for primary alkyl halides with strong nucleophiles/bases? (a) SN​¹ (b) SN​² (c) E₁ (d) AE​

Explanation: The correct answer is (b). Primary alkyl halides undergo SN​² reactions very effectively because of minimal steric hindrance around the carbon atom attacked by the nucleophile. Strong nucleophiles also favour SN​².

25. A reaction that involves the removal of atoms or groups from adjacent carbon atoms to form a multiple bond is called: (a) Substitution (b) Addition (c) Elimination (d) Rearrangement

Explanation: The correct answer is (c). Elimination reactions are characterized by the removal of two groups from adjacent carbon atoms, leading to the formation of a double or triple bond.

26. The rate-determining step in an SN​¹ reaction is: (a) Attack of the nucleophile (b) Formation of the carbocation (c) Deprotonation (d) Formation of the product

Explanation: The correct answer is (b). In SN^​1 reactions, the slowest and thus rate-determining step is the ionization of the substrate to form a carbocation.

27. Which of the following is an electron-withdrawing group by inductive effect (-I effect)? (a) −CH₃​ (b) −OH (c) −NH₂​ (d) −OCH₃​

Explanation: The correct answer is (b). The hydroxyl group (−OH) is more electronegative than carbon and pulls electron density away from the carbon chain through the inductive effect (-I effect). Alkyl groups are electron-donating (+I). Amino (−NH₂​) and alkoxy (−OCH₃​) groups are electron-donating by resonance (+M) but electron-withdrawing by induction (-I). However, usually, the more pronounced effect (resonance) is considered dominant for such groups when conjugated. But strictly by inductive effect, -OH is electron withdrawing.

28. What happens to the configuration of a chiral centre in an SN​¹ reaction? (a) Complete inversion (b) Complete retention (c) Racemization (partial inversion and retention) (d) No change

Explanation: The correct answer is (c). The carbocation intermediate formed in an SN​¹ reaction is planar. The nucleophile can attack from either side, leading to a mixture of enantiomers (racemic mixture), assuming no external factors bias the attack.

29. Nucleophilic addition reactions are characteristic of: (a) Alkanes (b) Alkenes (c) Aldehydes and ketones (d) Aromatic compounds

Explanation: The correct answer is (c). Aldehydes and ketones contain a polar carbonyl group (C=O), where the carbon is electrophilic. Nucleophiles readily attack this carbon, leading to nucleophilic addition reactions.

30. Which effect contributes to the extra stability of benzene? (a) Inductive effect (b) Hyperconjugation (c) Resonance (aromaticity) (d) Steric hindrance

Explanation: The correct answer is (c). The unusual stability of benzene and other aromatic compounds is due to the extensive delocalization of π-electrons within the cyclic, planar conjugated system, known as resonance or aromaticity (Huckel’s rule).

31. The intermediate formed in Hofmann bromamide degradation is a: (a) Carbene (b) Nitrene (c) Free radical (d) Carbanion

Explanation: The correct answer is (b). The Hofmann bromamide degradation (rearrangement) reaction proceeds through the formation of an acyl nitrene intermediate.

32. An E2 reaction typically occurs with: (a) Weak base in protic solvent (b) Strong base in aprotic solvent (c) Weak acid in protic solvent (d) Strong acid in aprotic solvent

Explanation: The correct answer is (b). E₂ reactions are favoured by strong bases, which efficiently abstract a β-hydrogen, and polar aprotic solvents, which do not solvate the base as much, keeping it reactive.

33. What is the stability order of carbanions? (a) 1∘>2∘>3∘ (b) 3∘>2∘>1∘ (c) CH₃−​>1∘>2∘>3∘ (d) All are equally stable

Explanation: The correct answer is (c). The stability of carbanions is inversely related to the number of alkyl groups attached to the negatively charged carbon. Alkyl groups are electron-donating (+I effect), which destabilizes the negative charge. Thus, methyl carbanion is the most stable.

34. The rearrangement of carbocations from less stable to more stable is often achieved by: (a) Addition of nucleophile (b) Hydride or alkyl shifts (c) Oxidation (d) Reduction

Explanation: The correct answer is (b). Carbocations can undergo rearrangement to form more stable carbocations (e.g., from 1∘ to 2∘ or 2∘ to 3∘) through 1,2-hydride shifts or 1,2-alkyl shifts.

35. Electrophilic aromatic substitution is a characteristic reaction of: (a) Alkanes (b) Alkenes (c) Arenes (aromatic compounds) (d) Alkyl halides

Explanation: The correct answer is (c). Aromatic compounds (arenes) are characterized by their tendency to undergo electrophilic substitution reactions, preserving their aromaticity.

36. Which of the following functional groups is a strong electron-withdrawing group by resonance (-M effect)? (a) −OH (b) −CH₃​ (c) −NO₂​ (d) −NH₂​

Explanation: The correct answer is (c). The nitro group (−NO₂​) strongly pulls electron density from the conjugated system (e.g., benzene ring) through the resonance effect, making it a powerful electron-withdrawing group.

37. The anti-Markovnikov’s addition of HBr to an alkene occurs in the presence of: (a) Acid catalyst (b) Peroxides (c) UV light (d) Lewis acid

Explanation: The correct answer is (b). The anti-Markovnikov’s addition of HBr to alkenes (e.g., propene) occurs via a free radical mechanism in the presence of peroxides (Kharasch effect), where the bromine radical adds first.

38. Which type of reaction involves the replacement of an atom or group by another atom or group? (a) Addition (b) Elimination (c) Substitution (d) Rearrangement

Explanation: The correct answer is (c). In substitution reactions, one atom or group in a molecule is replaced by another atom or group.

39. The most reactive alkyl halide towards SN​2 reactions is: (a) Tertiary alkyl halide (b) Secondary alkyl halide (c) Primary alkyl halide (d) Methyl halide

Explanation: The correct answer is (d). SN​² reactions are highly sensitive to steric hindrance. Methyl halides (CH₃​X) have the least steric hindrance, making them the most reactive, followed by primary, secondary, and tertiary alkyl halides (which are usually unreactive towards SN​²).

40. A nucleophile can also be termed as a: (a) Lewis acid (b) Brønsted-Lowry acid (c) Lewis base (d) Electrophile

Explanation: The correct answer is (c). Nucleophiles are electron-rich species that can donate a pair of electrons. According to Lewis acid-base theory, electron pair donors are Lewis bases.