Reactions of Haloarenes; Nucleophilic Substitution and Electrophilic Substitution Reactions MCQ Quiz - Objective Question with Answer for Reactions of Haloarenes; Nucleophilic Substitution and Electrophilic Substitution Reactions - Download Free PDF

CONCEPT:

SN1 Reaction (Unimolecular Nucleophilic Substitution)

• The SN1 reaction is a two-step process where the first step is the rate-determining step.
• The first step involves the formation of a carbocation intermediate by the loss of the leaving group.
• The rate of the SN1 reaction depends on the stability of the carbocation intermediate formed.
• More stable carbocations form faster, leading to a faster SN1 reaction.

EXPLANATION:

• Comparing the given compounds:
  1. Chloromethane (CH3Cl) - Forms a methyl carbocation, which is highly unstable.
     
  2. 2-bromo-3-methylbutane - Forms a secondary carbocation, which is relatively more stable than a methyl carbocation.
     
  3. 2-chloro-3-methylbutane - Also forms a secondary carbocation, but chlorine is a poorer leaving group compared to bromine.
  4. 
  5. 2-bromo-2-methylpropane - Forms a tertiary carbocation, which is highly stable.
     
• Among the given options, the 2-bromo-2-methylpropane forms a tertiary carbocation which is the most stable.
• Because the stability of the carbocation intermediate is the key factor in the SN1 reaction, the compound that forms the most stable carbocation will undergo the SN1 reaction the fastest.

Therefore, the compound that would undergo the SN1 reaction faster is 2-bromo-2-methylpropane (option 4).

CONCEPT:

Ortho- and Para-Directing Groups in Electrophilic Aromatic Substitution

• In electrophilic aromatic substitution reactions, substituents already present on the benzene ring influence the position where the new electrophile will attack.
• Substituents can either be activating or deactivating. Activating groups typically direct electrophiles to the ortho and para positions, while deactivating groups direct to the meta position.
• Chlorine is unique because it is an electron withdrawing group through inductive effect, yet it is ortho- and para-directing due to its resonance effect.

EXPLANATION:

• Chlorine withdraws electrons through its inductive effect due to its high electronegativity:
  • Chlorine is more electronegative than carbon, leading to a withdrawal of electron density from the benzene ring through the sigma bonds (inductive effect).
• Despite this, chlorine releases electrons through resonance, which stabilizes the intermediate carbocation formed during electrophilic substitution:
  • Chlorine has lone pairs of electrons that can participate in resonance with the benzene ring. This delocalization of electrons can stabilize the positively charged intermediate formed during the substitution.
  • Resonance donation of electron density to the ortho and para positions makes these positions more reactive towards electrophilic attack.

Therefore, the correct answer is (A), (B), and (D) only.

CONCEPT:

SN1 Reaction (Unimolecular Nucleophilic Substitution)

• The SN1 reaction is a two-step process where the first step is the rate-determining step.
• The first step involves the formation of a carbocation intermediate by the loss of the leaving group.
• The rate of the SN1 reaction depends on the stability of the carbocation intermediate formed.
• More stable carbocations form faster, leading to a faster SN1 reaction.

EXPLANATION:

• Comparing the given compounds:
  1. Chloromethane (CH3Cl) - Forms a methyl carbocation, which is highly unstable.
     
  2. 2-bromo-3-methylbutane - Forms a secondary carbocation, which is relatively more stable than a methyl carbocation.
     
  3. 2-chloro-3-methylbutane - Also forms a secondary carbocation, but chlorine is a poorer leaving group compared to bromine.
  4. 
  5. 2-bromo-2-methylpropane - Forms a tertiary carbocation, which is highly stable.
     
• Among the given options, the 2-bromo-2-methylpropane forms a tertiary carbocation which is the most stable.
• Because the stability of the carbocation intermediate is the key factor in the SN1 reaction, the compound that forms the most stable carbocation will undergo the SN1 reaction the fastest.

Therefore, the compound that would undergo the SN1 reaction faster is 2-bromo-2-methylpropane (option 4).

CONCEPT:

SN₂ Reactivity and Steric/Electronic Effects

• The rate of SN₂ reactions depends on the electrophilicity of the carbon and the steric hindrance around it.
• Methyl halides > Primary alkyl halides > Secondary alkyl halides for SN₂ reactivity.
• Electron withdrawing groups near the reaction center increase reactivity by increasing electrophilicity.

EXPLANATION:

• S: Has a benzoyl (PhCO) group, which is strongly electron-withdrawing → increases electrophilicity → maximum SN₂ reactivity
• P: Methyl chloride, very low steric hindrance → high SN₂ reactivity
• R: Primary halide → moderate reactivity
• Q: Secondary halide → highest steric hindrance → least reactive

Reactivity order: S > P > R > Q

Relative rates: S = 1000, P = 200, R = 79, Q = 0.02

Therefore, the correct answer is: Option 2) S > P > R > Q

CONCEPT:

Substitution Reactions with Methane

• Substitution reactions involve the replacement of a hydrogen atom in a molecule with another atom or group of atoms.
• In the case of methane (CH₄), the hydrogen atoms can be substituted by halogens (Cl, Br, I, F).
• Different halogens require different conditions and reagents for the substitution reaction to occur.

EXPLANATION:

• In the case of iodination of methane:

  CH₄ + I₂ → CH₃I + HI

  • Iodination of methane is less favorable compared to chlorination or bromination because the C-I bond is weaker than C-Cl or C-Br bonds.
  • To make the iodination reaction feasible, an oxidizing agent like HIO₃ (iodic acid) is used.
• HIO₃ helps in the following way:
  • It oxidizes the byproduct HI (hydrogen iodide) back to I₂ (iodine), allowing the reaction to proceed.
  • This prevents the accumulation of HI, which would otherwise reverse the reaction.

Therefore, the substitution reaction of methane that requires HIO₃ as an oxidizing agent is iodination (Option 3).

The ccorrect answer is option 1 that is Benzyl chloride.

Concept:

• Compounds that can produce a stable carbocation intermediate are more likely to undergo a nucleophilic substitution reaction that exclusively proceeds via an SN1 mechanism.

Explanation:

• The chemical that, in the list of choices, can produce a stable benzylic carbocation when it loses the chloride ion is benzyl chloride (C₆H₅CH₂Cl). This is because of its resonance stabilization with the phenyl group.
• Ethyl chloride (CH₃CH₂Cl) often undergoes an SN₂ mechanism because it creates a less stable primary carbocation, which makes it less appropriate for an SN₁ reaction.
• Due to the aromatic ring's resonance stabilization, chlorobenzene (C₆H₅Cl) typically does not undergo nucleophilic substitution.
• In order to increase the benzene ring's resistance to both the SN1 and SN2 reactions, the reaction would necessitate breaking its aromaticity, which is energetically unfavorable.

CONCEPT:

Ortho- and Para-Directing Groups in Electrophilic Aromatic Substitution

• In electrophilic aromatic substitution reactions, substituents already present on the benzene ring influence the position where the new electrophile will attack.
• Substituents can either be activating or deactivating. Activating groups typically direct electrophiles to the ortho and para positions, while deactivating groups direct to the meta position.
• Chlorine is unique because it is an electron withdrawing group through inductive effect, yet it is ortho- and para-directing due to its resonance effect.

EXPLANATION:

• Chlorine withdraws electrons through its inductive effect due to its high electronegativity:
  • Chlorine is more electronegative than carbon, leading to a withdrawal of electron density from the benzene ring through the sigma bonds (inductive effect).
• Despite this, chlorine releases electrons through resonance, which stabilizes the intermediate carbocation formed during electrophilic substitution:
  • Chlorine has lone pairs of electrons that can participate in resonance with the benzene ring. This delocalization of electrons can stabilize the positively charged intermediate formed during the substitution.
  • Resonance donation of electron density to the ortho and para positions makes these positions more reactive towards electrophilic attack.

Therefore, the correct answer is (A), (B), and (D) only.

The ccorrect answer is option 1 that is Benzyl chloride.

Concept:

• Compounds that can produce a stable carbocation intermediate are more likely to undergo a nucleophilic substitution reaction that exclusively proceeds via an SN1 mechanism.

Explanation:

• The chemical that, in the list of choices, can produce a stable benzylic carbocation when it loses the chloride ion is benzyl chloride (C₆H₅CH₂Cl). This is because of its resonance stabilization with the phenyl group.
• Ethyl chloride (CH₃CH₂Cl) often undergoes an SN₂ mechanism because it creates a less stable primary carbocation, which makes it less appropriate for an SN₁ reaction.
• Due to the aromatic ring's resonance stabilization, chlorobenzene (C₆H₅Cl) typically does not undergo nucleophilic substitution.
• In order to increase the benzene ring's resistance to both the SN1 and SN2 reactions, the reaction would necessitate breaking its aromaticity, which is energetically unfavorable.

CONCEPT:

SN1 Reaction (Unimolecular Nucleophilic Substitution)

• The SN1 reaction is a two-step process where the first step is the rate-determining step.
• The first step involves the formation of a carbocation intermediate by the loss of the leaving group.
• The rate of the SN1 reaction depends on the stability of the carbocation intermediate formed.
• More stable carbocations form faster, leading to a faster SN1 reaction.

EXPLANATION:

• Comparing the given compounds:
  1. Chloromethane (CH3Cl) - Forms a methyl carbocation, which is highly unstable.
     
  2. 2-bromo-3-methylbutane - Forms a secondary carbocation, which is relatively more stable than a methyl carbocation.
     
  3. 2-chloro-3-methylbutane - Also forms a secondary carbocation, but chlorine is a poorer leaving group compared to bromine.
  4. 
  5. 2-bromo-2-methylpropane - Forms a tertiary carbocation, which is highly stable.
     
• Among the given options, the 2-bromo-2-methylpropane forms a tertiary carbocation which is the most stable.
• Because the stability of the carbocation intermediate is the key factor in the SN1 reaction, the compound that forms the most stable carbocation will undergo the SN1 reaction the fastest.

Therefore, the compound that would undergo the SN1 reaction faster is 2-bromo-2-methylpropane (option 4).

CONCEPT:

SN1 Reaction (Unimolecular Nucleophilic Substitution)

• The SN1 reaction is a two-step process where the first step is the rate-determining step.
• The first step involves the formation of a carbocation intermediate by the loss of the leaving group.
• The rate of the SN1 reaction depends on the stability of the carbocation intermediate formed.
• More stable carbocations form faster, leading to a faster SN1 reaction.

EXPLANATION:

• Comparing the given compounds:
  1. Chloromethane (CH3Cl) - Forms a methyl carbocation, which is highly unstable.
     
  2. 2-bromo-3-methylbutane - Forms a secondary carbocation, which is relatively more stable than a methyl carbocation.
     
  3. 2-chloro-3-methylbutane - Also forms a secondary carbocation, but chlorine is a poorer leaving group compared to bromine.
  4. 
  5. 2-bromo-2-methylpropane - Forms a tertiary carbocation, which is highly stable.
     
• Among the given options, the 2-bromo-2-methylpropane forms a tertiary carbocation which is the most stable.
• Because the stability of the carbocation intermediate is the key factor in the SN1 reaction, the compound that forms the most stable carbocation will undergo the SN1 reaction the fastest.

Therefore, the compound that would undergo the SN1 reaction faster is 2-bromo-2-methylpropane (option 4).

CONCEPT:

SN₂ Reactivity and Steric/Electronic Effects

• The rate of SN₂ reactions depends on the electrophilicity of the carbon and the steric hindrance around it.
• Methyl halides > Primary alkyl halides > Secondary alkyl halides for SN₂ reactivity.
• Electron withdrawing groups near the reaction center increase reactivity by increasing electrophilicity.

EXPLANATION:

• S: Has a benzoyl (PhCO) group, which is strongly electron-withdrawing → increases electrophilicity → maximum SN₂ reactivity
• P: Methyl chloride, very low steric hindrance → high SN₂ reactivity
• R: Primary halide → moderate reactivity
• Q: Secondary halide → highest steric hindrance → least reactive

Reactivity order: S > P > R > Q

Relative rates: S = 1000, P = 200, R = 79, Q = 0.02

Therefore, the correct answer is: Option 2) S > P > R > Q

CONCEPT:

Substitution Reactions with Methane

• Substitution reactions involve the replacement of a hydrogen atom in a molecule with another atom or group of atoms.
• In the case of methane (CH₄), the hydrogen atoms can be substituted by halogens (Cl, Br, I, F).
• Different halogens require different conditions and reagents for the substitution reaction to occur.

EXPLANATION:

• In the case of iodination of methane:

  CH₄ + I₂ → CH₃I + HI

  • Iodination of methane is less favorable compared to chlorination or bromination because the C-I bond is weaker than C-Cl or C-Br bonds.
  • To make the iodination reaction feasible, an oxidizing agent like HIO₃ (iodic acid) is used.
• HIO₃ helps in the following way:
  • It oxidizes the byproduct HI (hydrogen iodide) back to I₂ (iodine), allowing the reaction to proceed.
  • This prevents the accumulation of HI, which would otherwise reverse the reaction.

Therefore, the substitution reaction of methane that requires HIO₃ as an oxidizing agent is iodination (Option 3).

CONCEPT:

Effect of Base Size on Substitution and Elimination Reactions

• The base CH₃CH₂O⁻ (ethoxide ion) is a smaller and less hindered base, favoring the substitution (S_N2) reaction over elimination due to easier access to the electrophilic carbon.
• The bulkier base, tert-butoxide ((CH₃)₃CO⁻), has steric hindrance that makes elimination (E2) more favorable than substitution as it cannot approach the electrophilic center easily.
• Given that μ = k_s/k_e, a higher rate constant for substitution (k_s) relative to elimination (k_e) will increase μ, favoring the smaller base (A) for substitution reactions and leading to μ_A > μ_B.

EXPLANATION:

• Since CH₃CH₂O⁻ (ethoxide ion) is smaller, it predominantly undergoes substitution, making k_s(A) > k_e(A).
• The bulkier tert-butoxide ((CH₃)₃CO⁻) favors elimination, so k_e(B) > k_s(B).
• Thus, μ_A > μ_B and ke(B) > ke(A), which aligns with the option stating μ_A > μ_B and k_e(B) > k_e(A).

Answer: 3) μ_A > μ_B and ke(B) > ke(A)

S_N2 rate ∝  EWG

The Correct Answer is CF₃ > CH₃ > OCH₃.

Concept:-

Nucleophilic Aromatic Substitution- Nucleophilic Aromatic Substitution is a reaction where a nucleophile replaces a leaving group on an aromatic ring.

Mechanism:
Nucleophilic Attack:

• A nucleophile (Nu⁻) attacks an electron-deficient carbon in the aromatic ring.
• The aromatic ring's pi electrons are pushed towards the carbon bearing the leaving group (LG), creating a resonance-stabilized intermediate.

Leaving Group Departure:

• The leaving group (LG) leaves, assisted by the nucleophile.
• This step reforms the aromaticity of the ring.

Product Formation:

• The nucleophile is now attached to the aromatic ring in place of the leaving group.
• The final product is a substitution of the leaving group by the nucleophile.

Explanation:-

In nucleophilic aromatic substitution reactions (S_NAr), the rate of reaction is significantly influenced by the electron-withdrawing or electron-donating nature of the substituents on the aromatic ring. Substituents that can stabilize the negative charge in the intermediate transition state by withdrawing electrons through inductive or resonance effects generally increase the rate of the reaction.

• CF₃ (Trifluoromethyl): The CF₃ group is a strong electron-withdrawing group due to the high electronegativity of fluorine atoms combined with their inductive effects. This makes the aromatic ring more susceptible to nucleophilic attack because it stabilizes the negative charge in the intermediate transition state.
• CH₃ (Methyl): The CH₃ group is electron-donating by hyperconjugation but has a relatively small effect. It does not withdraw electrons and thus makes the aromatic ring less reactive towards nucleophilic attack compared to CF_3.
• OCH₃ (Methoxy): The OCH₃ group is an electron-donating group via resonance, which increases the electron density on the aromatic ring, making it less susceptible to nucleophilic attack because it destabilizes the negative charge in the intermediate transition state.

Based on the above explanation, the correct order of the rate constants for nucleophilic aromatic substitution reactions, considering the electron-withdrawing or donating nature of the substituents, would indeed be:

CF₃ > CH₃ > OCH₃

Given reaction is an example of nucleophilic aromatic substitution. The presence of better electron withdrawing group will increase the rate of reaction. The order of better electron withdrawing group is CF₃ > CH₃ > OCH₃. 

The order of rate constant is also the same.

Conclusion:- 

The correct order of the rate constants for the given series of reactions CF₃ > CH₃ > OCH₃.