Chapter: Nucleophilic Substitution at Saturated Carbon

1. Introduction to Nucleophilic Substitution

• Definition: Nucleophilic substitution reactions are a fundamental class of organic reactions where a nucleophile (an electron-rich species) replaces a leaving group (an atom or group that departs with its bonding electrons) on an electrophilic carbon atom. The carbon atom undergoing substitution is typically sp3 hybridized (saturated).
• General Equation:
  Nu−+R-L→R-Nu+L− Where:
  • Nu− = Nucleophile (electron-rich, typically a Lewis base)
  • R-L = Substrate (R is the alkyl group, L is the leaving group)
  • L− = Leaving Group (departs with the electron pair)
• Importance: These reactions are crucial for interconverting functional groups and building molecular complexity in organic synthesis.

2. The SN​2 Mechanism (Substitution Nucleophilic Bimolecular)

• Definition: A concerted (one-step) reaction where bond breaking (C-L) and bond forming (C-Nu) occur simultaneously. It is a bimolecular process because the rate-determining step involves both the nucleophile and the substrate.
• Key Characteristics:
  • Concerted: No intermediates are formed.
  • Backside Attack: The nucleophile approaches the electrophilic carbon from the side opposite to the leaving group. This maximizes orbital overlap and minimizes steric repulsion.
  • Inversion of Configuration: Due to backside attack, if the carbon atom undergoing substitution is chiral, its configuration is inverted (Walden inversion). The product will have the opposite configuration (R becomes S, S becomes R) if the priority of groups does not change.
• Rate Law: Rate=k[Substrate][Nucleophile]
  • This is a second-order reaction (first-order with respect to substrate, first-order with respect to nucleophile).
• Energy Diagram: A single transition state (high-energy point) where the carbon is momentarily pentacoordinate (has five groups partially bonded to it).

2.1. Factors Affecting SN​2 Rate:

• Substrate Structure (Steric Hindrance): The most critical factor.
  • Primary alkyl halides (e.g., CH3​X, RCH2​X): React fastest. Less steric hindrance allows easy backside attack.
  • Secondary alkyl halides (e.g., R2​CHX): React slower than primary.
  • Tertiary alkyl halides (e.g., R3​CX): Do not undergo SN​2 reactions due to severe steric hindrance.
  • Order of Reactivity: Methyl > Primary > Secondary >> Tertiary
  • Allylic and Benzylic halides: React faster than saturated alkyl halides of similar substitution due to resonance stabilization of the transition state.
• Nucleophile Strength: Stronger nucleophiles lead to faster SN​2 rates.
  • Trends:
    • Charge: Negatively charged nucleophiles are generally stronger than their neutral conjugate acids (e.g., OH−>H2​O).
    • Across a Period: Nucleophilicity generally decreases with increasing electronegativity (e.g., CH3−​>NH2−​>OH−>F−) in protic solvents.
    • Down a Group (in protic solvents): Nucleophilicity generally increases with increasing size/polarizability because larger anions are less solvated and their electron density is more diffuse, making them better able to overlap with the electrophile (e.g., I−>Br−>Cl−>F−).
    • Down a Group (in polar aprotic solvents): Nucleophilicity follows basicity (e.g., F−>Cl−>Br−>I−) because anions are “naked” and not extensively solvated.
• Leaving Group Ability: Good leaving groups lead to faster rates. A good leaving group is a weak base (stable anion) that can readily depart with the bonding electrons.
  • Weak Bases: Strong conjugate acids have weak conjugate bases, which are good leaving groups (e.g., I−, Br−, Cl−, TsO− (tosylate), H2​O).
  • Poor Leaving Groups: Strong bases are poor leaving groups (e.g., OH−, NH2−​, OR−). Alcohols are often protonated to make H2​O a good leaving group.
  • Order of Reactivity: I−>Br−>Cl−≫F−.
• Solvent Effects: Polar aprotic solvents favor SN​2 reactions.
  • Polar Aprotic Solvents (e.g., DMSO, DMF, Acetone, Acetonitrile): Have high dielectric constants (polar) but cannot hydrogen bond to anions (aprotic). They effectively solvate cations, but leave anions “naked” or poorly solvated, making them more nucleophilic and reactive.

3. The SN​1 Mechanism (Substitution Nucleophilic Unimolecular)

• Definition: A stepwise reaction that proceeds in two or more steps, with the formation of a carbocation intermediate in the rate-determining step. It is a unimolecular process because the rate-determining step involves only the substrate.
• Key Characteristics:
  • Stepwise: Involves at least one intermediate (carbocation).
  • Carbocation Intermediate: Planar (sp2 hybridized) carbocation is formed.
  • Racemization: If the carbon atom undergoing substitution is chiral, the nucleophile can attack from either face of the planar carbocation, leading to a racemic mixture (equal amounts of R and S enantiomers) if a new chiral center is formed.
• Rate Law: Rate=k[Substrate]
  • This is a first-order reaction (dependent only on substrate concentration). The nucleophile concentration does not affect the rate.
• Energy Diagram: Involves two transition states (one for leaving group departure, one for nucleophile attack) and one energy minimum for the carbocation intermediate. The first transition state (carbocation formation) is typically higher in energy and is the rate-determining step.

3.1. Factors Affecting SN​1 Rate:

• Substrate Structure (Carbocation Stability): The most critical factor.
  • Tertiary alkyl halides (e.g., R3​CX): React fastest because tertiary carbocations are the most stable (due to hyperconjugation and inductive effects from alkyl groups).
  • Secondary alkyl halides (e.g., R2​CHX): React slower than tertiary.
  • Primary alkyl halides (e.g., RCH2​X) & Methyl halides: Do not undergo SN​1 reactions because primary and methyl carbocations are highly unstable.
  • Order of Reactivity: Tertiary > Secondary >> Primary > Methyl
  • Allylic and Benzylic halides: React faster than saturated alkyl halides due to resonance stabilization of the carbocation intermediate.
  • Rearrangements: Carbocations can rearrange (e.g., hydride or alkyl shifts) to form more stable carbocations before the nucleophile attacks, leading to rearranged products.
• Nucleophile Strength: Nucleophile strength does NOT affect the rate of an SN​1 reaction (it’s not in the rate-determining step). However, a weaker nucleophile will favor SN​1 over SN​2 when competition exists. Often, the solvent acts as the nucleophile (solvolysis).
• Leaving Group Ability: Good leaving groups lead to faster rates (same as SN​2).
  • Order of Reactivity: I−>Br−>Cl−≫F−.
• Solvent Effects: Polar protic solvents favor SN​1 reactions.
  • Polar Protic Solvents (e.g., H2​O, CH3​OH, CH3​CH2​OH): Have high dielectric constants and can hydrogen bond to both cations and anions. They stabilize the transition state leading to the carbocation and the carbocation intermediate itself through solvation, thus lowering the activation energy for carbocation formation.

4. Factors Affecting Nucleophilicity (Revisited for Clarity)

• Charge: Anionic nucleophiles (R−, RO−, CN−, Cl−) are stronger than neutral nucleophiles (H2​O, CH3​OH, NH3​).
• Electronegativity (across a row): Decreases as electronegativity increases (e.g., OH−>F− in water). This is related to the basicity of the atom.
• Size/Polarizability (down a group):
  • In protic solvents: Nucleophilicity increases down a group (I−>Br−>Cl−>F−). Larger, more polarizable atoms are less solvated and their electron cloud is more easily distorted to overlap with the electrophilic carbon.
  • In aprotic solvents: Nucleophilicity follows basicity (i.e., decreases down a group, F−>Cl−>Br−>I−). Smaller, more concentrated charge in F- makes it more reactive when ‘naked’.
• Steric Hindrance: Bulky nucleophiles are weaker nucleophiles (but might be stronger bases, favoring E2).

5. Factors Affecting Leaving Group Ability (Revisited for Clarity)

• Stability of the Leaving Group: A good leaving group is a stable anion once it departs.
• Basicity: Weak bases are good leaving groups. Strong acids have very weak conjugate bases.
  • Order of Leaving Group Ability: TsO−(pKa of TsOH≈−3)>I−(pKa of HI≈−10)>Br−(pKa of HBr≈−9)>Cl−(pKa of HCl≈−7)>H2​O(pKa of H3​O+≈−1.7).
  • Very poor leaving groups: OH−, NH2−​, OR−. These are strong bases.
• Resonance Stabilization: A leaving group whose negative charge can be delocalized by resonance is a better leaving group (e.g., tosylate).

6. Solvent Effects (Revisited for Clarity)

• Polar Protic Solvents: (e.g., water, methanol, ethanol, acetic acid). Contain -OH or -NH groups that can form hydrogen bonds.
  • Stabilize nucleophiles: By hydrogen bonding, which can reduce their nucleophilicity (especially for smaller, more concentrated anions like F−).
  • Stabilize transition states/carbocations: By solvating both positive and negative ends of the transition state or the carbocation intermediate. This lowers the activation energy for carbocation formation, favoring SN​1.
• Polar Aprotic Solvents: (e.g., DMSO, DMF, Acetone, Acetonitrile, HMPA). Have a high dielectric constant but no acidic hydrogens to form hydrogen bonds.
  • Solvate cations well: Through dipole interactions.
  • Leave anions “naked”: Do not effectively solvate anions, making them more reactive/nucleophilic. This enhances the rate of SN​2 reactions.

7. Competition Between SN​1 and SN​2 (and Elimination)

• Substrate is Key:
  • Methyl & Primary: Almost exclusively SN​2.
  • Tertiary: Almost exclusively SN​1 (if good leaving group & polar protic solvent) OR E2 (if strong, bulky base). No SN​2.
  • Secondary: Most complicated. Can undergo SN​1, SN​2, E1, or E2 depending on nucleophile/base strength, solvent, and temperature.
• Nucleophile/Base Strength:
  • Strong nucleophile (weak base): Favors SN​2.
  • Weak nucleophile (weak base): Favors SN​1 (often solvolysis).
  • Strong, bulky base: Favors E2 over SN​2.
• Temperature: Higher temperatures generally favor elimination (E1/E2) over substitution (due to entropy).

8. Summary Table: SN​1 vs. SN​2

Feature	SN​1 (Substitution Nucleophilic Unimolecular)	SN​2 (Substitution Nucleophilic Bimolecular)
Mechanism	Stepwise (2+ steps)	Concerted (1 step)
Rate-determining step	Formation of carbocation	Backside attack of nucleophile
Intermediate	Carbocation	None
Rate Law	Rate=k[Substrate]	Rate=k[Substrate][Nu]
Order	First-order overall	Second-order overall
Substrate Reactivity	3∘>2∘>1∘ (methyl inert)	Methyl >1∘>2∘ (3∘ inert)
Stereochemistry	Racemization (attack from both faces of planar carbocation)	Inversion of configuration (Walden inversion)
Nucleophile Strength	Not important for rate, but weaker Nu favors SN​1	Important for rate; stronger Nu favors SN​2
Leaving Group	Good leaving group (weak base)	Good leaving group (weak base)
Solvent	Polar Protic (stabilizes carbocation)	Polar Aprotic (activates Nu)
Rearrangements	Common	Not possible

Multiple Choice Questions (MCQ) on Nucleophilic Substitution at Saturated Carbon

Instructions: Choose the best answer for each question.

1. In a nucleophilic substitution reaction, what type of carbon atom is typically attacked by the nucleophile? a) sp hybridized b) sp2 hybridized c) sp3 hybridized d) Aromatic

2. Which of the following best describes the SN​2 mechanism? a) A two-step process involving a carbocation intermediate. b) A concerted reaction with backside attack and inversion of configuration. c) A reaction whose rate depends only on the substrate concentration. d) A reaction that typically leads to racemization.

3. What is the rate law for an SN​2 reaction? a) Rate=k[Substrate] b) Rate=k[Nucleophile] c) Rate=k[Substrate][Nucleophile] d) Rate=k[Substrate]2

4. If the carbon atom undergoing SN​2 reaction is chiral, what is the stereochemical outcome? a) Racemization b) Retention of configuration c) Inversion of configuration (Walden inversion) d) Epimerization

5. Which type of substrate reacts fastest in an SN​2 reaction? a) Tertiary alkyl halide b) Secondary alkyl halide c) Primary alkyl halide d) Aryl halide

6. Which of the following is considered a strong nucleophile? a) H2​O b) CH3​OH c) F− (in protic solvent) d) CN−

7. Which of the following is the best leaving group? a) OH− b) NH2−​ c) Cl− d) CH3−​

8. What type of solvent favors SN​2 reactions by making anions more reactive? a) Polar protic solvents b) Non-polar solvents c) Polar aprotic solvents d) Acidic solvents

9. Which of the following is a common polar aprotic solvent? a) Water (H2​O) b) Ethanol (CH3​CH2​OH) c) Dimethyl sulfoxide (DMSO) d) Acetic acid (CH3​COOH)

10. What is the definition of the SN​1 mechanism? a) A concerted reaction forming a pentacoordinate transition state. b) A stepwise reaction involving a carbocation intermediate. c) A reaction whose rate depends on both substrate and nucleophile. d) A reaction exclusive to primary alkyl halides.

11. What is the rate-determining step in an SN​1 reaction? a) Nucleophilic attack on the carbocation. b) Formation of the carbocation intermediate. c) Departure of the nucleophile. d) Proton transfer.

12. If the carbon atom undergoing SN​1 reaction is chiral, what is the typical stereochemical outcome? a) Inversion of configuration b) Retention of configuration c) Racemization d) Complete inversion or retention, depending on the nucleophile.

13. Which type of substrate reacts fastest in an SN​1 reaction? a) Methyl halide b) Primary alkyl halide c) Secondary alkyl halide d) Tertiary alkyl halide

14. Why do tertiary alkyl halides favor the SN​1 mechanism? a) They have minimal steric hindrance. b) They form the most stable carbocation intermediate. c) They are highly reactive towards strong nucleophiles. d) They are easily solvated by aprotic solvents.

15. In an SN​1 reaction, how does the strength of the nucleophile affect the reaction rate? a) Stronger nucleophiles increase the rate. b) Weaker nucleophiles increase the rate. c) Nucleophile strength does not affect the rate. d) It depends on the leaving group.

16. Which type of solvent favors SN​1 reactions by stabilizing the carbocation intermediate? a) Non-polar solvents b) Polar aprotic solvents c) Polar protic solvents d) Basic solvents

17. What is a common characteristic of carbocation intermediates in SN​1 reactions? a) They are tetrahedral. b) They are sp hybridized. c) They are planar and sp2 hybridized. d) They are very stable.

18. What is a common side reaction that can occur with carbocation intermediates in SN​1 reactions? a) Nucleophilic attack by a strong nucleophile. b) SN​2 attack. c) Rearrangements (e.g., hydride or alkyl shifts). d) Reduction.

19. Which of the following is a poor leaving group? a) I− b) Br− c) TsO− (tosylate) d) OR− (alkoxide)

20. What generally determines a good leaving group? a) It is a strong base. b) It is a weak base (stable as an anion). c) It is a strong acid. d) It is bulky.

21. Order the following nucleophiles from strongest to weakest in a polar protic solvent: F−, Cl−, Br−, I−. a) F−>Cl−>Br−>I− b) I−>Br−>Cl−>F− c) Cl−>Br−>I−>F− d) Br−>Cl−>F−>I−

22. Order the following nucleophiles from strongest to weakest in a polar aprotic solvent: F−, Cl−, Br−, I−. a) F−>Cl−>Br−>I− b) I−>Br−>Cl−>F− c) Cl−>Br−>I−>F− d) Br−>Cl−>F−>I−

23. Which of the following substrates is most likely to undergo an SN​2 reaction? a) 2-bromo-2-methylpropane b) Bromobenzene c) Bromoethane d) 2-bromopropane

24. Which of the following conditions would favor an SN​1 reaction over an SN​2 reaction for a secondary alkyl halide? a) Strong, unhindered nucleophile. b) Polar aprotic solvent. c) Weak nucleophile/base in a polar protic solvent. d) High concentration of nucleophile.

25. When an alcohol is reacted with HBr, the actual leaving group is: a) OH− b) Br− c) H2​O d) H3​O+

26. Why are primary alkyl halides generally unreactive towards SN​1 reactions? a) They are too sterically hindered. b) They form unstable primary carbocations. c) They have poor leaving groups. d) They only react via SN​2.

27. A reaction with the rate law Rate=k[RX] suggests which mechanism is operating? a) SN​2 b) E2 c) SN​1 d) Radical

28. What type of solvent stabilizes carbocations by hydrogen bonding, lowering the activation energy for their formation? a) Non-polar solvent b) Polar aprotic solvent c) Polar protic solvent d) Basic solvent

29. The intermediate formed in an SN​1 reaction is a(n): a) Alkyl radical b) Carbonyl c) Carbocation d) Carbanion

30. Which of the following nucleophiles is most likely to favor an SN​2 reaction rather than elimination (E2) when reacting with a secondary alkyl halide? a) Sodium tert-butoxide ((CH3​)3​CO−Na+) b) Potassium hydroxide (KOH) c) Methoxide (CH3​O−) d) Methyl sulfide (CH3​S−)

31. If an SN​2 reaction proceeds, the transition state involves: a) A fully formed carbocation. b) The nucleophile and leaving group fully detached. c) Partial bond breaking and bond forming at the carbon center. d) A radical intermediate.

32. What is the effect of increasing the concentration of a strong nucleophile on the rate of an SN​1 reaction? a) The rate increases. b) The rate decreases. c) The rate remains unchanged. d) The reaction mechanism changes to SN​2.

33. Which of the following is the order of reactivity for alkyl halides in SN​1 reactions? a) Methyl > Primary > Secondary > Tertiary b) Tertiary > Secondary > Primary > Methyl c) Primary > Secondary > Tertiary > Methyl d) Secondary > Primary > Tertiary > Methyl

34. Why do polar aprotic solvents increase the rate of SN​2 reactions? a) They stabilize the transition state. b) They solvate the leaving group more effectively. c) They don’t solvate the nucleophile as much, making it more reactive. d) They increase the concentration of the substrate.

35. A primary alkyl halide with a good leaving group and a strong, unhindered nucleophile in a polar aprotic solvent will predominantly undergo which reaction? a) SN​1 b) E1 c) SN​2 d) E2

36. Allylic and benzylic halides often show enhanced reactivity in both SN​1 and SN​2 reactions due to: a) Their higher electronegativity. b) The resonance stabilization of their carbocation intermediate (for SN​1) or transition state (for SN​2). c) Their ability to form stable radicals. d) Their saturated nature.

37. Which of the following is a weak base and therefore a good leaving group? a) OH− b) NH2−​ c) Br− d) CH3​CH2​O−

38. If the nucleophile for an SN​1 reaction is also the solvent, the reaction is called: a) Hydrolysis b) Solvolysis c) Transesterification d) Saponification

39. What is the typical effect of increasing temperature on the competition between substitution and elimination reactions? a) Favors substitution. b) Favors elimination. c) Has no effect. d) Favors SN​1 only.

40. Which statement about the energy diagrams of SN​1 and SN​2 reactions is true? a) SN​1 has one transition state, SN​2 has two. b) SN​1 has an intermediate, SN​2 does not. c) Both mechanisms have the same activation energy. d) SN​2 has an intermediate, SN​1 does not.

Answer Key with Explanations

1. c) sp3 hybridized.
   • Explanation: Nucleophilic substitution reactions, as discussed in this chapter, specifically occur at saturated (all single bonds) carbon atoms, which are sp3 hybridized.
2. b) A concerted reaction with backside attack and inversion of configuration.
   • Explanation: The SN​2 mechanism is a single-step (concerted) process where the nucleophile attacks from the opposite side of the leaving group, leading to an inversion of stereochemistry at the reacting carbon.
3. c) Rate=k[Substrate][Nucleophile].
   • Explanation: The SN​2 reaction is bimolecular, meaning its rate depends on the concentrations of both the substrate and the nucleophile in the rate-determining step.
4. c) Inversion of configuration (Walden inversion).
   • Explanation: The backside attack of the nucleophile in an SN​2 reaction causes the inversion of configuration at a chiral center.
5. c) Primary alkyl halide.
   • Explanation: SN​2 reactions are highly sensitive to steric hindrance. Primary alkyl halides (and methyl halides) are least sterically hindered at the electrophilic carbon, allowing for the fastest backside attack.
6. d) CN−.
   • Explanation: Negatively charged species are generally stronger nucleophiles than neutral ones. Among the options, CN− is a well-known strong nucleophile. F− is strong as a base, but its nucleophilicity is hindered by solvation in protic solvents.
7. c) Cl−.
   • Explanation: Good leaving groups are weak bases. Cl− is the conjugate base of HCl (a strong acid), making Cl− a relatively weak base and therefore a good leaving group. OH−, NH2−​, and CH3−​ are all strong bases and poor leaving groups.
8. c) Polar aprotic solvents.
   • Explanation: Polar aprotic solvents solvate cations well but do not form strong hydrogen bonds with anions, leaving the anions “naked” and highly reactive as nucleophiles, thus favoring SN​2.
9. c) Dimethyl sulfoxide (DMSO).
   • Explanation: DMSO is a classic example of a polar aprotic solvent, along with DMF, acetone, and acetonitrile. Water, ethanol, and acetic acid are polar protic.
10. b) A stepwise reaction involving a carbocation intermediate.
    • Explanation: The SN​1 mechanism proceeds in at least two steps, with the initial departure of the leaving group forming a carbocation intermediate, which is then attacked by the nucleophile.
11. b) Formation of the carbocation intermediate.
    • Explanation: This first step, the dissociation of the leaving group to form a carbocation, is typically the slowest (highest energy) step and therefore the rate-determining step in an SN​1 reaction.
12. c) Racemization.
    • Explanation: Because the carbocation intermediate is planar, the nucleophile can attack from either face with roughly equal probability, leading to a mixture of R and S enantiomers (a racemic mixture) if a new chiral center is formed.
13. d) Tertiary alkyl halide.
    • Explanation: SN​1 reactions are favored by substrates that can form stable carbocations. Tertiary carbocations are the most stable (due to hyperconjugation and inductive effects), making tertiary alkyl halides react fastest.
14. b) They form the most stable carbocation intermediate.
    • Explanation: The stability of the carbocation intermediate is the primary factor determining the rate of an SN​1 reaction. Tertiary carbocations are highly stabilized.
15. c) Nucleophile strength does not affect the rate.
    • Explanation: The nucleophile is not involved in the rate-determining step (carbocation formation), so its concentration and strength do not appear in the rate law and do not affect the reaction rate.
16. c) Polar protic solvents.
    • Explanation: Polar protic solvents stabilize the highly charged transition state leading to the carbocation and the carbocation intermediate itself through solvation (especially hydrogen bonding), thereby lowering the activation energy for the rate-determining step.
17. c) They are planar and sp2 hybridized.
    • Explanation: Carbocations are sp2 hybridized at the positively charged carbon, which adopts a trigonal planar geometry.
18. c) Rearrangements (e.g., hydride or alkyl shifts).
    • Explanation: Carbocations can undergo rearrangements to form a more stable carbocation (e.g., from secondary to tertiary), leading to unexpected or rearranged products.
19. d) OR− (alkoxide).
    • Explanation: Alkoxide ions (OR−) are very strong bases (conjugate bases of very weak acids, alcohols) and thus are very poor leaving groups. I−, Br−, and TsO− are good leaving groups.
20. b) It is a weak base (stable as an anion).
    • Explanation: A good leaving group is one that can depart with the bonding pair of electrons and exist as a stable anion. Stable anions are typically weak bases.
21. b) I−>Br−>Cl−>F−.
    • Explanation: In polar protic solvents, nucleophilicity increases down a group because larger anions are less solvated and more polarizable, making them better able to overlap with the electrophilic carbon.
22. a) F−>Cl−>Br−>I−.
    • Explanation: In polar aprotic solvents, anions are “naked” or poorly solvated. Nucleophilicity then generally follows basicity, meaning smaller, more concentrated charges (like F−) are more reactive as nucleophiles.
23. c) Bromoethane.
    • Explanation: Bromoethane is a primary alkyl halide, which is the most favored substrate for SN​2 reactions due to minimal steric hindrance. 2-bromo-2-methylpropane is tertiary (favors SN​1/E2). 2-bromopropane is secondary. Bromobenzene is an aryl halide, which does not undergo simple SN​1 or SN​2.
24. c) Weak nucleophile/base in a polar protic solvent.
    • Explanation: For secondary alkyl halides, these conditions (weak nucleophile, polar protic solvent) specifically disfavor SN​2 (which needs a strong nucleophile and polar aprotic solvent) and favor SN​1 (which is promoted by carbocation stabilization in protic solvents).
25. c) H2​O.
    • Explanation: Alcohols are protonated by HBr to form an oxonium ion (R-OH2+​). Water (H2​O) is then the actual leaving group, as it is a much weaker base and more stable than OH−.
26. b) They form unstable primary carbocations.
    • Explanation: Primary carbocations are highly unstable and very difficult to form, making SN​1 reactions generally unfavorable for primary alkyl halides.
27. c) SN​1.
    • Explanation: A rate law that only depends on the concentration of the substrate (RX) is characteristic of a first-order reaction, specifically the SN​1 mechanism.
28. c) Polar protic solvent.
    • Explanation: Polar protic solvents (like water, alcohols) stabilize carbocations (positive charge) through hydrogen bonding and dipole interactions, thereby facilitating their formation.
29. c) Carbocation.
    • Explanation: The defining intermediate in an SN​1 reaction is a carbocation, which is formed after the departure of the leaving group.
30. d) Methyl sulfide (CH3​S−).
    • Explanation: CH3​S− (methylthiolate) is a good nucleophile but a relatively weak base (due to the large sulfur atom’s polarizability), which favors substitution over elimination. The other options are strong bases and would favor E2. Sodium tert-butoxide is a strong, bulky base, highly favoring E2.
31. c) Partial bond breaking and bond forming at the carbon center.
    • Explanation: The transition state in an SN​2 reaction involves the nucleophile and leaving group simultaneously forming and breaking bonds with the central carbon, making it momentarily pentacoordinate.
32. c) The rate remains unchanged.
    • Explanation: The rate of an SN​1 reaction is determined only by the concentration of the substrate, as the nucleophile is not involved in the rate-determining step.
33. b) Tertiary > Secondary > Primary > Methyl.
    • Explanation: This order reflects the increasing stability of the carbocation intermediate, which is the key factor for SN​1 reactivity.
34. c) They don’t solvate the nucleophile as much, making it more reactive.
    • Explanation: Polar aprotic solvents do not form strong hydrogen bonds with anions, leaving the anionic nucleophile “naked” and more accessible/reactive for attack, thereby increasing the SN​2 rate.
35. c) SN​2.
    • Explanation: This set of conditions (primary substrate, good leaving group, strong unhindered nucleophile, polar aprotic solvent) are all optimal for the SN​2 mechanism.
36. b) The resonance stabilization of their carbocation intermediate (for SN​1) or transition state (for SN​2).
    • Explanation: Both allylic and benzylic systems benefit from resonance stabilization. In SN​1, it stabilizes the carbocation. In SN​2, it stabilizes the developing partial charges in the transition state. This makes them more reactive than simple alkyl halides.
37. c) Br−.
    • Explanation: Br− is the conjugate base of HBr (a strong acid), making it a weak base and an excellent leaving group. OH−, NH2−​, and CH3​CH2​O− are all strong bases and poor leaving groups.
38. b) Solvolysis.
    • Explanation: A solvolysis reaction is a substitution (or sometimes elimination) reaction where the solvent also acts as the nucleophile (or base).
39. b) Favors elimination.
    • Explanation: Elimination reactions often have a higher entropy of activation (more molecules or freedom in the transition state) and are therefore generally favored by higher temperatures. Substitution reactions are less affected or can be disfavored at high temperatures.
40. b) SN​1 has an intermediate, SN​2 does not.
    • Explanation: The SN​1 mechanism is stepwise and involves a carbocation intermediate. The SN​2 mechanism is concerted and proceeds through a single transition state without forming an intermediate.