Chapter: Elimination Reactions

1. Introduction to Elimination Reactions

• Definition: Elimination reactions are a class of organic reactions in which two substituents are removed from a molecule, typically from adjacent carbon atoms, to form a new π bond (usually a double bond, but can be a triple bond).
• General Equation:
  R-CH2​-CH2​-L+Base→R-CH=CH2​+Base-H+L− Where:
  • L = Leaving Group (atom or group that departs with its bonding electrons)
  • Base = Lewis base that abstracts a proton (H) from the β-carbon
• Relationship with Substitution: Elimination reactions often compete with nucleophilic substitution reactions (SN​1 and SN​2) at saturated carbon atoms. The conditions (substrate, base/nucleophile, solvent, temperature) determine the favored pathway.

2. The E2 Mechanism (Elimination Bimolecular)

• Definition: A concerted (one-step) reaction where bond breaking (C-H and C-L) and bond forming (C=C) occur simultaneously. It is a bimolecular process because the rate-determining step involves both the substrate and the base.
• Key Characteristics:
  • Concerted: No intermediates are formed.
  • Stereospecific: Requires an anti-periplanar arrangement of the leaving group and the β-hydrogen. This means they must be in the same plane but on opposite sides of the C$\alpha$-C$\beta$ bond.
  • Strong Base Required: A strong base is necessary to deprotonate the β-hydrogen in the single, concerted step.
  • Primary Kinetic Isotope Effect: If the β-hydrogen is replaced by deuterium, the reaction rate decreases significantly, indicating that the C-H bond breaking is involved in the rate-determining step.
• Rate Law: Rate=k[Substrate][Base]
  • This is a second-order reaction (first-order with respect to substrate, first-order with respect to base).
• Energy Diagram: A single transition state.

2.1. Factors Affecting E2 Rate:

• Substrate Structure: Increased substitution around the α and β carbons generally favors E2 over SN​2 due to increased steric hindrance for SN​2 and increased stability of the alkene product.
  • Order of Reactivity: Tertiary > Secondary > Primary
  • Note: While primary alkyl halides can undergo E2, SN​2 is often competitive or dominant unless a very strong, hindered base is used.
• Base Strength: Stronger bases lead to faster E2 rates.
  • Common strong bases for E2: NaOCH3​ (sodium methoxide), NaOCH2​CH3​ (sodium ethoxide), NaOH, KOH, NaH, LDA (lithium diisopropylamide), KOtBu (potassium tert-butoxide).
  • Bulky bases like KOtBu preferentially favor E2 (especially Hofman elimination) and disfavor SN​2 due to steric hindrance.
• Leaving Group Ability: Good leaving groups lead to faster E2 rates (same as SN​2 and SN​1).
  • Order of Reactivity: I−>Br−>Cl−≫F−. Tosylates (TsO−) are excellent leaving groups.
• Solvent Effects: Polar aprotic solvents generally favor E2 due to increased basicity of the nucleophile/base. However, polar protic solvents can also be used, especially with very strong bases.
• Temperature: Higher temperatures significantly favor E2 (and E1) over substitution reactions (SN​2 and SN​1) because elimination reactions typically have a higher activation entropy (more molecules formed in the products, higher disorder).

2.2. Regioselectivity (Zaitsev’s vs. Hofmann’s Rule):

• Zaitsev’s Rule (Saytzeff’s Rule): When multiple β-hydrogens are available, the major product is the more substituted (more stable) alkene. This is generally the thermodynamic product. Favored by smaller, less hindered bases and/or less sterically hindered substrates.
• Hofmann’s Rule: When a bulky base or a sterically hindered substrate is used, the major product is the less substituted (less stable) alkene. This is often the kinetic product due to steric hindrance preventing the base from reaching the more hindered β-hydrogens. Also observed with poor leaving groups or nitrogen-containing leaving groups.

2.3. Stereoselectivity and Stereospecificity:

• Stereospecific: E2 is stereospecific because it requires the anti-periplanar geometry of the H and L.
  • If there are multiple anti-periplanar hydrogens, multiple stereoisomers (cis/trans) can be formed.
  • For cyclic systems (e.g., cyclohexanes): The β-hydrogen and the leaving group must both be in axial positions and anti to each other for E2 to occur from a chair conformation. This can limit the number of possible products.

3. The E1 Mechanism (Elimination 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.
  • Weak Base Required: A weak base (often the solvent itself, solvolysis) is sufficient for the second step (deprotonation).
  • Not Stereospecific: The carbocation is planar, allowing deprotonation from multiple directions.
  • No Primary Kinetic Isotope Effect: C-H bond breaking is not in the rate-determining step.
• Rate Law: Rate=k[Substrate]
  • This is a first-order reaction (dependent only on substrate concentration). The base concentration does not affect the rate.
• Energy Diagram: Involves two transition states (one for leaving group departure, one for deprotonation) 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 E1 Rate:

• Substrate Structure (Carbocation Stability): The most critical factor.
  • Tertiary alkyl halides: React fastest because tertiary carbocations are the most stable.
  • Secondary alkyl halides: React slower than tertiary.
  • Primary alkyl halides & Methyl halides: Do not undergo E1 reactions because primary and methyl carbocations are highly unstable.
  • Order of Reactivity: Tertiary > Secondary >> Primary > Methyl
  • Allylic and Benzylic halides: React faster due to resonance stabilization of the carbocation intermediate.
  • Rearrangements: Carbocations can rearrange (e.g., hydride or alkyl shifts) to form more stable carbocations, leading to rearranged products.
• Base Strength: Base strength does NOT affect the rate of an E1 reaction (it’s not in the rate-determining step). A weaker base will favor E1 over E2 when competition exists.
• Leaving Group Ability: Good leaving groups lead to faster rates (same as SN​1 and SN​2).
• Solvent Effects: Polar protic solvents favor E1 reactions. They stabilize the transition state leading to the carbocation and the carbocation intermediate itself through solvation.
• Temperature: Higher temperatures favor E1 over SN​1.

3.2. Regioselectivity (Zaitsev’s Rule):

• E1 reactions typically follow Zaitsev’s Rule, leading to the more substituted (more stable) alkene as the major product because its formation from the carbocation is usually more facile.

4. Competition Between Substitution (S_N1/S_N2) and Elimination (E1/E2)

The outcome of a reaction involving an alkyl halide and a nucleophile/base depends on a delicate balance of factors:

• Substrate Structure:
  • Methyl & Primary: Favor SN​2. E2 can occur with very strong, hindered bases.
  • Tertiary: Favor SN​1 (with weak nucleophile/base, polar protic solvent) or E2 (with strong base). No SN​2.
  • Secondary: Most complex. Can undergo all four reactions.
• Nucleophile/Base Strength and Steric Hindrance:
  • Strong, unhindered nucleophile/base: Favors SN​2 (if primary) or E2 (if secondary/tertiary, or if base is stronger than nucleophile).
  • Strong, bulky base: Favors E2 (often Hofmann product) over SN​2 or E1.
  • Weak nucleophile/base: Favors SN​1/E1 (especially in polar protic solvents).
• Solvent:
  • Polar Protic: Favors SN​1/E1 (stabilizes carbocation).
  • Polar Aprotic: Favors SN​2/E2 (activates nucleophile/base).
• Temperature: Higher temperatures favor elimination (E1/E2) over substitution (SN​1/SN​2).

5. Summary Table: E1 vs. E2

Feature	E1 (Elimination Unimolecular)	E2 (Elimination Bimolecular)
Mechanism	Stepwise (2+ steps)	Concerted (1 step)
Rate-determining step	Formation of carbocation	C-H and C-L bond breaking
Intermediate	Carbocation	None
Rate Law	Rate=k[Substrate]	Rate=k[Substrate][Base]
Order	First-order overall	Second-order overall
Substrate Reactivity	3∘>2∘≫1∘ (methyl inert)	3∘>2∘>1∘
Stereospecificity	Not stereospecific	Stereospecific (anti-periplanar H and L required)
Base Strength	Weak base (often solvent)	Strong base required
Leaving Group	Good leaving group (weak base)	Good leaving group (weak base)
Solvent	Polar Protic (stabilizes carbocation)	Polar Aprotic favored (activates base); protic can work
Rearrangements	Common	Not possible
Regioselectivity	Zaitsev (more substituted alkene)	Zaitsev (unless bulky base/Hofmann)
Temperature	Favored by high temp (over SN​1)	Favored by high temp (over SN​2)

Multiple Choice Questions (MCQ) on Elimination Reactions

Instructions: Choose the best answer for each question.

1. What is the defining characteristic of an elimination reaction? a) Two substituents are replaced by new ones. b) Two substituents are removed to form a new π bond. c) A molecule undergoes rearrangement. d) A radical intermediate is formed.

2. Which of the following best describes the E2 mechanism? a) A two-step process involving a carbocation. b) A concerted reaction requiring an anti-periplanar arrangement. c) A reaction whose rate depends only on the substrate. d) A reaction that typically leads to racemization.

3. What is the rate law for an E2 reaction? a) Rate=k[Substrate] b) Rate=k[Base] c) Rate=k[Substrate][Base] d) Rate=k[Substrate]2

4. What is the required geometric arrangement of the leaving group and β-hydrogen for an E2 reaction to occur? a) Syn-periplanar b) Gauche c) Anti-periplanar d) Eclipsed

5. Which type of substrate reacts fastest in an E2 reaction? a) Methyl halide b) Primary alkyl halide c) Secondary alkyl halide d) Tertiary alkyl halide

6. Which of the following is a characteristic feature of the E2 mechanism but not E1? a) Carbocation intermediate b) Rate-determining step c) Primary kinetic isotope effect d) Zaitsev regioselectivity

7. Which type of base is typically required for an E2 reaction? a) Weak base b) Neutral nucleophile c) Strong base d) Lewis acid

8. According to Zaitsev’s Rule, the major product of an elimination reaction is the: a) Less substituted alkene. b) More substituted alkene. c) Alkane. d) Alkyne.

9. What is the definition of the E1 mechanism? a) A concerted reaction with backside attack. b) A stepwise reaction involving a carbocation intermediate. c) A reaction whose rate depends on both substrate and base. d) A reaction exclusive to primary alkyl halides.

10. What is the rate-determining step in an E1 reaction? a) Deprotonation of the β-hydrogen. b) Formation of the carbocation intermediate. c) Attack of the base on the substrate. d) Rearrangement of the product.

11. Which type of substrate reacts fastest in an E1 reaction? a) Methyl halide b) Primary alkyl halide c) Secondary alkyl halide d) Tertiary alkyl halide

12. Why do tertiary alkyl halides favor the E1 mechanism? a) They have minimal steric hindrance. b) They form the most stable carbocation intermediate. c) They are highly reactive towards strong bases. d) They are easily solvated by aprotic solvents.

13. In an E1 reaction, how does the strength of the base affect the reaction rate? a) Stronger bases increase the rate. b) Weaker bases increase the rate. c) Base strength does not affect the rate. d) It depends on the leaving group.

14. Which type of solvent favors E1 reactions by stabilizing the carbocation intermediate? a) Non-polar solvents b) Polar aprotic solvents c) Polar protic solvents d) Basic solvents

15. What is a common side reaction that can occur with carbocation intermediates in E1 reactions? a) Nucleophilic attack (SN​1). b) E2 elimination. c) Rearrangements (e.g., hydride or alkyl shifts). d) Reduction.

16. Which of the following is a good leaving group for elimination reactions? a) OH− b) NH2−​ c) Br− d) CH3−​

17. What type of alkene product is generally favored by E1 reactions, following Zaitsev’s Rule? a) The less substituted alkene. b) The more substituted alkene. c) An alkyne. d) A rearranged alkane.

18. How does increasing the temperature typically affect the competition between substitution and elimination reactions? a) Favors substitution. b) Favors elimination. c) Has no effect. d) Favors SN​1 only.

19. Which of the following bases would most likely lead to the Hofmann product (less substituted alkene) in an E2 reaction? a) Sodium methoxide (NaOCH3​) b) Potassium hydroxide (KOH) c) Potassium tert-butoxide (KOtBu) d) Sodium ethoxide (NaOCH2​CH3​)

20. For an E2 reaction in a cyclohexane ring, what must be the relative orientation of the β-hydrogen and the leaving group? a) Both equatorial. b) Both axial and anti-periplanar. c) Both axial and syn-periplanar. d) One axial, one equatorial.

21. A substrate that is a primary alkyl halide with a strong, bulky base would most likely undergo which reaction? a) SN​1 b) E1 c) SN​2 d) E2

22. What is the order of reactivity for alkyl halides in E1 reactions? a) Methyl > Primary > Secondary > Tertiary b) Tertiary > Secondary > Primary > Methyl c) Primary > Secondary > Tertiary > Methyl d) Secondary > Primary > Tertiary > Methyl

23. Which statement is true about the stereochemistry of E1 reactions? a) They are strictly stereospecific. b) They often lead to a mixture of cis and trans alkenes. c) They always result in retention of configuration. d) They always result in inversion of configuration.

24. The rate law Rate=k[Substrate] is characteristic of which elimination mechanism? a) E2 b) E1 c) Both E1 and E2 d) Neither

25. Which type of solvent would favor an E2 reaction for a secondary alkyl halide with a strong base? a) Water b) Methanol c) DMSO (Dimethyl sulfoxide) d) Acetic acid

26. What does the term “solvolysis” refer to in the context of elimination reactions? a) Elimination in a non-polar solvent. b) Elimination where the solvent acts as the base. c) Elimination at very high temperatures. d) Elimination that produces a soluble product.

27. Which of the following is NOT a factor that influences the outcome of a reaction involving an alkyl halide and a nucleophile/base? a) Temperature b) Color of the reaction mixture c) Solvent type d) Substrate structure

28. Why are primary alkyl halides generally unreactive towards E1 reactions? a) They are too sterically hindered. b) They form unstable primary carbocations. c) They have poor leaving groups. d) They only react via E2.

29. The loss of H2​O from an alcohol after protonation to form an alkene is an example of which mechanism? a) SN​1 b) SN​2 c) E1 d) E2 (if strong base used) / E1 (if weak base/acidic conditions) Correction: For alcohols, dehydration is often acid-catalyzed, which typically proceeds via an E1 mechanism for 2∘ and 3∘ alcohols.

30. Which elimination product is typically favored when using a strong, bulky base? a) Zaitsev product b) Hofmann product c) Markovnikov product d) Only highly substituted products

31. A reaction that involves a concerted mechanism and requires anti-periplanar geometry is: a) E1 b) SN​1 c) E2 d) SN​2

32. The stability of the carbocation intermediate is most crucial for which elimination mechanism? a) E2 b) E1 c) Both E1 and E2 d) Neither

33. If an E2 reaction of 2-bromobutane results in a mixture of cis-2-butene and trans-2-butene, this indicates: a) Lack of regioselectivity. b) Lack of stereospecificity. c) Stereoselectivity (favoring one geometric isomer over another). d) A side reaction occurred.

34. In the E2 mechanism, if the β-hydrogen is replaced by deuterium, what happens to the reaction rate? a) It increases. b) It remains unchanged. c) It decreases significantly (primary kinetic isotope effect). d) The reaction stops.

35. What is the typical effect of steric hindrance around the β-hydrogens on the regioselectivity of E2 reactions? a) Favors the Zaitsev product. b) Favors the Hofmann product. c) Has no effect. d) Prevents the reaction.

36. A reaction involves a substrate and a weak base in a polar protic solvent at high temperature. Which pathway is most likely to be favored? a) SN​2 b) E2 c) SN​1 d) E1

37. Which of the following is true about the carbocation formed in an E1 reaction? a) It is sp hybridized and linear. b) It is sp2 hybridized and planar. c) It is sp3 hybridized and tetrahedral. d) It is generally stable and non-reactive.

38. What is the approximate pKa of the conjugate acid of a strong base typically used for E2 reactions (e.g., CH3​O−)? a) 5 b) 10 c) 16 d) 35

39. For 1-bromopropane, which type of elimination reaction would be most likely favored with NaOCH2​CH3​ in ethanol? a) E1 b) E2 c) E1cb d) Neither, it would be SN​2.

40. Why do higher temperatures generally favor elimination over substitution reactions? a) Elimination reactions are always more exothermic. b) Elimination reactions have a lower activation energy. c) Elimination reactions have a more positive change in entropy (greater disorder). d) Substitution reactions are reversible at high temperatures.

Answer Key with Explanations

1. b) Two substituents are removed to form a new π bond.
   • Explanation: Elimination reactions are characterized by the removal of two groups (typically a hydrogen and a leaving group) from adjacent carbons to form a new double (or triple) bond.
2. b) A concerted reaction requiring an anti-periplanar arrangement.
   • Explanation: The E2 mechanism is a single-step (concerted) reaction where the abstraction of the β-hydrogen and the departure of the leaving group occur simultaneously, and this requires a specific anti-periplanar geometry.
3. c) Rate=k[Substrate][Base].
   • Explanation: The E2 reaction is bimolecular; its rate depends on the concentrations of both the substrate and the base in the rate-determining step.
4. c) Anti-periplanar.
   • Explanation: For an E2 reaction to occur efficiently, the β-hydrogen and the leaving group must be anti-periplanar, meaning they are in the same plane but on opposite sides of the C$\alpha$-C$\beta$ bond.
5. d) Tertiary alkyl halide.
   • Explanation: The reactivity order for E2 is 3∘>2∘>1∘ because more substituted alkenes (the products) are generally more stable.
6. c) Primary kinetic isotope effect.
   • Explanation: The primary kinetic isotope effect (slower rate when C-H is replaced by C-D) is observed in E2 reactions because the C-H bond breaking occurs in the rate-determining step. It is not observed in E1 as C-H bond breaking occurs after the RDS.
7. c) Strong base.
   • Explanation: E2 reactions require a strong base to effectively abstract the β-hydrogen in the concerted mechanism.
8. b) More substituted alkene.
   • Explanation: Zaitsev’s Rule states that in an elimination reaction, the major product will be the more substituted (more stable) alkene, which is generally the thermodynamic product.
9. b) A stepwise reaction involving a carbocation intermediate.
   • Explanation: The E1 mechanism is a two-step process. The first step involves the spontaneous departure of the leaving group to form a carbocation intermediate, followed by deprotonation by a base.
10. 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 E1 reaction.
11. d) Tertiary alkyl halide.
    • Explanation: E1 reactions are favored by substrates that can form stable carbocations. Tertiary carbocations are the most stable, making tertiary alkyl halides react fastest in E1.
12. b) They form the most stable carbocation intermediate.
    • Explanation: The stability of the carbocation intermediate is the primary factor determining the rate of an E1 reaction. Tertiary carbocations are highly stabilized.
13. c) Base strength does not affect the rate.
    • Explanation: The base is not involved in the rate-determining step (carbocation formation) of an E1 reaction, so its concentration and strength do not appear in the rate law and do not affect the reaction rate.
14. c) Polar protic solvents.
    • Explanation: Polar protic solvents (like water, methanol) stabilize the highly charged transition state leading to the carbocation and the carbocation intermediate itself through solvation, thereby lowering the activation energy for the rate-determining step in E1.
15. c) Rearrangements (e.g., hydride or alkyl shifts).
    • Explanation: Since E1 reactions involve carbocation intermediates, these highly reactive species can undergo rearrangements to form a more stable carbocation before the deprotonation step.
16. c) Br−.
    • Explanation: Good leaving groups are weak bases. Br− is the conjugate base of HBr (a strong acid), making Br− a relatively weak base and therefore a good leaving group. OH−, NH2−​, and CH3−​ are all strong bases and poor leaving groups.
17. b) The more substituted alkene.
    • Explanation: E1 reactions typically follow Zaitsev’s Rule, favoring the formation of the more stable (more substituted) alkene.
18. b) Favors elimination.
    • Explanation: Higher temperatures generally favor elimination (E1/E2) over substitution reactions (SN​1/SN​2) because elimination reactions have a more positive change in entropy (increase in disorder from forming more molecules or a more open transition state).
19. c) Potassium tert-butoxide (KOtBu).
    • Explanation: Potassium tert-butoxide is a strong, bulky base. Bulky bases preferentially abstract the most accessible β-hydrogen (often leading to the less substituted, or Hofmann, product) due to steric hindrance.
20. b) Both axial and anti-periplanar.
    • Explanation: For E2 to occur efficiently in a cyclohexane ring, the β-hydrogen and the leaving group must be in a trans-diaxial orientation (i.e., both axial and anti-periplanar).
21. d) E2.
    • Explanation: For primary alkyl halides, SN​2 is usually favored. However, using a strong, bulky base like KOtBu will sterically hinder the SN​2 pathway and strongly favor E2.
22. b) Tertiary > Secondary > Primary > Methyl.
    • Explanation: This order reflects the increasing stability of the carbocation intermediate, which is the key factor for E1 reactivity.
23. b) They often lead to a mixture of cis and trans alkenes.
    • Explanation: Since the carbocation intermediate is planar, deprotonation can occur from either side, leading to the formation of both cis and trans alkene isomers if possible. E1 is not stereospecific.
24. b) E1.
    • Explanation: A rate law that only depends on the concentration of the substrate is characteristic of a first-order reaction, specifically the E1 mechanism (or SN​1).
25. c) DMSO (Dimethyl sulfoxide).
    • Explanation: DMSO is a polar aprotic solvent. Polar aprotic solvents do not strongly solvate anions, leaving the strong base more reactive and thus favoring E2 (and SN​2) pathways.
26. b) Elimination where the solvent acts as the base.
    • Explanation: Solvolysis refers to a reaction where the solvent acts as the nucleophile or base. In an E1 solvolysis, the solvent abstracts the β-hydrogen.
27. b) Color of the reaction mixture.
    • Explanation: The color of the reaction mixture is generally not a fundamental factor that determines the mechanism or outcome of substitution/elimination reactions. The other factors are critical.
28. b) They form unstable primary carbocations.
    • Explanation: Primary carbocations are highly unstable and very difficult to form, making E1 reactions generally unfavorable for primary alkyl halides.
29. c) E1.
    • Explanation: The dehydration of secondary and tertiary alcohols to alkenes, especially under acidic conditions, typically proceeds via an E1 mechanism involving a carbocation intermediate.
30. b) Hofmann product.
    • Explanation: Bulky bases (like KOtBu) preferentially remove the most accessible β-hydrogen, which is often a hydrogen on the less substituted carbon, leading to the less substituted (Hofmann) alkene product.
31. c) E2.
    • Explanation: The E2 mechanism is a concerted reaction (single step) that has a strict stereochemical requirement for an anti-periplanar alignment of the leaving group and the β-hydrogen.
32. b) E1.
    • Explanation: The stability of the carbocation intermediate is the most crucial factor determining the feasibility and rate of an E1 reaction, as its formation is the rate-determining step.
33. c) Stereoselectivity (favoring one geometric isomer over another).
    • Explanation: If both cis and trans isomers can be formed from an E2 reaction (due to multiple anti-periplanar hydrogens), and one is formed in a greater amount than the other, it indicates stereoselectivity. It does not imply a lack of stereospecificity (E2 is always stereospecific for the anti-elimination).
34. c) It decreases significantly (primary kinetic isotope effect).
    • Explanation: In an E2 reaction, the C-H bond is broken in the rate-determining step. Replacing hydrogen with the heavier deuterium isotope slows down this bond breaking, leading to a primary kinetic isotope effect.
35. b) Favors the Hofmann product.
    • Explanation: If the β-hydrogens leading to the Zaitsev product are sterically hindered, bulky bases will preferentially abstract the more accessible β-hydrogens, resulting in the less substituted (Hofmann) product.
36. d) E1.
    • Explanation: High temperature favors elimination over substitution. A weak base/nucleophile disfavors E2 and SN​2. A polar protic solvent favors carbocation formation (SN​1/E1). Combining these, E1 is the most likely favored pathway.
37. b) It is sp2 hybridized and planar.
    • Explanation: Carbocations are trigonal planar at the carbon with the positive charge, and this carbon is sp2 hybridized.
38. c) 16.
    • Explanation: Strong bases like methoxide (CH3​O−) are the conjugate bases of weak acids (alcohols). Methanol (CH3​OH) has a pKa of approximately 16, so its conjugate base is strong.
39. d) Neither, it would be SN​2.
    • Explanation: 1-bromopropane is a primary alkyl halide. NaOCH2​CH3​ (sodium ethoxide) is a strong base and a strong, relatively unhindered nucleophile. For a primary substrate, SN​2 is highly favored over E2, E1, or SN​1.
40. c) Elimination reactions have a more positive change in entropy (greater disorder).
    • Explanation: Elimination reactions typically result in an increase in the number of molecules (e.g., one reactant molecule forming two product molecules: alkene and H-Base), leading to a greater increase in entropy (ΔS>0). At higher temperatures, the TΔS term in ΔG=ΔH−TΔS becomes more significant, thus favoring reactions with a positive ΔS.