Organic Chemistry 2 Quiz: Alcohol Substitution And Elimination
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Alcohol Substitution And EliminationQuestion 1 of 20

The reaction of (R)-2-butanol with thionyl chloride (SOCl₂) in an inert solvent (e.g., ether) without pyridine is known to proceed via an Sₙi mechanism. What is the stereochemical outcome of this reaction?

Formation of (S)-2-chlorobutane (inversion)
Formation of (R)-2-chlorobutane (retention)
Formation of a racemic mixture of 2-chlorobutane
Formation of 2-butene as the major product
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Alcohol Substitution And Elimination

Practice Alcohol Substitution And Elimination in Organic Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Alcohol Substitution And Elimination, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The reaction of (R)-2-butanol with thionyl chloride (SOCl₂) in an inert solvent (e.g., ether) without pyridine is known to proceed via an Sₙi mechanism. What is the stereochemical outcome of this reaction?

  1. Formation of (S)-2-chlorobutane (inversion)
  2. Formation of (R)-2-chlorobutane (retention) (correct answer)
  3. Formation of a racemic mixture of 2-chlorobutane
  4. Formation of 2-butene as the major product
Explanation: In the absence of a base like pyridine, the reaction of an alcohol with SOCl₂ proceeds through an Sₙi (substitution nucleophilic internal) mechanism. The alcohol first forms a chlorosulfite ester intermediate. This intermediate then collapses in a concerted step where the chloride is delivered from the same face as the departing SO₂ group. This internal delivery of the nucleophile results in overall retention of configuration at the stereocenter. Thus, (R)-2-butanol yields (R)-2-chlorobutane. If pyridine were present, it would intercept the intermediate and promote an Sₙ2 reaction, leading to inversion of configuration (Choice A).

Question 2

Which condition would most likely shift 2-butanol dehydration toward substitution instead of elimination?

  1. Lower temperature, high [Br−^-] (correct answer)
  2. Higher temperature, no nucleophile
  3. Bulky base, heat
  4. Concentrated H2SO4 only, heat
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; conditions can shift dehydration outcomes. In this scenario, lower temperatures and high nucleophile concentrations favor substitution over elimination. The correct answer is choice A because it accurately reflects the shift toward substitution. Choice B fails because it suggests elimination-favoring heat, a common misconception in competition control. To help students avoid these errors, emphasize temperature and concentration effects on pathways. Practice predicting ratios with variable condition examples.

Question 3

Which intermediate is characteristic of SN1 when a tertiary alcohol reacts with HX?

  1. Carbocation (correct answer)
  2. Carbanion
  3. Radical
  4. Carbene
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; tertiary ones form carbocations easily with HX. In this scenario, SN1 proceeds via a planar carbocation intermediate. The correct answer is choice A because it accurately reflects the key SN1 intermediate. Choice B fails because it suggests anionic species, a common misconception from base-catalyzed reactions. To help students avoid these errors, emphasize intermediate stability and mechanism steps. Practice drawing SN1 mechanisms with energy profiles.

Question 4

Which step makes an alcohol a better leaving group for E1 dehydration in acid?

  1. Deprotonation to alkoxide
  2. Protonation to form H2O leaving group (correct answer)
  3. Oxidation to aldehyde
  4. Reduction to alkane
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; acid protonation converts them to water. In this scenario, protonation facilitates carbocation formation for E1 dehydration. The correct answer is choice B because it accurately reflects the activation step. Choice A fails because it suggests base-catalyzed processes, a common error confusing E1 and E2. To help students avoid these errors, emphasize acid's role in leaving group improvement and mechanism initiation. Practice drawing stepwise mechanisms with protonation emphasis.

Question 5

Which product is major for E2 of 2-bromobutane made from 2-butanol using strong base (small)?

  1. 1-butene
  2. cis-2-butene
  3. trans-2-butene (correct answer)
  4. 2-butanol
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; conversion to halides enables E2 with bases. In this scenario, small strong bases yield Zaitsev products like trans-2-butene. The correct answer is choice C because it accurately reflects the major stable alkene. Choice A fails because it suggests Hofmann product, a common error ignoring stability. To help students avoid these errors, emphasize Zaitsev's rule and isomer stability. Practice drawing E2 products with conformational analysis.

Question 6

Which of the following sequences is the most plausible synthesis of 1-methoxybutane from 1-butanol?

    1. Concentrated H₂SO₄, heat; 2. CH₃OH, H⁺
    1. CH₃MgBr in ether; 2. H₂O
  3. CH₃OH, catalytic H₂SO₄
    1. NaH in THF; 2. CH₃I
     (correct answer)
Explanation: This transformation is a Williamson Ether Synthesis. It requires converting the alcohol into a nucleophilic alkoxide, which then attacks an alkyl halide in an Sₙ2 reaction. Sequence D correctly accomplishes this. First, sodium hydride (NaH), a strong base, deprotonates 1-butanol to form the sodium butoxide salt. Second, the butoxide ion attacks methyl iodide, an excellent Sₙ2 substrate, to form 1-methoxybutane. Choice A would cause elimination to form butenes, followed by addition of methanol to likely form 2-methoxybutane. Choice B would result in an acid-base reaction where the Grignard reagent deprotonates the alcohol to form butane and methane. Choice C would result in a slow equilibrium and potential for side reactions, not a clean synthesis of the asymmetric ether.

Question 7

A student attempts to synthesize tert-butyl methyl ether by adding sodium methoxide (NaOCH₃) to tert-butyl bromide. This method fails to produce a significant amount of the ether. What is the major product and the reason for this outcome?

  1. 2-methylpropene, because methoxide acts as a base promoting E2 elimination. (correct answer)
  2. tert-butanol, because methoxide acts as a nucleophile in an Sₙ1 reaction with water.
  3. 1-methoxy-2-methylpropane, because a carbocation rearrangement occurs.
  4. No reaction, because tert-butyl bromide is too sterically hindered to react.
Explanation: This is a classic example of the limitations of the Williamson ether synthesis. The reaction involves a strong nucleophile/strong base (methoxide) and a tertiary alkyl halide. Due to the significant steric hindrance at the tertiary carbon, the Sₙ2 pathway is blocked. Instead, the methoxide acts as a base and removes a β-proton from one of the methyl groups. This initiates an E2 elimination reaction, where the bromide ion is ejected, forming the alkene 2-methylpropene as the major product. Sₙ1/E1 could also occur, but with a strong base like methoxide, E2 is dominant.

Question 8

The reaction of 3,3-dimethyl-2-butanol with concentrated H₂SO₄ yields 2,3-dimethyl-2-butene as the major product. What would be the major product if 3,3-dimethyl-2-butanol were instead treated with POCl₃ in pyridine?

  1. 2,3-dimethyl-2-butene
  2. 2,3-dimethyl-1-butene
  3. 3,3-dimethyl-1-butene (correct answer)
  4. No reaction would occur.
Explanation: The reaction with H₂SO₄ proceeds via an E1 mechanism involving a carbocation intermediate, which allows for a 1,2-methyl shift to form a more stable tertiary carbocation, leading to the rearranged product 2,3-dimethyl-2-butene. In contrast, the reaction with phosphorus oxychloride (POCl₃) in pyridine is an E2 dehydration. The E2 mechanism is concerted and does not involve a carbocation intermediate, so rearrangements are not possible. Elimination occurs on the original carbon skeleton. The only β-hydrogens available for elimination are on the methyl group at C1, leading to the formation of 3,3-dimethyl-1-butene.

Question 9

A solution of tert-butanol in aqueous sulfuric acid reacts to form both substitution and elimination products. If the temperature of the reaction is significantly increased from 0 °C to 50 °C, what change in the product distribution is expected?

  1. The rate of substitution will increase, but the rate of elimination will decrease.
  2. The proportion of 2-methylpropene will increase relative to the substitution products. (correct answer)
  3. The reaction will switch to an Sₙ2 pathway due to the increased kinetic energy.
  4. A carbocation rearrangement will occur, leading to the formation of n-butane derivatives.
Explanation: Both Sₙ1 and E1 reactions proceed through a common carbocation intermediate. Elimination reactions are generally more favored by higher temperatures than substitution reactions. This is because elimination reactions typically have a larger positive entropy change (ΔS) since the number of moles of products is greater than the moles of reactants. According to the Gibbs free energy equation (ΔG = ΔH - TΔS), the -TΔS term becomes more negative at higher T, making ΔG more favorable for elimination. Thus, increasing the temperature increases the E1/Sₙ1 ratio. Choice C is incorrect as tertiary substrates do not undergo Sₙ2 reactions. Choice D is incorrect as the tert-butyl carbocation is already the most stable C4 carbocation and will not rearrange.

Question 10

In converting a primary alcohol like 1-butanol to 1-bromobutane, PBr₃ is often preferred over concentrated HBr. What is the best chemical justification for this preference in a synthesis requiring high purity?

  1. HBr is a gas, making it harder to handle, whereas PBr₃ is a liquid.
  2. The reaction with HBr has a lower yield due to the formation of a phosphonic acid byproduct.
  3. HBr's strong acidity can catalyze a competing E1 elimination to form butenes as side products. (correct answer)
  4. The reaction with HBr proceeds via an Sₙ1 mechanism, which allows for unwanted rearrangements.
Explanation: The primary reason for preferring PBr₃ is to avoid the strongly acidic conditions of concentrated HBr. While the reaction of a primary alcohol with HBr proceeds via an Sₙ2 mechanism, the presence of a strong acid catalyst (H⁺) and heat can promote a competing dehydration (E1 or E2) pathway, leading to the formation of butene impurities. The mechanism with PBr₃ does not involve free protons or carbocations, thus minimizing elimination side reactions and ensuring a cleaner conversion to the alkyl bromide. Choice D is incorrect because primary alcohols react with HBr via an Sₙ2, not Sₙ1, mechanism, so rearrangements are not a concern for 1-butanol itself, but elimination is.

Question 11

What is the major product when 5-bromo-1-pentanol is treated with sodium hydride (NaH)?

  1. Tetrahydropyran (correct answer)
  2. 1,5-pentanediol
  3. Pent-4-en-1-ol
  4. 2-methyltetrahydrofuran
Explanation: This is an intramolecular Williamson ether synthesis. Sodium hydride (NaH) is a strong, non-nucleophilic base that deprotonates the alcohol to form a sodium alkoxide. The resulting alkoxide ion is a potent nucleophile. It attacks the electrophilic carbon bearing the bromine atom at the other end of the molecule in an intramolecular Sₙ2 reaction. This cyclization reaction forms a six-membered ring containing an oxygen atom, which is named tetrahydropyran. This 6-exo-tet cyclization is favorable according to Baldwin's rules. Choice B would require substitution with hydroxide. Choice C is an E2 product, which is kinetically slower than the intramolecular Sₙ2 reaction in this case. Choice D is an incorrect ring size.

Question 12

How does polar protic solvent affect SN2 on a primary alkyl halide made from an alcohol?

  1. Speeds SN2 by unsolvating nucleophile
  2. Slows SN2 by solvating nucleophile (correct answer)
  3. Forces SN1 by forming carbocation
  4. Has no effect on rate
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; derived halides react under solvent-influenced conditions. In this scenario, polar protic solvents solvate nucleophiles, reducing SN2 rates on primary halides. The correct answer is choice B because it accurately reflects the solvent's inhibitory effect. Choice A fails because it suggests acceleration, a common error confusing protic and aprotic effects. To help students avoid these errors, emphasize solvation and nucleophilicity trends. Practice solvent classification with rate impact examples.

Question 13

Which alcohol class reacts fastest by SN1 with HCl (ZnCl2) to form an alkyl chloride?

  1. Primary alcohol
  2. Secondary alcohol
  3. Tertiary alcohol (correct answer)
  4. Methanol
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; reactivity varies by class in SN1 conditions. In this scenario, tertiary alcohols form stable carbocations fastest with HCl/ZnCl2. The correct answer is choice C because it accurately reflects the fastest-reacting class. Choice A fails because it suggests primary alcohols, a common error ignoring carbocation stability. To help students avoid these errors, emphasize alkyl group effects and rate orders. Practice ranking reactivities with structural examples.

Question 14

Which solvent best promotes SN2 when converting an alcohol-derived tosylate with NaCN?

  1. DMSO (correct answer)
  2. Methanol
  3. Water
  4. tert-Butanol
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; tosylates pair well with nucleophiles in optimal solvents. In this scenario, DMSO enhances NaCN nucleophilicity for SN2 cyanation. The correct answer is choice A because it accurately reflects the polar aprotic solvent's promotion of SN2. Choice B fails because it suggests protic solvents that slow reactions, a common misconception in solvent choice. To help students avoid these errors, emphasize aprotic advantages and solvent-nucleophile synergy. Practice selecting solvents with mechanism-specific examples.

Question 15

Which reagent is commonly used to convert an alcohol to a mesylate (good leaving group) without changing C stereochemistry?

  1. MsCl, Et3N (correct answer)
  2. NaOH, H2O
  3. H2SO4, 94
  4. KMnO4
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; mesylates provide good alternatives without stereochemical change. In this scenario, MsCl with Et3N activates alcohols retaining carbon configuration. The correct answer is choice A because it accurately reflects the reagent for stereochemistry-preserving conversion. Choice C fails because it suggests elimination conditions, a common misconception in activation versus reaction. To help students avoid these errors, emphasize sulfonate formation and stereochemical retention. Practice comparing sulfonates with halides in mechanisms.

Question 16

Which reagent best converts 1-propanol to 1-bromopropane via SN2 under mild conditions?

  1. HBr (reflux)
  2. PBr3 (correct answer)
  3. H2SO4, 94
  4. AgNO3 in ethanol
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; mild reagents are preferred for primary alcohols to avoid side reactions. In this scenario, the choice of reagent such as PBr3 enables SN2 substitution under mild conditions without rearrangement. The correct answer is choice B because it accurately reflects the conditions necessary for the SN2 mechanism to occur efficiently. Choice A fails because it suggests harsher conditions that may lead to elimination, which is a common error when confusing primary and secondary alcohol reactivities. To help students avoid these errors, emphasize the importance of selecting reagents based on substrate type and understanding temperature effects on mechanism selection. Practice mapping reagents to mechanisms using flowcharts and example reactions.

Question 17

Which base is most likely to give E2 over SN2 with a secondary alkyl tosylate from an alcohol?

  1. I−^-
  2. t-BuO−^- (correct answer)
  3. H2O
  4. HS−^-
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; tosylates allow base-promoted eliminations. In this scenario, bulky bases like t-BuO- favor E2 over SN2 on secondary substrates. The correct answer is choice B because it accurately reflects the base promoting elimination. Choice A fails because it suggests good nucleophiles for substitution, a common misconception in base selection. To help students avoid these errors, emphasize steric effects and mechanism competition. Practice predicting products with base type comparisons.

Question 18

Which first step best turns an -OH into a good leaving group for SN2 by I−^- on a primary carbon?

  1. NaOH
  2. TsCl, pyridine (correct answer)
  3. H2O
  4. NaOEt
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; conversion to tosylates is common for SN2 reactions on primary carbons. In this scenario, TsCl with pyridine forms a good leaving group for subsequent iodide displacement. The correct answer is choice B because it accurately reflects the first step necessary for the SN2 mechanism. Choice A fails because it suggests deprotonation without activation, a common misconception about base roles. To help students avoid these errors, emphasize multi-step transformations and leaving group quality. Practice sequencing reactions with flowcharts and stereochemistry considerations.

Question 19

Which condition most favors SN2 when converting a primary alcohol-derived tosylate to a bromide?

  1. Polar protic solvent, weak nucleophile
  2. Polar aprotic solvent, strong nucleophile (correct answer)
  3. High temperature, bulky base
  4. Carbocation-stabilizing solvent, heat
Explanation: This question tests intermediate Organic Chemistry skills specifically related to alcohol substitution and elimination mechanisms. Alcohols often act as poor leaving groups; tosylates facilitate SN2 with optimized conditions. In this scenario, polar aprotic solvents enhance strong nucleophile activity for bromide formation. The correct answer is choice B because it accurately reflects the conditions necessary for the SN2 mechanism. Choice A fails because it suggests protic solvents that hinder SN2, a common misconception about solvent effects. To help students avoid these errors, emphasize aprotic solvent advantages and nucleophile strength. Practice classifying conditions with mechanism maps and rate comparisons.

Question 20

Which of the following reaction sequences would be most effective for converting 2-butanol into 1-butene as the major product, minimizing the formation of 2-butene?

  1. Concentrated H₂SO₄, heat
    1. TsCl, pyridine; 2. NaOEt
    1. PBr₃; 2. KOtBu
    1. TsCl, pyridine; 2. KOtBu
     (correct answer)
Explanation: The goal is to synthesize the Hofmann (less substituted) alkene. This requires an E2 reaction with a sterically hindered base. Sequence D first converts the alcohol into a good leaving group (tosylate) and then uses potassium tert-butoxide (KOtBu), a bulky base, to promote E2 elimination. The bulky base preferentially removes the less sterically hindered proton from C1, yielding 1-butene. Choice A (acid-catalyzed dehydration) and Choice B (E2 with a non-bulky base) would both favor the Zaitsev (more substituted) product, 2-butene. Choice C converts the alcohol to a bromide, and then KOtBu would also give the Hofmann product, but the two-step tosylation route is a very common and effective method.