Organic Chemistry 2 Quiz: Epoxide Opening Reactions
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Epoxide Opening ReactionsQuestion 1 of 13

When 1-hexene is treated with m-CPBA and the resulting product is subsequently treated with sodium azide (NaN₃) in ethanol, what is the final major product?

2-ethoxyhexan-1-ol
2-azidohexan-1-ol
1,2-diazidohexane
1-azidohexan-2-ol
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Epoxide Opening Reactions

Practice Epoxide Opening Reactions 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 Epoxide Opening Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

How to use this quiz

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.

All questions

Question 1

When 1-hexene is treated with m-CPBA and the resulting product is subsequently treated with sodium azide (NaN₃) in ethanol, what is the final major product?

  1. 2-ethoxyhexan-1-ol
  2. 2-azidohexan-1-ol
  3. 1,2-diazidohexane
  4. 1-azidohexan-2-ol (correct answer)
Explanation: This question tests your understanding of epoxide formation and ring-opening reactions, a key sequence in organic chemistry. When you see m-CPBA (meta-chloroperoxybenzoic acid) followed by a nucleophile, think epoxide formation then nucleophilic attack. The reaction proceeds in two steps. First, m-CPBA converts 1-hexene into 1,2-epoxyhexane through electrophilic addition across the double bond, creating a three-membered ring with oxygen. Second, when this epoxide encounters sodium azide (NaN3\text{NaN}_3) in ethanol, the azide ion acts as a nucleophile and attacks the epoxide ring. Epoxide ring-opening follows specific regiochemical rules. Under basic conditions (like NaN3\text{NaN}_3 in ethanol), the nucleophile attacks the less substituted carbon to minimize steric hindrance. In 1,2-epoxyhexane, the less substituted carbon is C1 (primary), while C2 is secondary. Therefore, azide attacks at C1, and the ring opens with the hydroxyl group forming at C2, giving 1-azidohexan-2-ol. Looking at the wrong answers: Choice A (2-ethoxyhexan-1-ol) incorrectly suggests ethanol acts as the nucleophile instead of azide. Choice B (2-azidohexan-1-ol) places the azide at the wrong carbon, ignoring the regioselectivity rules. Choice C (1,2-diazidohexane) incorrectly suggests both carbons receive azide groups, which doesn't happen in simple epoxide opening. Remember: epoxide ring-opening under basic conditions follows anti-Markovnikov selectivity—the nucleophile attacks the less hindered carbon. This regioselectivity is crucial for predicting products correctly.

Question 2

An unknown epoxide with molecular formula C₄H₈O is treated with sodium methoxide in methanol. The ¹H NMR spectrum of the product consists of three singlets at δ 1.3 (6H), δ 3.3 (3H), and δ 3.4 (2H). What was the structure of the starting epoxide?

  1. 1,2-epoxybutane
  2. cis-2,3-epoxybutane
  3. 1,2-epoxy-2-methylpropane (correct answer)
  4. tetrahydrofuran
Explanation: The ¹H NMR data indicates a high degree of symmetry in the product. A singlet at δ 1.3 integrating to 6H suggests two equivalent methyl groups attached to a quaternary carbon. A singlet at δ 3.3 (3H) is characteristic of a methoxy group (OCH₃). A singlet at δ 3.4 (2H) corresponds to a CH₂ group with no adjacent protons. This pattern matches the structure of 1-methoxy-2-methylpropan-2-ol, (CH₃)₂C(OH)CH₂OCH₃. This product is formed by the Sₙ2 attack of methoxide (CH₃O⁻) on the less sterically hindered primary carbon (C1) of 1,2-epoxy-2-methylpropane (isobutylene oxide).

Question 3

What is the major product when (S)-styrene oxide is treated with lithium aluminum deuteride (LiAlD₄), followed by an aqueous (H₂O) workup?

  1. (S)-2-deuterio-1-phenylethanol (correct answer)
  2. (R)-1-deuterio-1-phenylethanol
  3. (R)-2-deuterio-1-phenylethanol
  4. (S)-1-deuterio-2-phenylethanol
Explanation: Lithium aluminum deuteride (LiAlD₄) is a source of the strong nucleophile deuteride (D⁻). Under these basic conditions, the nucleophile attacks the less sterically hindered carbon of the epoxide via an Sₙ2 mechanism. In styrene oxide (phenyloxirane), the less hindered carbon is C2 (the CH₂ group). The benzylic carbon (C1) is the stereocenter. Since the attack occurs at C2, the configuration of the stereocenter at C1 is unaffected. Thus, the (S)-styrene oxide yields (S)-1-phenylethanol with a deuterium atom at C2, which is (S)-2-deuterio-1-phenylethanol.

Question 4

In the acid-catalyzed ring-opening of 1,2-epoxy-1-methylcyclohexane with methanol, the nucleophile attacks the more substituted carbon (C1). Which statement provides the best mechanistic explanation for this regioselectivity?

  1. The transition state has significant Sₙ1 character, with partial positive charge better stabilized by the tertiary carbon. (correct answer)
  2. The reaction proceeds through a discrete, fully-formed tertiary carbocation intermediate which is then trapped by methanol.
  3. Steric hindrance from the axial protons on the cyclohexane ring directs the nucleophile to the more substituted carbon.
  4. Protonation of the epoxide makes it a better leaving group, allowing a standard Sₙ2 attack at the less hindered secondary carbon.
Explanation: In the acid-catalyzed opening of an epoxide with a tertiary carbon, the reaction pathway has a transition state with significant Sₙ1 character. After protonation of the epoxide oxygen, the C-O bonds begin to break. The bond to the more substituted carbon (tertiary C1) breaks more easily because that carbon can better stabilize the resulting partial positive charge. This leads to preferential attack by the weak nucleophile (methanol) at the more substituted position. It is not a full carbocation (choice B), and choices C and D describe incorrect reasoning or outcomes.

Question 5

The acid-catalyzed hydrolysis of 2,2,3,3-tetramethyloxirane is significantly slower than that of 2,2-dimethyloxirane. What is the best explanation for this rate difference?

  1. The four methyl groups in 2,2,3,3-tetramethyloxirane electronically destabilize the protonated intermediate.
  2. Backside attack by water is severely sterically hindered at both tertiary carbons of 2,2,3,3-tetramethyloxirane. (correct answer)
  3. The C-O bonds in 2,2,3,3-tetramethyloxirane are stronger due to hyperconjugation with the methyl groups.
  4. A full carbocation intermediate cannot be formed from 2,2,3,3-tetramethyloxirane due to excessive ring strain.
Explanation: Acid-catalyzed epoxide opening, even at tertiary centers, requires a backside approach by the nucleophile. In 2,2-dimethyloxirane, water can attack the tertiary carbon from the side opposite the C-O bond. However, in 2,2,3,3-tetramethyloxirane, each tertiary carbon is adjacent to another carbon bearing two bulky methyl groups. These groups create severe steric hindrance that blocks the trajectory for backside attack, dramatically slowing the reaction rate.

Question 6

A sample of (R)-2-phenyloxirane is divided. Portion A is treated with NaSH. Portion B is treated with ethanol and catalytic H₂SO₄. Which statement accurately compares the major products from these two reactions?

  1. The products are constitutional isomers; the stereocenter from Portion A is retained while that from Portion B is inverted. (correct answer)
  2. The products are enantiomers; Portion A yields the (R)-enantiomer and Portion B yields the (S)-enantiomer.
  3. Both reactions yield the same regioisomer, but the stereochemistry is inverted in Portion B and retained in Portion A.
  4. The products are diastereomers because a new stereocenter is formed in the reaction with ethanol.
Explanation: Portion A (NaSH, basic conditions): The strong nucleophile SH⁻ attacks the less hindered carbon (C2), leaving the stereocenter at C1 untouched. The product is (R)-2-mercapto-1-phenylethanol. Portion B (EtOH/H⁺, acidic conditions): The weak nucleophile EtOH attacks the more substituted benzylic carbon (C1) with inversion of configuration. The product is (S)-2-ethoxy-1-phenylethanol. The products have different connectivity (mercapto vs. ethoxy group, and at different positions), so they are constitutional isomers. The stereochemistry is retained in A and inverted in B.

Question 7

The reaction of propylene oxide with a large excess of ammonia (NH₃), followed by a mild workup, is a key step in the synthesis of certain amino alcohols. Which statement correctly describes the outcome of this reaction?

  1. Attack occurs primarily at the more substituted carbon, yielding 2-aminopropan-1-ol.
  2. Attack occurs primarily at the less substituted carbon, yielding 1-aminopropan-2-ol. (correct answer)
  3. The ammonia acts as a base, causing elimination to form allyl alcohol.
  4. The reaction forms a mixture of both regioisomers because ammonia is a weak nucleophile.
Explanation: Ammonia (NH₃) is a nucleophile. While it is not as strong as an alkoxide, it does not require acid catalysis to open an epoxide. Therefore, the reaction proceeds under neutral/basic conditions via an Sₙ2 mechanism. The nucleophile attacks the less sterically hindered carbon of propylene oxide (2-methyloxirane), which is the primary carbon (C1). This results in the formation of 1-aminopropan-2-ol. A large excess of ammonia is used to prevent the product amino alcohol from reacting with another molecule of the epoxide.

Question 8

A student aims to synthesize (R)-1-chloro-2-butanol. Which combination of epoxide and nucleophile under appropriate conditions would be the most effective route?

  1. (R)-1,2-epoxybutane treated with HCl
  2. (S)-1,2-epoxybutane treated with LiCl in THF
  3. (R)-1,2-epoxybutane treated with LiCl in THF
  4. (S)-1,2-epoxybutane treated with HCl (correct answer)
Explanation: The target molecule, (R)-1-chloro-2-butanol, has a stereocenter at C2. To form this product, we need to open an epoxide. Using HCl (acidic conditions) on 1,2-epoxybutane will lead to nucleophilic attack by Cl⁻ at the more substituted carbon, C2, with inversion of configuration. To obtain an (R) product at C2, we must start with an (S) stereocenter at C2 in the epoxide. Therefore, reacting (S)-1,2-epoxybutane with HCl is the correct choice. Basic conditions (LiCl) would cause attack at the less hindered C1, which would not affect the stereocenter at C2 and would produce 2-chloro-1-butanol (the wrong regioisomer).

Question 9

Which of the following reaction schemes is the most suitable for synthesizing racemic trans-2-methoxycyclohexanol?

  1. Reacting 2-methoxycyclohexene with cold, dilute KMnO₄.
  2. Reacting cyclohexene with methanol in the presence of H₂SO₄.
  3. Reacting cyclohexene oxide with sodium methoxide in methanol. (correct answer)
  4. Reacting 2-chlorocyclohexanol with NaH, followed by addition of methyl iodide.
Explanation: The target product has a trans relationship between the hydroxyl and methoxy groups, which is characteristic of the ring-opening of an epoxide. Cyclohexene oxide is a symmetrical epoxide. Reacting it with the strong nucleophile sodium methoxide (CH₃O⁻) will proceed via an Sₙ2 backside attack, yielding the trans product. Acid-catalyzed opening with methanol/H⁺ would also yield the same trans product. Choice A would give a cis diol. Choice B would likely yield a mixture of products from carbocation intermediates. Choice D describes a Williamson ether synthesis which might compete with elimination.

Question 10

Which set of conditions would selectively produce (S)-2-ethoxy-1-phenylethanol starting from (R)-styrene oxide?

  1. Sodium ethoxide (NaOEt) in ethanol
    1. Mercury(II) acetate in ethanol; 2. NaBH₄
    1. Ethylmagnesium bromide (EtMgBr); 2. H₃O⁺
  4. Ethanol (EtOH) with catalytic sulfuric acid (H₂SO₄) (correct answer)
Explanation: When you encounter epoxide ring-opening reactions, the key is understanding how different conditions affect both regioselectivity (which carbon gets attacked) and stereochemistry (configuration at stereocenters). Epoxides can open under either nucleophilic or electrophilic conditions, leading to different outcomes. The correct answer is D because acid-catalyzed epoxide opening with ethanol proceeds through an SN2S_N2-like mechanism. The ethanol attacks the less substituted carbon of the (R)-styrene oxide, causing inversion of configuration at the stereocenter and placing the ethoxy group on the less hindered carbon. This gives the desired (S)-2-ethoxy-1-phenylethanol with the correct regiochemistry and stereochemistry. A is incorrect because sodium ethoxide is a strong base that would attack the more substituted, benzylic carbon through an SN2S_N2 mechanism, giving the wrong regioisomer (1-ethoxy-2-phenylethanol). B is wrong because oxymercuration-demercuration is used for alkene hydration, not epoxide opening. This reaction sequence wouldn't work with an epoxide starting material. C fails because Grignard reagents attack the less hindered carbon of epoxides, but this would give 1-phenyl-1-butanol after protonation, not an ethoxy-containing product. The Grignard adds a two-carbon chain, not an ethoxy group. Study tip: Remember that acid-catalyzed epoxide opening favors attack at the more substituted carbon (carbocation-like), while base-catalyzed opening favors the less substituted carbon (SN2S_N2-like). Always check both regiochemistry and stereochemistry in epoxide problems.

Question 11

A student attempts to hydrolyze 2-methyloxirane. Which statement correctly compares the mechanism of this reaction under acidic (H₃O⁺) versus basic (NaOH, H₂O) conditions?

  1. Both reactions proceed via attack at the more substituted carbon, but the acidic reaction is faster.
  2. The acidic reaction involves attack at the more substituted carbon, while the basic reaction involves attack at the less substituted carbon. (correct answer)
  3. Both reactions proceed via attack at the less substituted carbon, but the basic reaction is faster.
  4. The basic reaction involves attack at the more substituted carbon, while the acidic reaction involves attack at the less substituted carbon.
Explanation: This question addresses the fundamental difference in regioselectivity for unsymmetrical epoxides. Under acidic conditions (H₃O⁺), the weak nucleophile (H₂O) attacks the more substituted carbon (C2), which better stabilizes the partial positive charge of the protonated epoxide. Under basic conditions (OH⁻), the strong nucleophile attacks the less sterically hindered carbon (C1) via an Sₙ2 mechanism. The final product (1,2-propanediol) is the same in both cases, but the position of nucleophilic attack differs.

Question 12

Which sequence of reagents is most appropriate to convert 1-methylcyclohexene into cis-1-methyl-1,2-cyclohexanediol?

    1. m-CPBA; 2. H₃O⁺
    1. OsO₄, NMO; 2. NaHSO₃/H₂O
     (correct answer)
    1. KMnO₄, NaOH, heat; 2. H₃O⁺
    1. H₃O⁺; 2. m-CPBA
Explanation: The target product is a cis-diol (syn-dihydroxylation). The sequence of epoxidation (m-CPBA) followed by acid-catalyzed ring opening (H₃O⁺) results in anti-dihydroxylation, yielding a trans-diol. In contrast, osmium tetroxide (OsO₄) with a co-oxidant like NMO performs syn-dihydroxylation, adding both hydroxyl groups to the same face of the double bond, which produces the desired cis-diol. Cold, dilute KMnO₄ also produces cis-diols, but the heated conditions in choice C would cause oxidative cleavage. Choice D has the reagents in the wrong order.

Question 13

A chemist prepares a product by treating cyclohexene oxide with methylmagnesium bromide (CH₃MgBr), followed by an acidic workup. What are the key structural features of the purified major product?

  1. It is a racemic mixture of trans-2-methylcyclohexanol. (correct answer)
  2. It is a racemic mixture of cis-2-methylcyclohexanol.
  3. It is the achiral compound 1-methylcyclohexanol.
  4. It is the achiral compound 2-methylcyclohexanone.
Explanation: The Grignard reagent (CH₃MgBr) provides a strong nucleophile (CH₃⁻) that opens the epoxide under basic conditions. The reaction is an Sₙ2 backside attack. In a cyclic system like cyclohexene oxide, this backside attack results in anti-addition. The incoming methyl group and the resulting hydroxyl group will be on opposite faces of the ring, leading to a trans product. Since the starting epoxide is achiral and attack can occur with equal probability at either carbon of the epoxide, a racemic mixture of the two enantiomeric trans products is formed.