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An SN2 reaction is a type of nucleophilic substitution reaction where the nucleophile attacks the substrate from the backside, leading to inversion of configuration. This topic appears in exams to test your understanding of reaction mechanisms, rate laws, stereochemistry, and substrate requirements.
This topic is frequently tested in organic chemistry exams, including those for undergraduate courses, MCAT, and GRE Chemistry. It typically carries moderate to high marks and tests your ability to apply mechanistic knowledge, understand kinetics, and predict reaction outcomes based on molecular structure.
Intermediate
Question: Identify the product of the following SN2 reaction: CH3CH2Br + NaOH-? Step-by-Step:1. Identify the nucleophile (OH-) and the substrate (CH3CH2Br).2. The nucleophile attacks the substrate from the backside.3. The leaving group (Br-) departs, resulting in inversion of configuration.4. The product is CH3CH2OH. Answer: CH3CH2OH Key Rule: Inversion of configuration
Question: Calculate the rate of the following SN2 reaction if [NaOH] = 0.1 M and [CH3CH2Br] = 0.2 M, given k = 0.05 M^-1 s^-1. Step-by-Step:1. Use the rate law: Rate = k[NaOH][CH3CH2Br]2. Substitute the values: Rate = 0.05 M^-1 s^-1 * 0.1 M * 0.2 M3. Calculate: Rate = 0.001 M s^-1 Answer: 0.001 M s^-1 Key Rule: Rate law
Question: Predict the stereochemical outcome of the following SN2 reaction: (R)-2-bromobutane + NaCN-? Step-by-Step:1. Identify the nucleophile (CN-) and the substrate ((R)-2-bromobutane).2. The nucleophile attacks the substrate from the backside.3. The leaving group (Br-) departs, resulting in inversion of configuration.4. The product is (S)-2-cyanobutane. Answer: (S)-2-cyanobutane Key Rule: Inversion of configuration
Question: What is the product of the SN2 reaction between CH3CH2Br and NaOH? Options: A. CH3CH2OH B. CH3CH2Br C. CH3CH2OCH3 D. CH3CH2ONa Correct Answer: A. CH3CH2OH Explanation: The nucleophile (OH-) attacks the substrate (CH3CH2Br) from the backside, resulting in inversion of configuration and the product CH3CH2OH. Why the Distractors Are Tempting: B is the starting material, C is a possible product but not from SN2, D is incorrect as it suggests a different mechanism.
Question: What is the rate of the SN2 reaction if [NaOH] = 0.1 M and [CH3CH2Br] = 0.2 M, given k = 0.05 M^-1 s^-1? Options: A. 0.001 M s^-1 B. 0.01 M s^-1 C. 0.05 M s^-1 D. 0.1 M s^-1 Correct Answer: A. 0.001 M s^-1 Explanation: Use the rate law: Rate = k[NaOH][CH3CH2Br] = 0.05 M^-1 s^-1 * 0.1 M * 0.2 M = 0.001 M s^-1. Why the Distractors Are Tempting: B, C, and D are plausible rates but incorrect due to misapplication of the rate law.
Question: What is the stereochemical outcome of the SN2 reaction between (R)-2-bromobutane and NaCN? Options: A. (R)-2-cyanobutane B. (S)-2-cyanobutane C. (R)-2-bromobutane D. (S)-2-bromobutane Correct Answer: B. (S)-2-cyanobutane Explanation: The nucleophile (CN-) attacks the substrate ((R)-2-bromobutane) from the backside, resulting in inversion of configuration and the product (S)-2-cyanobutane. Why the Distractors Are Tempting: A suggests retention of configuration, C is the starting material, D is incorrect as it suggests a different mechanism.
Question: Which of the following substrates will NOT undergo an SN2 reaction? Options: A. CH3CH2Br B. (CH3)3CBr C. CH3CH2Cl D. CH3CH2I Correct Answer: B. (CH3)3CBr Explanation: Tertiary halides are sterically hindered and do not undergo SN2 reactions. Why the Distractors Are Tempting: A, C, and D are primary or secondary halides that can undergo SN2 reactions.
Question: Which of the following is a strong nucleophile that favors SN2 reactions? Options: A. H2O B. CH3OH C. CN- D. NH3 Correct Answer: C. CN- Explanation: CN- is a strong nucleophile that favors SN2 reactions. Why the Distractors Are Tempting: A, B, and D are weaker nucleophiles that may not favor SN2 reactions.
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