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Ch.6 - Alkyl Halides; Nucleophilic Substitution
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh.6 - Alkyl Halides; Nucleophilic SubstitutionProblem 18
Chapter 6, Problem 18

Rank the following compounds in decreasing order of their reactivity toward the SN2 reaction with sodium ethoxide (Na+ –OCH2CH3) in ethanol.
methyl chloride
tert-butyl iodide 
neopentyl bromide
isopropyl bromide
methyl iodide
ethyl chloride

Verified step by step guidance
1
Step 1: Understand the SN2 reaction mechanism. SN2 reactions are bimolecular nucleophilic substitution reactions where the nucleophile attacks the electrophilic carbon in a single step, leading to the inversion of configuration. Reactivity in SN2 reactions is influenced by steric hindrance and the leaving group ability.
Step 2: Analyze steric hindrance for each compound. The less sterically hindered the electrophilic carbon is, the more reactive the compound will be in an SN2 reaction. Methyl halides (e.g., methyl chloride and methyl iodide) are the least hindered, followed by primary alkyl halides (e.g., ethyl chloride), secondary alkyl halides (e.g., isopropyl bromide), and tertiary alkyl halides (e.g., tert-butyl iodide). Neopentyl bromide is a primary alkyl halide but has significant steric hindrance due to the bulky neopentyl group.
Step 3: Evaluate the leaving group ability. The better the leaving group, the more reactive the compound will be in an SN2 reaction. Iodide (I⁻) is a better leaving group than bromide (Br⁻), which is better than chloride (Cl⁻). This means compounds with iodide as the leaving group will generally be more reactive than those with bromide or chloride.
Step 4: Combine steric hindrance and leaving group ability to rank the compounds. Methyl iodide will likely be the most reactive due to minimal steric hindrance and an excellent leaving group. Methyl chloride will follow due to minimal steric hindrance but a weaker leaving group. Ethyl chloride, being a primary alkyl halide, will be less reactive than methyl halides. Isopropyl bromide, a secondary alkyl halide, will be less reactive due to increased steric hindrance. Neopentyl bromide will be even less reactive due to significant steric hindrance. Tert-butyl iodide will be the least reactive because tertiary alkyl halides are highly hindered and do not favor SN2 reactions.
Step 5: Final ranking based on decreasing reactivity toward SN2: Methyl iodide > Methyl chloride > Ethyl chloride > Isopropyl bromide > Neopentyl bromide > Tert-butyl iodide.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

SN2 Reaction Mechanism

The SN2 (substitution nucleophilic bimolecular) reaction is a type of nucleophilic substitution where the nucleophile attacks the electrophile simultaneously as the leaving group departs. This mechanism involves a single transition state and results in an inversion of configuration at the carbon center. The rate of the reaction depends on the concentration of both the nucleophile and the substrate, making it a second-order reaction.
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Steric Hindrance

Steric hindrance refers to the prevention of chemical reactions due to the spatial arrangement of atoms within a molecule. In SN2 reactions, bulky groups around the electrophilic carbon can hinder the approach of the nucleophile, thus decreasing reactivity. Compounds with less steric hindrance, such as methyl and primary alkyl halides, are more reactive in SN2 reactions compared to tertiary alkyl halides, which are significantly less reactive.
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Leaving Group Ability

The ability of a leaving group to depart from a substrate is crucial in determining the reactivity of compounds in nucleophilic substitution reactions. Good leaving groups, such as iodide and bromide, stabilize the negative charge after leaving, facilitating the reaction. In contrast, poor leaving groups, like chloride, are less favorable, making the substrate less reactive in SN2 reactions.
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Related Practice
Textbook Question

For each pair of compounds, state which compound is the better SN2 substrate.

a. 2-methyl-1-iodopropane or tert-butyl iodide.

b. cyclohexyl bromide or 1-bromo-1-methylcyclohexane

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Textbook Question

For each pair, predict the stronger nucleophile in the SN2 reaction (using an alcohol as the solvent). Explain your prediction.

e. (CH3)3N or (CH3)2O

f. CH3COO– or CF3COO–

1343
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Textbook Question

For each pair of compounds, state which compound is the better SN2 substrate.

c. 2-bromobutane or isopropyl bromide

d. 1-chloro-2,2-dimethylbutane or 2-chlorobutane

e. 1-iodobutane or 2-iodopropane

1102
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Textbook Question

Draw a perspective structure or a Fischer projection for the products of the following SN2 reactions.

(a) trans-1-bromo-3-methylcyclopentane + KOH

(b) (R)-2-bromopentane + KCN

1045
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Textbook Question

For each pair, predict the stronger nucleophile in the SN2 reaction (using an alcohol as the solvent). Explain your prediction.

g. (CH3)2CHO– or CH3CH2CH2O–

h. I– or Cl–

831
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Textbook Question

When diethyl ether (CH3CH2OCH2CH3) is treated with concentrated HBr, the initial products are CH3CH2Br and CH3CH2OH. Propose a mechanism to account for this reaction.

2811
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