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Ch. 12 - Substitution and Elimination: Reactions of Haloalkanes
Mullins - Organic Chemistry: A Learner Centered Approach 1st Edition
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Tutti i libri di testoMullins 1st EditionCh. 12 - Substitution and Elimination: Reactions of HaloalkanesProblema 50a
Capitolo 11, Problema 50a

For which of the following reactions would you expect elimination to be more favored than substitution?
(a)

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Step 1: Analyze the reaction conditions for both cases. In the first reaction, NaOEt (sodium ethoxide) in EtOH (ethanol) is used, which is a small, strong base. In the second reaction, NaOt-Bu (sodium tert-butoxide) in t-BuOH (tert-butanol) is used, which is a bulky, strong base.
Step 2: Understand the role of the base size in determining the reaction pathway. Small bases like NaOEt can easily approach the electrophilic carbon and favor substitution (S_N2 mechanism). Bulky bases like NaOt-Bu, due to steric hindrance, are less likely to approach the electrophilic carbon and instead favor elimination (E2 mechanism).
Step 3: Consider the structure of the alkyl halide. Both alkyl halides are secondary bromides, which can undergo both substitution and elimination. However, the steric hindrance around the electrophilic carbon plays a role in determining the favored pathway.
Step 4: Evaluate the solvent effects. Ethanol (EtOH) is a protic solvent that can stabilize the transition state for substitution reactions. Tert-butanol (t-BuOH), being bulky, does not stabilize substitution as effectively and further promotes elimination.
Step 5: Conclude that elimination is more favored in the second reaction (NaOt-Bu/t-BuOH) due to the bulky base and solvent, while substitution is more competitive in the first reaction (NaOEt/EtOH) due to the smaller base and solvent effects.

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Elimination vs. Substitution Reactions

Elimination reactions involve the removal of a leaving group and a hydrogen atom, resulting in the formation of a double bond, while substitution reactions replace a leaving group with a nucleophile. The preference for one mechanism over the other depends on factors such as the structure of the substrate, the strength of the nucleophile/base, and the reaction conditions.
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Intro to Substitution/Elimination Problems

Zaitsev's Rule

Zaitsev's Rule states that in elimination reactions, the more substituted alkene is typically the major product. This is due to the stability of more substituted alkenes, which are favored thermodynamically. Understanding this rule helps predict the outcome of elimination reactions, especially when considering steric hindrance and the nature of the base used.
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Defining Zaitsev’s Rule

Sterics and Base Strength

The steric hindrance of the base and the substrate plays a crucial role in determining whether elimination or substitution will be favored. Bulky bases, like t-BuOK, tend to favor elimination due to their inability to effectively approach the substrate for substitution. Conversely, smaller bases can facilitate substitution reactions more readily, highlighting the importance of base choice in reaction pathways.
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Understanding steric effects.
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