Question

Which of the following most readily undergoes \(\mathrm{E} 2\) elimination with a strong base? (a) 2 - bromo - 3 - methylbutane (b) 2 - bromopentane (c) 2 - bromo-2-methylbutane (d) 1 - bromo- 2,2 - dimethylpropane

Short answer

2-bromo-2-methylbutane (c) undergoes E2 elimination most readily.

Step-by-step solution

Understanding E2 Elimination

The E2 elimination reaction involves the removal of a leaving group (such as a halide) and a hydrogen on an adjacent carbon to form a double bond. This process is bimolecular, meaning that the reaction rate depends on the concentration of both the substrate and the base. The reaction is favored by a strong base and typically occurs in a single, concerted step.

Analyzing Substrate Structure

The substrate that undergoes E2 elimination most readily will typically be the one with the most accessible hydrogen atom adjacent to the leaving group, as well as the most stable possible alkene product. Substrates with more highly substituted alkenes as products are generally more stable.

Evaluating Each Compound

Let's consider each provided compound: (a) 2-bromo-3-methylbutane, (b) 2-bromopentane, (c) 2-bromo-2-methylbutane, and (d) 1-bromo-2,2-dimethylpropane. We check the structure for the presence of β-hydrogens and the stability of the resulting alkene. More substituted alkenes are more stable.

Analyzing 2-Bromo-2-Methylbutane

For compound (c), 2-bromo-2-methylbutane, the bromine is on a tertiary carbon which favors formation of a stable, more substituted alkene. Removal of a β-hydrogen from an adjacent methyl group would form a tertiary alkene. Due to the highly substituted nature of the potential double bond, this is the most favorable for E2 elimination.

Conclusion

Compound (c), 2-bromo-2-methylbutane, will most readily undergo E2 elimination due to the presence of a highly substituted, and hence more stable, possible alkene product. Tertiary carbon compounds facilitate E2 elimination effectively under strong base conditions.

Key concepts

Organic Chemistry

Organic chemistry is the branch of chemistry that studies the structure, properties, and reactions of compounds primarily made of carbon atoms. It's a broad field since carbon forms the backbone of various molecules, including those found in living organisms.
These organic compounds can have a variety of functional groups, such as alcohols, alkanes, alkenes, and halides. Each group influences how a compound reacts.
In the context of the E2 elimination reaction, understanding these various functional groups helps us predict how an organic molecule might react under specific conditions, such as the presence of a strong base. Having a grasp of organic chemistry is essential to navigate through different reaction mechanisms and outcomes.

Elimination Reaction

Elimination reactions involve the removal of atoms or groups from a molecule, leading to the formation of a multiple bond, usually a double or triple bond. Among these is the E2 mechanism which stands for bimolecular elimination.

• Bimolecular: The '2' indicates that two molecules are involved in the rate-determining step. Specifically, the substrate and the base both affect the reaction rate.
• Strong Base: A key feature of E2 eliminations is their reliance on strong bases to deprotonate a hydrogen atom adjacent to the leaving group.
• One Step Process: Unlike E1 eliminations, E2 reactions occur in a single concerted step involving simultaneous removal of the leaving group and the formation of a pi bond.

These features make E2 eliminations important when synthesizing alkenes, especially when a strong, non-polarized base is available.

Reaction Mechanism

Understanding the reaction mechanism of an E2 elimination is crucial for predicting the outcome of the process. This mechanism relies on a concerted step where:

• The base removes a hydrogen atom adjacent to the carbon bonded to the leaving group (often a halogen).
• The electrons from the hydrogen-carbon bond help form the new double bond while the leaving group departs, simultaneously creating an alkene.
• No intermediates are formed during this process, as it happens all at once, which makes it distinct from the E1 mechanism.

This concerted mechanism means the orientation of the substrate matters, and anti-periplanar orientation (where the leaving group and hydrogen are 180° apart) is often required for E2 eliminations.

Substrate Stability

Substrate stability plays a critical role in how readily an E2 elimination occurs. The factors influencing stability include:

• Leaving Group Quality: A good leaving group, such as a halide ion, makes it easier for the elimination to occur.
• Substituent Effects: More substituted carbon atoms, especially tertiary structures, can lead to more stable alkenes due to hyperconjugation and electron-donating effects.
• Alkene Stability: Alkenes that result from the elimination are more stable if they are more substituted.

For the example given, 2-bromo-2-methylbutane undergoes E2 elimination readily due to its tertiary carbon structure, which allows for the formation of a highly substituted and hence more stable alkene product. This understanding helps in selecting appropriate substrates for desired reactions.