Which alkyl halide in each pair is more reactive in an E2 reaction with hydroxide ion?
c.
d.

검증된 단계별 안내

Step 1: Understand the E2 reaction mechanism. E2 reactions are bimolecular elimination reactions where a base (such as hydroxide ion) removes a proton from a β-carbon, and a leaving group (such as Br or Cl) departs simultaneously, forming a double bond.

Step 2: Analyze the structure of the alkyl halides. The reactivity in E2 reactions depends on steric hindrance and the ability of the β-hydrogens to be abstracted. Less steric hindrance around the β-hydrogens generally increases reactivity.

Step 3: Compare the first pair of alkyl halides (bromine-containing compounds). The compound on the left has a secondary bromine atom, while the compound on the right has a tertiary bromine atom. Tertiary alkyl halides are more sterically hindered, making it harder for the base to access the β-hydrogens, thus less reactive in E2 reactions.

Step 4: Compare the second pair of alkyl halides (chlorine-containing compounds). The compound on the left has a secondary chlorine atom, while the compound on the right has a tertiary chlorine atom. Similar to the bromine pair, the tertiary alkyl halide is more sterically hindered, reducing its reactivity in E2 reactions.

Step 5: Conclude that in both pairs, the secondary alkyl halides (left structures) are more reactive in E2 reactions with hydroxide ion due to reduced steric hindrance and better accessibility of β-hydrogens.

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주요 개념

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E2 Reaction Mechanism

The E2 (bimolecular elimination) reaction is a concerted process where a base removes a proton from a β-carbon while a leaving group departs from the α-carbon, resulting in the formation of a double bond. The reaction rate depends on the concentration of both the alkyl halide and the base, making it a second-order reaction. The steric hindrance and the nature of the leaving group significantly influence the reactivity of the alkyl halide.

Leaving Group Ability

The ability of a leaving group to depart from the substrate is crucial in determining the reactivity of alkyl halides in E2 reactions. Good leaving groups, such as iodide (I-) and bromide (Br-), stabilize the negative charge better than poorer leaving groups like chloride (Cl-). Therefore, alkyl halides with better leaving groups will generally react more readily in elimination reactions.

Steric Hindrance

Steric hindrance refers to the crowding around a reactive center in a molecule, which can impede the approach of the base in an E2 reaction. Tertiary alkyl halides are typically more reactive than primary ones due to less steric hindrance, allowing the base to effectively abstract a proton. In the context of the provided pairs, the structure of the alkyl halide influences its reactivity based on steric factors.

Which alkyl halide in each pair is more reactive in an E2 reaction with hydroxide ion?c. d.