2-Bromopentane is treated with alcoholic KOH solution. What will be the major product formed in this reaction and what is the type of elimination) called? (a) Pent-1-ene, \(\beta\)-Elimination (b) Pent-2-ene, \(\beta\)-Elimination (c) Pent-1-ene, Nucleophilic substitution (d) Pent-2-ene, Nucleophilic substitution.

Short Answer

Expert verified
The major product formed is (b) Pent-2-ene, and the type of elimination is \beta-Elimination.

Step by step solution

01

Understand the Reaction

Alcoholic KOH is a strong base and is commonly used to promote elimination reactions in organic chemistry. When 2-bromopentane reacts with alcoholic KOH, a beta-elimination reaction occurs, specifically an E2 mechanism.
02

Identify the Elimination Mechanism

In the E2 mechanism, a proton adjacent to the carbon with the leaving group (bromine in this case) is abstracted by the base (KOH), causing the electrons from the C-H bond to form a double bond as the leaving group exits. This results in the formation of an alkene.
03

Determine the Major Product

For 2-bromopentane, there are two adjacent protons that could potentially be abstracted. However, the more substituted alkene is generally the more stable product and thus formed preferentially. This favors the formation of pent-2-ene over pent-1-ene.
04

Identify the Type of Reaction

Since a base is removing a proton to form a double bond while a leaving group is simultaneously removed, this is a beta-elimination reaction, also known as E2 elimination.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

E2 Mechanism
The E2 mechanism is a pivotal type of elimination reaction in organic chemistry. It's a concise acronym for the full name - Elimination Bimolecular reaction. This mechanism characteristically involves the simultaneous removal of a proton (a hydrogen atom) and a leaving group, typically a halide ion like bromide or chloride, from adjacent carbon atoms, hence the 'Elimination' in E2.

The 'Bimolecular' part indicates that the rate-determining step involves two molecular entities, the substrate (such as 2-bromopentane) and the base (like alcoholic KOH). During the reaction, the base attacks a hydrogen atom that is directly adjacent to the carbon bearing the leaving group. This process forms a double bond between the two carbons as the leaving group is expelled.

One of the defining features of the E2 mechanism is that it proceeds via a single, concerted step. Both the abstraction of the proton by the base and the departure of the leaving group happen simultaneously. This concerted action often results in the creation of more stable alkenes through the formation of the new double bonds, which is a subtle dance of electrons and atomic positioning that underpins organic transformations.
Beta-Elimination
Beta-elimination refers to a specific type of elimination reaction where a molecule loses atoms or groups of atoms from two adjacent carbon atoms (known otherwise as alpha and beta carbons). In organic chemistry, the alpha carbon is the one directly attached to the functional group or atom in question, and the beta carbon is the one next to the alpha.

During a beta-elimination, a hydrogen on the beta carbon and a leaving group on the alpha carbon are removed, resulting in the formation of a double bond between these two carbons. The beauty of beta-elimination is in its selectivity and its trend towards creating more thermodynamically favored products. These reactions are guided by certain rules, such as Zaitsev's rule, which predicts that the more substituted, stabilized alkene will be the preferred product. This is due to its greater number of alkyl groups which provide hyperconjugation and inductive effects, stabilizing the double bond.
2-bromopentane Reactions
2-bromopentane is a suitable and common substrate for examining elimination reactions in educational settings. Its structure contains a bromine atom attached to the second carbon atom of pentane, which defines it as a halogenated aliphatic compound. In organic chemistry, when 2-bromopentane is subject to a reactant like alcoholic KOH, it primarily undergoes elimination reactions.

Given its configuration, the 2-bromopentane molecule has the potential to form two different alkenes when the bromine, a good leaving group, is eliminated. The creation of these alkenes depends on which adjacent hydrogen atom the basic reactant decides to abstract. In most scenarios, the most stable alkene formed is favored. This specificity aligns with Zaitsev's rule, and hence, in the case of 2-bromopentane, the major product is typically pent-2-ene, which is more substituted, rather than pent-1-ene, which would be less substituted and less stable.
Alcoholic KOH Reaction
Alcoholic KOH (potassium hydroxide in alcohol) is a strong, bulky base used to promote elimination reactions over substitution reactions. When we dissolve KOH in alcohol, such as ethanol, the resulting solution is highly effective in abstracting protons from organic substrates like 2-bromopentane due to the presence of the alkoxide ions.

In reactions with alcoholic KOH, the base attacks and removes a proton, creating a double bond as the leaving group departs from the molecule. The strength and structure of this base make it particularly adept at eliminating beta hydrogens, thereby favoring the E2 mechanism over other potential reaction pathways, making it a trusted reagent for inducing beta-elimination in organic synthesis.

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Most popular questions from this chapter

The correct trend of acidic nature of the following alkynes is (a) \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3}\) (b) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3}>\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}\) (d) \(\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3}>\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\)

Which alkane is produced when sodium salt of butanoic acid is heated with soda lime? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{4}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Identify the reagent from the following list which can easily distinguish between 1 -butyne and 2 -butyne. (a) Bromine water (b) Baeyer's reagent (c) Dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{HgSO}_{4}\) (d) Ammoniacal \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\)

When 1-butyne undergoes oxymercuration with the help of \(\mathrm{HgSO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}\), the product(s) formed is/are (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{HCOOH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\)

Ozonolysis products of 2-pentyne after decomposition of ozonide with water and subsequent oxidation are (a) ethanoic acid and propanoic acid (b) ethanoic acid and propanone (c) ethanoic acid (d) formic acid and glyoxal.

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