Question

The major product formed when 2-bromo-3-methyl-1-phenyl butane is treated with alcoholic KOH is

Short answer

Answer: The major product formed in the reaction is 1-phenyl-2-methylbut-1-ene.

Step-by-step solution

Write down the given compound and the reagent

Draw the structure of 2-bromo-3-methyl-1-phenyl butane, the starting compound, and alcoholic KOH, the reagent. It will help visualize the reaction and the elimination process.

Identify the alpha and beta carbons

The alpha carbon is the carbon atom that is directly attached to the bromine atom (the leaving group). The beta carbons are the carbon atoms adjacent to the alpha carbon. In this case, there are two beta carbons: one is attached to a phenyl group and the other is attached to a methyl group.

Determine which beta carbon will lose a hydrogen

According to Zaitsev's rule, the more substituted alkene will be the major product. Compare the two possible products resulting from the elimination of a hydrogen atom from the two beta carbons.

Eliminate hydrogen bromide

The hydrogen is eliminated from the beta carbon with the greater number of carbon substituents (the one attached to the phenyl group). The remaining electrons from the broken C-H bond form a double bond between the alpha and beta carbons, and the bromine atom leaves as bromide ion (Br-). The final product is formed with a double bond between the alpha and beta carbons.

Write the major product's structure and name

Draw the structure of the major product, which is an alkene: 1-phenyl-2-methylbut-1-ene.

Key concepts

Zaitsev's Rule

Understanding Zaitsev's rule is essential for predicting the outcome of elimination reactions in organic chemistry. This empirical rule states that, when an alkene is formed during an elimination reaction, the most substituted alkene will be the predominant product. In simpler terms, this means that the hydrogen atom is more likely to be removed from the beta carbon that already has more carbon atoms attached to it.

Applying Zaitsev's rule to the problem at hand, the elimination of hydrogen would predominantly occur at the beta carbon connected to the phenyl group over the one connected to a mere methyl group. This is because the resulting alkene will have more alkyl groups attached to it, which stabilizes the double bond through hyperconjugation and inductive effects, leading to a more stable alkene product.

Elimination Reactions

Elimination reactions are a type of organic chemical reaction where two substituents are removed from a molecule, resulting in the formation of a pi bond. Most commonly, these reactions involve removing a halogen and a hydrogen from adjacent carbons (alpha and beta carbons), as seen in our exercise. With alcoholic KOH as the base, the reaction fits into the broad category of beta-elimination, specifically E2 (bimolecular elimination), where the reaction rate depends on the concentration of both the substrate and the base.

This process occurs in one concerted step: as the base abstracts the proton from the beta carbon, the electrons left behind form the double bond while the leaving group (often a halide) is expelled, thus forming an alkene.

Alkene Formation

The creation of alkenes is a central theme in organic synthesis and is often achieved through elimination reactions, where double bonds are formed between carbon atoms. Alkene formation is not a random process, but one that is governed by specific rules and patterns, with Zaitsev's rule being particularly important in determining the position of the double bond.

The exercise illustrates this through the transformation of 2-bromo-3-methyl-1-phenyl butane into 1-phenyl-2-methylbut-1-ene. The selectivity of the reaction towards the more substituted alkene is a direct result of the stability conferred by the additional alkyl substituents adjacent to the double bond, which is an essential concept for students to grasp.

Substitution Patterns in Hydrocarbons

Hydrocarbons can undergo a variety of reactions including substitution, where an atom or group of atoms is replaced by another atom or group. The substitution patterns in hydrocarbons influence the chemical reactivity and stability of the molecules. In terms of alkene formation, the stability of the potential products is often assessed by the substitution pattern of the alkene.

In the given exercise, we have a secondary alpha carbon bonded to a tertiary and a secondary carbon. The elimination reaction preferentially occurs in such a way as to result in a more substituted, and hence more stable, alkene. Recognition of substitution patterns is crucial in predicting not only the outcome of elimination reactions but also the stability and reactivity of organic molecules in a variety of chemical contexts.