Question

Predict the major product(s) for the reaction between \((E)-3\)-methyl-2-pentene and the following reagents: (a) \(\mathrm{H}_{3} \mathrm{O}^{+} / \mathrm{H}_{2} \mathrm{O}\) (b) \(1 . \mathrm{BH}_{3} / \mathrm{THF} ; 2 . \mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{NaOH}\) (c) \(\mathrm{HCl}\) (d) \(\mathrm{HBr}\) (e) \(\mathrm{H}_{3} \mathrm{O}^{+} / \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

Short answer

(a) 3-methyl-3-pentanol(b) 3-methyl-2-pentanol(c) 3-chloro-3-methylpentane(d) 3-bromo-3-methylpentane(e) 3-ethoxy-3-methylpentane

Step-by-step solution

Reaction with \(\text{H}_3\text{O}^+/ \text{H}_2\text{O}\)

This reaction undergoes acid-catalyzed hydration. The major product is formed by adding a \(\text{-OH}\) group to the more substituted carbon atom (Markovnikov addition). \( (E)-3-methyl-2-pentene \rightarrow 3-methyl-3-pentanol \)

Reaction with \(1. \text{BH}_3 / \text{THF}; 2. \text{H}_2\text{O}_2 / \text{NaOH}\)

This reaction undergoes hydroboration-oxidation, leading to anti-Markovnikov addition of \(\text{-OH}\). The major product forms with \(\text{-OH}\) on the less substituted carbon. \( (E)-3-methyl-2-pentene \rightarrow 3-methyl-2-pentanol \)

Reaction with \(\text{HCl}\)

This reaction undergoes hydrohalogenation. \(\text{Cl}\) is added to the more substituted carbon atom (Markovnikov addition). \( (E)-3-methyl-2-pentene \rightarrow 3-chloro-3-methylpentane \)

Reaction with \(\text{HBr}\)

This reaction undergoes hydrohalogenation. \(\text{Br}\) is added to the more substituted carbon atom (Markovnikov addition). \( (E)-3-methyl-2-pentene \rightarrow 3-bromo-3-methylpentane \)

Reaction with \(\text{H}_3\text{O}^+ / \text{CH}_3\text{CH}_2\text{OH}\)

This reaction undergoes acid-catalyzed addition of ethanol. The \(\text{-OCH}_2\text{CH}_3\) group is added to the more substituted carbon atom (Markovnikov addition). \( (E)-3-methyl-2-pentene \rightarrow 3-ethoxy-3-methylpentane \)

Key concepts

Markovnikov Addition

Markovnikov addition explains how atoms add to the different carbons in a double bond. According to Markovnikov’s rule, when a protic acid (like HCl or HBr) is added to an alkene, the hydrogen atom attaches to the carbon with more hydrogen atoms already attached (the less substituted carbon). Correspondingly, the additional group (like Cl or Br) goes to the more substituted carbon. This is because the more substituted carbocation intermediate is more stable.

Markovnikov addition ensures the dominant product is formed by stabilizing the carbocation through neighboring carbon atoms, often observed in hydrohalogenation reactions (e.g., HCl and HBr). These additions result in Markovnikov products, such as 3-chloro-3-methylpentane and 3-bromo-3-methylpentane in hydrohalogenation reactions.

Anti-Markovnikov Addition

Anti-Markovnikov addition occurs when the incoming group attaches to the less substituted carbon atom of the double bond. A common method to achieve this is through hydroboration-oxidation. During this reaction sequence, borane (BH3) reacts with the alkene in a syn addition fashion, where hydrogen and boron add to the same side.

Subsequently, oxidation with hydrogen peroxide (H2O2) and a base like sodium hydroxide (NaOH) replaces the boron with a hydroxyl group (-OH). This positioning flips the usual Markovnikov rule, thus leading to anti-Markovnikov products. For instance, (E)-3-methyl-2-pentene reacts via hydroboration-oxidation to yield 3-methyl-2-pentanol, with the -OH group on the less substituted carbon.

Acid-Catalyzed Hydration

Acid-catalyzed hydration involves adding water to an alkene in the presence of a strong acid, often yielding Markovnikov products. This reaction proceeds through the formation of a carbocation intermediate. The alkene reacts with the acid (like H3O^+), leading to the creation of a carbocation on the more substituted carbon.

Water (H2O) then attacks the carbocation, and deprotonation occurs to form the alcohol. For example, in the hydration of (E)-3-methyl-2-pentene with H3O^+/H2O, the product is 3-methyl-3-pentanol, placing the hydroxyl group on the more substituted carbon.

Hydroboration-Oxidation

Hydroboration-oxidation is a two-step reaction that converts alkenes to alcohols with anti-Markovnikov selectivity. In the first stage, borane (BH3) adds across the alkene in a syn addition, placing hydrogen and boron on the same side of the molecule.

This is followed by treatment with hydrogen peroxide (H2O2) and sodium hydroxide (NaOH), which replaces the boron with an OH group. This method avoids carbocation rearrangement, ensuring the hydroxyl group ends up on the less substituted carbon. For example, treating (E)-3-methyl-2-pentene results in 3-methyl-2-pentanol.

Hydrohalogenation

Hydrohalogenation is the addition of a hydrogen halide (like HCl or HBr) to an alkene, forming an alkyl halide. The hydrogen attaches to the carbon with more hydrogen atoms (less substituted carbon), while the halide attaches to the more substituted carbon. This results in Markovnikov products driven by the stability of the carbocation intermediate.

For instance, when (E)-3-methyl-2-pentene reacts with HCl, 3-chloro-3-methylpentane is produced. Similarly, the reaction with HBr forms 3-bromo-3-methylpentane. These reactions are straightforward as they follow Markovnikov’s rule, often without rearrangements.