Question

Identify \(Z\) in the following series: \(\mathrm{CH}_{2}=\mathrm{CH}_{2} \stackrel{\mathrm{HBr}}{\longrightarrow} \mathrm{X} \stackrel{\text { Hydrolysis }}{\longrightarrow} \mathrm{Y} \stackrel{\mathrm{I}_{2}, \text { Excess }}{\longrightarrow} \mathrm{Z}\) (a) \(\mathrm{CHI}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\)

Short answer

Intermediate \(Z\) is \(\mathrm{CH}_3\mathrm{CHO}\), option (b).

Step-by-step solution

Identify Intermediate X

The first step involves the reaction of ethene (\(\mathrm{CH}_2=\mathrm{CH}_2\)) with HBr. This is an electrophilic addition reaction that produces bromoethane as the major product due to the addition of the hydrogen atom to the less substituted carbon (Markovnikov's Rule). Thus, \(\mathrm{X} = \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}\).

Hydrolysis to Yield Y

The intermediate\(\mathrm{X}, \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}\), is then hydrolyzed to form ethanol (\(\mathrm{Y} = \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\)). This is a nucleophilic substitution reaction where the bromine atom in bromoethane is replaced by a hydroxyl group.

Iodination to Yield Z

In the final step, ethanol (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\)) is treated with an excess of iodine (\(\mathrm{I}_2\)). However, ethanol does not readily form a simple iodinated compound under these conditions. Instead, excess \(\mathrm{I}_2\) oxidizes ethanol to form acetaldehyde (\(\mathrm{CH}_3\mathrm{CHO}\)). Therefore, intermediate \(\mathrm{Z}\) is acetaldehyde.

Key concepts

Electrophilic Addition Reaction

Electrophilic addition reactions are a vital concept in organic chemistry, especially when understanding how alkenes, such as ethene (\(\mathrm{CH}_2=\mathrm{CH}_2\)), react with hydrogen halides like HBr. These reactions involve the addition of an electrophile across a double bond.

In the initial step, an electron-poor electrophile, in this case, the hydrogen ion from HBr, approaches the electron-rich double bond in ethene. The double bond is broken, and a new \(\mathrm{C-H}\) bond is formed. This temporarily creates a carbocation.

• The bromide ion (\(\mathrm{Br}^-\)) then attacks the positively charged carbon in the carbocation, completing the addition reaction by forming bromoethane (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}\)).
• The key here is understanding that the reaction proceeds through a two-step mechanism—formation of a carbocation followed by nucleophilic attack.

The choice of where the hydrogen atom adds (to which carbon of the double bond) follows a principle known as Markovnikov's Rule.

Markovnikov's Rule

In the realm of addition reactions, Markovnikov's Rule is a guiding principle that helps predict the major product. This rule states that when a hydrogen halide, such as HBr, is added to an alkene, the hydrogen atom will attach to the less substituted carbon atom.

This happens because the formation of the more stable carbocation intermediate is favored. A stable carbocation typically has more alkyl groups attached.

• In the case of ethene, as it reacts with HBr, the hydrogen attaches itself to the first carbon atom, leading to the formation of the more stable ethyl carbocation.
• The bromide ion then attacks this carbocation, resulting in bromoethane as the major product.

Understanding Markovnikov's Rule is crucial as it allows chemists to anticipate the products of electrophilic addition reactions accurately.

Nucleophilic Substitution Reaction

Nucleophilic substitution reactions are common in organic chemistry, especially for converting a compound like bromoethane into ethanol. These reactions involve the replacement of one group in a molecule by a nucleophile.

In the hydrolysis of bromoethane, the nucleophile is the hydroxide ion (\(\mathrm{OH}^-\)).

• The \(\mathrm{OH}^-\) ion approaches the carbon atom bonded to the bromine, allowing for the substitution of bromine by the hydroxyl group.
• This process is known as substitution, specifically SN2 in this reaction, indicating a single-step mechanism involving two reactants.
• The bromine leaves as a bromide ion, and ethanol (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\)) is formed.

Such a transformation elucidates how functional groups can be altered in organic molecules through nucleophilic substitution reactions.

Oxidation Reactions

Oxidation reactions in organic chemistry often involve the conversion of alcohols into aldehydes or ketones. In this scenario, ethanol undergoes oxidation when treated with excess iodine (\(\mathrm{I}_2\)).

Here's how it happens:

• Excess \(\mathrm{I}_2\) leads to the formation of iodoform (\(\mathrm{CHI}_3\)) as a side product while the \(\mathrm{OH}\) group of ethanol is oxidized to an aldehyde group, yielding acetaldehyde (\(\mathrm{CH}_3\mathrm{CHO}\)).
• This transformation is a type of oxidation reaction where the increase in the oxidation state of the molecule is observed, and it is facilitated by oxidizing agents, such as \(\mathrm{I}_2\)

These reactions are significant because they allow for the synthesis of different functional groups and the understanding of organic transformations.