Problem 94 - A Complete Resource Book in Chemistry for JEE Main

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) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

Step-by-Step Solution

Answer

The correct answer is (a) CHI₃.

Step 1: Recognize the Initial Reaction

The series begins with ethene ( CH_2=CH_2) reacting with hydrogen bromide (HBr). HBr adds across the double bond in an electrophilic addition reaction, forming bromoethane (CH_3CH_2Br). The structure X is CH_3CH_2Br.

Step 2: Understand Hydrolysis Process

The next step is the hydrolysis of bromoethane (CH_3CH_2Br). During hydrolysis, the bromine atom is replaced by a hydroxyl group, forming ethanol (CH_3CH_2OH). Thus, the structure Y in the series is CH_3CH_2OH.

Step 3: Interpret the Iodination Reaction

In the final step, the product from hydrolysis, ethanol (CH_3CH_2OH), is treated with excess iodine (I_2). Excess iodine typically causes oxidation or halogenation reactions. Here, it leads to the formation of iodoform (CHI_3) and other by-products. Therefore, the structure Z is CHI_3.

Key Concepts

Electrophilic Addition Reaction

In the realm of organic chemistry, the electrophilic addition reaction is a fundamental process, especially crucial for alkenes like ethene. When ethene (\(\mathrm{CH}_2 = \mathrm{CH}_2\)) encounters hydrogen bromide (HBr), an interesting dance of electrons occurs. The double bond in ethene is like a high-energy site, rich in electrons and ready to react.

HBr is composed of a hydrogen and a bromine atom. The hydrogen acts as an electrophile, meaning it seeks electrons to complete its outer shell. It approaches the electron-rich double bond in ethene. The bond between the two carbon atoms in ethene grabs the hydrogen, while the remaining bromine goes to the more electronegative part of the molecule.

As a result, the double bond breaks, and bromoethane (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}\)) forms. This process is called an electrophilic addition because the electrophile, hydrogen, adds itself to the molecule.

Hydrolysis

Hydrolysis involves breaking a bond in a molecule using water. In the context of the exercise, hydrolysis affects bromoethane (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}\)) and transforms it into ethanol (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\)).

The key player here is the hydroxyl group, (-OH). When bromoethane undergoes hydrolysis, water supplies the hydroxyl group, replacing the bromine atom. This kind of chemical reaction is a nucleophilic substitution. In such reactions:

• The water molecule acts as a nucleophile by donating a pair of electrons to form a new bond.
• The bromine leaves, taking its electrons with it, which results in ethanol formation.

Understanding hydrolysis is essential because it often serves as a crucial step in converting alkyl halides into alcohols.

Iodination Reaction

The iodination reaction involves introducing iodine into a compound, typically for oxidation or substitution purposes. In this exercise, the substance ethanol (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}\)) is treated with excess iodine (\(\mathrm{I}_2\)).

In the presence of iodine and sometimes a base, ethanol undergoes a complex series of steps resulting in the formation of iodoform (\(\mathrm{CHI}_3\)). This specific reaction is a well-known example of iodination where the methyl group of ethanol is oxidized and transformed.

Iodoform forms as the iodine causes the substitution of hydrogen atoms in ethanol, a process promoting the formation of the yellow crystalline solid, \(\mathrm{CHI}_3\). This reaction is useful in organic synthesis and analytical chemistry. It's another route to track the transformations of organic molecules using iodine.