Problem 80 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify \(\mathrm{Z}\) in the following series: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I} \stackrel{\text { Alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{X} \stackrel{\mathrm{Br}_{2}}{\longrightarrow} \mathrm{Y} \stackrel{\mathrm{KCN}}{\longrightarrow} \mathrm{Z}\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}\) (b) \(\mathrm{NCCH}_{2}-\mathrm{CH}_{2} \mathrm{CN}\) (c) \(\mathrm{BrCH}_{2}-\mathrm{CH}_{2} \mathrm{CN}\) (d) \(\mathrm{BrCH}=\mathrm{CHCN}\)

Step-by-Step Solution

Answer

The compound \(\mathrm{Z}\) is (b) \(\mathrm{NCCH}_{2}-\mathrm{CH}_{2} \mathrm{CN}\).

Step 1: Formation of Compound X

The initial compound, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\), is treated with alcoholic \(\mathrm{KOH}\), leading to an elimination reaction. This reaction forms \(\mathrm{CH}_{2} = \mathrm{CH}_{2}\) (ethene) as compound \(\mathrm{X}\), where iodide is eliminated and a double bond is formed.

Step 2: Formation of Compound Y

Compound \(\mathrm{X}\), ethene (\(\mathrm{CH}_{2} = \mathrm{CH}_{2}\)), is then reacted with \(\mathrm{Br}_{2}\). This forms 1,2-dibromoethane (\(\mathrm{BrCH}_{2}-\mathrm{CH}_{2}\mathrm{Br}\)) as compound \(\mathrm{Y}\) through an addition reaction across the double bond.

Step 3: Formation of Compound Z

Finally, compound \(\mathrm{Y}\), 1,2-dibromoethane (\(\mathrm{BrCH}_{2}-\mathrm{CH}_{2}\mathrm{Br}\)), is reacted with \(\mathrm{KCN}\), which replaces both bromine atoms with cyano (\(\mathrm{CN}\)) groups in a nucleophilic substitution reaction. This results in the formation of compound \(\mathrm{Z}\), which is \(\mathrm{NCCH}_{2}-\mathrm{CH}_{2} \mathrm{CN}\).

Key Concepts

Elimination Reactions

In organic chemistry, elimination reactions are essential for forming double bonds by removing atoms or groups from adjacent carbon atoms. They involve the removal of a small molecule, typically a halide ion and a hydrogen ion. This process results in the formation of alkenes, compounds characterized by carbon-carbon double bonds.

The exercise begins with an elimination reaction. When ethyl iodide (\(\text{C}_2\text{H}_5\text{I}\)), a compound containing the iodine atom, is treated with alcoholic potassium hydroxide (alc. KOH), the iodine atom and a hydrogen atom are removed.

• The iodide ion exits as a leaving group.
• Potassium hydroxide provides a hydroxide ion, which acts as a base to deprotonate the nearby carbon atom.
• A double bond forms between the two carbon atoms, creating ethene (\(\text{CH}_2 = \text{CH}_2\)), also known as compound X.

Elimination reactions like this one are crucial for introducing unsaturation into molecules, making them more reactive for further chemical transformations. Understanding them is key to grasping more complex reaction mechanisms later in organic chemistry.

Addition Reactions

Addition reactions play a significant role in organic chemistry, particularly when converting alkenes or alkynes through the addition of atoms across double or triple bonds. These reactions result in the saturation of a molecule, transforming alkenes by breaking the double bond and adding new elements across these formerly bonded carbons.

In the exercise, after ethene (\(\text{CH}_2 = \text{CH}_2\)) is formed (compound X), it undergoes an addition reaction with bromine (\(\text{Br}_2\)).

• The double bond in ethene acts as a reactive site for bromine addition.
• Bromine molecules add across the double bond, each attaching to one of the two carbons involved in the original bond.
• This yields 1,2-dibromoethane (\(\text{BrCH}_2-\text{CH}_2\text{Br}\)), referred to as compound Y.

Addition reactions, like this one, are used to transform and modify the structure of hydrocarbon chains, facilitating the stepwise buildup of more complex molecules from simpler precursors.

Nucleophilic Substitution

Nucleophilic substitution reactions are key transformations in organic chemistry where a nucleophile replaces a leaving group in a molecule. The nucleophile is typically an electron-rich substance that donates an electron pair to form a new covalent bond. Conversely, the leaving group detaches and exits.

In the given exercise, compound Y, 1,2-dibromoethane (\(\text{BrCH}_2-\text{CH}_2\text{Br}\)), undergoes nucleophilic substitution with potassium cyanide (KCN).

• The CN ion, acting as a nucleophile, is rich in electrons.
• It attacks the carbon atoms bonded with the bromine atoms in 1,2-dibromoethane.
• As the CN groups replace the bromine atoms, they form cyanide groups (\(\text{CN}\)), resulting in a new compound, which is compound Z: \(\text{NCCH}_2-\text{CH}_2\text{CN}\).

Nucleophilic substitution is foundational in many synthesis routes, particularly when introducing functional groups such as nitriles, aldehydes, or ketones. Understanding this reaction helps learners appreciate how reactive intermediates can drive the synthesis of diverse organic compounds.