Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify the final product (C) in the following sequence of reactions. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \stackrel{\text { Alc. } \mathrm{KOH}, \Delta}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{Br}_{2}}{\longrightarrow}(\mathrm{B}) \stackrel{\mathrm{KCN}}{\longrightarrow}(\mathrm{C})\) (a) \(\mathrm{NCCH}_{2} \mathrm{CH}_{2} \mathrm{CN}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2}^{2} \mathrm{CN}\) (d) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CN}\)

Step-by-Step Solution

Answer

The final product (C) is \(\mathrm{NCCH}_2\mathrm{CH}_2\mathrm{CN}\) .

Step 1: Dehydrohalogenation Reaction

The first step involves the reaction of ethyl iodide (\(\mathrm{CH}_3\mathrm{CH}_2\mathrm{I}\)) with alcoholic KOH under heat. This is a dehydrohalogenation reaction where the iodine is removed, along with a hydrogen from the adjacent carbon, forming ethylene (\(\mathrm{C}_2\mathrm{H}_4\)), which is compound (A).

Step 2: Halogenation Reaction

In the second step, the ethylene (\(\mathrm{C}_2\mathrm{H}_4\)) from step 1 undergoes a halogenation reaction with bromine (\(\mathrm{Br}_2\)). The double bond in ethylene reacts with bromine, resulting in the formation of 1,2-dibromoethane (\(\mathrm{BrCH}_2\mathrm{CH}_2\mathrm{Br}\)), which is compound (B).

Step 3: Nucleophilic Substitution Reaction

The final step involves 1,2-dibromoethane (\(\mathrm{BrCH}_2\mathrm{CH}_2\mathrm{Br}\)) reacting with potassium cyanide (\(\mathrm{KCN}\)). A nucleophilic substitution occurs where both bromine atoms are replaced by cyanide groups, leading to the formation of 1,2-dicyanoethane (\(\mathrm{NCCH}_2\mathrm{CH}_2\mathrm{CN}\)), which is compound (C).

Key Concepts

Dehydrohalogenation

In organic chemistry, dehydrohalogenation is a crucial reaction for forming alkenes. This process involves the elimination of a hydrogen halide (such as HCl or HI) from an alkyl halide. By using alcoholic potassium hydroxide (KOH) as a strong base and applying heat, we remove both a halogen atom and a hydrogen atom from adjacent carbon atoms. This elimination results in the creation of a carbon-carbon double bond, forming an alkene.

• Starting compound: Alkyl halide (e.g., ethyl iodide, \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{I}\)).
• Reaction conditions: Alcoholic KOH and heat.
• Product: Ethylene (\(\mathrm{C}_{2}\mathrm{H}_{4}\)), an alkene.

This reaction is not only straightforward but also significant for the production of alkenes, which are key intermediates in many synthetic processes. The resulting compound from this specific exercise is ethylene or compound (A). Understanding this reaction paves the way for transforming simple alkanes into more complex molecules.

Halogenation

Halogenation is a fundamental reaction in organic chemistry where halogen atoms are added to organic compounds. When ethylene (\(\mathrm{C}_{2}\mathrm{H}_{4}\)) undergoes halogenation, it reacts with a halogen like bromine (\(\mathrm{Br}_{2}\)). This reaction targets the double bond in ethylene, breaking it and substituting each carbon atom with one bromine atom, resulting in 1,2-dibromoethane (\(\mathrm{BrCH}_{2}\mathrm{CH}_{2}\mathrm{Br}\)).

• Starting compound: Ethylene, \(\mathrm{C}_{2}\mathrm{H}_{4}\).
• Reagent: Bromine, \(\mathrm{Br}_{2}\).
• Product: 1,2-dibromoethane, \(\mathrm{BrCH}_{2}\mathrm{CH}_{2}\mathrm{Br}\).

This reaction is particularly useful in synthesizing organohalogen compounds, which can be employed in various subsequent chemical transformations, including nucleophilic substitutions. Here, the product from this transformation is identified as compound (B). Halogenation not only modifies the physical and chemical properties of compounds but often is a preparatory step for further chemical reactions.

Nucleophilic Substitution

Nucleophilic substitution is a critical type of substitution reaction where a nucleophile replaces a leaving group, typically a halide ion. In this exercise, 1,2-dibromoethane (\(\mathrm{BrCH}_{2}\mathrm{CH}_{2}\mathrm{Br}\)) serves as the starting compound, and potassium cyanide (\(\mathrm{KCN}\)) is the nucleophile. The strong nucleophilicity of the cyanide ion (\(\mathrm{CN}^-\)) allows it to effectively replace the bromine atoms in the compound.

• Starting compound: 1,2-dibromoethane, \(\mathrm{BrCH}_{2}\mathrm{CH}_{2}\mathrm{Br}\).
• Nucleophile: Cyanide ion from \(\mathrm{KCN}\).
• Product: 1,2-dicyanoethane, \(\mathrm{NCCH}_{2}\mathrm{CH}_{2}\mathrm{CN}\).

The reaction results in converting the dibromoethane to a dicyanoethane, known as compound (C). This process is crucial in organic synthesis for creating nitriles, which are valuable intermediates in producing pharmaceuticals and other fine chemicals. Nucleophilic substitution reactions expand the diversity of possible organic molecules significantly.