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

Question

Identify \(Z\) in the following series: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I} \stackrel{\mathrm{Alc} . \mathrm{KOH}}{\longrightarrow} \mathrm{X} \stackrel{\mathrm{Br}_{2}}{\longrightarrow \mathrm{Y}} \stackrel{\mathrm{KCN}}{\longrightarrow} \mathrm{Z}\) (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 \(Z\) is \(\mathrm{CH}_{3}\mathrm{CH_{2}CN}\) which corresponds to option (a).

Step 1: Understand the Reaction Setup

The exercise presents a sequence of chemical reactions involving a starting material \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\). This compound undergoes a series of transformations to form product \(Z\). We will break down each stage to determine this final product.

Step 2: First Transformation

The first transformation involves the reaction of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) with alcoholic \(\mathrm{KOH}\). Here, a dehydrohalogenation occurs, resulting in the elimination of \(\mathrm{HI}\) to form \(\mathrm{C}_{2} \mathrm{H}_{4}\) (ethylene, \(\mathrm{X}\)).

Step 3: Second Transformation

In the second step, \(\mathrm{X} = \mathrm{C}_{2} \mathrm{H}_{4}\) reacts with \(\mathrm{Br}_{2}\) via an addition reaction across the double bond, forming 1,2-dibromoethane (\(\mathrm{BrCH}_{2}-\mathrm{CH}_{2}\mathrm{Br}\)). This compound is \(\mathrm{Y}\).

Step 4: Final Transformation

In the last step, \(\mathrm{Y} = \mathrm{BrCH}_{2}-\mathrm{CH}_{2}\mathrm{Br}\) reacts with \(\mathrm{KCN}\). This substitution reaction replaces a bromine atom with a cyano group \(\mathrm{CN}\), yielding the final product \(Z = \mathrm{CH}_{3}\mathrm{CH_{2}CN}\).

Step 5: Identify the Correct Option

Now that we have identified \(Z\) as \(\mathrm{CH}_{3}\mathrm{CH_{2}CN}\), we match it with the provided options. The correct answer is option (a) \(\mathrm{CH}_{3}\mathrm{CH_{2}CN}\).

Key Concepts

Dehydrohalogenation

Dehydrohalogenation is a fundamental transformation in organic chemistry where a hydrogen halide (such as HCl, HBr, HI) is eliminated from an alkyl halide, resulting in the formation of an alkene. This process effectively removes a hydrogen atom and a halogen atom from adjacent carbon atoms. The reaction typically requires a strong base, such as potassium hydroxide (KOH).
In the given exercise, dehydrohalogenation occurs in the first step, where ethyl iodide ( C_2H_5I ) reacts with alcoholic KOH. The base abstracts a hydrogen atom, causing the iodine atom to depart, thereby producing ethylene ( C_2H_4 ), which contains a carbon-carbon double bond.
This reaction is particularly useful because it converts more reactive alkyl halides into the less reactive, yet synthetically valuable, alkenes.

Addition Reaction

Addition reactions involve the addition of atoms or groups of atoms to a double or triple bond. This type of reaction results in the saturation of the unsaturated bond, converting compounds like alkenes into more complex molecules.
In the exercise, addition occurs in the second step, where ethylene ( C_2H_4 ), formed from the dehydrohalogenation, reacts with bromine ( Br_2 ). This reaction takes place across the double bond of ethylene, breaking the C=C bond and adding two bromine atoms, resulting in the formation of 1,2-dibromoethane ( BrCH_2-CH_2Br ).

• Double bonds are reactive sites for addition reactions due to the electron-rich environment, which attracts electrophiles like Br₂.
• The product becomes more saturated, increasing its utility in further chemical transformations.

Nucleophilic Substitution

Nucleophilic substitution is a reaction where a nucleophile (an electron-rich species) displaces a leaving group from a carbon atom. This reaction is a cornerstone of organic synthesis, providing a pathway to introduce diverse functional groups into organic molecules.
In the final step of the problem, the 1,2-dibromoethane ( BrCH_2-CH_2Br ) undergoes nucleophilic substitution with potassium cyanide ( KCN ). Cyanide ion, CN^- , acts as a nucleophile, attacking one of the carbon-bromine bonds and replacing the bromine atom with a cyano group ( CN ).
This transformation is essential for synthesizing nitriles, as seen in the formation of CH_3CH_2-CN , which becomes the product Z . Nucleophilic substitution opens doors to create various organic compounds by modifying the starting material through strategic introduction of functional groups.

Synthesis of Nitriles

Nitriles are a class of organic compounds containing a cyano group ( CN ) attached to a carbon atom. They are immensely valuable in organic chemistry due to their versatility and ability to undergo further transformations, such as hydrolysis and reduction.
Synthesizing nitriles often involves nucleophilic substitution reactions, where halogens in alkyl halides are replaced by cyano groups. In the exercise, this strategy is employed in the last reaction step, converting 1,2-dibromoethane with KCN into CH_3CH_2-CN .

• Nitriles serve as important intermediates in the formation of carboxylic acids and amines, expanding their use in pharmaceutical and chemical industries.
• The presence of the cyano group in nitriles can contribute to stability and reactivity, advantageous for designing molecular architectures.

By understanding the synthesis of nitriles, one can exploit their potential in creating various chemical products, highlighting their significance in both academic and commercial settings.