Problem 89 - A Complete Resource Book in Chemistry for JEE Main

Question

An alkyl chloride produces a single alkene on reaction with sodium ethoxide and ethanol. The alkene further undergoes hydrogenation to yield 2 -methylbutane. Identify the alkyl chloride from amongst the following. (a) \(\mathrm{ClCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{ClCH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{Cl})\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Step-by-Step Solution

Answer

The correct alkyl chloride is (a) \(\mathrm{ClCH}_{2}\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_3\).

Step 1: Analyze the Reaction

The reaction given is an elimination reaction where an alkyl chloride reacts with sodium ethoxide in ethanol to form an alkene. Given that the alkene undergoes hydrogenation to yield 2-methylbutane, we need to deduce which alkene results in this product after hydrogenation.

Step 2: Understand the Product

2-Methylbutane is a pentane derivative with the structure: \( \text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_3 \). The alkene precursor to this would likely be a double bond between the carbon that becomes the methyl-substituted carbon and one adjacent carbon, reflecting cis- or trans-2-pentene.

Step 3: Identify Possible Alkenes

From the options, the alkyl chloride that can form an alkene that hydrogenates to 2-methylbutane: Consider each chloride reacting to lose HCl and form the alkene.

Step 4: Evaluate Each Option

(a) \(\mathrm{ClCH}_{2}\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_3\): Forms 2-pentene upon elimination (correct).(b) \(\mathrm{ClCH}_{2}\mathrm{C}(\mathrm{CH}_3)_{2}\mathrm{CH}_3\): Forms 2-methylpropene (incorrect).(c) \(\mathrm{ClCH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}\): Forms 1-butene (incorrect).(d) \(\mathrm{CH}_{3}\mathrm{C}(\mathrm{Cl})(\mathrm{CH}_{3})\mathrm{CH}_2\mathrm{CH}_3\): Forms 2-methyl-1-butene (incorrect).

Step 5: Identify the Correct Alkyl Chloride

Option (a) can form 2-pentene which upon hydrogenation yields 2-methylbutane. Therefore, the alkyl chloride in option (a) is the correct precursor that results in the desired alkene.

Key Concepts

Elimination Reactions

Elimination reactions are a vital part of organic chemistry where a molecule loses atoms or groups, usually forming a double bond or an alkene. They often involve the loss of a hydrogen atom and a leaving group, such as a chlorine atom in alkyl chlorides, hence they're called dehydrohalogenation reactions. These reactions are classified mainly into two types: E1 and E2.
- **E1 Reaction:** This is a two-step process usually involving a carbocation intermediate. It's a unimolecular reaction, meaning the rate depends only on the concentration of the substrate. - **E2 Reaction:** This is a single-step bimolecular reaction where the loss of the leaving group and hydrogen occurs simultaneously. The rate of this reaction depends on both the substrate and the base, such as sodium ethoxide.
In your exercise, the alkyl chloride undergoes an E2 elimination when treated with sodium ethoxide in ethanol, resulting in the formation of an alkene. Understanding these basic reaction mechanisms is vital as they frequently appear in synthesis and various organic chemistry processes.

Alkenes

Alkenes, unsaturated hydrocarbons with one or more carbon-carbon double bonds, play a crucial role in organic chemistry. The presence of a double bond introduces unique chemical behavior, such as reactivity in addition reactions. Alkenes are characterized by the general formula \\(C_nH_{2n}\).
The double bond in alkenes consists of a sigma bond and a pi bond. The pi bond is primarily responsible for the reactivity of alkenes. Due to the presence of the pi bond's electrons, alkenes can undergo reactions like hydrogenation and halogenation comparatively easily.
In our exercise, the alkyl chloride forms an alkene, specifically 2-pentene, through an elimination reaction. This 2-pentene can exist as either a cis or trans isomer, affecting its physical properties and reactivity. Understanding the structure and behavior of alkenes helps when examining their transformation through additional reactions like hydrogenation, leading to saturated hydrocarbons.

Hydrogenation

Hydrogenation is the process of adding hydrogen (\(H_2\)) to an organic compound, typically converting alkenes into alkanes. This reaction is widely used in the chemical industry to produce saturated fats, oils, and other chemicals. Hydrogenation usually requires a metal catalyst such as platinum, palladium, or nickel.
The mechanism of hydrogenation involves the adsorption of hydrogen and the alkene onto the metal catalyst's surface. The pi bond in the alkene is then broken, allowing the addition of hydrogen atoms across the former double bond, resulting in an alkane. This process effectively "saturates" the double bond.
In your exercise, the alkene 2-pentene undergoes hydrogenation to form 2-methylbutane. This illustrates hydrogenation's ability to convert alkenes into more stable and less reactive alkanes, which are crucial for various applications in organic synthesis and industrial chemistry.

Alkyl Chlorides

Alkyl chlorides are organic compounds featuring a chlorine atom bonded to an alkane chain. They're a subclass of alkyl halides and are pivotal in organic synthesis due to their reactivity. The presence of a highly electronegative chlorine atom makes these compounds good electrophiles, prone to nucleophilic substitution and elimination reactions.
Alkyl chlorides can be formed via chlorination of alkanes and have various industrial and synthetic uses. They can undergo elimination reactions to form alkenes when reacting with bases like sodium ethoxide, as in the exercise.
In the given problem, we examine different alkyl chlorides to determine which one forms an alkene that hydrogenates to 2-methylbutane. Option (a) forms 2-pentene, which upon hydrogenation, results in the desired 2-methylbutane. Recognizing the behavior of alkyl chlorides in elimination reactions helps in predicting product formation and understanding their key role in transforming into useful chemical intermediates.