Question

Number of alkenes obtained on dehydrohalogenation of 2-chloro-3-methylpentane with caustic potash is (A) 2 (B) 3 (C) 4 (D) 5

Short answer

Upon dehydrohalogenation of 2-chloro-3-methylpentane with caustic potash, there are 3 possible alkenes that can be formed. Therefore, the correct answer is (B) 3.

Step-by-step solution

Draw 2-chloro-3-methylpentane structure

Draw the structure of 2-chloro-3-methylpentane as shown below: ``` CH3-CH(Cl)-CH(CH3)-CH2-CH3 ```

Identify the β-carbon

In dehydrohalogenation, a proton is removed from a β-carbon (carbon adjacent to the carbon containing the halogen). Identify all possible β-carbons. There are two β-carbons in this molecule, as shown below: ``` CH3-CH(1)(Cl)-CH(2)(CH3)-CH2-CH3 ```

Determine potential alkenes

Remove one hydrogen atom from each β-carbon and remove the halogen atom (Cl). Evaluate the different alkenes that can form: 1. Removing a hydrogen from position 1: ``` CH2=CH-CH(CH3)-CH2-CH3 ``` 2. ``` CH2=CH-CH(CH3)-CH=CH3 ``` 3. ``` CH3-CH(Cl)-CH(CH3)-CH=CH3 ``` 4. ``` CH3-CH(Cl)-CH(CH3)-CH=CH3 ``` Since options 3 and 4 are identical, then there are only three unique potential alkenes that can form.

Choose correct answer

Based on the previous steps, we found that there are 3 possible alkenes that can be formed upon dehydrohalogenation of 2-chloro-3-methylpentane. Therefore, the correct answer is: (B) 3

Key concepts

Alkene Formation

Alkene formation through the process called dehydrohalogenation is an essential concept in organic chemistry. This process involves the elimination of a halogen atom alongside a hydrogen atom from an alkyl halide. The hydrogen atom is specifically removed from a carbon atom that is adjacent to the carbon bonded to the halogen, known as the beta carbon. During this reaction, a double bond is created, leading to the formation of an alkene.

The reaction typically requires the presence of a strong base, such as caustic potash (potassium hydroxide, KOH), which facilitates the abstraction of the beta hydrogen. When studying for competitive exams like JEE MAIN and ADVANCED, understanding the mechanism behind this reaction is crucial as it not only aids in predicting the products but also helps in unraveling more complex reaction pathways in synthesis.

• Dehydrohalogenation: A base abstracts a β-hydrogen, the halide leaves, forming a double bond.
• Alkene: The product of dehydrohalogenation, characterized by a carbon-carbon double bond.
• Reagent: Strong base like KOH (caustic potash) is often used.
• Regioselectivity: The major product is usually the more substituted alkene (according to Zaitsev's rule).

Beta-Carbon Identification

Identification of the beta-carbon(s) is a critical step when predicting the outcome of dehydrohalogenation reactions. The beta carbon is defined as a carbon atom that is directly adjacent to the carbon atom bonded to the halogen (the alpha carbon). For each halogen present in the molecule, there can be one or more beta carbons, depending on the structure of the molecule.

Identifying beta carbons correctly allows us to predict where double bonds might form, as the hydrogen atoms that are removed during the reaction will come from these beta-carbon atoms. This process is governed by certain rules and patterns:

• Zaitsev's Rule: Preferentially remove the hydrogen from the beta carbon that has fewer hydrogen atoms (more substituted) as it typically leads to the more stable, major alkene product.
• Number of Alkenes: The number of possible alkenes can often be deduced by considering the number of unique beta carbons.
• Reaction Conditions: The strength of the base and the reaction conditions (like temperature and solvent) can influence which hydrogen gets removed.

The ability to identify beta-carbons is not only useful for mechanisms involving dehydrohalogenation but is broadly applicable in organic chemistry, as various elimination reactions require beta-carbon considerations.

Organic Chemistry for JEE MAIN and ADVANCED

Mastering organic chemistry is a cornerstone for students preparing for competitive exams like JEE MAIN and ADVANCED. It requires a deep conceptual understanding and the ability to apply the principles to different problems. Organic reactions, such as the dehydrohalogenation of alkyl halides, are often tested due to the intricacies involved in their mechanisms and the variety of products they can produce.

For students aiming to score high on these exams, it’s not enough to memorize reactions; they must also be able to critically analyze the molecular structure and understand the factors affecting these reactions. This includes:

• Identifying reaction types and the reagents that drive them.
• Appreciating the role of stereochemistry in the reactants and products.
• Predicting reaction outcomes based on electronic and steric considerations.
• Understanding mechanisms at a granular level to solve complex synthesis problems.

In the context of dehydrohalogenation, a solid grasp of beta-carbon identification and alkene formation, along with Zaitsev's rule, are critical tools in the arsenal of a JEE candidate. By working through numerous practice questions and applying concepts in different scenarios, students enhance their problem-solving skills and prepare themselves for success on these challenging exams.