Problem 28 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

1-Chlorobutane on reaction with alcoholic potash gives (a) 1-butene (b) 1 -butanol (c) 2-butene (d) 2-butanol

Step-by-Step Solution

Answer

The reaction yields (c) 2-butene.

Step 1: Identify the Reaction

The reaction of 1-chlorobutane with alcoholic potash (KOH) is typically an elimination reaction, known as dehydrohalogenation. This reaction involves the removal of a hydrogen atom and a halide (chlorine in this case) to form a double bond, leading to an alkene.

Step 2: Determine the Alkenes Formed

During the dehydrohalogenation of 1-chlorobutane, a hydrogen atom is usually removed from the carbon adjacent to the carbon bearing the halide. In 1-chlorobutane, Chlorine is on the first carbon, so the hydrogen is removed from the second carbon, leading to the formation of 1-butene.

Step 3: Apply Zaitsev's Rule

Zaitsev's rule states that in elimination reactions, the more substituted alkene is preferred. However, since the possible positions are limited (from the primary carbon with chlorine to secondary carbons with hydrogen), the Zaitsev's rule directly applies here and prefers 2-butene over 1-butene.

Step 4: Conclusion

Based on the steps, the primary product formed from the reaction of 1-chlorobutane with alcoholic KOH is 2-butene, following Zaitsev's rule for more substituted alkene preference.

Key Concepts

Dehydrohalogenation

Dehydrohalogenation is a type of elimination reaction. It involves the removal of a halogen atom and a hydrogen atom from adjacent carbon atoms in an organic molecule. This reaction leads to the formation of a double bond, resulting in the creation of an alkene.
In the case of 1-chlorobutane, reacting it with alcoholic potash (KOH) induces this process. The chlorine atom from the one carbon atom and a hydrogen atom from an adjacent carbon atom are removed. This specific change transforms the single-bonded structure into one with a double bond, characteristic of alkenes.
These reactions are fundamental in organic chemistry as they help convert saturated hydrocarbons into unsaturated hydrocarbons, opening the door to many further reactions and product possibilities.

Zaitsev's Rule

Zaitsev's rule is a guideline used in predicting the outcome of elimination reactions like dehydrohalogenation. It states that in most cases, the most substituted alkene will be the major product. The more substituted an alkene, the more stable it typically is.
When 1-chlorobutane undergoes dehydrohalogenation, potential products include both 1-butene and 2-butene. However, due to Zaitsev's rule, 2-butene becomes the favored product as it is more substituted compared to 1-butene. This means that 2-butene, having more alkyl groups attached to the carbon-carbon double bond, stabilizes those bonds better than 1-butene.
Understanding Zaitsev's rule helps chemists and students predict the structure of the major product in elimination reactions before actual experimentation.

Alkenes Formation

Alkenes are hydrocarbons characterized by a carbon-carbon double bond. Forming alkenes through dehydrohalogenation process is a common method in organic synthesis. This process is prominently known for converting alkyl halides into alkenes with the aid of a base.
In our example with 1-chlorobutane, the elimination of a hydrogen and chlorine atom results in the formation of a double bond. As mentioned before, due to the preferences outlined by Zaitsev's rule, 2-butene is the major product formed.
The production of alkenes is significant in organic chemistry as these molecules serve as important intermediates in various chemical reactions, including polymerizations and epoxidations. Therefore, understanding their formation and the factors affecting their stability and predominance is essential for students and professionals in the field.