Problem 102 - A Complete Resource Book in Chemistry for JEE Main

Question

n-propyl bromide on treatment with ethanolic potassium hydroxide produces (a) propane (b) propene (c) propyne (d) propanol

Step-by-Step Solution

Answer

(b) propene

Step 1: Understanding the Reaction

N-propyl bromide is a type of haloalkane, and the reaction with ethanolic potassium hydroxide typically leads to a dehydrohalogenation process. Dehydrohalogenation results in the removal of a hydrogen halide (HBr in this case) from the molecule.

Step 2: Identifying the Function of Ethanolic KOH

Ethanolic potassium hydroxide (KOH) acts as a base, and under these conditions, it promotes an elimination reaction rather than nucleophilic substitution. This means that a double bond will be formed in the resulting molecule.

Step 3: Determining the Elimination Product

In dehydrohalogenation of n-propyl bromide, once HBr is removed, a double bond is formed between the adjacent carbon atoms, resulting in the formation of propene. The reaction can be represented as: \( \text{C}_3\text{H}_7\text{Br} + \text{KOH (ethanol)} \rightarrow \text{C}_3\text{H}_6 + \text{KBr} + \text{H}_2\text{O} \).

Step 4: Conclusion of Reaction Type

The result of the reaction due to the influence of ethanolic KOH is the formation of an alkene, which in the case of n-propyl bromide, is propene due to the created double bond.

Key Concepts

n-propyl bromide

N-propyl bromide is a haloalkane, which means it consists of an alkane group with a halogen atom attached. In this case, the bromine is the halogen. N-propyl bromide is derived from propane, and it possesses a three-carbon chain structure. When talking about haloalkanes, such as n-propyl bromide, they are typically reactive due to the presence of the halogen. The carbon-bromine bond in n-propyl bromide is polarized. This happens because bromine is more electronegative than carbon, making the bond more susceptible to chemical reactions. In reactions, n-propyl bromide can undergo various transformations, and one such transformation is the dehydrohalogenation process. This is where n-propyl bromide's nature and reactivity come into play, as they dictate the reaction pathways and products that can be formed. By understanding n-propyl bromide, you can predict how it will behave when introduced to different chemical environments.

ethanolic potassium hydroxide

Ethanolic potassium hydroxide (KOH in ethanol) plays a significant role both in lab settings and industrial applications. KOH is a strong base, and when dissolved in ethanol, it creates an environment conducive to elimination reactions, specifically dehydrohalogenation. Here's why ethanolic KOH is essential in dehydrohalogenation of n-propyl bromide:

• Potassium hydroxide, being a strong base, facilitates the removal of a hydrogen atom adjacent to the carbon that holds the halogen (bromine), leading to the formation of a double bond.
• Ethanolic conditions are preferred because ethanol is less polar than water, minimizing unwanted side reactions, such as nucleophilic substitution, which are more likely in aqueous conditions.
• By using ethanolic KOH, the reaction favors the formation of an alkene, which is the key target in the reaction mechanism.

These features ensure that ethanolic KOH is tailored for such elimination reactions, making it a critical component in transforming n-propyl bromide into propene.

alkene formation

Alkenes are a class of hydrocarbons characterized by at least one carbon-carbon double bond. In the context of the dehydrohalogenation of n-propyl bromide, the alkene formed is propene. Let's break down this process: Propene, a three-carbon alkene, results from the removal of hydrogen bromide (HBr) from n-propyl bromide when treated with ethanolic KOH. The mechanism works as follows:

• An ethoxide ion (from ethanolic KOH) abstracts a hydrogen from the carbon adjacent to the bromine-bearing carbon.
• The bond between this hydrogen and its carbon breaks, while simultaneously, the electrons from this bond form a new double bond with the carbon previously bearing the bromine.
• The result is the release of HBr and the formation of a double bond between the two central carbon atoms in the chain, producing propene.

This reaction is a classic example of an elimination reaction, specifically E2 (bimolecular elimination), which is crucial for synthesizing alkenes like propene in organic chemistry.