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: Identify the Reaction Type

Ethanolic potassium hydroxide (KOH in ethanol) is used to carry out elimination reactions, specifically dehydrohalogenation. This implies that we are expecting the removal of hydrogen (H) and halogen (Br) from n-propyl bromide.

Step 2: Analyze n-Propyl Bromide

n-Propyl bromide has the chemical structure CH3CH2CH2Br. In the presence of ethanolic KOH, a hydrogen atom adjacent to the bromine will be removed along with the bromine atom.

Step 3: Identify the Products of Elimination

When a hydrogen from the beta carbon (next to the carbon bonded to bromine) is removed and the bromine is eliminated, a double bond forms between the alpha and beta carbons, resulting in the formation of an alkene.

Step 4: Determine the Specific Alkene Formed

The elimination process will form propene with the structure CH3CH=CH2. This is due to the creation of a double bond between the carbons after elimination of H and Br.

Key Concepts

Elimination Reaction

An elimination reaction is a fundamental chemical reaction where elements are removed from a molecule, forming a new structure. This reaction is typical when you want to create a double bond between atoms. In the case of n-propyl bromide, treating it with ethanolic potassium hydroxide leads to a specific type of elimination reaction called dehydrohalogenation. Here's why elimination reactions are important:

• They transform saturated compounds (single bonds only) into unsaturated compounds (such as alkenes with double bonds).
• They are vital for synthesizing a wide range of organic compounds.

Elimination reactions can produce various products depending on the conditions and the specific molecules involved. Understanding the type of molecules involved and the reaction conditions helps predict and control the reaction outcome.

Dehydrohalogenation

Dehydrohalogenation is a specific type of elimination reaction where a hydrogen atom and a halogen are removed from an organic molecule. With n-propyl bromide, the term "dehydrohalogenation" indicates that one hydrogen and one bromine atom are eliminated. Here's a quick rundown of the process:

• The base, potassium hydroxide (KOH), often in an ethanolic (alcoholic) solution, helps remove the bromine atom.
• A hydrogen atom is removed from the adjacent carbon atom, often referred to as the beta (β) carbon, to the carbon bonded to the bromine.

Once these atoms are removed, the result is the formation of a double bond between the carbon atoms that are left, creating a more stable, unsaturated compound. This reaction type is widely used in organic chemistry to convert simple alkyl halides into more complex alkenes.

Alkene Formation

Alkene formation through elimination reactions like dehydrohalogenation is a crucial reaction in organic chemistry. In our specific case, n-propyl bromide, upon elimination, forms propene. Here's how the process works:

• After the removal of a bromine and hydrogen, a double bond is formed between two carbon atoms.
• This transformation from a saturated to an unsaturated molecule involves the formation of a pi-bond, which contributes to the alkene's chemical reactivity.

The final product, propene (\[\text{CH}_3\text{CH}=\text{CH}_2\]), is a simple alkene. This kind of transformation is essential from an application standpoint, allowing the synthesis of important precursors for polymers, pharmaceuticals, and other organic materials.