Problem 101 - A Complete Resource Book in Chemistry for JEE Main

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

(a) 1-butene

Step 1: Understand the Reaction Context

The reaction involves 1-chlorobutane reacting with alcoholic potash. You need to identify it as one involving elimination (dehydrohalogenation), where KOH in alcohol as a solvent leads to the removal of a hydrogen and a halogen from adjacent carbon atoms, forming a double bond, i.e., creating an alkene.

Step 2: Identify Relevant Mechanism

1-chlorobutane, an alkyl halide, will undergo elimination when treated with alcoholic KOH. This is known as a dehydrohalogenation reaction, specifically an E2 elimination, because alcoholic KOH favors elimination over substitution.

Step 3: Select the Correct Product Formation

In the E2 reaction of 1-chlorobutane, the hydrogen is removed from the β-carbon next to the carbon bonded to the chlorine. Since 1-chlorobutane is used, the most straightforward elimination produces 1-butene, as there is only one β-hydrogen available for abstraction.

Key Concepts

1-chlorobutane

1-chlorobutane is a type of alkyl halide, specifically, a chlorinated hydrocarbon. It is made up of a butane chain, which consists of four carbon atoms, and one of these carbon atoms is bonded to a chlorine atom.

The presence of the chlorine atom makes 1-chlorobutane reactive and an important compound in organic chemistry. It can undergo different types of reactions, but, in the context of our exercise, we focus on elimination reactions.

• Elimination reactions involve the removal of two groups from a molecule, typically forming a double bond in the process.
• In 1-chlorobutane, the chlorine atom and a hydrogen atom are removed during the reaction with alcoholic potash.

Understanding its structure and reactivity helps to predict the products formed when 1-chlorobutane is exposed to certain chemical conditions, such as those involved in our exercise.

alkenes

Alkenes are a major class of hydrocarbons that contain at least one carbon-carbon double bond in their molecular structure. They are unsaturated, meaning they have fewer hydrogen atoms compared to alkanes, which are saturated hydrocarbons with only single bonds.

When 1-chlorobutane undergoes a reaction with alcoholic potash, it forms an alkene known as 1-butene. This reaction is an example of a dehydrohalogenation reaction, which is a common method for synthesizing alkenes from alkyl halides.

• The elimination reaction removes halogen and hydrogen atoms to create a double bond, resulting in an alkene.
• In the case of 1-chlorobutane, the elimination leads to the formation of 1-butene, highlighting the creation of a double bond.

This concept is essential because it explains how simple reactions can be used to convert saturated compounds into more chemically reactive unsaturated compounds.

alcoholic potash

Alcoholic potash is a commonly used reagent in organic chemistry. It consists of potassium hydroxide (KOH) dissolved in an alcohol, typically ethanol. This combination is particularly effective for carrying out elimination reactions.

When alcoholic potash reacts with an alkyl halide like 1-chlorobutane, it promotes the elimination process rather than substitution. This results in the formation of alkenes. Here are a few key points about alcoholic potash:

• It provides a strong base, KOH, which is necessary for the deprotonation step in elimination reactions.
• The alcoholic environment helps in favoring the elimination pathway by improving the solubility of organic reactants.

The role of alcoholic potash in reactions like the one with 1-chlorobutane is crucial, as it effectively facilitates the transformation from an alkyl halide to an alkene through an E2 elimination mechanism.